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  • Please Solve RD Sharma class 12 chapter 7 Solution of Simultaneous Linear Exercise Very short answer Question 2 Maths Textbook solution

Please Solve RD Sharma class 12 chapter 7 Solution of Simultaneous Linear  Exercise Very short answer  Question 2 Maths Textbook solution.

Answers (1)

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Answer \rightarrow x=1, y=0, z=-1

Given
             \rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]

Explanation
            \rightarrow\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]

                    \left[\begin{array}{l} x \times 1+0 \times y+0 \times z \\ 0 \times x-1 \times y+0 \times z \\ 0 \times x+0 \times y-1 \times z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]

                    \left[\begin{array}{c} x \\ -y \\ -z \end{array}\right]=\left[\begin{array}{l} 1 \\ 0 \\ 1 \end{array}\right]

Comparing both sides we get  x=1,y=0,z=-1

 

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