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  • Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.1 question 47

Explain solution for rd sharma class 12 chapter Algebra of matrices exercise 4.1 question 47

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Answer: 

\left[\begin{array}{ll}x & y \\ z & \mu\end{array}\right]=\left[\begin{array}{ll}1 & -4 \\ 3 & -2\end{array}\right]

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

\left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right]\left[\begin{array}{cc}x & y \\ z & \mu\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]

Firstly, we will multiply both matrices in LHS

\left[\begin{array}{cc}5 x-7 z & 5 y-7 \mu \\ -2 x+3 z & -2 y+3 \mu\end{array}\right]=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right]

Since, corresponding entries of equal matrices are equal, so

\begin {array }{ll} 5 x-7 z=-16 \quad \ldots(i) \\\\ -2 x+3 z=7 \quad \ldots(ii) \\\\ 5 y-7 \mu=-6 \quad \ldots(iii)\\\\ -2 y+3 \mu=2 \quad \cdots (iv) \end{}

First solving equation (i) & ii

Multiply equation i by 2 and equation ii by 5 and then add both equations

\begin {array}{ll} 10 x-14 z=-32\\\\ -10 x+15 z=35\\\\ z=3 \end{}

Put the value of z in equation i

\begin {array}{ll} 5 x-7(3)=-16\\\\ 5 x-21=-16\\\\\ 5 x=-16+21 \\\\ 5 x=5 \\\\ x=1 \end{}

Solving equation iii & iv

Multiply equation iii by 2 and equation iv by 5 and then add both equation

\begin{array}{l} 10 y-14 \mu=-12\\\\ -10 y+15 \mu=10\\\\ \mu=-2\\ \end{array}

\begin {array}{ll} 5 y-7 \mu=-6\\\\ 5 y-7(-2)=-6\\\\ 5 y+14=-6\\\\ 5 y=-20\\\\ y=-4\\ \end{}

Therefore,y=-4, \mu=-2, x=1\ \ \ \text { and } \ \ \ z=3 y=-4\\

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