Question 410 of 472MediumMCQ
Let \[A=\left(\[1−22−5\]\right)\]. Let \[\alpha, \beta \in \mathbb{R}\] be such that \[\alpha A^{2}+\beta A=2 I\]. Then \[\alpha+\beta\] is equal to Solutions · 1
Official solution
\[{A^2} = \left[ {(1−22−5) } \right]\left[ {(1−22−5) } \right] = \left[ {(−38−821) } \right]\]
\[\alpha {A^2} + \beta A = \left[ {(−3α8α−8α21α) } \right] + \left[ {(β−2β2β−5β) } \right]\]
\[ = \left[ {(−3α+β8α−2β−8α+2β21α−5β) } \right] = \left[ {(2002) } \right]\]
On Comparing
\[8\alpha = 2\beta ,\, - 3\alpha + \beta = 2,\,21\alpha - 5\beta = 2\]
\[ \Rightarrow \alpha = 2,\,\beta = 8\]
So, \[\alpha + \beta = 10\] Answer the question to unlock