If the matrix A=1 0 00 2 03 0 −1 satisfies the matrix equation A20+αA19+βA=1 0 00 4 00 0 1, then the value of α+β is

We have, A=1 0 00 2 03 0 −1∴ A2=1 0 00 4 00 0 1,A3=1 0 00 8 03 0 −1 and A4=1000160001⇒ A20=10002200001 and A19=1000219030−1∴ A20+αA19+βA=100040001⇒ 1+α+β000220+α219+2β03+3β01−α−β=100040001⇒1+α+β=1,220+α⋅219+2β=4,3α+3β=0and 1−α−β=1⇒α+β=0,220+α⋅219+2β=4⇒β=−α and 220+α⋅219+2β=4⇒220+α⋅219−2α=4⇒(α+2)219−2(α+2)=0⇒(α+2)219−2=0⇒α+2=0⇒α=−2∴β=−α⇒β=2Hence, α+β=0

If the matrix A=1 0 00 2 03 0 −1 satisfies the matrix equation A20+αA19+βA=1 0 00 4 00 0 1, then the value of α+β is

We have, A=1 0 00 2 03 0 −1∴ A2=1 0 00 4 00 0 1,A3=1 0 00 8 03 0 −1 and A4=1000160001⇒ A20=10002200001 and A19=1000219030−1∴ A20+αA19+βA=100040001⇒ 1+α+β000220+α219+2β03+3β01−α−β=100040001⇒1+α+β=1,220+α⋅219+2β=4,3α+3β=0and 1−α−β=1⇒α+β=0,220+α⋅219+2β=4⇒β=−α and 220+α⋅219+2β=4⇒220+α⋅219−2α=4⇒(α+2)219−2(α+2)=0⇒(α+2)219−2=0⇒α+2=0⇒α=−2∴β=−α⇒β=2Hence, α+β=0

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If the matrix $A = [\begin{array}{rrr} 1 0 0 \\ 0 2 0 \\ 3 0 - 1 \end{array}]$ satisfies the matrix equation $A^{20} + α A^{19} + β A = [\begin{array}{lll} 1 0 0 \\ 0 4 0 \\ 0 0 1 \end{array}]$ , then the value of $α + β$ is

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Detailed Solution

We have, $A = [\begin{array}{l} 1 0 0 \\ 0 2 0 \\ 3 0 - 1 \end{array}]$

$∴ A^{2} = [\begin{array}{lll} 1 0 0 \\ 0 4 0 \\ 0 0 1 \end{array}], A^{3} = [\begin{array}{rrr} 1 0 0 \\ 0 8 0 \\ 3 0 - 1 \end{array}] and A^{4} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1 \end{array}]$

$\Rightarrow A^{20} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1 \end{array}]$ and $A^{19} = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & - 1 \end{array}]$

$\begin{array}{l} ∴ A^{20} + α A^{19} + β A = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow [\begin{array}{ccc} 1 + α + β & 0 & 0 \\ 0 & 2^{20} + α 2^{19} + 2 β & 0 \\ 3 + 3 β & 0 & 1 - α - β \end{array}] = [\begin{array}{lll} 1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{array}] \\ \Rightarrow 1 + α + β = 1, 2^{20} + α \cdot 2^{19} + 2 β = 4, 3 α + 3 β = 0 \end{array}$

and $1 - α - β = 1$

$\Rightarrow α + β = 0, 2^{20} + α \cdot 2^{19} + 2 β = 4$

$\Rightarrow β = - α$ and $2^{20} + α \cdot 2^{19} + 2 β = 4$

$\begin{array}{l} \Rightarrow 2^{20} + α \cdot 2^{19} - 2 α = 4 \\ \Rightarrow (α + 2) 2^{19} - 2 (α + 2) = 0 \\ \Rightarrow (α + 2) (2^{19} - 2) = 0 \Rightarrow α + 2 = 0 \Rightarrow α = - 2 \\ ∴ β = - α \Rightarrow β = 2 \end{array}$