If the matrix $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{array}\right]$ satisfies the equation $A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]$ for some real numbers $\alpha$ and $\beta$, then $\beta-\alpha$ is equal to ____

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Matrices

If the matrix

A = 103020 00 - 1

satisfies the equation

A^{20} + α A^{19} + β A = 100040001

for some real numbers

α

and

β

, then

β - α

is equal to ____

Answer: 4

Multiplying out,

A^{2} = 100040001

which is precisely the right side, so the condition is

A^{20} + α A^{19} + β A = A^{2}

. Since

A^{2}

is diagonal,

A^{18} = (A^{2})^{9} = diag (1, 2^{18}, 1), A^{20} = diag (1, 2^{20}, 1), A^{19} = diag (1, 2^{18}, 1) A = 103 0 2^{19} 0 00 - 1

Comparing the

(1, 1)

entries gives

1 + α + β = 1

, so

β = - α

, and the

(3, 1)

and

(3, 3)

entries repeat the same relation. Comparing the

(2, 2)

entries,

2^{20} + α 2^{19} + 2 β = 4 \Rightarrow α (2^{19} - 2) = 4 - 2^{20} = - 4 (2^{18} - 1)

and

2^{19} - 2 = 2 (2^{18} - 1)

, so

α = - 2

and

β = 2

. Hence

β - α = 4

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer