Question

For the matrix $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right],$ the values of ${}^{\prime }{a}^{\prime }$ and ${}^{\prime }{b}^{\prime }$ such that $A^2+aA+bI=0$ are

• A. $2,3$
• B. $1,4$
• C. $-4,1$
• D. $-2,3$

Answer 1

The correct option is C $-4,1$

$A^2=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]$
Now,
$A^2+aA+bI=0$
Putting values
$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+a\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]+b\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$
$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+\left[\begin{array}{cc}3a& 2a\\ a& a\end{array}\right]+\left[\begin{array}{cc}b& 0\\ 0& b\end{array}\right]=0$
$\left[\begin{array}{cc}11+3a+b& 8+2a+0\\ 4+a+0& 3+a+b\end{array}\right]=0$
$\left[\begin{array}{cc}3a+b+11& 2a+8\\ 4+a& a+b+3\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
Since the matrices are equal,
Comparing corresponding elements
$4+a=0\dots \left(1\right)$
$a+b+3=0\dots \left(2\right)$
Solving $\left(1\right)$ and $\left(2\right)$
$a=-4\Rightarrow b=1$