Let A=(\beginarraycc1 & 2 -2 & -5\endarray) . Let \alpha, \beta \in \mathbbR be such that \alpha A^2+\beta A=2 I . Then \alpha+\beta is equal to

Matrices and Determinants PYQ — JEE Main 2022 · 27 Jul · Shift 1 · Q27 with Solution

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Correct answer: D

We are given $A=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}$ and need to find real numbers $\alpha,\beta$ such that $\alpha A^2+\beta A=2I.$ Then we must compute $\alpha+\beta$ .
First compute $A^2$ : $A^2=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}\begin{pmatrix}1&2\\-2&-5\end{pmatrix}.$ Now multiply:

1\cdot 1+2\cdot(-2) & 1\cdot 2+2\cdot(-5)\\ (-2)\cdot 1+(-5)\cdot(-2) & (-2)\cdot 2+(-5)\cdot(-5) \end{pmatrix} =\begin{pmatrix}-3&-8\\8&21\end{pmatrix}.$$ 3. Use the relation $$\alpha A^2+\beta A=2I.$$ Substitute $A^2$ and $A$: $$\alpha\begin{pmatrix}-3&-8\\8&21\end{pmatrix}+\beta\begin{pmatrix}1&2\\-2&-5\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ So, $$\begin{pmatrix}-3\alpha+\beta & -8\alpha+2\beta\\8\alpha-2\beta & 21\alpha-5\beta\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ 4. Equate corresponding entries: - From the off-diagonal entries, $$-8\alpha+2\beta=0$$ $$8\alpha-2\beta=0$$ which both give $$\beta=4\alpha.$$ 5. Use the $(1,1)$ entry: $$-3\alpha+\beta=2.$$ Substitute $\beta=4\alpha$: $$-3\alpha+4\alpha=2$$ $$\alpha=2.$$ Hence, $$\beta=4\alpha=8.$$ 6. Therefore, $$\alpha+\beta=2+8=10.$$ 7. So the correct option is $$\boxed{10}$$ which is option $\boxed{D}$. 8. Comparison with stored answer: Stored correct answer is $D$, which matches our result.

View written solutionFree

Correct answer: D

We are given $A=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}$ and need to find real numbers $\alpha,\beta$ such that $\alpha A^2+\beta A=2I.$ Then we must compute $\alpha+\beta$ .
First compute $A^2$ : $A^2=\begin{pmatrix}1&2\\-2&-5\end{pmatrix}\begin{pmatrix}1&2\\-2&-5\end{pmatrix}.$ Now multiply:

1\cdot 1+2\cdot(-2) & 1\cdot 2+2\cdot(-5)\\ (-2)\cdot 1+(-5)\cdot(-2) & (-2)\cdot 2+(-5)\cdot(-5) \end{pmatrix} =\begin{pmatrix}-3&-8\\8&21\end{pmatrix}.$$ 3. Use the relation $$\alpha A^2+\beta A=2I.$$ Substitute $A^2$ and $A$: $$\alpha\begin{pmatrix}-3&-8\\8&21\end{pmatrix}+\beta\begin{pmatrix}1&2\\-2&-5\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ So, $$\begin{pmatrix}-3\alpha+\beta & -8\alpha+2\beta\\8\alpha-2\beta & 21\alpha-5\beta\end{pmatrix}=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ 4. Equate corresponding entries: - From the off-diagonal entries, $$-8\alpha+2\beta=0$$ $$8\alpha-2\beta=0$$ which both give $$\beta=4\alpha.$$ 5. Use the $(1,1)$ entry: $$-3\alpha+\beta=2.$$ Substitute $\beta=4\alpha$: $$-3\alpha+4\alpha=2$$ $$\alpha=2.$$ Hence, $$\beta=4\alpha=8.$$ 6. Therefore, $$\alpha+\beta=2+8=10.$$ 7. So the correct option is $$\boxed{10}$$ which is option $\boxed{D}$. 8. Comparison with stored answer: Stored correct answer is $D$, which matches our result.