Question

If $A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$ and $A^2-\lambda A-I_2=0$, then $\lambda =$

• A. $-4$
• B. $-2$
• C. $2$
• D. $4$

Answer 1

Answer: (D)

$4$

$A=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

$A^2=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

$A^2=\left[\begin{array}{cc}1+4& 2+6\\ 2+6& 4+9\end{array}\right]$

$A^2=\left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]$

$\lambda A=\lambda \left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right]$

$=\left[\begin{array}{cc}\lambda & 2\lambda \\ 2\lambda & 3\lambda \end{array}\right]$

Now,$A^2-\lambda A-I_2=0$,

$\Rightarrow \left[\begin{array}{cc}5& 8\\ 8& 13\end{array}\right]-\left[\begin{array}{cc}\lambda & 2\lambda \\ 2\lambda & 3\lambda \end{array}\right]-\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{cc}5-\lambda -1& 8-2\lambda -0\\ 8-2\lambda -0& 13-3\lambda -1\end{array}\right]=0$

$\Rightarrow \left[\begin{array}{cc}4-\lambda & 8-2\lambda \\ 8-2\lambda & 12-3\lambda \end{array}\right]=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right]$

Equating , we get

$4-\lambda =0\Rightarrow \lambda =4$

$8-2\lambda =0\Rightarrow 2\lambda =8$ or $\lambda =\frac{8}{2}=4$

$12-3\lambda =0\Rightarrow 3\lambda =12$ or $\lambda =\frac{12}{3}=4$

$\therefore \lambda =4$