Medium INTEGER +4 / -1 PYQ · JEE Mains 2021

If the matrix $$A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right]$$ satisfies the equation

$${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ for some real numbers $\alpha$ and $\beta$, then $\beta$ $-$ $\alpha$ is equal to ___________.

Answer (integer) 4

Solution

$${A^2} = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 2 &amp; 0 \cr 3 &amp; 0 &amp; { - 1} \cr } } \right]\left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 2 &amp; 0 \cr 3 &amp; 0 &amp; { - 1} \cr } } \right] = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 4 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]$$<br><br>$${A^3} = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 4 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]\left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 2 &amp; 0 \cr 3 &amp; 0 &amp; { - 1} \cr } } \right] = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 8 &amp; 0 \cr 3 &amp; 0 &amp; { - 1} \cr } } \right]$$<br><br>$${A^4} = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 4 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]\left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 4 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right] = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; {16} &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]$$<br><br>$$\eqalign{ &amp; . \cr &amp; . \cr &amp; . \cr &amp; . \cr &amp; . \cr &amp; . \cr} $$<br><br>$${A^{19}} = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; {{2^{19}}} &amp; 0 \cr 3 &amp; 0 &amp; { - 1} \cr } } \right],{A^{20}} = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; {{2^{20}}} &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right]$$<br><br>$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A =$ <br><br>$$\left[ {\matrix{ {1 + \alpha + \beta } &amp; 0 &amp; 0 \cr 0 &amp; {{2^{20}} + \alpha {2^{19}} + 2\beta } &amp; 0 \cr {3\alpha + 3\beta } &amp; 0 &amp; {1 - \alpha - \beta } \cr } } \right]$$<br><br>$$R.H.S. = \left[ {\matrix{ 1 &amp; 0 &amp; 0 \cr 0 &amp; 4 &amp; 0 \cr 0 &amp; 0 &amp; 1 \cr } } \right] $$ <br><br>$\Rightarrow \alpha + \beta = 0$ and ${2^{20}} + \alpha {2^{19}} + 2\beta = 4$<br><br>$\Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$<br><br>$\Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$<br><br>$\Rightarrow \beta = 2$<br><br>$\therefore$ $\beta - \alpha = 4$

About this question

Subject: Mathematics · Chapter: Matrices and Determinants · Topic: Types of Matrices and Operations

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