Question

Let a matrix $A=\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right],$ such that $A^2+aA+bI=O$. Then $(a,b)$ is

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

Answer 1

The correct option is C $(-4,1)$

$A^2=\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]$
Now,
$A^2+aA+bI=O$
Putting values
$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+a\left[\begin{array}{cc}3& 2\\ 1& 1\end{array}\right]+b\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=O$

$\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]+\left[\begin{array}{cc}3a& 2a\\ a& a\end{array}\right]+\left[\begin{array}{cc}b& 0\\ 0& b\end{array}\right]=O$

$\left[\begin{array}{cc}11+3a+b& 8+2a+0\\ 4+a+0& 3+a+b\end{array}\right]=O$

$\left[\begin{array}{cc}3a+b+11& 2a+8\\ 4+a& a+b+3\end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$
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 $\left(2\right),$ we get
$a=-4\Rightarrow b=1$ which also satisfy the above equation.