Question 23

If $$A = \begin{bmatrix}1 & 2 \\3 & a \end{bmatrix}$$ where as is a real number and det $$(A^{3} − 3A^{2} − 5A) =0$$ then one of the value of a can be

Solution

Given, det $$(A^{3} − 3A^{2} − 5A) =0$$

So, det $$(A^3-3A^2-5A)$$ = det $$\left((A)\left(A^2-3A-5I\right)\right)$$=det $$A$$ $$\times\ $$ det $$(A^2-3A-5I)$$ =0

So, det $$A$$=0

or, $$\begin{vmatrix}1&2 \\ 3&a\end{vmatrix}$$=0

or, $$1\times\ a-2\times\ 3=0$$

or, $$a=6$$

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