If the matrix A = [ 1 0 0 0 2 0 3 0 − 1 ] satisfies the equation A20 = αA19 + βA = [ 1 0 0 0 4 0 0 0 1 ] for some real numbers α and β, then β - α is equal to…………

A 2 = [ 1 0 0 0 2 0 3 0 − 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 2 0 0 0 1 ] A 3 = [ 1 0 0 0 2 2 0 0 0 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 3 0 3 0 − 1 ] A 4 [ 1 0 0 0 2 3 0 3 0 − 1 ] [ 1 0 0 0 2 0 3 0 − 1 ] = [ 1 0 0 0 2 4 0 0 0 1 ] Similarly we get A19 = = [ 1 0 0 0 2 1 9 0 3 0 − 1 ] & A 2 0 = [ 1 0 0 0 2 2 0 0 0 0 1 ] = [ 1 0 0 0 4 0 0 0 1 ] ⇒ 1 + α + β = 1 g i v e s α + β = 0 . . . . . . . . ( i ) 2 2 0 + ( 2 1 9 − 2 ) α = 4 f r o m ( i ) ⇒ α = 4 − 2 2 0 2 1 9 − 2 = 4 ( 1 − 2 1 8 ) − 2 ( 1 − 2 1 8 ) = − 2 So, β = 2 Hence β - α = 4

If the matrix A = $[\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1 \end{matrix}]$ satisfies the equation A²⁰ = αA¹⁹ + βA = $[\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{matrix}]$ for some real numbers α and β, then β - α is equal to…………

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Vishal Baghel | Contributor-Level 10

5 months ago

$A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{2} & 0 \\ 0 & 0 & 1 \end{matrix}]$

$A^{3} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{2} & 0 \\ 0 & 0 & 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{3} & 0 \\ 3 & 0 & - 1 \end{matrix}]$
$A^{4} [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{3} & 0 \\ 3 & 0 & - 1 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{4} & 0 \\ 0 & 0 & 1 \end{matrix}]$

Similarly we get A¹⁹ = $= [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & - 1 \end{matrix}] & A^{20} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{matrix}]$

= $[\begin{matrix} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{matrix}]$

$\Rightarrow 1 + α + β = 1 g i v e s α + β = 0 . . . . . . . . (i)$

$2^{20} + (2^{19} - 2) α = 4 f r o m (i)$

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

So, β = 2

Hence β - α = 4

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