Question

(i) $2i-j$ is the $(i,j)$th element of a $2\times 2$ matrix $A$.
(ii) Find the matrix $B$ such that $3A-2B=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
(iii) Prove that $A^2+2I=3A$

Answer 1

$\left(i\right)2i-j$ is the $(i,j)$th element of a $2\times 2$ matrix $A$

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

$\Rightarrow A=\left[\begin{array}{cc}2\left(1\right)-1& 2\left(1\right)-2\\ 2\left(2\right)-1& 2\left(2\right)-2\end{array}\right]$

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

$\left(ii\right)3A-2B=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow 2B=3\left[\begin{array}{cc}1& 0\\ 3& 2\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 9& 6\end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$

$\Rightarrow B=\frac{1}{2}\left[\begin{array}{cc}2& 0\\ 9& 5\end{array}\right]=\left[\begin{array}{cc}1& 0\\ \frac{9}{2}& \frac{5}{2}\end{array}\right]$

$\left(iii\right)A^2+2I=\left[\begin{array}{cc}1& 0\\ 3& 2\end{array}\right]\left[\begin{array}{cc}1& 0\\ 3& 2\end{array}\right]+2\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 9& 4\end{array}\right]+\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]=\left[\begin{array}{cc}3& 0\\ 9& 6\end{array}\right]=3\left[\begin{array}{cc}1& 0\\ 3& 2\end{array}\right]=3A$

$\Rightarrow A^2+2I=3A$