Question

find the matrix c such that ac=b​

Answer 1

$\textbf{Given:}$

$\mathsf{A=\left(\begin{array}{cc}3&2\\-1&1\end{array}\right)}$

$\mathsf{B=\left(\begin{array}{cc}14&3\\2&4\end{array}\right)}$

$\mathsf{and\;AC=B}$

$\textbf{To find:}$

$\textsf{Matrix C}$

$\textbf{Solution:}$

$\mathsf{Consider,}$

$\mathsf{A=\left(\begin{array}{cc}3&2\\-1&1\end{array}\right)}$

$\mathsf{adjA=\left(\begin{array}{cc}1&-2\\1&3\end{array}\right)}$

$\mathsf{|A|=\left|\begin{array}{cc}3&2\\-1&1\end{array}\right|=3+2=5{\neq}0}$

$\mathsf{Now}$

$\mathsf{A^{-1}=\dfrac{1}{|A|}adjA}$

$\mathsf{A^{-1}=\dfrac{1}{5}\left(\begin{array}{cc}1&-2\\1&3\end{array}\right)}$

$\mathsf{Also}$

$\mathsf{AC=B}$

$\implies\mathsf{C=A^{-1}B}$

$\implies\mathsf{C=\dfrac{1}{5}\left(\begin{array}{cc}1&-2\\1&3\end{array}\right)\left(\begin{array}{cc}14&3\\2&4\end{array}\right)}$

$\implies\mathsf{C=\dfrac{1}{5}\left(\begin{array}{cc}14-4&3-8\\14+6&3+12\end{array}\right)}$

$\implies\mathsf{C=\dfrac{1}{5}\left(\begin{array}{cc}10&-5\\20&15\end{array}\right)}$

$\implies\boxed{\mathsf{C=\left(\begin{array}{cc}2&-1\\4&3\end{array}\right)}}$

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