RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices

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Here we provide RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter5
Exercise5.2
CategoryRD Sharma Solutions

Table of Contents

RD Sharma Class 12 Ex 5.2 Solutions Chapter 5 Algebra of Matrices

Question 1(i): Compute the following sum:\begin{bmatrix} 3 & -2\\ 1 & 4 \end{bmatrix} + \begin{bmatrix} -2 & 4\\ 1 & 3 \end{bmatrix} .

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 2×2.

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

=>\begin{bmatrix} 1 & 2 \\ 2 & 7 \end{bmatrix}

Question 1(ii): Compute the following sum:\begin{bmatrix} 2 & 1 & 3 \\ 0 & 3 & 5 \\ -1 & 2 & 5 \\ \end{bmatrix} + \begin{bmatrix} 1 & -2 & 3 \\ 2 & 6 & 1 \\ 0 & -3 & 1 \\ \end{bmatrix} .

Solution:

As the matrices are of the same dimensions, we can add them to get a matrix of the same dimensions which is 3×3.

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

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

Question 2: Let A =\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} , B =\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} and C =\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} . Find each of the following:

(i): 2A – 3B

Solution:

Both the matrices A and B are of the same order which is 2×2, hence the operation can be performed.

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

=> 3B =3\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 3 & 9 \\ -6 & 15 \end{bmatrix}

=> 2A – 3B =\begin{bmatrix} 4-3 & 8-9 \\ 6+6 & 4-15 \end{bmatrix}

=> 2A – 3B =\begin{bmatrix} 1 & -1 \\ 12 & -11 \end{bmatrix}

(ii): B – 4C

Solution:

Both the matrices B and C are of the same order which is 2×2, hence the operation can be performed.

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

=> 4C =4\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -8 & 20 \\ 12 & 16 \end{bmatrix}

=> B – 4C =\begin{bmatrix} 1+8 & 3-20 \\ -2-12 & 5-16 \end{bmatrix}

=> B – 4C =\begin{bmatrix} 9 & -17 \\ -14 & -11 \end{bmatrix}

(iii): 3A – C

Solution:

Both the matrices A and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}

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

=> 3A – C =\begin{bmatrix} 6+2 & 12-5 \\ 9-3 & 6-4 \end{bmatrix}

=> 3A – C =\begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}

(iv): 3A -2B + 3C

Solution:

The matrices A, B and C are of the same order which is 2×2, hence the operation can be performed.

=> 3A =3\begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}

=> 2B =2\begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix} = \begin{bmatrix} 2 & 6 \\ -4 & 10 \end{bmatrix}

=> 3C =3\begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix} = \begin{bmatrix} -6 & 15 \\ 9 & 12 \end{bmatrix}

=> 3A – 2B + 3C =\begin{bmatrix} 6-2-6 & 12-6+15 \\ 9+4+9 & 6-10+12 \end{bmatrix}

=> 3A – 2B + 3C =\begin{bmatrix} -2 & 21 \\ 22 & 8 \end{bmatrix}

Question 3: If A =\begin{bmatrix} 2 & 3 \\ 5 & 7 \end{bmatrix} , B =\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix} , C =\begin{bmatrix} -1 & 2 & 3\\ 2 & 1 & 0 \end{bmatrix} , find:

(i): A + B and B + C

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3.

B+C can be computed and is solved as follows:

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

=> B + C =\begin{bmatrix} -2 & 2 & 5 \\ 5 & 5 & 1 \end{bmatrix}

(ii): 2B + 3A and 3C – 4B

Solution:

A and B can not be added since A’s order is 2×2 which is different from B’s order which is 2×3, and thus 2B + 3A can not be calculated.

3C – 4B can be computed and is solved as follows:

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

=> 4B =4\begin{bmatrix} -1 & 0 & 2\\ 3 & 4 & 1 \end{bmatrix} = \begin{bmatrix} -4 & 0 & 8\\ 12 & 16 & 4 \end{bmatrix}

=> 3C – 4B =\begin{bmatrix} -3+4 & 6+0 & 9-8 \\ 6-12 & 3-16 & 10-4 \end{bmatrix}

=> 3C – 4B =\begin{bmatrix} 1 & 6 & 1 \\ -6 & -13 & 6 \end{bmatrix}

Question 4: Let A =\begin{bmatrix} -1 & 0 & 2 \\ 3 & 1 & 4 \end{bmatrix} , B =\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} and C =\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix} . Compute 2A – 3B + 4C.

Solution:

The result can be computed since A, B and C are of the same order which is 2×3.

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

=> 3B =3\begin{bmatrix} 0 & -2 & 5 \\ 1 & -3 & 1 \end{bmatrix} = \begin{bmatrix} 0 & -6 & 15 \\ 3 & -9 & 3 \end{bmatrix}

=> 4C =4\begin{bmatrix} 1 & -5 & 2 \\ 6 & 0 & -4 \end{bmatrix} = \begin{bmatrix} 4 & -20 & 8 \\ 24 & 0 & -16 \end{bmatrix}

=> 2A – 3B + 4C =\begin{bmatrix} -2-0+4 & 0+6-20 & 4-15+8 \\ 6-3+24 & 2+9+0 & 8-3-16 \end{bmatrix}

=> 2A – 3B + 4C =\begin{bmatrix} 2 & -14 & -3 \\ 27 & 11 & -11 \end{bmatrix}

Question 5: If A =diag(2, -5, 9), B = diag(1, 1, -4) and C = diag(-6, 3, 4), find:

(i): A – 2B

Solution:

In the given question A and B are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> A = diag(2, -5, 9)

=> 2B = 2. diag(1, 1, -4) = diag(2, 2, -8)

=> A – 2B = diag(2-2, -5-2, 9+8)

=> A – 2B = diag(0, -7, 17)

(ii): B + C – 2A

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> B = diag(1, 1, -4)

=> C = diag(-6, 3, 4)

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> B + C – 2A = diag(1-6-4, 1+3+10, -4+4-18)

=> B + C – 2A = diag(-9, 14, -18)

(iii): 2A + 3B – 5C

Solution:

In the given question A, B and C are diagonal matrices of the order 3×3, thus the only non-zero elements are present in the diagonal.

=> 2A = 2. diag(2, -5, 9) = diag(4, -10, 18)

=> 3B = 3. diag(1, 1, -4) = diag(3, 3, -12)

=> 5C = 5. diag(-6, 3, 4) = diag(-30, 15, 20)

=> 2A + 3B – 5C = diag(4+3+30, -10+3-15, 18-12-20)

=> 2A + 3B – 5C = diag(37, -22, -14)

Question 6: Given the matrices A =\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} , B =\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix} and C =\begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix} . Verify that (A + B) + C = A + (B + C).

Solution:

Given L.H.S :

=> (A + B) = \begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix}

=> (A + B) =\begin{bmatrix} 2+9 & 1+7 & 1-1 \\ 3+3 & -1+5 & 0+4 \\ 0+2 & 2+1 & 4+6 \\ \end{bmatrix}

=> (A + B) =\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 11 & 8 & 0 \\ 6 & 4 & 4 \\ 2 & 3 & 10 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 11+2 & 8-4 & 0+3 \\ 6+1 & 4-1 & 4+0 \\ 2+9 & 3+4 & 10+5 \\ \end{bmatrix}

=> (A + B) + C =\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}

Given R.H.S :

=> (B + C) =\begin{bmatrix} 9 & 7 & -1 \\ 3 & 5 & 4 \\ 2 & 1 & 6 \\ \end{bmatrix} + \begin{bmatrix} 2 & -4 & 3 \\ 1 & -1 & 0 \\ 9 & 4 & 5 \\ \end{bmatrix}

=> (B + C) =\begin{bmatrix} 9+2 & 7-4 & -1+3 \\ 3+1 & 5-1 & 4+0 \\ 2+9 & 1+4 & 6+5 \\ \end{bmatrix}

=> (B + C) =\begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 2 & 1 & 1 \\ 3 & -1 & 0 \\ 0 & 2 & 4 \\ \end{bmatrix} + \begin{bmatrix} 11 & 3 & 2 \\ 4 & 4 & 4 \\ 11 & 5 & 11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 2+11 & 1+3 & 1+2 \\ 3+4 & -1+4 & 0+4 \\ 0+11 & 2+5 & 4+11 \\ \end{bmatrix}

=> A + (B + C) =\begin{bmatrix} 13 & 4 & 3 \\ 7 & 3 & 4 \\ 11 & 7 & 15 \\ \end{bmatrix}

Hence R.H.S = L.H.S has been verified.

Question 7: Find the matrices X and Y, if X + Y =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} and X – Y =\begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} .

Solution:

We know that (X + Y) + (X – Y) = 2X.

=> (X + Y) + (X – Y) =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}

=> 2X =\begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9-1 \end{bmatrix}

=> 2X =\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}

=> X =\dfrac {1}{2}\begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}

=> X =\begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}

Now Y = (X + Y) – X

=> Y =\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}

=> Y =\begin{bmatrix} 5-4 & 2-4 \\ 0-0 & 9-4 \end{bmatrix}

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

Question 8: Find X, if Y =\begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix} and 2X + Y =\begin{bmatrix} 1 & 0 \\ -3 & 2 \end{bmatrix} .

Solution:

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

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

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

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

=> 2X =\begin{bmatrix} -2 & -2 \\ -4 & -2 \end{bmatrix}

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

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

Question 9: Find matrices X and Y, if 2X – Y =\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix} and X + 2Y =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix} .

Solution:

We know that 2 (2X – Y) + (X + 2Y) = 4X – 2Y + X + 2Y = 5X .

=> 2 (2X – Y) =2\begin{bmatrix} 6 & -6 & 0\\ -4 & 2 & 1\end{bmatrix}

=> 2 (2X -Y) =\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix}

=> 2 (2X – Y) + (X + 2Y) =\begin{bmatrix} 12 & -12 & 0\\ -8 & 4 & 2\end{bmatrix} + \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=> 5X =\begin{bmatrix} 12+3 & -12+2 & 0+5\\ -8-2 & 4+1 & 2-7\end{bmatrix}

=> 5X =\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}

=> X =\dfrac{1}{5}\begin{bmatrix} 15 & -10 & 5\\ -10 & 5 & -5\end{bmatrix}

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

As (X + 2Y) =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=>\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix} + 2Y = \begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix}

=> 2Y =\begin{bmatrix} 3 & 2 & 5\\ -2 & 1 & -7\end{bmatrix} -\begin{bmatrix} 3 & -2 & 1\\ -2 & 1 & -1\end{bmatrix}

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

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

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

Question 10: If X – Y =\begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix} and X + Y =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} , find X and Y.

Solution:

We know that (X + Y) + (X – Y) = 2X.

=> 2X =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}

=> 2X =\begin{bmatrix} 1+3 & 1+5 & 1+1\\ 1-1 & 1+1 & 0+4 \\1+11 & 0+8 & 0+0 \end{bmatrix}

=> 2X =\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}

=> X =\dfrac {1}{2}\begin{bmatrix} 4 & 6 & 2\\ 0 & 2 & 4 \\12 & 8 & 0 \end{bmatrix}

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

Also (X + Y) – (X -Y) = 2Y.

=> 2Y =\begin{bmatrix} 3 & 5 & 1\\ -1 & 1 & 4 \\11 & 8 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 1\\ 1 & 1 & 0 \\1 & 0 & 0 \end{bmatrix}

=> 2Y =\begin{bmatrix} 3-1 & 5-1 & 1-1\\ -1-1 & 1-1 & 4-0 \\11-1 & 8-0 & 0-0 \end{bmatrix}

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

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

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

Question 11: Find matrix A, if\begin{bmatrix} 1 & 2 & -1\\ 0 & 4 & 9 \end{bmatrix} + A =\begin{bmatrix} 9 & -1 & 4\\ -2 & 1 & 3 \end{bmatrix} .

Solution:

Given,\begin{bmatrix} 1 & 2 & -1\\ 0 & 4 & 9 \end{bmatrix} + A =\begin{bmatrix} 9 & -1 & 4\\ -2 & 1 & 3 \end{bmatrix} .

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

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

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

Question 12: If A =\begin{bmatrix} 9 & 1\\ 7 & 8 \end{bmatrix} , B =\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix} , find C such that 5A + 3B + 2C is a null matrix.

Solution:

Given 5A + 3B + 2C =O, where O is the null matrix.

=> 2C = O – 5A – 3B.

=> 5A =5\begin{bmatrix} 9 & 1\\ 7 & 8 \end{bmatrix} = \begin{bmatrix} 45 & 5\\ 35 & 40 \end{bmatrix}

=> 3B =3\begin{bmatrix} 1 & 5\\ 7 & 12 \end{bmatrix} = \begin{bmatrix} 3 & 15\\ 21 & 36 \end{bmatrix}

=> 2C =\begin{bmatrix} 0 & 0\\ 0 & 0 \end{bmatrix} - \begin{bmatrix} 45 & 5\\ 35 & 40 \end{bmatrix} - \begin{bmatrix} 27 & 3\\ 21 & 24 \end{bmatrix}

=> 2C =\begin{bmatrix} 0-45-3 & 0-5-15\\ 0-35-21 & 0-40-36 \end{bmatrix}

=> 2C =\begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}

=> C =\dfrac{1}{2}\begin{bmatrix} -48 & -20\\ -56 & -76 \end{bmatrix}

=> C =\begin{bmatrix} -24 & -10\\ -28 & -38 \end{bmatrix}

Question 13: If A =\begin{bmatrix} 2 & -2\\ 4 & 2 \\ -5 & 1 \end{bmatrix} , B =\begin{bmatrix} 8 & 0\\ 4 & -2 \\ 3 & 6 \end{bmatrix} , find matrix X such that 2A + 3X = 5B.

Solution:

Given 2A + 3X = 5B.

=> 3X = 5B – 2A.

=> 5B =5\begin{bmatrix} 8 & 0\\ 4 & -2 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 40 & 0\\ 20 & -10 \\ 15 & 30 \end{bmatrix}

=> 2A =2\begin{bmatrix} 2 & -2\\ 4 & 2 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} 4 & -4\\ 8 & 4 \\ -10 & 2 \end{bmatrix}

=> 3X =\begin{bmatrix} 40 & 0\\ 20 & -10 \\ 15 & 30 \end{bmatrix} - \begin{bmatrix} 4 & -4\\ 8 & 4 \\ -10 & 2 \end{bmatrix}

=> 3X =\begin{bmatrix} 40-4 & 0+4\\ 20-8 & -10-4 \\ 15+10 & 30-2 \end{bmatrix}

=> 3X =\begin{bmatrix} 36 & 4\\ 12 & -14 \\ 25 & 28 \end{bmatrix}

=> X =\dfrac{1}{3}\begin{bmatrix} 36 & 4\\ 12 & -14 \\ 25 & 28 \end{bmatrix}

=> X =\begin{bmatrix} 12 & \dfrac{4}{3}\\\\ 4 & \dfrac{-14}{3} \\\\ \dfrac{25}{3} & \dfrac{28}{3} \end{bmatrix}

Question 14: If A =\begin{bmatrix} 1 & -3 & 2\\ 2 & 0 & 2\\ \end{bmatrix} and B =\begin{bmatrix} 2 & -1 & -1\\ 1 & 0 & -1\\ \end{bmatrix} , find the matrix C such that A + B + C is a zero matrix.

Solution:

Given that A + B + C = O, where O is a null matrix.

=> C = O – A – B.

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

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

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

Question 15(i): Find x, y satisfying the matrix equation\begin{bmatrix} x-y & 2 & -2\\ 4 & x & 6\\ \end{bmatrix} + \begin{bmatrix} 3 & -2 & 2\\ 1 & 0 & -1\\ \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0\\ 5 & 2x+y & 5\\ \end{bmatrix}

Solution:

Given that,

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

We can arrive at 2 equations from the above matrix equation.

=> x – y + 3 = 6

=> x – y = 3 ……(eq.1)

=> x + 0 = 2x + y

=> -x = y ……….(eq.2)

Solving (eq.1) and (eq.2) for x and y.

=> 2x = 3

=> x = 3/2

Substitute x in (eq.2)

=> y = -3/2

Question 15(ii): Find x, y and z satisfying the matrix equation\begin{bmatrix} x & y+2 & z-3\\ \end{bmatrix} + \begin{bmatrix} y & 4 & 5\\ \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12\\ \end{bmatrix}

Solution:

Given that,

\begin{bmatrix} x & y+2 & z-3\\ \end{bmatrix} + \begin{bmatrix} y & 4 & 5\\ \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12\\ \end{bmatrix}

=>\begin{bmatrix} x+y & y+2+4 & z-3+5\\ \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12\\ \end{bmatrix}

=>\begin{bmatrix} x+y & y+6 & z+2\\ \end{bmatrix} = \begin{bmatrix} 4 & 9 & 12\\ \end{bmatrix}

We can arrive at 3 equations from the above matrix equation.

=> x + y = 4 ……(eq.1)

=> y + 6 = 9 ……(eq.2)

=> z + 2 = 12 ….(eq.3)

From (eq.2),

=> y = 9 – 6

=> y = 3

From (eq.3),

=> z = 12 – 2

=> z = 10

Substitute the value of y in (eq.1),

=> x + 3 = 4

=> x = 4 – 3

=> x = 1

Question 15(iii): Find x and y satisfying the matrix equation x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = O.

Solution:

Given that,

x\begin{bmatrix} 2 \\ 1 \end{bmatrix} + y\begin{bmatrix} 3 \\ 5 \end{bmatrix} + \begin{bmatrix} -8 \\ -11 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}

We can arrive at 2 equations from the above matrix equation.

=> 2x + 3y – 8 = 0

=> 2x + 3y = 8 …….(eq.1)

=> x + 5y -11 = 0

=> x + 5y = 11 …….(eq.2)

Solving for x and y , (eq.1) – 2.(eq.2),

=> 2x -2x + 3y – 10y = 8 – 22

=> -7y = -14

=> y = 2

Substitute y in (eq.2),

=> x + 5(2) = 11

=> x = 11 – 10

=> x = 1

Question 16: If2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix} , find x and y.

Solution:

Given that,

2\begin{bmatrix} 3 & 4 \\ 5 & x \end{bmatrix} + \begin{bmatrix} 1 & y \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 10 & 5 \end{bmatrix}

We can arrive at 2 equations from the above matrix equation.

=> 2x + 1 = 5…….(eq.1)

=> 8 + y = 0……..(eq.2)

Solving for x,

=> 2x = 5 – 1

=> 2x = 4

=> x = 2

Solving for y,

=> y = -8

Question 17: Find the value of\lambda , a non-zero scalar, if\lambda \begin{bmatrix} 1 & 0 & 2\\ 3 & 4 & 5 \end{bmatrix} + 2 \begin{bmatrix} 1 & 2 & 3\\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10\\ 4 & 2 & 14 \end{bmatrix}

Solution:

Given that,

\lambda \begin{bmatrix} 1 & 0 & 2\\ 3 & 4 & 5 \end{bmatrix} + 2 \begin{bmatrix} 1 & 2 & 3\\ -1 & -3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 4 & 10\\ 4 & 2 & 14 \end{bmatrix}

=> We can arrive at several equations to solve for\lambda however lets take one.

=>\lambda + 2 = 4

=>\lambda = 2

If we substitute\lambda = 2 , in the matrix we see that the equation remains consistent.

Hence,\lambda = 2 .

Question 18(i): Find a matrix X such that 2A + B + X = O, where A =\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix} , B =\begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix} .

Solution:

Given that, 2A + B + X = O.

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

=> X = O – 2A – B

=> X =\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} - \begin{bmatrix} -2 & 4 \\ 6 & 8 \end{bmatrix} - \begin{bmatrix} 3 & -2 \\ 1 & 5 \end{bmatrix}

=> X =\begin{bmatrix} 0+2-3 & 0-4+2 \\ 0-6-1 & 0-8-5 \end{bmatrix}

=> X =\begin{bmatrix} -1 & -2 \\ -7 & -13 \end{bmatrix}

Question 18(ii): If A =\begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} and B =\begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} , then find the matrix X of order 3×2 such that 2A + 3X = 5B.

Solution:

Given that 2A + 3X = 5B.

=> 3X = 5B – 2A.

=> 5B =5\begin{bmatrix} 2 & -2 \\ 4 & 2 \\ -5 & 1 \end{bmatrix} = \begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix}

=> 2A =2\begin{bmatrix} 8 & 0 \\ 4 & -2 \\ 3 & 6 \end{bmatrix} = \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}

=> 3X =\begin{bmatrix} 10 & -10 \\ 20 & 10 \\ -25 & 5 \end{bmatrix} - \begin{bmatrix} 16 & 0 \\ 8 & -4 \\ 6 & 12 \end{bmatrix}

=> 3X =\begin{bmatrix} 10-16 & -10-0 \\ 20-8 & 10+4 \\ -25-6 & 5-12 \end{bmatrix}

=> 3X =\begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}

=> X =\dfrac{1}{3}\begin{bmatrix} -6 & -10 \\ 12 & 14 \\ -31 & -7 \end{bmatrix}

=> X =\begin{bmatrix} -2 & \dfrac{-10}{3} \\\\ 4 & \dfrac{14}{3} \\\\ \dfrac{-31}{3} & \dfrac{-7}{3} \end{bmatrix}

Question 19(i): Find x, y, z and t, if 3\begin{bmatrix} x & y \\ z & t \end{bmatrix} = \begin{bmatrix} x & 6 \\ -1 & 2t \end{bmatrix} + \begin{bmatrix} 4 & x+y \\ z+t & 3 \end{bmatrix} .

Solution:

Given that,

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

We can arrive at 4 different equations from the above matrix equation,

=> 3x = x + 4 …………(eq.1)

=> 3y = 6 + x + y ….(eq.2)

=> 3z = -1 + z + t …(eq.3)

=> 3t = 2t + 3 ………..(eq.4)

From (eq.1),

=> 2x = 4

=> x = 2

Substitute x=2 in (eq.2),

=> 3y = 6 + 2 + y

=> 2y = 8

=> y = 4

From (eq.4),

=> t = 3

Substitute t=3 in (eq.3),

=> 3z = -1 + z + 3

=> 2z = 2

=> z = 1

Question 19(ii): Find x, y, z and t, if 2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ 15 & 14 \end{bmatrix} .

Solution:

Given that,

2\begin{bmatrix} x & 5 \\ 7 & y-3 \end{bmatrix} + \begin{bmatrix} 3 & 4 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 7 & 14 \\ 15 & 14 \end{bmatrix}

We can arrive at 2 equations from the above matrix equation,

=> 2x + 3 = 7 ………………..(eq.1)

=> 2 (y – 3) + 2 = 14 ….(eq.2)

From (eq.1),

=> 2x = 7 – 3

=> 2x = 4

=> x = 2

From (eq.2),

=> 2y – 6 + 2 = 14

=> 2y = 14 + 4

=> 2y = 18

=> y = 9

Question 20: If X and Y are 2×2 matrices, then solve the following matrix equations for X and Y, 2X + 3Y =\begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} , 3X + 2Y =\begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} .

Solution:

Let 2X + 3Y =\begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} be (eq.1) and let 3X + 2Y =\begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} , be (eq.2) .

=> 2(2X + 3Y) – 3(3X + 2Y) = 4X + 6Y – 9X – 6Y = -5X.

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

=> -5X =\begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix} - \begin{bmatrix} -6 & 6 \\ 3 & -15 \end{bmatrix}

=> -5X =\begin{bmatrix} 4+6 & 6-6 \\ 8-3 & 0+15 \end{bmatrix}

=> -5X =\begin{bmatrix} 10 & 0 \\ 5 & 15 \end{bmatrix}

=> 5X =\begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix}

=> X =\dfrac{1}{5}\begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix}

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

Substitute the matrix X in (eq.1),

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

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

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

=> 3Y =\begin{bmatrix} 6 & 3 \\ 6 & 6 \end{bmatrix}

=> Y =\dfrac{1}{3}\begin{bmatrix} 6 & 3 \\ 6 & 6 \end{bmatrix}

=> Y =\begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}

Question 21: In a certain city there are 30 colleges. Each college has 15 peons, 6 clerks, 1 typist and 1 section officer. Express the given information as a column matrix. Using scalar multiplication, find the total number of posts of each kind of in all the colleges.

Solution:

Let the different posts in each college be represented as :\begin{bmatrix} 15 \\ 6\\ 1\\ 1 \end{bmatrix}

Now the total posts will be computed as follows:30\begin{bmatrix} 15 \\ 6\\ 1\\ 1 \end{bmatrix} = \begin{bmatrix} 450 \\ 180\\ 30\\ 30 \end{bmatrix}

Question 22: The monthly incomes of Aryan and Babban are in the ratio 3 : 4 and their monthly expenditures are in the ratio 5 : 7. If each saves 15000 per month, find their monthly incomes using the matrix method.

Solution:

The problem can be solved by considering two matrices, one for expenditure and one for income.

=> The income matrix is: \begin{bmatrix} 3x \\ 4x\\ \end{bmatrix} where x is a constant.

=> The expenditure matrix is:\begin{bmatrix} 5y \\ 7y \end{bmatrix} where y is a constant.

=>\begin{bmatrix} 3x \\ 4x\\ \end{bmatrix}-\begin{bmatrix} 5y \\ 7y \end{bmatrix} = \begin{bmatrix} 15000\\15000 \end{bmatrix}

We arrive at 2 equations from the above matrix equation.

=> 3x – 5y = 15000……..(eq.1)

=> 4x – 7y = 15000……..(eq.2)

Solving for y by 4(eq.1) – 3(eq.2),

=> 12x – 20y – 12x + 21y = 4(15000) – 3(15000)

=> y = 15000

Substitute the value of y in (eq.1),

=> 3x = 15000 + 5(15000)

=> 3x = 15000 + 75000

=> 3x = 90000

=> x = 30000

=> Their incomes and expenditures are,

=> 3x = 3(30000) = 90000 and 5y = 5(15000) = 75000

=> 4x = 4(30000) = 120000 and 7y = 7(15000) = 105000

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