Let A be a 2 × 2 symmetric matrix such that A [ 1 1 ] = [ 3 7 ] and the determinant of A be 1. If A - 1 = A + I , where I is an identity matrix of order 2 × 2 , then equals _______ . [2024]

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Solution

Q.
Let $A$ be a $2 \times 2$ symmetric matrix such that $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$ and the determinant of $A$ be 1. If $A^{- 1} = α A + β I,$ where $I$ is an identity matrix of order $2 \times 2,$ then $α + β$ equals _______ . [2024]

Ans.
(5)

Let $A = [\begin{matrix} a & b \\ b & c \end{matrix}]$

$| A | = 1 \Rightarrow a c - b^{2} = 1$ ...(i)

Given, $A [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}] \Rightarrow [\begin{matrix} a & b \\ b & c \end{matrix}] [\begin{matrix} 1 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 7 \end{matrix}]$

$\Rightarrow a + b = 3$ ...(ii)

and $b + c = 7$

On solving (i), (ii) and (iii), we get

$a = 1, b = 2, c = 5$

$∴$ $A = [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}] \Rightarrow A^{- 1} = [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}]$

Given, $A^{- 1} = α A + β I$

$\Rightarrow [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}] = α [\begin{matrix} 1 & 2 \\ 2 & 5 \end{matrix}] + β [\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}]$

$\Rightarrow [\begin{matrix} α + β & 2 α \\ 2 α & 5 α + β \end{matrix}] = [\begin{matrix} 5 & - 2 \\ - 2 & 1 \end{matrix}]$

On comparing, we get

$α = - 1 and β = 6$

Hence, $α + β = - 1 + 6 = 5$

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