Question
Find matrices A and B, where
$2 A-B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$ and $A+3 B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$
$2 A-B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$ and $A+3 B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$
$\begin{aligned} & 2 A-B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \ldots \ldots(i) \\ & \text { and } A+3 B=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \text { By (i) } \times 3+\text { (ii), we get } \\ & 7 A=3\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \therefore \quad 7 \mathrm{~A}=\left[\begin{array}{cc}3 & -3 \\ 0 & 3\end{array}\right]+\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & \therefore \quad 7 \mathrm{~A}=\left[\begin{array}{cc}4 & -4 \\ 0 & 4\end{array}\right] \\ & \therefore \quad \mathrm{A}=\frac{1}{7}\left[\begin{array}{cc}4 & -4 \\ 0 & 4\end{array}\right] \\ & \end{aligned}$
By (i) - (ii) $\times 2$, we get
$\begin{aligned}-7 B & =\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]-2\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] \\ & =\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]-\left[\begin{array}{cc}2 & -2 \\ 0 & 2\end{array}\right] \\ \therefore \quad-7 B & =\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right] \\ \therefore \quad B & =\frac{1}{7}\left[\begin{array}{cc}-1 & 1 \\ 0 & -1\end{array}\right]\end{aligned}$
$\therefore \quad B=\frac{1}{7}\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.