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provide solution for rd sharma maths class 12 chapter Solution of simultaneous linear equation exercise 7.2 question 2

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Answer:

            x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k

Hint:

If \left | A \right |= 0, then the system of equation has non-trivial solution.

Given:

                2x-y+2z=0

                5x+3y-z=0

                x+5y-5z=0

Explanation:

The given homogeneous system can be written in matrix form

i.e.      \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}

Let     A=\begin{bmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{bmatrix},X=\begin{bmatrix} x\\ y\\ z\end{bmatrix},B=\begin{bmatrix} 0\\ 0\\ 0\end{bmatrix}

i.e.     AX=B

           \left | A \right |=\begin{vmatrix} 2& -1& 2\\ 5 & 3& -1\\ 1& 5& -5 \end{vmatrix}

                 =2(-15+5)+1(-25+1)+2(25-3)

                 =-20-24+ 44

                 =0

            So, the given system has non-trivial solution.

            To find the solutions we write

                        2x-y=-2z

                        5x+3y=z

           let          z=k

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

                       

                        \left | A \right |=\begin{vmatrix} 2 &-1 \\ 5 & 3 \end{vmatrix}=11\neq 0

            So A-1 exists

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

                            A^{-1}=\frac{1}{\left | A \right |}adjA

                            A^{-1}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}

        Now,           X=A^{-1}B

                       \begin{bmatrix} x\\ y\end{bmatrix}=\frac{1}{11}\begin{bmatrix} 3 & 1\\ -5 &2 \end{bmatrix}\begin{bmatrix} -2k\\ k\end{bmatrix}

                                =\frac{1}{11}\begin{bmatrix} -5k\\ 12k\end{bmatrix}

                        x=\frac{-5k}{11},y=\frac{12k}{11}

            x,z and y satisfy the third equation

                        \frac{-5k}{11}+5\left ( \frac{12k}{11} \right )-5z=0

                        5z=\frac{55k}{11}

                        z=k

 

Hence x=\frac{-5k}{11}, y=\frac{12k}{11}, z=k, where k\in R

 

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