Let A = x 1 1 0 , x ε R and A 4 = a i j . If a 11 = 109 , then a 22 is equal to_____________.

A 2 = x 1 1 0 × x 1 1 0 = x 2 + 1 x x 1 A 4 = x 2 + 1 x x 1 × x 2 + 1 x x 1 = x 4 + 3 x 2 + 1 x 3 + 2 x x 3 + 2 x x 2 + 1 Given a 11 = 109 ⇒ x 4 + 3 x 2 + 1 = 109 x 2 + 12 x 2 - 9 = 0 ⇒ x 2 = 9 ∴ a 22 = x 2 + 1 = 10

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MathematicsMatricesProduct of MatricesMedium2 minPYQ_2020

MathematicsMediumnumerical

Let $A = [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}], x ε R$ and $A^{4} = [a_{i j}] .$ If $a_{11} = 109,$ then $a_{22}$ is equal to_____________.

Answer:

10.00

Solution:

$A^{2} = [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}] \times [\begin{matrix} x & 1 \\ 1 & 0 \end{matrix}] = [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}]$

$A^{4} = [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}] \times [\begin{matrix} x^{2} + 1 & x \\ x & 1 \end{matrix}] = [\begin{matrix} x^{4} + 3 x^{2} + 1 & x^{3} + 2 x \\ x^{3} + 2 x & x^{2} + 1 \end{matrix}]$

Given $a_{11} = 109 \Rightarrow x^{4} + 3 x^{2} + 1 = 109$

$(x^{2} + 12) (x^{2} - 9) = 0 \Rightarrow x^{2} = 9$

$∴ a_{22} = x^{2} + 1 = 10$

Stream:JEESubject:MathematicsTopic:MatricesSubtopic:Product of Matrices

⏱ 2mℹ️ Source: PYQ_2020

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