Question

If $A=\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$, then $A^2$ equals to-

Answer 1

We have,

$A=\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$

Now,

$A^2=A.A$

Therefore,

$A^2=\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$$\left(\begin{array}{cc}a& 1\\ 0& a\end{array}\right)$

$A^2=\left(\begin{array}{cc}{a}^2+0& a+a\\ 0+0& 0+a^2\end{array}\right)$

$A^2=\left(\begin{array}{cc}{a}^2& 2a\\ 0& a^2\end{array}\right)$

Hence, this is the answer.