If the matrix A = 0 2 K - 1 satisfies A A 3 + 3 I = 2 I , then the value of K is

A A 3 + 3 I = 2 I A 2 = 0 2 K - 1 0 2 K - 1 = 2 K - 2 - K 2 K + 1 A 3 = 2 K - 2 - K 2 K + 1 0 - 1 K 2 = - 2 K 4 K + 2 2 K 2 + K - 2 K - 2 K - 1 A 4 = 2 K - 2 - K 2 K + 1 2 K - 2 - K 2 K + 1 = 4 K 2 + 2 K - 4 K - 4 K - 2 - 2 K 2 - 2 K 2 - K 2 K + 4 K 2 + 1 + 4 K A 4 + 3 AI = 2 I 2 I = 4 K 2 + 2 K - 8 K - 2 - 4 K 2 - K 4 K 2 + 6 K + 1 + 3 0 2 K - 1 1 0 0 1 2 I = 4 K 2 + 2 K - 8 K - 2 - 4 K 2 - K 4 K 2 + 6 K + 1 + 3 0 2 K - 1 2 I = 4 K 2 + 2 K 6 - 8 K - 2 3 K - K - 4 K 2 4 K 2 + 6 K + 1 - 3 4 - 8 K = 0 K = 1 2

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MathematicsMatricesProduct of MatricesMedium2 minPYQ_2021

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If the matrix $A = [\begin{matrix} 0 & 2 \\ K & - 1 \end{matrix}]$ satisfies $A (A^{3} + 3 I) = 2 I,$ then the value of $K$ is

Options:

Answer:

Solution:

$A (A^{3} + 3 I) = 2 I$

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

$= [\begin{matrix} 2 K & - 2 \\ - K & 2 K + 1 \end{matrix}]$

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

$= [\begin{matrix} - 2 K & 4 K + 2 \\ 2 K^{2} + K & - 2 K - 2 K - 1 \end{matrix}]$

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

$= [\begin{matrix} 4 K^{2} + 2 K & - 4 K - 4 K - 2 \\ - 2 K^{2} - 2 K^{2} - K & 2 K + 4 K^{2} + 1 + 4 K \end{matrix}]$

$A^{4} + 3 AI = 2 I$

$2 I = [\begin{matrix} 4 K^{2} + 2 K & - 8 K - 2 \\ - 4 K^{2} - K & 4 K^{2} + 6 K + 1 \end{matrix}] + 3 [\begin{matrix} 0 & 2 \\ K & - 1 \end{matrix}] [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$2 I = [\begin{matrix} 4 K^{2} + 2 K & - 8 K - 2 \\ - 4 K^{2} - K & 4 K^{2} + 6 K + 1 \end{matrix}] + 3 [\begin{matrix} 0 & 2 \\ K & - 1 \end{matrix}]$

$2 I = [\begin{matrix} 4 K^{2} + 2 K & 6 - 8 K - 2 \\ 3 K - K - 4 K^{2} & 4 K^{2} + 6 K + 1 - 3 \end{matrix}]$

$4 - 8 K = 0$

$K = \frac{1}{2}$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2021

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