Question

If $A=\left[\begin{array}{cc}4& -1\\ -1& k\end{array}\right]$ such that $A^2-6A+7I=0$, then $k=$

• A. $1$
• B. $3$
• C. $2$
• D. $4$

Answer 1

The correct option is A $1$

$A=\left[\begin{array}{cc}4& -1\\ -1& k\end{array}\right]$

$A^2=\left[\begin{array}{cc}4& -1\\ -1& k\end{array}\right]\times \left[\begin{array}{cc}4& -1\\ -1& k\end{array}\right]$

$\Rightarrow A^2=\left[\begin{array}{cc}16+1& -4-k\\ -4-k& 1+k^2\end{array}\right]=\left[\begin{array}{cc}17& -(4+k)\\ -(4+k)& 1+k^2\end{array}\right]$

$6A=6\times \left[\begin{array}{cc}4& -1\\ -1& k\end{array}\right]=\left[\begin{array}{cc}24& -6\\ -6& 6k\end{array}\right]$

$7I=7\times \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

Therefore,

$A^2-6A+7I=0$

$\Rightarrow A^2=6A-7I$

$\left[\begin{array}{cc}17& -(4+k)\\ -(4+k)& 1+k^2\end{array}\right]=\left[\begin{array}{cc}24& -6\\ -6& 6k\end{array}\right]-\left[\begin{array}{cc}7& 0\\ 0& 7\end{array}\right]$

$\left[\begin{array}{cc}17& -(4+k)\\ -(4+k)& 1+k^2\end{array}\right]=\left[\begin{array}{cc}17& -6\\ -6& 6k-7\end{array}\right]$

Comparing both sides, we get

$-(4+k)=-6$

$\Rightarrow k=6-4=2$