RD Sharma Class 12 Ex 8.2 Solutions Chapter 8 Solution of Simultaneous Linear Equations

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Here we provided RD Sharma Class 12 Ex 8.2 Solutions Chapter 8 Solution of Simultaneous Linear Equations for English medium students, Which will very helpful for every student in their exams. Students can download the RD Sharma Class 12 Ex 8.2 Solutions Chapter 8 Solution of Simultaneous Linear Equations book pdf download. Now you will get step-by-step solutions to each question.

TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter8
Exercise8.2
CategoryRD Sharma Solutions

RD Sharma Class 12 Ex 8.2 Solutions Chapter 8 Solution of Simultaneous Linear Equations

Solve the following systems of homogeneous linear equations by matrix method:

Question 1. 

2x – y + z = 0

3x + 2y – z = 0

x + 4y + 3z = 0

Solution:

Given

2x – y + z = 0

3x + 2y – z = 0

X + 4y + 3z = 0

The system can be written as

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

A X = 0

Now, |A| = 2(6 + 4) + 1(9 + 1) + 1(12 – 2)

|A| = 2(10) + 10 + 10

|A| = 40 ≠ 0

Since, |A|≠ 0, hence x = y = z = 0 is the only solution of this homogeneous equation.

Question 2. 

2x – y + 2z = 0

5x + 3y – z = 0

X + 5y – 5z = 0

Solution:

Given 2x – y + 2z = 0

5x + 3y – z = 0

X + 5y – 5z = 0

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

A X = 0

Now, |A| = 2(– 15 + 5) + 1(– 25 + 1) + 2(25 – 3)

|A| = – 20 – 24 + 44

|A| = 0

Thus, the system has infinite solutions

Let z = k

2x – y = – 2k

5x + 3y = k

\begin{bmatrix} 2 & -1\\ 3 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -2k\\ k \end{bmatrix}\\ AX=B\\ |A|=6+5=11\neq0 \ \ So,A^{-1}\ exist\\ Now\ adj\ A=\begin{bmatrix} 3&-5\\ 1&2 \end{bmatrix}^T=\begin{bmatrix} 3&1\\ -5&2 \end{bmatrix}\\ X=A^{-1}B=\frac{1}{|A|}(adjA)B=\frac{1}{11}\begin{bmatrix} 3&1\\ -5&2 \end{bmatrix}\begin{bmatrix} -2k\\ k\end{bmatrix}\\ X=\begin{bmatrix}\frac{-5k}{11}\\{\frac{12k}{11}}\end{bmatrix} \\Hence, X=\frac{-5k}{11},Y=\frac{12k}{11}and\ z=k

Question 3.

3x – y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z = 0

Solution:

Given: 

3x – y + 2z = 0

4x + 3y + 3z = 0

5x + 7y + 4z = 0

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

A X = 0

Now, |A| = 3(12 – 21) + 1(16 – 15) + 2(28 – 15)

|A| = – 27 + 1 + 26

|A| = 0

Hence, the system has infinite solutions

Let z = k

3x – y = – 2k

4x + 3y = – 3k

\begin{bmatrix} 3 & -1\\ 4 & 3 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -2k\\ -3k \end{bmatrix}\\ AX=B\\ |A|=9+4=13\neq0 \ \ So,A^{-1}\ exist\\ Now\ adj\ A=\begin{bmatrix} 3&-1\\ 4&3 \end{bmatrix}^T=\begin{bmatrix} 3&1\\ -4&3 \end{bmatrix}\\ X=A^{-1}B=\frac{1}{|A|}(adjA)B=\frac{1}{12}\begin{bmatrix} 3&1\\ -4&3 \end{bmatrix}\begin{bmatrix} -2k\\ -3k\end{bmatrix}\\ X=\begin{bmatrix}\frac{-9k}{13}\\{\frac{-k}{13}}\end{bmatrix}\\Hence, X=\frac{-9k}{13},Y=\frac{-k}{13}and\ z=k

Question 4. 

x + y – 6z = 0

x – y + 2z = 0

– 3x + y + 2z = 0

Solution:

Given: 

x + y – 6z = 0

x – y + 2z = 0

– 3x + y + 2z = 0

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

A X = 0

Now, |A| = 1(– 2 – 2) – 1(2 + 6) – 6(1 – 3)

|A| = – 4 – 8 + 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = 6k

x – y = – 2k

\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} 6k\\ -2k \end{bmatrix}\\ AX=B\\ |A|=-1-1=-2\neq0 \ \ So,A^{-1}\ exist\\ Now\ adj\ A=\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}^T=\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}\\ X=A^{-1}B=\frac{1}{|A|}(adjA)B=\frac{1}{-2}\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}\begin{bmatrix} 6k\\ -2k\end{bmatrix}\\ X=\frac{1}{-2}\begin{bmatrix}-6k+2k\\-6k-2k\end{bmatrix}\\ X=\begin{bmatrix}-4k\\-8k\end{bmatrix}\\ Hence, X=2k,\ Y=4k\ and\ Z=k

Question 5. Solve the system of homogeneous linear equations by matrix method:

x + y + z = 0

x – y – 5z = 0

x + 2y + 4z = 0

Solution:

Given:

x + y + z = 0

x – y – 5z = 0

x + 2y + 4z = 0

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

A X = 0

Now, |A| = 1(6) – 1(9) + 1(3)

|A| = 9 – 9

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = –k

x – y = 5k

\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} -k\\ 5k \end{bmatrix}\\ AX=B\\ |A|=-1-1=-2\neq0 \ \ So,A^{-1}\ exist\\ Now\ adj\ A=\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}^T=\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}\\ X=A^{-1}B=\frac{1}{|A|}(adjA)B=\frac{1}{-2}\begin{bmatrix} -1&-1\\ -1&1 \end{bmatrix}\begin{bmatrix} -k\\ 5k\end{bmatrix}\\ \begin{bmatrix}X\\Y\end{bmatrix}=\frac{1}{-2}\begin{bmatrix}k-5k\\k+5k\end{bmatrix}=\begin{bmatrix}2k\\-3k\end{bmatrix}\\ Hence, X=2k,\ Y=-3k\ and\ Z=k

Question 6. Solve the system of homogeneous linear equations by matrix method:

x + y – z = 0

x – 2y + z = 0

3x + 6y –5z = 0

Solution:

Given:

x + y – z = 0

x – 2y + z = 0

3x + 6y –5z = 0

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

A X = 0

Now, |A| = 1(4) – 1(–8) – 1(12)

|A| = 4 + 8  – 12

|A| = 0

Hence, the system has infinite solutions

Let z = k

x + y = –k

x – 2y = –k

\begin{bmatrix} 1 & 1\\ 1 & -2 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix}=\begin{bmatrix} k\\ -k \end{bmatrix}\\ AX=B\\ |A|=-2\neq0 \ \ So,A^{-1}\ exist\\ Now\ adj\ A=\begin{bmatrix} -2&-1\\ -1&1 \end{bmatrix}^T=\begin{bmatrix} -2&-1\\ -1&1 \end{bmatrix}\\ X=A^{-1}B=\frac{1}{|A|}(adjA)B=\frac{1}{-3}\begin{bmatrix} -2&-1\\ -1&1 \end{bmatrix}\begin{bmatrix} k\\ -k\end{bmatrix}\\ =\frac{-1}{3}\begin{bmatrix}-2k+k\\-2k\end{bmatrix}\\=\frac{-1}{3}\begin{bmatrix}-k\\-2k\end{bmatrix}=\begin{bmatrix}\frac{k}{3}\\\frac{2k}{3}\end{bmatrix}\\ Hence, X=\frac{k}{3},\ Y=\frac{2k}{3}\ and\ Z=k

Question 7. Solve the system of homogeneous linear equations by matrix method:

3x + y – 2z = 0

x + y + z = 0

x – 2y + z =0

Solution:

Given:

3x + y – 2z = 0

x + y + z = 0

x – 2y + z =0

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

A X = 0

Now, |A| = 3(3) – 1(0) – 2(–3)

|A| = 9 – 0  + 6

|A| = 15 ≠ 0,

Hence, the given system has only trivial solutions given by x = y = z = 0.

Question 8. Solve the system of homogeneous linear equations by matrix method:

2x + 3y – z =0

x – y – 2z = 0

3x + y + 3z = 0

Solution:

Given:

2x + 3y – z =0

x – y – 2z = 0

3x + y + 3z = 0

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

A X = 0

Now, |A| = 2(–3 + 2) – 3(3 + 6) – 1(4)

|A| = –2 – 27  – 4

|A| = –33 ≠ 0,

Hence, the given system has only trivial solutions given by x = y = z = 0.

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