Question

For the matrix A=\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\),find the numbers a and b such that A²+ aA+bI=O. 

Answer 1

Let  A=\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)

A²=\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)=\(\begin{bmatrix}9+2&6+2\\3+1&2+1\end{bmatrix}\)

=\(\begin{bmatrix}11&8\\4&3\end{bmatrix}\)
Now, A²+aA+bI=0   [post multiplying by A^-1 as IAI≠0]
\(\Rightarrow\) (AA)A^-1+aAA^-1+bAA^-1=0
\(\Rightarrow\) A(AA^-1)+aI+b(IA^-1)=0
AI+aI+bA^-1=0
\(\Rightarrow\) A+aI=-bA^-1
\(\Rightarrow\) A^-1=-\(\frac{1}{b}\)(A+aI)
Now, A^-1=\(\frac{1}{\mid A\mid}\)adj A=\(\frac{1}{1}\)\(\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\)
we have:
\(\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\)=-\(\frac{1}{b}\)\(\bigg(\)\(\begin{bmatrix}3&2\\1&1\end{bmatrix}\)+\(\begin{bmatrix}a&0\\0&a\end{bmatrix}\)\(\bigg)\)=-\(\frac{1}{b}\)\(\begin{bmatrix}3+a&2\\1&1+a\end{bmatrix}\)=\(\begin{bmatrix}\frac{-3-a}{b}&\frac{-2][b}2\\\frac{-1}{b}&\frac{-1-a}{b}\end{bmatrix}\)
Comparing the corresponding elements of the two matrices have:
-\(\frac{1}{b}=-1\geq b=1\)
\(\frac{-3-a}{b}=1\geq -3-a=1\)
\(\Rightarrow\) a=-4

Hence, −4 and 1 are the required values of a and b respectively.