JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 16)
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 ___________.
$${A^{20}} + \alpha {A^{19}} + \beta A = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]$$ for some real numbers $$\alpha$$ and $$\beta$$, then $$\beta$$ $$-$$ $$\alpha$$ is equal to ___________.
Answer
4
Explanation
$${A^2} = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right]\left[ {\matrix{
1 & 0 & 0 \cr
0 & 2 & 0 \cr
3 & 0 & { - 1} \cr
} } \right] = \left[ {\matrix{
1 & 0 & 0 \cr
0 & 4 & 0 \cr
0 & 0 & 1 \cr
} } \right]$$
$${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right]$$
$${A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {16} & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$\eqalign{ & . \cr & . \cr & . \cr & . \cr & . \cr & . \cr} $$
$${A^{19}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{19}}} & 0 \cr 3 & 0 & { - 1} \cr } } \right],{A^{20}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{20}}} & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A = $$
$$\left[ {\matrix{ {1 + \alpha + \beta } & 0 & 0 \cr 0 & {{2^{20}} + \alpha {2^{19}} + 2\beta } & 0 \cr {3\alpha + 3\beta } & 0 & {1 - \alpha - \beta } \cr } } \right]$$
$$R.H.S. = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] $$
$$\Rightarrow \alpha + \beta = 0$$ and $${2^{20}} + \alpha {2^{19}} + 2\beta = 4$$
$$ \Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$$
$$ \Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$$
$$ \Rightarrow \beta = 2$$
$$ \therefore $$ $$\beta - \alpha = 4$$
$${A^3} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 2 & 0 \cr 3 & 0 & { - 1} \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 8 & 0 \cr 3 & 0 & { - 1} \cr } } \right]$$
$${A^4} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right]\left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {16} & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$\eqalign{ & . \cr & . \cr & . \cr & . \cr & . \cr & . \cr} $$
$${A^{19}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{19}}} & 0 \cr 3 & 0 & { - 1} \cr } } \right],{A^{20}} = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & {{2^{20}}} & 0 \cr 0 & 0 & 1 \cr } } \right]$$
$$L.H.S. = {A^{20}} + \alpha {A^{19}} + \beta A = $$
$$\left[ {\matrix{ {1 + \alpha + \beta } & 0 & 0 \cr 0 & {{2^{20}} + \alpha {2^{19}} + 2\beta } & 0 \cr {3\alpha + 3\beta } & 0 & {1 - \alpha - \beta } \cr } } \right]$$
$$R.H.S. = \left[ {\matrix{ 1 & 0 & 0 \cr 0 & 4 & 0 \cr 0 & 0 & 1 \cr } } \right] $$
$$\Rightarrow \alpha + \beta = 0$$ and $${2^{20}} + \alpha {2^{19}} + 2\beta = 4$$
$$ \Rightarrow {2^{20}} + \alpha ({2^{19}} - 2) = 4$$
$$ \Rightarrow \alpha = {{4 - {2^{20}}} \over {{2^{19}} - 2}} = - 2$$
$$ \Rightarrow \beta = 2$$
$$ \therefore $$ $$\beta - \alpha = 4$$
Comments (0)
