Question

If $A = \begin{bmatrix} {1}&{-3}\\ {2}&{K} \\ \end{bmatrix}$ and $A^2-14A+10I=A$,then $K$ = is equal to :

• A. 1 or 4
• B. 4 and not 1
• C. -4 .
• D. 0

Answer 1

The Correct Option is B

$A = \begin{bmatrix}1&-3\\ 2&k\end{bmatrix} $ $ \therefore A^{2} -4A +10I = A $ $\Rightarrow \begin{bmatrix}1&-3\\ 2&k\end{bmatrix} \begin{bmatrix}1&-3\\ 2&k\end{bmatrix}-4 \begin{bmatrix}1&-3\\ 2&k\end{bmatrix} $ $ + 10 \begin{bmatrix}1&0\\ 0&1\end{bmatrix} = \begin{bmatrix}1&-3\\ 2&k\end{bmatrix} $ $\Rightarrow \begin{bmatrix}-5&-3-3k\\ 2+2k& -6+k^{2}\end{bmatrix} - \begin{bmatrix}4&-12\\ 8&4k\end{bmatrix} + \begin{bmatrix}10&0\\ 0&10\end{bmatrix} = \begin{bmatrix}1&-3\\ 2&k\end{bmatrix} $ $\Rightarrow \begin{bmatrix}1&9-3k\\ -6+2k&4+k^{2}-4k\end{bmatrix} =\begin{bmatrix}1&-3\\ 2&k\end{bmatrix} $ $ \Rightarrow 9 -3 k = -3, - 6+2k = 2 \,\,\,\,\dots(i)$ and $ 4 + k^{2} - 4k = k $ $\Rightarrow \, k^{2}- 5k + 4 = 0 $ $\Rightarrow\, \left(k - 4\right) \left(k - 1\right) = 0$ $\Rightarrow \, k = 4,1 $ But k = 1 is not satisfied the E (i)