Question

Find the value of $a$ and $b$ if $\left[\begin{array}{cc}a+3& b^2+2\\ 0& -6\end{array}\right]=\left[\begin{array}{cc}2a+1& 3b\\ 0& b^2-5b\end{array}\right]$.

Answer 1

Comparing both sides

$\Rightarrow a+3=2a+1$

$a-2a=1-3$

$a=2$

$b^2+2=3b{b}^2-3b+2=0b^2-2b-b+2=0b(b-2)-1(b-2)=0(b-1)(b-2)=0b=1,2$

Also $-6=b^2-5b$

$b^2-5b+6=0b^2-3b-2b+6=0b(b-3)-2(b-3)=0(b-2)(b-3)=0b=2,3\therefore a=2,b=2$