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Find the matrix A such that \left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] \cdot \mathrm{A}=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right].
CBSE | Class 12 | Exercise 5C | Maths | Matrices | RS Aggarwal
Find the matrix A such that \left[\begin{array}{cc}5 & -7 \\ -2 & 3\end{array}\right] \cdot \mathrm{A}=\left[\begin{array}{cc}-16 & -6 \\ 7 & 2\end{array}\right].

Solution:

We have \boldsymbol{B}=\left(\begin{array}{cc}\mathbf{5} & -\mathbf{7} \\ -\mathbf{2} & \mathbf{3}\end{array}\right) and \boldsymbol{C}=\left(\begin{array}{cc}-\mathbf{1 6} & -\mathbf{6} \\ \mathbf{7} & \mathbf{2}\end{array}\right).
We need to find matrix \boldsymbol{A} such that \boldsymbol{B} \boldsymbol{A}=\boldsymbol{C}.
We’ll use the inverse application here by modification of the equation we have as \boldsymbol{A}=\boldsymbol{B}^{-1} \boldsymbol{C}

(i) The inverse of \boldsymbol{B} is computed as follow:
We have \boldsymbol{B}=\left(\begin{array}{cc}\mathbf{5} & -7 \\ -\mathbf{2} & \mathbf{3}\end{array}\right). Then \operatorname{det}(\boldsymbol{B})=\left|\begin{array}{cc}\mathbf{5} & -7 \\ -\mathbf{2} & \mathbf{3}\end{array}\right|=\mathbf{1 5}- 14=1 \neq 0
Thus inverse of \boldsymbol{B} exist.
Then \boldsymbol{C}_{B}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{2} \\ \mathbf{7} & \mathbf{5}\end{array}\right)
\operatorname{Now} \boldsymbol{\operatorname { a d } \boldsymbol { j }}(\boldsymbol{B})=\boldsymbol{C}_{\boldsymbol{B}}^{T}=\left(\begin{array}{ll}\mathbf{3} & \mathbf{7} \\ \mathbf{2} & \mathbf{5}\end{array}\right)
Thus the inverse of the given matrix is \boldsymbol{B}^{\mathbf{- 1}}=\frac{\mathbf{1}}{\operatorname{det}(\boldsymbol{B})} \boldsymbol{a d j}(\boldsymbol{B})= \left(\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right)

(ii)Finally we have \boldsymbol{A}=\boldsymbol{B}^{-1} \boldsymbol{C}
\begin{array}{l} =\left(\begin{array}{ll} 3 & 7 \\ 2 & 5 \end{array}\right)\left(\begin{array}{cc} -16 & -6 \\ 7 & 2 \end{array}\right) \\ =\left(\begin{array}{ll} 1 & -4 \\ 3 & -6 \end{array}\right) \end{array}

i

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