Question

Find a matrix A such that

$\left[\begin{array}{cc}2& 3\\ 1& 4\end{array}\right]\mathit{A}\mathbf{=}\left[\begin{array}{cc}3& 0\\ 1& 2\end{array}\right]$

[Finding A requires that you solve system of linear equation]

Short answer

Required matrix is

$A=\left[\begin{array}{cc}\frac{9}{5}& -\frac{6}{5}\\ -\frac{1}{5}& \frac{4}{5}\end{array}\right]$

Step-by-step solution

Multiplication of matrices

Let$\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$$A=\left[\begin{array}{cc}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]$

data-custom-editor="chemistry" $\begin{array}{r}\left[\begin{array}{ll}2& 3\\ 1& 4\end{array}\right]A=\left[\begin{array}{ll}3& 0\\ 1& 2\end{array}\right]\\ \left[\begin{array}{ll}2& 3\\ 1& 4\end{array}\right]\left[\begin{array}{ll}{a}_{11}& a_{12}\\ a_{21}& a_{22}\end{array}\right]=\left[\begin{array}{ll}3& 0\\ 1& 2\end{array}\right]\\ \left[\begin{array}{cc}2a_{11}+3a_{21}& 2a_{12}+3a_{22}\\ a_{11}+4a_{21}& a_{12}+4a_{22}\end{array}\right]=\left[\begin{array}{ll}3& 0\\ 1& 2\end{array}\right]\end{array}$

On comparing both equations, we get

role="math" localid="1668436656954" $\begin{array}{l}2a_{11}+3a_{21}=3\\ 2a_{12}+3a_{22}=0\\ a_{11}+4a_{21}=1\\ a_{12}+4a_{22}=2\end{array}$

Multiply the third and fourth equation by 2 on each side

$\begin{array}{l}2a_{11}+3a_{21}=3\\ 2a_{12}+3a_{22}=0\\ 2a_{21}+8a_{21}=2\\ 2a_{22}+8a_{22}=4\end{array}$

Solving equations

Subtract the first equation from the third equation

Subtract the second equation from the fourth equation

Divide each side of each equation by 5

$\begin{array}{r}5a_{21}=-1\\ 5a_{22}=4\\ a_{21}=-\frac{1}{5}\\ a_{22}=\frac{4}{5}\\ a_{11}=1-4a_{21}\\ =1-4\left(-\frac{1}{5}\right)\\ =1+\frac{4}{5}\\ =\frac{9}{5}\\ a_{12}=2-4a_{12}\\ =2-4\left(\frac{4}{5}\right)\\ =2-\frac{16}{5}\\ =-\frac{6}{5}\end{array}$

Hence, we get

$A=\left[\begin{array}{cc}\frac{9}{5}& -\frac{6}{5}\\ -\frac{1}{5}& \frac{4}{5}\end{array}\right]$