RD Sharma Class 12 Ex 5.3 Solutions Chapter 5 Algebra of Matrices  

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter5
Exercise5.3
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 5.3 Solutions Chapter 5 Algebra of Matrices  

Question 1. Compute the indicated products:

(i) \begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}

Solution:

We have,

\begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}=\begin{bmatrix}a(a) +b(b) & a(-b) +b(a) \\-b(a) +a(b) & (-b) (-b) +a(a) \end{bmatrix}

=\begin{bmatrix}a^2+b^2& -ab+ab\\-ab+ab & a^2-b^2\end{bmatrix}

=\begin{bmatrix}a^2+b^2& 0\\0 & a^2-b^2\end{bmatrix}

(ii) \begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\-3 & 2 & -1\end{bmatrix}

Solution:

We have,

\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\-3 & 2 & -1\end{bmatrix}=\begin{bmatrix}1(1) +(-2) (-3) & 1(2) +(-2) (2) & 1(3) +(-2) (-1) \\2(1) +3(-3) & 2(2) +3(2) & 2(3) +3(-1) \end{bmatrix}

=\begin{bmatrix}1+6 & 2-4 & 3+2\\2-9 & 4+6 & 6-3\end{bmatrix}

=\begin{bmatrix}7 & -2 & 5\\-7 & 10 & 3\end{bmatrix}

(iii) \begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}\begin{bmatrix}1 & -3 & 5\\0 & 2 & 4\\3 & 0 & 5\end{bmatrix}

Solution:

We have,

\begin{bmatrix}2 & 3 & 4\\3 & 4 & 5\\4 & 5 & 6\end{bmatrix}\begin{bmatrix}1 & -3 & 5\\0 & 2 & 4\\3 & 0 & 5\end{bmatrix}=\begin{bmatrix}2(1) +3(0) +4(3) & 2(-3) +3(2) +4(0) & 2(5) +3(4) +4(5) \\3(1) +4(0) +5(3) & 3(-3) +4(2) +5(0) & 3(5) +4(4) +5(5) \\4(1) +5(0) +6(3) & 4(-3) +5(2) +6(0) & 4(5) +5(4) +6(5) \end{bmatrix}

=\begin{bmatrix}2+0+12 & -6+6+0 & 10+12+20\\3+0+15 & -9+8+0 & 15+16+25\\4+0+18 & -12+10+0 & 20+20+30\end{bmatrix}

=\begin{bmatrix}14 & 0 & 42\\18 & -1 & 56\\22 & -2 & 70\end{bmatrix}

Question 2. Show that AB ≠ BA in each of the following cases:

(i) A=\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}   and B=\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}   and B =\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}

AB =\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}

=\begin{bmatrix}10-3 & 5-4\\12+21 & 6+28\end{bmatrix}

=\begin{bmatrix}7 & 1\\33 & 34\end{bmatrix}

And we have,

BA =\begin{bmatrix}2 & 1\\3 & 4\end{bmatrix}\begin{bmatrix}5 & -1\\6 & 7\end{bmatrix}

=\begin{bmatrix}10+6 & -2+7\\15+24 & -3+28\end{bmatrix}

=\begin{bmatrix}16 & 5\\39 & 25\end{bmatrix}

Therefore, AB ≠ BA.

Hence, proved.

(ii) A=\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

AB =\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

=\begin{bmatrix}-1+0+0 & -2+1+0 & -3+0+0\\0+0+1 & 0-1+1 & 0+0+0\\2+0+4 & 4+3+4 & 6+0+0\end{bmatrix}

=\begin{bmatrix}-1 & -1 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}1 & 2 & 3\\0 & 1 & 0\\1 & 1 & 0\end{bmatrix}\begin{bmatrix}-1 & 1 & 0\\0 & -1 & 1\\2 & 3 & 4\end{bmatrix}

=\begin{bmatrix}-1+0+6 & 1-2+9 & 0+2+12\\0+0+0 & 0-1+0 & 0+1+0\\-1+0+0 & 1-1+0 & 0+1+0\end{bmatrix}

=\begin{bmatrix}5 & 8 & 14\\0 & -1 & 1\\-1 & 0 & 1\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

(iii) A=\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}   and B=\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}   and B =\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

AB =\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}

=\begin{bmatrix}0+3+0 & 1+0+0 & 0+0+0\\0+1+0 & 1+0+0 & 0+0+0\\0+1+0 & 4+0+0 & 0+0+0\end{bmatrix}

=\begin{bmatrix}3 & 1 & 0\\1 & 1 & 0\\1 & 4 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}0 & 1 & 0\\1 & 0 & 0\\0 & 5 & 1\end{bmatrix}\begin{bmatrix}1 & 3 & 0\\1 & 1 & 0\\4 & 1 & 0\end{bmatrix}

=\begin{bmatrix}0+1+0 & 0+1+0 & 0+0+0\\1+0+0 & 3+0+0 & 0+0+0\\0+5+4 & 0+5+1 & 0+0+0\end{bmatrix}

=\begin{bmatrix}1 & 1 & 0\\1 & 3 & 0\\9 & 6 & 0\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

Question 3. Compute the products AB and BA whichever exists in each of the following cases:

(i) A=\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

As A is of order 2 × 2 and B is of order 2 × 3, AB is possible but BA is not possible.

So, we get

AB =\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}

=\begin{bmatrix}1-4 & 2-6 & 3-2\\2+6 & 4+9 & 6+3\end{bmatrix}

=\begin{bmatrix}-3 & -4 & 1\\8 & 13 & 9\end{bmatrix}

(ii) A=\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}   and B=\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}   and B =\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}

As A is of order 3 × 2 and B is of order 2 × 3, AB and BA both are possible.

So, we get,

AB =\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}

=\begin{bmatrix}12+0 & 15+2 & 18+4\\-4+0 & -5+0 & -6+0\\-4+0 & -5+0 & -6+2\end{bmatrix}

=\begin{bmatrix}12 & 17 & 22\\-4 & -5 & -6\\-4 & -5 & -4\end{bmatrix}

Also we have,

BA =\begin{bmatrix}4 & 5 & 6\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}3 & 2\\-1 & 0\\-1 & 1\end{bmatrix}

=\begin{bmatrix}12-5-6 & 8+0+6\\0-1-2 & 0+0+2\end{bmatrix}

=\begin{bmatrix}1 & 14\\-3 & 2\end{bmatrix}

(iii) A=\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}   and B=\begin{bmatrix}0\\1\\3\\2\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}   and B =\begin{bmatrix}0\\1\\3\\2\end{bmatrix}

As A is of order 1 × 4 and B is of order 4 × 1, AB and BA both are possible.

So, we get,

AB =\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}\begin{bmatrix}0\\1\\3\\2\end{bmatrix}

=\begin{bmatrix}1(0) +(-1) (1) +2(3) +3(2) \end{bmatrix}

=\begin{bmatrix}0-1+6+6\end{bmatrix}

=\begin{bmatrix}11\end{bmatrix}

Also, we have,

BA =\begin{bmatrix}0\\1\\3\\2\end{bmatrix}\begin{bmatrix}1 & -1 & 2 & 3\end{bmatrix}

=\begin{bmatrix}0(1) & 0(-1) & 0(2) & 0(3) \\1(1) & 1(-1) & 1(2) & 1(3) \\3(1) & 3(-1) & 3(2) & 3(3) \\2(1) & 2(-1) & 3(2) & 2(3) \end{bmatrix}

=\begin{bmatrix}0 & 0 & 0 & 0\\1 & -1 & 2 & 3\\3 & -3 & 6 & 9\\2 & -2 & 6 & 6\end{bmatrix}

(iv) \begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a & b & c & d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}

Solution:

We have,

=\begin{bmatrix}a & b\end{bmatrix}\begin{bmatrix}c\\d\end{bmatrix}+\begin{bmatrix}a & b & c & d\end{bmatrix}\begin{bmatrix}a\\b\\c\\d\end{bmatrix}

=\begin{bmatrix}ac+bd\end{bmatrix}\begin{bmatrix}a^2+b^2+c^2+d^2\end{bmatrix}

=\begin{bmatrix}ac+bd+a^2+b^2+c^2+d^2\end{bmatrix}

Question 4. Show that AB ≠ BA in each of the following cases:

(i) A=\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}   and B=\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}   and B =\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}

AB =\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}

=\begin{bmatrix}-2-3+6 & 3+6-9 & -1-3+4\\-4+1+6 & 6-2-9 & -2+1+4\\-6+0+6 & 9+0-9 & -3+0+4\end{bmatrix}

=\begin{bmatrix}1 & 0 & 0\\3 & -5 & 3\\0 & 0 & 1\end{bmatrix}

And we have,

BA =\begin{bmatrix}-2 & 3 & -1\\1 & 2 & -1\\-6 & 9 & -4\end{bmatrix}\begin{bmatrix}1 & 3 & -1\\2 & -1 & -1\\3 & 0 & -1\end{bmatrix}

=\begin{bmatrix}-2+6-3 & -6-3+0 & 2-3+1\\-1+4-3 & -3-2+0 & 1-2+1\\-6+18-12 & -18-9+0 & 6-9+4\end{bmatrix}

=\begin{bmatrix}1 & -9 & 0\\0 & -5 & 0\\0 & -27 & 1\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

(ii) A=\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}   and B=\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}   and B =\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}

AB =\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}

=\begin{bmatrix}10-12-1 & 20-16-3 & 10-8-2\\-11+15+0 & -22+20+0 & -11+10+0\\9-15+1 & 18-20+3 & 9-10+2\end{bmatrix}

=\begin{bmatrix}-3 & 1 & 0\\4 & -2 & -1\\-5 & 1 & 1\end{bmatrix}

And we have,

BA =\begin{bmatrix}1 & 2 & 1\\3 & 4 & 2\\1 & 3 & 2\end{bmatrix}\begin{bmatrix}10 & -4 & -1\\-11 & 5 & 0\\9 & -5 & 1\end{bmatrix}

=\begin{bmatrix}10-22+9 & -4+10-5 & -9+0+1\\30-44+10 & -12+20-10 & -3+0+2\\10-33+18 & -4+15-10 & -1+0+2\end{bmatrix}

=\begin{bmatrix}-3 & 1 & 0\\4 & -2 & -1\\-5 & 1 & 1\end{bmatrix}

Therefore, AB ≠ BA.

Hence proved.

Question 5. Evaluate the following:

(i) \left(\begin{bmatrix}1 & 3\\-1 & -4\end{bmatrix}+\begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

Solution:

We have,

=\left(\begin{bmatrix}1 & 3\\-1 & -4\end{bmatrix}+\begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\left(\begin{bmatrix}1+3 & 3-2\\-1+1 & -4+1\end{bmatrix}\right) \begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\begin{bmatrix}4 & 1\\0 & -3\end{bmatrix}\begin{bmatrix}1 & 3 & 5\\2 & 4 & 6\end{bmatrix}

=\begin{bmatrix}4+2 & 12+4 & 20+6\\-2-6 & -6-12 & -10-18\\\end{bmatrix}

=\begin{bmatrix}6 & 16 & 26\\-8 & -18 & -28\\\end{bmatrix}

(ii) \begin{bmatrix}1 & 2 & 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

Solution:

We have,

=\begin{bmatrix}1 & 2 & 3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\\0 & 1 & 2\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}1+4+0 & 0+0+3 & 2+0+6\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}5 & 3 & 8\end{bmatrix}\begin{bmatrix}2\\4\\6\end{bmatrix}

=\begin{bmatrix}10+12+60\end{bmatrix}

=\begin{bmatrix}82\end{bmatrix}

(iii) \begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\left(\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\1 & 0 & 2\end{bmatrix}\right)

Solution:

We have,

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\left(\begin{bmatrix}1 & 0 & 2\\2 & 0 & 1\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\1 & 0 & 2\end{bmatrix}\right)

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\begin{bmatrix}1-0 & 0-1 & 2-2\\2-1 & 0-0 & 1-2\end{bmatrix}

=\begin{bmatrix}1 & -1\\0 & 2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & -1 & 0\\1 & 0 & -1\end{bmatrix}

=\begin{bmatrix}1-1 & -1+0 & 0+1\\0+2 & 0+0 & 0-2\\2+3 & -2+0 & 0-3\\\end{bmatrix}

=\begin{bmatrix}0 & -1 & 1\\2 & 0 & -2\\5 & -2 & -3\\\end{bmatrix}

Question 6. If A =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}   and C =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}   , then show that A2 = B2 = C2 = I2.

Solution:

We have,

A =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}   and C =

\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

A2 =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, A2 = I2

B2 =\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}\begin{bmatrix}1 & 0\\0 & -1\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, B2 = I2

C2 =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

=\begin{bmatrix}1+0 & 0+0\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

Therefore, C2 = I2

So, we get A2 = B2 = C2 = I2

Hence proved.

Question 7. If A =\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}   and B =\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}   , find 3A2 – 2B + I.

Solution:

We are given,

A =\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}   and B =\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}

So, we get,

3A2 – 2B + I =3\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}\begin{bmatrix}2 & -1\\3 & 2\end{bmatrix}-2\begin{bmatrix}0 & 4\\-1 & 7\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=3\begin{bmatrix}4-3 & -2-2\\6+6 & -3+4\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=3\begin{bmatrix}1 & -4\\12 & 1\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}3 & -12\\36 & 3\end{bmatrix}-\begin{bmatrix}0 & 8\\-2 & 14\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}3-0+1 & -12-8+0\\36+2+0 & 3-14+1\end{bmatrix}

=\begin{bmatrix}4 & -20\\38 & -10\end{bmatrix}

Question 8. If A =\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}   , prove that (A – 2I) (A – 3I) = 0.

Solution:

We are given,

A =\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}

L.H.S. = (A – 2I) (A – 3I)

=\left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right) \left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\right)

=\left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}\right) \left(\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}-\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}\right)

=\begin{bmatrix}4-2 & 2-0\\-1-0 & 1-2\end{bmatrix}\begin{bmatrix}4-3 & 2-0\\-1-0 & 1-3\end{bmatrix}

=\begin{bmatrix}2 & 2\\-1 & -1\end{bmatrix}\begin{bmatrix}1 & 2\\-1 & -2\end{bmatrix}

=\begin{bmatrix}2-2 & 4-4\\-1+1 & -2+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 9. If A =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}   , show that A2 =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}   and A3 =\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}   .

Solution:

We have,

A =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}

So, A2 =\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 1+1\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}

Hence, A3 = A2 . A

=\begin{bmatrix}1 & 2\\0 & 1\end{bmatrix}\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}

=\begin{bmatrix}1+0 & 1+2\\0+0 & 0+1\end{bmatrix}

=\begin{bmatrix}1 & 3\\0 & 1\end{bmatrix}

Hence proved.

Question 10. If A =\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}   , show that A2 = 0.

Solution:

We have,

A =\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}

So, we get

L.H.S. = A=\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}\begin{bmatrix}ab & b^2\\-a^2 & -ab\end{bmatrix}

=\begin{bmatrix}a^2b^2-a^2b^2 & ab^3-ab^3\\-a^3b+a^3b & -a^2b^2+a^2b^2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 11. If A =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}   , find A2.

Solution:

We have,

A =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}

So, we get

A2 =\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}\begin{bmatrix}\cos2\theta & \sin2\theta\\-\sin2\theta & \cos2\theta\end{bmatrix}

=\begin{bmatrix}\cos^22\theta-\sin^22\theta & \cos2\theta\sin2\theta+\cos2\theta\sin2\theta\\-\cos2\theta\sin2\theta-\cos2\theta\sin2\theta & \cos^22\theta-\sin^22\theta\end{bmatrix}

=\begin{bmatrix}\cos4\theta & 2\cos2\theta\sin2\theta\\-2\cos2\theta\sin2\theta & \cos4\theta\end{bmatrix}

=\begin{bmatrix}\cos4\theta & \sin4\theta\\-\sin4\theta & \cos4\theta\end{bmatrix}

Question 12. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}   , show that AB = BA = O3×3.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}

So, we get

AB =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 6+9-15 & 10+15+25\\1+4-5 & -3-12+15 & -5-20+25\\-1-3+4 & 3+9-12 & 5+15-20\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

And we have,

BA =\begin{bmatrix}-1 & 3 & 5\\1 & -3 & -5\\-1 & 3 & 5\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 3+12-15 & 5+15-20\\2+3-5 & -3-12+15 & -5-15+20\\-2-3+5 & 3+12-15 & 5+15-20\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Question 13. If A =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}   and B =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}   , show that AB = BA = O3×3.

Solution:

We have,

A =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}   and B =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}

So, we have,

AB =\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}

=\begin{bmatrix}0+abc-abc & 0+b^2c-b^2c & 0+bc^2-bc^2\\-a^2c+0+a^2c & -abc+0+abc & -ac^2+0+ac^2\\a^2b-a^2b+0 & ab^2-ab^2+0 & abc-abc+0\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

And we have,

BA =\begin{bmatrix}a^2 & ab & ac\\ab & b^2 & bc\\ac & bc & c^2\end{bmatrix}\begin{bmatrix}0 & c & -b\\-c & 0 & a\\b & -a & 0\end{bmatrix}

=\begin{bmatrix}0+abc-abc & 0+b^2c-b^2c & 0+bc^2-bc^2\\-a^2c+0+a^2c & -abc+0+abc & -ac^2+0+ac^2\\a^2b-a^2b+0 & ab^2-ab^2+0 & abc-abc+0\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= O3×3

Therefore, AB = BA = O3×3.

Hence proved.

Question 14. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}   , show that AB = A and BA = B.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   and B =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

AB =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

=\begin{bmatrix}4+3-5 & -4-9+10 & -8-12+15\\-2-4+5 & 2+12-10 & 4+16-15\\2+3-4 & -2-9+18 & -4-12+12\end{bmatrix}

=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

= A

And we have,

BA =\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}4+2-4 & -6-8+12 & -10-10+16\\-2-3+4 & 3+12-12 & 5+15-16\\2+2-3 & -3-8+9 & -5-10+12\end{bmatrix}

=\begin{bmatrix}2 & -2 & -4\\-1 & 3 & 4\\1 & -2 & -3\end{bmatrix}

= B

Hence proved.

Question 15. If A =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}   and B =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}   , compute A2 – B2.

Solution:

We have,

A =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}   and B =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}

A2 =\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}\begin{bmatrix}-1 & 1 & -1\\3 & -3 & 3\\5 & 5 & 5\end{bmatrix}

=\begin{bmatrix}4+2-4 & -6-8+12 & -10-10+16\\-2-3+4 & 3+12-12 & 5+15-16\\2+2-3 & -3-8+9 & -5-10+12\end{bmatrix}

=\begin{bmatrix}-1 & -9 & -1\\3 & 27 & 3\\35 & 15 & 35\end{bmatrix}

And we have,

B2 =\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}\begin{bmatrix}0 & 4 & 3\\1 & -3 & -3\\-1 & 4 & 4\end{bmatrix}

=\begin{bmatrix}-2-3+5 & 3+12-15 & 5+15-20\\2+3-5 & -3-12+15 & -5-15+20\\-2-3+5 & 3+12-15 & 5+15-20\end{bmatrix}

=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

So, we get

A2 – B2 =\begin{bmatrix}-1 & -9 & -1\\3 & 27 & 3\\35 & 15 & 35\end{bmatrix}-\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}-1-1 & -9-0 & -1-0\\3-0 & 27-1 & 3-0\\35-0 & 15-0 & 35-1\end{bmatrix}

=\begin{bmatrix}-2 & -9 & -1\\3 & 26 & 3\\35 & 15 & 34\end{bmatrix}

Question 16. For the following matrices verify the associativity of matrix multiplication i.e. (AB) C = A (BC).

(i) A =\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}   , C =\begin{bmatrix}1\\-1\end{bmatrix}

Solution:

We are given,

A =\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}   , C =\begin{bmatrix}1\\-1\end{bmatrix}

L.H.S. = (AB) C

=\left(\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}\right) \begin{bmatrix}1\\-1\end{bmatrix}

=\begin{bmatrix}1-2+0 & 0+4+0\\-1+0+0 & 0+0+3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}

=\begin{bmatrix}-1 & 4\\-1 & 3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}

=\begin{bmatrix}-5\\-4\end{bmatrix}

And R.H.S. = A (BC)

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\left(\begin{bmatrix}1 & 0\\-1 & 2\\0 & 3\end{bmatrix}\begin{bmatrix}1\\-1\end{bmatrix}\right)

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1+0\\-1-2\\0-3\end{bmatrix}

=\begin{bmatrix}1 & 2 & 0\\-1 & 0 & 1\end{bmatrix}\begin{bmatrix}1\\-3\\-3\end{bmatrix}

=\begin{bmatrix}1-6+0\\-1+0-3\end{bmatrix}

=\begin{bmatrix}-5\\-4\end{bmatrix}

= L.H.S.

Hence proved.

(ii) A =\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

Solution:

We are given,

A =\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}   , B =\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

L.H.S. = (AB) C

=\left(\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}\right) \begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}4+0+6 & -4+2-3 & 4+4+3\\1+0+4 & -1+1-2 & 1+2+2\\3+0+2 & -3+0-1 & 3+0+1\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}10 & -5 & 11\\5 & -2 & 5\\5 & -4 & 4\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}10-15+0 & 20+0+0 & -10+5+11\\5-6+0 & 10+0+0 & -5-2+5\\5-12+0 & 10+0+0 & -5-4+4\end{bmatrix}

=\begin{bmatrix}-5 & 20 & -4\\-1 & 10 & -2\\-7 & 10 & -5\end{bmatrix}

And R.H.S. = A (BC)

=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix} \left(\begin{bmatrix}1 & -1 & 1\\0 & 1 & 2\\2 & -1 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & -1\\3 & 0 & 1\\0 & 0 & 1\end{bmatrix}\right)

=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}1-3+0 & 2+0+0 & -1-1+1\\0+3+0 & 0+0+0 & 0+1+2\\2-3+0 & 4+0+0 & -2-1+1\end{bmatrix}

=\begin{bmatrix}4 & 2 & 3\\1 & 1 & 2\\3 & 0 & 1\end{bmatrix}\begin{bmatrix}-2 & 2 & -1\\3 & 0 & 3\\-1 & 4 & -2\end{bmatrix}

=\begin{bmatrix}-8+6-3 & 8+12+0 & -4+6-6\\-2+3-2 & 2+0+8 & -1+3-4\\-6+0-1 & 6+0+4 & -3-0-2\end{bmatrix}

=\begin{bmatrix}-5 & 20 & -4\\-1 & 10 & -2\\-7 & 10 & -5\end{bmatrix}

= L.H.S.

Hence proved.

Question 17. For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A (B + C) = AB + AC.

(i) A =\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}   , B =\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}   , C =\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}   , B =\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}   , C =\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

L.H.S. = A (B + C)

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix} \left(\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}\right)

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1+0 & 0+1\\2+1 & 1-1\end{bmatrix}

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1 & 1\\3 & 0\end{bmatrix}

=\begin{bmatrix}-1-3 & 1+0\\0+6 & 0+0\end{bmatrix}

=\begin{bmatrix}-4 & 1\\6 & 0\end{bmatrix}

R.H.S. = AB + AC

=\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}-1 & 0\\2 & 1\end{bmatrix}+\begin{bmatrix}1 & -1\\0 & 2\end{bmatrix}\begin{bmatrix}0 & 1\\1 & -1\end{bmatrix}

=\begin{bmatrix}-1-2 & 0-1\\0+4 & 0+2\end{bmatrix}+\begin{bmatrix}0-1 & 1+1\\0+2 & 0-2\end{bmatrix}

=\begin{bmatrix}-3 & -1\\4 & 2\end{bmatrix}+\begin{bmatrix}-1 & 2\\2 & -2\end{bmatrix}

=\begin{bmatrix}-3-1 & -1+2\\4+2 & 2-2\end{bmatrix}

=\begin{bmatrix}-4 & 1\\6 & 0\end{bmatrix}

= L.H.S.

Hence proved.

(ii) A =\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}   , B =\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

Solution:

We have,

A =\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}   , B =\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}   , C =\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

L.H.S. = A (B + C)

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix} \left(\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}\right)

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}0+1 & 1-1\\1+0 & 1+1\end{bmatrix}

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}1 & 0\\1 & 2\end{bmatrix}

=\begin{bmatrix}2-1 & 0+2\\1+1 & 0+2\\-1+2 & 0+4\end{bmatrix}

=\begin{bmatrix}1 & 2\\2 & 2\\1 & 4\end{bmatrix}

R.H.S. = AB + AC

=\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 1\end{bmatrix}+\begin{bmatrix}2 & -1\\1 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}1 & -1\\0 & 1\end{bmatrix}

=\begin{bmatrix}0+1 & 2-1\\0+1 & 1+1\\0+2 & -1+2\end{bmatrix}+\begin{bmatrix}2+0 & -2-1\\1+0 & -1+1\\-1+0 & 1+2\end{bmatrix}

=\begin{bmatrix}-1 & 1\\1 & 2\\2 & 1\end{bmatrix}+\begin{bmatrix}2 & -3\\1 & 0\\-1 & 3\end{bmatrix}

=\begin{bmatrix}-1+2 & 1-3\\1+1 & 2+0\\2-1 & 1+3\end{bmatrix}

=\begin{bmatrix}1 & -2\\2 & 2\\1 & 4\end{bmatrix}

= L.H.S.

Hence proved.

Question 18. If A =\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}   , B =\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}   and C =\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}   , show that A (B – C) = AB – AC.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}   , B =\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}   and C =\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}

L.H.S. = A (B – C)

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\left(\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}-\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}\right)

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}0-1 & 5-5 & -4-2\\-2+1 & 1-1 & 3-0\\-1-0 & 0+1 & 2-1\end{bmatrix}

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}-1 & 0 & -6\\-1 & 0 & 3\\-1 & 1 & 1\end{bmatrix}

=\begin{bmatrix}-1+0+2 & 0-0-2 & -6+0-2\\-3+1+0 & 0+0+0 & -18-3+0\\2-1-1 & 0+0+1 & 12+3+1\end{bmatrix}

=\begin{bmatrix}1 & -2 & -8\\-2 & 0 & -21\\0 & 1 & 16\end{bmatrix}

R.H.S. = AB – AC

=\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}0 & 5 & -4\\-2 & 1 & 3\\-1 & 0 & 2\end{bmatrix}-\begin{bmatrix}1 & 0 & -2\\3 & -1 & 0\\-2 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 5 & 2\\-1 & 1 & 0\\0 & -1 & 1\end{bmatrix}

=\begin{bmatrix}0+0+2 & 5+0+0 & -4+0-4\\0+2+0 & 15-1+0 & -12-3+0\\0-2-1 & -10+1+0 & 8+3+2\end{bmatrix}-\begin{bmatrix}1+0+0 & -5+0+2 & 2+0-2\\3+1+0 & 15-1+0 & 6+0+0\\0-2+1 & -10+1+1 & -4+0+1\end{bmatrix}

=\begin{bmatrix}2 & 5 & -8\\2 & 14 & -15\\-3 & -9 & 13\end{bmatrix}-\begin{bmatrix}1 & 7 & 0\\4 & 14 & 6\\-3 & -10 & -3\end{bmatrix}

=\begin{bmatrix}2-1 & 5-7 & -8-0\\2-4 & 14-14 & -15-6\\-3+3 & -9+10 & 13+3\end{bmatrix}

=\begin{bmatrix}1 & -2 & -8\\-2 & 0 & -21\\0 & 1 & 16\end{bmatrix}

Question 19. Compute the elements a43 and a22 of the matrix:

A =\begin{bmatrix}0 & 1 & 0\\2 & 0 & 2\\0 & 3 & 2\\4 & 0 & 4\end{bmatrix}\begin{bmatrix}2 & -1\\-3 & 2\\4 & 3\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

Solution:

We are given,

A =\begin{bmatrix}0 & 1 & 0\\2 & 0 & 2\\0 & 3 & 2\\4 & 0 & 4\end{bmatrix}\begin{bmatrix}2 & -1\\-3 & 2\\4 & 3\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}0-3+0 & 0+2+0\\4+0+8 & -2+0+6\\0-9+8 & 0+6+6\\8+0+16 & -4+0+12\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}-3 & 2\\12 & 4\\-1 & 12\\24 & 8\end{bmatrix}\begin{bmatrix}0 & 1 & -1 & 2 & -2\\3 & -3 & 4 & -4 & 0\end{bmatrix}

=\begin{bmatrix}0+6 & -3-6 & 3+8 & -6-8 & 6+0\\0+12 & 12-12 & -12+16 & 24-16 & 24+0\\0+36 & -1-36 & 1+48 & -2-48 & 2+0\\0+24 & -24+24 & -24+34 & 48-32 & -48+0\end{bmatrix}

=\begin{bmatrix}6 & -9 & 11 & -14 & 6\\12 & 0 & 4 & 8 & -24\\36 & -37 & 49 & -50 & 2\\24 & 0 & 8 & 16 & -48\end{bmatrix}

Therefore, a43 = 8 and a22 = 0.

Question 20. If A =\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}   and I is the identity matrix of order 3, show that A3 = pI + qA + rA2.

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

L.H.S. = A3

=\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0+0+0 & 0+0+0 & 0+1+0\\0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0 & 0 & 1\\p & q & r\\pr & p+qr & q+r^2\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\\0+0+pq+pr^2 & pr+0+q^2+qr^2 & 0+p+qr+qr+r^2\end{bmatrix}

=\begin{bmatrix}p & q & r\\pr & p+qr & q+r^2\\pq+pr^2 & pr+q^2+qr^2 & p+2qr+r^2\end{bmatrix}

And R.H.S. = pI + qA + rA2

=p\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}+q\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}+r\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}\begin{bmatrix}0 & 1 & 0\\0 & 0 & 1\\p & q & r\end{bmatrix}

=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+r\begin{bmatrix}0+0+0 & 0+0+0 & 0+1+0\\0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\end{bmatrix}

=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+r\begin{bmatrix}0 & 0 & 1\\p & q & r\\pr & p+qr & q+r^2\end{bmatrix}

=\begin{bmatrix}p & 0 & 0\\0 & p & 0\\0 & 0 & p\end{bmatrix}+\begin{bmatrix}0 & q & 0\\0 & 0 & q\\pq & q^2 & qr\end{bmatrix}+\begin{bmatrix}0 & 0 & r\\pr & qr & r^2\\pr^2 & pr+qr^2 & qr+r^3\end{bmatrix}

=\begin{bmatrix}0+0+p & 0+0+q & 0+0+r\\0+0+pr & p+0+qr & 0+q+r^2\\0+0+pq+pr^2 & pr+0+q^2+qr^2 & 0+p+qr+qr+r^2\end{bmatrix}

=\begin{bmatrix}p & q & r\\pr & p+qr & q+r^2\\pq+pr^2 & pr+q^2+qr^2 & p+2qr+r^2\end{bmatrix}

= L.H.S.

Hence proved.

Question 21. If ω is a complex cube root of unity, show that

\left(\begin{bmatrix}1 & ω & ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix}+\begin{bmatrix}ω & ω^2 & 1\\ω^2 & 1 & ω\\ω & ω^2 & 1\end{bmatrix}\right) \begin{bmatrix}1\\ω\\ω^2\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}

Solution:

We have,

L.H.S. =\left(\begin{bmatrix}1 & ω & ω^2\\ω & ω^2 & 1\\ω^2 & 1 & ω\end{bmatrix}+\begin{bmatrix}ω & ω^2 & 1\\ω^2 & 1 & ω\\ω & ω^2 & 1\end{bmatrix}\right) \begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}1+ω & ω+ω^2 & ω^2+1\\ω+ω^2 & ω^2+1 & 1+ω\\ω^2+ω & 1+ω^2 & ω+1\end{bmatrix}\begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}-ω^2 & -1 & -ω\\-1 & -ω & ω^2\\-1 & -ω & -ω^2\end{bmatrix}\begin{bmatrix}1\\ω\\ω^2\end{bmatrix}

=\begin{bmatrix}-ω^2-ω-ω^3\\-1-ω^2-ω^4\\-1-ω^2-ω^4\end{bmatrix}

=\begin{bmatrix}-ω(1+ω+ω^2) \\-1-ω^2-ω\\-1-ω^2-ω\end{bmatrix}

=\begin{bmatrix}-ω(0) \\-(1+ω^2+ω) \\-(1+ω^2+ω) \end{bmatrix}

=\begin{bmatrix}0\\0\\0\end{bmatrix}

= R.H.S.

Hence proved.

Question 22. If A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}   , prove that A2 = A.

Solution:

We have,

A =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

So, A2 =\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

=\begin{bmatrix}4+3-5 & -6-12+15 & -10-15+20\\-2-4+5 & 3+16-15 & 5+20-20\\2+3-4 & -3-12+12 & -5-15+16\end{bmatrix}

=\begin{bmatrix}2 & -3 & -5\\-1 & 4 & 5\\1 & -3 & -4\end{bmatrix}

= A

Hence proved.

Question 23. If A =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}   , show that A2 = I3.

Solution:

We have,

A =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}

So, A2 =\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}\begin{bmatrix}4 & -1 & -4\\3 & 0 & -4\\3 & -1 & -3\end{bmatrix}

=\begin{bmatrix}16-3-12 & -4+4+0 & -16+4+12\\12+0-12 & -3+0+4 & -12+0+12\\12-3-9 & -3+3+0 & -12+4+9\end{bmatrix}

=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

= I3

Hence proved.

Question 24.

(i) If\begin{bmatrix}1 & 1 & x\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0, find x.

Solution:

We have,

=>\begin{bmatrix}1 & 1 & x\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 1 & 0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+0+2x & 0+2+x & 2+1+0\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+2x & 2+x & 3\end{bmatrix}\begin{bmatrix}1\\1\\1\end{bmatrix}   = 0

=>\begin{bmatrix}1+2x+2+x+3\end{bmatrix}   = 0

=> [3x + 6] = 0

=> 3x = –6

=> x = –6/3

=> x = –2

Therefore, the value of x is –2.

(ii) If\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}\begin{bmatrix}1 & -3\\-2 & 4\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}   = 0, find x.

Solution:

We have,

=>\begin{bmatrix}2 & 3\\5 & 7\end{bmatrix}\begin{bmatrix}1 & -3\\-2 & 4\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

=>\begin{bmatrix}2-6 & -6+12\\5-14 & -15+28\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

=>\begin{bmatrix}-4 & 6\\-9 & 13\end{bmatrix}=\begin{bmatrix}-4 & 6\\-9 & x\end{bmatrix}

On comparing the above matrix we get,

x = 13

Therefore, the value of x is –13.

Question 25. If\begin{bmatrix}x & 4 & 1\end{bmatrix}\begin{bmatrix}2 & 1 & 2\\1 & 0 & 2\\0 & 2 & -4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0   , find x.

Solution:

We have,

=>\begin{bmatrix}x & 4 & 1\end{bmatrix}\begin{bmatrix}2 & 1 & 2\\1 & 0 & 2\\0 & 2 & -4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x+4+0 & x+0+2 & 2x+8-4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x+4 & x+2 & 2x+4\end{bmatrix}\begin{bmatrix}x\\4\\-1\end{bmatrix}=0

=>\begin{bmatrix}2x^2+4x+4x+8-2x-4\end{bmatrix}=0

=> 2x2 + 4x + 4x + 8 – 2x – 4 = 0

=> 2x2 + 6x + 4 = 0

=> 2x2 + 2x + 4x + 4 = 0

=> 2x (x + 1) + 4 (x + 1) = 0

=> (x + 1) (2x + 4) = 0

=> x = –1 or x = –2

Therefore, the value of x is –1 or –2.

Question 26. If\begin{bmatrix}1 & -1 & x\end{bmatrix}\begin{bmatrix}0 & 1 & -1\\2 & 1 & 3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix} = 0, find x.

Solution:

We have,

=>\begin{bmatrix}1 & -1 & x\end{bmatrix}\begin{bmatrix}0 & 1 & -1\\2 & 1 & 3\\1 & 1 & 1\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix} = 0

=>\begin{bmatrix}0-2+x & 1-1+x & -1-3+x\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0

=>\begin{bmatrix}-2+x & x & -4+x\end{bmatrix}\begin{bmatrix}0\\1\\1\end{bmatrix}=0

=>\begin{bmatrix}0+x-4+x\end{bmatrix}=0

=> 2x – 4 = 0

=> 2x = 4

=> x = 2

Therefore, the value of x is 2.

Question 27. If A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , then prove that A2 – A + 2I = 0.

Solution:

We have,

A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

A2 =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

=\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}

L.H.S. = A2 – A + 2I

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}-\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}+2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}-\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}1-3 & -2+2\\4-4 & -4+2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}-2 & 0\\0 & -2\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=\begin{bmatrix}-2+2 & 0+0\\0+0 & -2+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 28. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , then find λ so that A2 = 5A + λI.

Solution:

We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

We are given,

=> A2 = 5A + λI

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+λ\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}λ & 0\\0 & λ\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15+λ & 5+0\\-5+0 & 10+λ\end{bmatrix}

=>\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}=\begin{bmatrix}15+λ & 5\\-5 & 10+λ\end{bmatrix}

On comparing both sides, we get

=> 8 = 15 + λ

=> λ = –7

Therefore, the value of λ is –7.

Question 29. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} , show that A2 – 5A + 7I2 = 0.

Solution:

We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

L.H.S. = A2 – 5A + 7I2

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}

=\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 30. If A =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix} , show that A2 – 2A + 3I2 = 0.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}

A2 =\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}

=\begin{bmatrix}4-3 & 6+0\\-2+0 & -3+0\end{bmatrix}

=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}

L.H.S. = A2 – 2A + 3I2

=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}-2\begin{bmatrix}2 & 3\\-1 & 0\end{bmatrix}+3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1 & 6\\-2 & -3\end{bmatrix}-\begin{bmatrix}4 & 6\\-2 & 0\end{bmatrix}+\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}

=\begin{bmatrix}1-4+3 & 6-6+0\\-2+2+0 & -3+0+3\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 31. Show that the matrix A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} satisfies the equation A3 – 4A2 + A = 0.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

A2 =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix}

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}14+12 & 21+24\\8+7 & 12+14\end{bmatrix}

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}

L.H.S. = A3 – 4A2 + A

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}-4\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}+\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}26 & 45\\15 & 26\end{bmatrix}-\begin{bmatrix}28 & 48\\16 & 28\end{bmatrix}+\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}26-28+2 & 45-48+3\\15-16+1 & 26-28+2\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 32. Show that the matrix A =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix} is root of the equation A– 12A – I = 0

Solution:

We have,

A =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}

A2 =\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}

=\begin{bmatrix}25+36 & 15+21\\60+84 & 36+49\end{bmatrix}

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}

L.H.S. = A2 – 12A – I

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}-12\begin{bmatrix}5 & 3\\12 & 7\end{bmatrix}-\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}61 & 36\\144 & 85\end{bmatrix}-\begin{bmatrix}60 & 36\\144 & 84\end{bmatrix}-\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}61-60-1 & 36-36-0\\144-144-0 & 85-84-1\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 33. If A =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix} find A2 – 5A – 14I.

Solution:

We have,

A =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}

=\begin{bmatrix}9+20 & -15-10\\-12-8 & 20+4\end{bmatrix}

=\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}

A2 – 5A – 14I =\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}-5\begin{bmatrix}3 & -5\\-4 & 2\end{bmatrix}-14\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}29 & -25\\-20 & 24\end{bmatrix}-\begin{bmatrix}15 & -25\\-20 & 10\end{bmatrix}-\begin{bmatrix}14 & 0\\0 & 14\end{bmatrix}

=\begin{bmatrix}29-15-14 & -25+25+0\\-20+20+0 & 24-10-14\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

Question 34. If A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix} , find A2 – 5A + 7I = 0. Use this to find A4.

Solution:

We have,

A =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

A2 =\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}

=\begin{bmatrix}9-1 & 3+2\\-3-2 & -1+4\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}

L.H.S. = A2 – 5A + 7I = 0

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-5\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}+7\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}8 & 5\\-5 & 3\end{bmatrix}-\begin{bmatrix}15 & 5\\-5 & 10\end{bmatrix}+\begin{bmatrix}7 & 0\\0 & 7\end{bmatrix}

=\begin{bmatrix}8-15+7 & 5-5+0\\-5+5+0 & 3-10+7\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Now we have A2 – 5A + 7I = 0

=> A2 = 5A – 7I

=> A4 = (5A – 7I) (5A – 7I)

=> A4 = 25A2 – 35AI – 35AI + 49I

=> A4 = 25A2 – 70AI + 49I

=> A4 = 25 (5A – 7I) – 70AI + 49I

=> A4 = 125A – 175I – 70A + 49I

=> A4 = 55A – 126I

=> A4 =55\begin{bmatrix}3 & 1\\-1 & 2\end{bmatrix}-126\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=> A4 =\begin{bmatrix}165 & 55\\-55 & 110\end{bmatrix}-\begin{bmatrix}126 & 0\\0 & 126\end{bmatrix}

=> A4 =\begin{bmatrix}165-126 & 55-0\\-55-0 & 110-126\end{bmatrix}

=> A4 =\begin{bmatrix}39 & 55\\-55 & -16\end{bmatrix}

Question 35. If A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix} , find k such that A= kA – 2I2.

Solution:

We have,

A =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

A2 =\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}

=\begin{bmatrix}9-8 & -6+4\\12-8 & -8+4\end{bmatrix}

=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}

We are given,

=> A= kA – 2I2

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = k\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}-2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = \begin{bmatrix}3k & -2k\\4k & -2k\end{bmatrix}-\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

=>\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix} = \begin{bmatrix}3k-2 & -2k-0\\4k-0 & -2k-2\end{bmatrix}

On comparing both sides, we get

=> 3k – 2 = 1

=> 3k = 3

=> k = 1

Therefore, the value of k is 1.

Question 36. If A =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix} , find k such that A2 – 8A + kI = 0.

Solution:

We have,

A =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}

A2 =\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}

=\begin{bmatrix}1-0 & 0+0\\-1-7 & 0+49\end{bmatrix}

=\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}

We are given,

=> A2 – 8A + kI = 0

=>\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}-8\begin{bmatrix}1 & 0\\-1 & 7\end{bmatrix}+k\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=0

=>\begin{bmatrix}1 & 0\\-8 & 49\end{bmatrix}-\begin{bmatrix}8 & 0\\-8 & 56\end{bmatrix}+\begin{bmatrix}k & 0\\0 & k\end{bmatrix}=0

=>\begin{bmatrix}1-8+k & 0-0+0\\-8+8+0 & 49-56+k\end{bmatrix} = 0

=>\begin{bmatrix}-7+k & 0\\0 & -7+k\end{bmatrix} = 0

On comparing both sides, we get

=> –k + 7 = 0

=> k = 7

Therefore, the value of k is 7.

Question 37. If A =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix} and f(x) = x2 – 2x – 3, show that f(A) = 0.

Solution:

We have,

A =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix} and f(x) = x2 – 2x – 3

A2 =\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}

=\begin{bmatrix}1+4 & 2+2\\2+2 & 4+1\end{bmatrix}

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}

L.H.S. = f(A) = A2 – 2A – 3I2

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}-2\begin{bmatrix}1 & 2\\2 & 1\end{bmatrix}+3\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}5 & 4\\4 & 5\end{bmatrix}-\begin{bmatrix}2 & 4\\4 & 2\end{bmatrix}+\begin{bmatrix}3 & 0\\0 & 3\end{bmatrix}

=\begin{bmatrix}5-2-3 & 4-4-0\\4-4-0 & 5-2-3\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 38. If A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} and I =\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix} , find λ, μ so that A2 = λA + μI.

Solution:

We have,

A =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

A2 =\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix}

=\begin{bmatrix}4+3 & 6+6\\2+2 & 3+4\end{bmatrix}

=\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}

We are given,

=> A2 = λA + μI

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=λ\begin{bmatrix}2 & 3\\1 & 2\end{bmatrix} + μ\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ & 3λ\\λ & 2λ\end{bmatrix} + \begin{bmatrix}μ & 0\\0 & μ\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ+μ & 3λ+0\\λ+0 & 2λ+μ\end{bmatrix}

=>\begin{bmatrix}7 & 12\\4 & 7\end{bmatrix}=\begin{bmatrix}2λ+μ & 3λ\\λ & 2λ+μ\end{bmatrix}

On comparing both sides, we get,

=> 2λ + μ = 7 and λ = 4

=> 2(4) + μ = 7

=> μ = 7 – 8

=> μ = –1

Therefore, the value of λ is 4 and μ is –1.

Question 39. Find the value of x for which the matrix product\begin{bmatrix}2 & 0 & 7\\0 & 1 & 0\\1 & -2 & 1\end{bmatrix}\begin{bmatrix}-x & 14x & 7x\\0 & 1 & 0\\x & -4x & -2x\end{bmatrix} equals an identity matrix.

Solution:

We have,

=>\begin{bmatrix}2 & 0 & 7\\0 & 1 & 0\\1 & -2 & 1\end{bmatrix}\begin{bmatrix}-x & 14x & 7x\\0 & 1 & 0\\x & -4x & -2x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}-2x+0+7x & 28x+0-28x & 14x+0-14x\\0+0+0 & 0+1-0 & 0+0-0\\-x-0+x & 14x-2-4x & 7x-0-2x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}5x & 0 & 0\\0 & 1 & 0\\0 & 10x-2 & 5x\end{bmatrix}=\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

On comparing both sides, we get,

=> 5x = 1

=> x = 1/5

Therefore, the value of x is 1/5.

Question 40. Solve the following matrix equations:

(i)\begin{bmatrix}x & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}x & 1\end{bmatrix}\begin{bmatrix}1 & 0\\-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x-2 & 0-3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x-2 & -3\end{bmatrix}\begin{bmatrix}x \\5\end{bmatrix}=0

=>\begin{bmatrix}x^2-2x-15\end{bmatrix}=0

=> x2 – 2x – 15 = 0

=> x2 – 5x + 3x – 15 = 0

=> x (x – 5) + 3 (x – 5) = 0

=> (x – 5) (x + 3) = 0

=> x = 5 or –3

Therefore, the value of x is 5 or –3.

(ii)\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1&0&2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}1 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 0\\2 & 0 & 1\\1&0&2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}1+4+1 & 2+0+0 & 0+2+2\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}6 & 2 & 4\end{bmatrix}\begin{bmatrix}0 \\2\\x\end{bmatrix}=0

=>\begin{bmatrix}0+4+4x\end{bmatrix}=0

=> 4 + 4x = 0

=> 4x = –4

=> x = –1

Therefore, the value of x is –1.

(iii)\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}x & -5 & -1\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x-0-2 & 0-10-0 & 2x-5-3\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x-2 & -10 & 2x-8\end{bmatrix}\begin{bmatrix}x \\4\\1\end{bmatrix}=0

=>\begin{bmatrix}x^2-2x-40+2x-8\end{bmatrix}=0

=> x2 – 48 = 0

=> x2 = 48

=> x = ±4√3

Therefore, the value of x is ±4√3.

(iv)\begin{bmatrix}2x & 3\end{bmatrix}\begin{bmatrix}1 & 2\\-3 & 0\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

Solution:

We have,

=>\begin{bmatrix}2x & 3\end{bmatrix}\begin{bmatrix}1 & 2\\-3 & 0\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

=>\begin{bmatrix}2x-9 & 4x\end{bmatrix}\begin{bmatrix}x\\8 \end{bmatrix}=0

=>\begin{bmatrix}x(2x-9)+32x\end{bmatrix}=0

=>\begin{bmatrix}2x^2-9x+32x\end{bmatrix}=0

=> 2x2 + 23x = 0

=> x (2x + 23) = 0

=> x = 0 or x = –23/2

Therefore, the value of x is 0 or –23/2.

Question 41. If A =\begin{bmatrix}1 & 2 & 0\\3 & -4 & 5\\0&-1&3\end{bmatrix} , compute A2 – 4A + 3I3.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

A2 =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}1+6+0 & 2-8-0 & 0+10+0\\3-12+0 & 6+16-5 & 0-20+15\\0-3+0&0+4-3&0-5+9\end{bmatrix}

=\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}

So, A2 – 4A + 3I3 =\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}-4\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}+3\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0&0&1\end{bmatrix}

=\begin{bmatrix}7 & -6 & 10\\-9 & 17 & -5\\-3&1&4\end{bmatrix}-\begin{bmatrix}4 & 0 & 8\\0 & 8 & 4\\8&0&12\end{bmatrix}+\begin{bmatrix}3 & 0 & 0\\0 & 3 & 0\\0&0&3\end{bmatrix}

=\begin{bmatrix}7-4+3 & -6-8+0 & 10-0+0\\-9-12+0 & 17+16+3 & -5-20+0\\-3-0+0&1+4+0&4-12+3\end{bmatrix}

=\begin{bmatrix}6 & -14 & 10\\-21 & 36 & -25\\-3&5&-5\end{bmatrix}

Question 42. If f(x) = x2 – 2x, find f(A), where A =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix} .

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix} and f(x) = x2 – 2x

A2 =\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}

=\begin{bmatrix}0+4+0 & 0+5+4 & 0+0+6\\0+20+0 & 4+25+0 & 8+0+0\\0+8+0&0+10+6&0+0+9\end{bmatrix}

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}

So, f(A) = A2 – 2A

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}-2\begin{bmatrix}0 & 1 & 2\\4 & 5 & 0\\0&2&3\end{bmatrix}

=\begin{bmatrix}4 & 9 & 6\\20 & 29 & 8\\8&16&9\end{bmatrix}-\begin{bmatrix}0 & 2 & 4\\8 & 10 & 0\\0&4&6\end{bmatrix}

=\begin{bmatrix}4-0 & 9-2 & 6-4\\20-8 & 29-10 & 8-0\\8-0&16-4&9-6\end{bmatrix}

=\begin{bmatrix}4 & 7 & 2\\12 & 19 & 0\\8&12&3\end{bmatrix}

Question 43. If f(x) = x3 + 4x2 – x, find f(A) where A =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix} .

Solution:

We have,

A =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix} and f(x) = x3 + 4x2 – x

A2 =\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}0+2+2 & 0-3-2 & 0+0+0\\0-6+0 & 2+9-0 & 4-0+0\\0-2+0&1+3-0&2-0+0\end{bmatrix}

=\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}0-10+0 & 4+15-0 & 8-0+0\\0+22+4 & -6-33-4 & -12+0+0\\0+8+2&-2-12-2&-4+0+0\end{bmatrix}

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}

Now, f(A) = A3 + 4A2 – A

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}+4\begin{bmatrix}4 & -5 & 0\\-6 & 11 & 4\\-2&4&2\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}-10 & 19 & 8\\26 & -43 & -12\\10&-16&-4\end{bmatrix}+\begin{bmatrix}16 & -20 & 0\\-24 & 44 & 16\\-8&16&8\end{bmatrix}-\begin{bmatrix}0 & 1 & 2\\2 & -3 & 0\\1&-1&0\end{bmatrix}

=\begin{bmatrix}-10+16-0 & 19-20-1 & 8+0-2\\26-24-2 & -43+44+3 & -12+16-0\\10-8-1&-16+16+1&-4+8+0\end{bmatrix}

=\begin{bmatrix}6 & -2 & 6\\0 & 4 & 4\\1&1&4\end{bmatrix}

Question 44. If A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix} , then show that A is a root of the polynomial f(x) = x3 – 6x2 + 7x +2.

Solution:

We have,

A =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix} and f(x) = x3 – 6x2 + 7x +2.

A2 =\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}1+0+4 & 0+0+0 & 2+0+6\\0+0+2 & 0+4+0 & 0+2+3\\2+0+6&0+0+0&4+0+9\end{bmatrix}

=\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}

A3 = A2. A

=\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}

=\begin{bmatrix}5+0+16 & 0+0+0 & 10+0+24\\2+0+10 & 0+8+0 & 4+4+15\\8+0+26&0+0+0&16+0+39\end{bmatrix}

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}

In order to show that A is a root of above polynomial, we need to prove that f(A) = 0.

Now, f(A) = A3 – 6A2 + 7A + 2I

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}-6\begin{bmatrix}5 & 0 & 8\\2 & 4 & 5\\8&0&13\end{bmatrix}+7\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2&0&3\end{bmatrix}+2\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0&0&1\end{bmatrix}

=\begin{bmatrix}21 & 0 & 34\\12 & 8 & 23\\34&0&55\end{bmatrix}-\begin{bmatrix}30 & 0 & 48\\12 & 24 & 30\\48&0&78\end{bmatrix}+\begin{bmatrix}7 & 0 & 14\\0 & 14 & 7\\14&0&21\end{bmatrix}+\begin{bmatrix}2 & 0 & 0\\0 & 2 & 0\\0&0&2\end{bmatrix}

=\begin{bmatrix}21-30+7+2 & 0-0+0+0 & 34-48+14+0\\12-12+0+0 & 8-24+14+2 & 23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

Hence proved.

Question 45. If A =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix} , prove that A2 – 4A – 5I = 0.

Solution:

We have,

A =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}

A2 =\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}

=\begin{bmatrix}1+4+4 & 2+2+4 & 2+4+2\\2+2+4 & 4+1+4 & 4+2+2\\2+4+2 & 4+2+2 & 4+4+1\end{bmatrix}

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}

Now, L.H.S. = A2 – 4A – 5I

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}-4\begin{bmatrix}1 & 2 & 2\\2 & 1 & 2\\2 & 2 & 1\end{bmatrix}-5\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}9 & 8 & 8\\8 & 9 & 8\\8 & 8 & 9\end{bmatrix}-\begin{bmatrix}4 & 8 & 8\\8 & 4 & 8\\8 & 8 & 4\end{bmatrix}-\begin{bmatrix}5 & 0 & 0\\0 & 5 & 0\\0 & 0 & 5\end{bmatrix}

=\begin{bmatrix}9-4-5 & 8-8-0 & 8-8-0\\8-8-0 & 9-4-5 & 8-8-0\\8-8-0 & 8-8-0 & 9-4-5\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 46. If A =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix} , show that A2 – 7A + 10I3 = 0.

Solution:

We have,

A =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}

A2 =\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}

=\begin{bmatrix}9+2+0 & 6+8+0 & 0+0+0\\3+4+0 & 2+16+0 & 0+0+0\\0+0+0 & 0+0+0 & 0+0+25\end{bmatrix}

=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}

Now, L.H.S. = A2 – 7A + 10I3

=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}-7\begin{bmatrix}3 & 2 & 0\\1 & 4 & 0\\0 & 0 & 5\end{bmatrix}+10\begin{bmatrix}1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}

=\begin{bmatrix}11 & 14 & 0\\7 & 18 & 0\\0 & 0 & 25\end{bmatrix}-\begin{bmatrix}21 & 14 & 0\\7 & 28 & 0\\0 & 0 & 35\end{bmatrix}+\begin{bmatrix}10 & 0 & 0\\0 & 10 & 0\\0 & 0 & 10\end{bmatrix}

=\begin{bmatrix}11-21+10 & 14-14+0 & 0-0+0\\7-7+0 & 18-28+10 & 0-0+0\\0-0+0 & 0-0+0 & 25-35+10\end{bmatrix}

=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}

= 0

= R.H.S.

Hence proved.

Question 47. Without using the concept of inverse of a matrix, find the matrix\begin{bmatrix}x & y\\z & u\end{bmatrix} such that,\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix} \begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

Solution:

We have,

=>\begin{bmatrix}5 & -7\\-2 & 3\end{bmatrix} \begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

=>\begin{bmatrix}5x-7z & 5y-7u\\-2x+3z & -2y+3u\end{bmatrix}=\begin{bmatrix}-16 & -6\\7 & 2\end{bmatrix}

On comparing both sides, we get,

5x – 7z = –16

5y – 7u = –6

–2x + 3z = 7

–2y + 3u = 2

On solving the above equations, we get

=> x = 1, y = –4, z = 3 and u = –2.

So, we get\begin{bmatrix}x & y\\z & u\end{bmatrix} =\begin{bmatrix}1 & -4\\3 & -2\end{bmatrix} .

Question 48. Find the matrix A such that

(i)\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}A=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}

Given equation is,

=>\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}A=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}x+a & y+b & z+c\\a+0 & b+0 & c+0\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

=>\begin{bmatrix}x+a & y+b & z+c\\a & b & c\end{bmatrix}=\begin{bmatrix}3 & 3 & 5\\1 & 0 & 1\end{bmatrix}

On comparing both sides, we get, a = 1, b = 0 and c = 1.

And x + 1 = 3 => x = 2

Also, y = 3 and

z + 1 = 5 => z = 4

So, we have A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}2 & 3 & 4\\1 & 0 & 1\end{bmatrix}

(ii)A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

Solution:

Let A =\begin{bmatrix}w & x\\y & z\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

=>\begin{bmatrix}w+4x & 2w+5x & 3w+6x\\y+4z & 2y+5z & 3y+6z\end{bmatrix}=\begin{bmatrix}7 & -8 & -9\\2 & 4 & 6\end{bmatrix}

On comparing both sides, we get,

w + 4x = 7

2w + 5x = –6

y + 4z = 2

2y + 5z = 4

On solving the above equations, we get

=> x = –2, y = 2, w = 1 and z = 0.

So, we get A =\begin{bmatrix}w & x\\y & z\end{bmatrix}=\begin{bmatrix}1 & -2\\2 & 0\end{bmatrix}

(iii)\begin{bmatrix}4\\1\\3\end{bmatrix}A=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\end{bmatrix}

Given equation is,

=>\begin{bmatrix}4\\1\\3\end{bmatrix}A=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

=>\begin{bmatrix}4\\1\\3\end{bmatrix}\begin{bmatrix}x & y & z\end{bmatrix}=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

=>\begin{bmatrix}4x & 4y & 4z\\x & y & z\\3x & 3y & 3z\end{bmatrix}=\begin{bmatrix}-4 & 8 & 4\\-1 & 2 & 1\\-3 & 6 & 3\end{bmatrix}

On comparing both sides, we get,

=> 4x = – 4, 4y = 8 and 4z = 4.

=> x = –1, y = 2 and z = 1.

So, we get A =\begin{bmatrix}-1 & 2 & 1\end{bmatrix}

(iv)A=\begin{bmatrix}2 & 1 & 3\end{bmatrix}\begin{bmatrix}-1 & 0 & -1\\-1 & 1 & 0\\0 & 1 &1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

Solution:

We have,

A=\begin{bmatrix}2 & 1 & 3\end{bmatrix}\begin{bmatrix}-1 & 0 & -1\\-1 & 1 & 0\\0 & 1 &1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-2-1+0 & 0+1+3 & -2+0+3\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-3 & 4 & 1\end{bmatrix}\begin{bmatrix}1\\0\\-1\end{bmatrix}

A =\begin{bmatrix}-3+0-1\end{bmatrix}

A =\begin{bmatrix}-4\end{bmatrix}

(v)\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}A=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}

Given equation is,

=>\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}A=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

=>\begin{bmatrix}2 & -1\\1 & 0\\-3 & -4\end{bmatrix}\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

=>\begin{bmatrix}2x-a & 2y-b & 2z-c\\x & y & z\\-3x+4a & -3y+4b & -3z+4c\end{bmatrix}=\begin{bmatrix}-1 & -8 & -10\\1 & -2 & -5\\9 & 22 & 15\end{bmatrix}

On comparing both sides, we get,

=> x = 1, y = –2 and z = –5

And also we have,

2x – a = –1

2y – b = –8

2z – c = –10

On solving these, we get,

=> a = 3, b = 4 and c = 0.

So, we get A =\begin{bmatrix}x & y & z\\a & b & c\end{bmatrix}=\begin{bmatrix}1 & -2 & -5\\3 & 4 & 0\end{bmatrix}

(vi)A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

Solution:

Let A =\begin{bmatrix}x & a\\y & b\\z & c\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

=>\begin{bmatrix}x & a\\y & b\\z & c\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

=>\begin{bmatrix}x+4a & 2x+5a & 3x+6a\\y+4b & 2y+5b & 3y+6b\\z+4c & 2z+5c & 3z+6c\end{bmatrix}=\begin{bmatrix}-7 & -8 & -9\\2 & 4 & 6\\11 & 10 & 9\end{bmatrix}

On comparing both sides, we get

x + 4a = –7 and 2x + 5a = –8

=> x = 1 and a = –2

y + 4b = 2 and 2y + 5b = 4

=> b = 0 and y = 2

z + 4c = 11 and 2z + 5c = 10

=> c = 4 and z = –5

So, we get A =\begin{bmatrix}1 & -2\\2 & 0\\-5 & 4\end{bmatrix}

Question 49. Find a 2 × 2 matrix A such thatA\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix} = 6I2.

Solution:

Let A =\begin{bmatrix}w & x\\y & z\end{bmatrix}

Given equation is,

=>A\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix} = 6I

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=6\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=>\begin{bmatrix}w & x\\y & z\end{bmatrix}\begin{bmatrix}1 & -2\\1 & 4\end{bmatrix}=\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix}

=>\begin{bmatrix}w+x & -2w+4x\\y+z & -2y+4z\end{bmatrix}=\begin{bmatrix}6 & 0\\0 & 6\end{bmatrix}

On comparing both sides, we get

w + x = 6 and –2w + 4x = 0

=> w = 4 and x = 2

y + z = 0 and –2y + 4z = 6

=> y = –1 and z = 1

So, we get A =\begin{bmatrix}w & x\\y & z\end{bmatrix}=\begin{bmatrix}4 & 2\\-1 & 1\end{bmatrix}

Question 50. If A =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix} , find A16.

Solution:

We have,

A =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}

A2 =\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\4 & 0\end{bmatrix}

=\begin{bmatrix}0+0 & 0+0\\0+0 & 0+0\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

A16 = A2 A2 A2 A2

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

=\begin{bmatrix}0 & 0\\0 & 0\end{bmatrix}

Question 51. If A =\begin{bmatrix}0 & -x\\x & 0\end{bmatrix} , B =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} and x2 = –1, then show that (A + B)2 = A2 + B2.

Solution:

We have,

A =\begin{bmatrix}0 & -x\\x & 0\end{bmatrix} , B =\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix} and x2 = –1

L.H.S. = (A + B)2

=\left(\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\right)^2

=\left(\begin{bmatrix}0+0 & -x+1\\x+1 & 0+0\end{bmatrix}\right)^2

=\left(\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}\right)^2

=\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}\begin{bmatrix}0 & -x+1\\x+1 & 0\end{bmatrix}

=\begin{bmatrix}0+(1-x)(1+x)& 0+0\\0+0 & (x+1)(1-x)+0\end{bmatrix}

=\begin{bmatrix}1-x^2 & 0\\0 & 1-x^2\end{bmatrix}

=\begin{bmatrix}1-(-1) & 0\\0 & 1-(-1)\end{bmatrix}

=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

R.H.S. = A2 + B2

=\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}\begin{bmatrix}0 & -x\\x & 0\end{bmatrix}+\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}

=\begin{bmatrix}0-x^2 & 0+0\\0+0 & -x^2+0\end{bmatrix}+\begin{bmatrix}0+1 & 0+0\\0+0 & 1+0\end{bmatrix}

=\begin{bmatrix}-x^2 & 0\\0 & -x^2\end{bmatrix}+\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}

=\begin{bmatrix}1-x^2 & 0\\0 & 1-x^2\end{bmatrix}

=\begin{bmatrix}1-(-1) & 0\\0 & 1-(-1)\end{bmatrix}

=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}

= L.H.S.

Hence proved.

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