If \(A=\begin{bmatrix}3 & 2\\2 & 1 \end{bmatrix}\) and $A^2 - λA - I = 0$, then the value of λ is:

4

Given

$A=\begin{bmatrix}3&2\\2&1\end{bmatrix}$

First find $A^2$

$A^2=\begin{bmatrix}3&2\\2&1\end{bmatrix} \begin{bmatrix}3&2\\2&1\end{bmatrix}$

$=\begin{bmatrix}9+4&6+2\\6+2&4+1\end{bmatrix}$

$=\begin{bmatrix}13&8\\8&5\end{bmatrix}$

Given

$A^2-\lambda A-I=0$

$A^2=\lambda A+I$

$\lambda A+I= \lambda\begin{bmatrix}3&2\\2&1\end{bmatrix} +\begin{bmatrix}1&0\\0&1\end{bmatrix}$

$=\begin{bmatrix}3\lambda+1&2\lambda\\2\lambda&\lambda+1\end{bmatrix}$

Equate with $A^2$

$\begin{bmatrix}13&8\\8&5\end{bmatrix} =\begin{bmatrix}3\lambda+1&2\lambda\\2\lambda&\lambda+1\end{bmatrix}$

$2\lambda=8$

$\lambda=4$

Check

$3\lambda+1=3(4)+1=13$ ✔

The value of $\lambda$ is $4$.