Let A be a matrix such that , , and . If det (A) = 1, then is equal to : [2026]

Question

Let A be a $3 \times 3$ matrix such that $A^{T} = [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 5 \\ 2 \\ 2 \end{matrix}]$ , $A^{T} = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 1 \\ 1 \end{matrix}]$ , $A = [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 4 \\ 4 \end{matrix}]$ and $A = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 1 \end{matrix}]$ . If det (A) = 1, then $\det (adj (A^{2} + A))$ is equal to : [2026]

Smart Achievers · Accepted Answer

(4)

Let $A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix}]$

$\Rightarrow A = [\begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 4 \\ 4 \end{matrix}]$

$\begin{matrix} \Rightarrow a_{1} + a_{3} = 3 \\ \Rightarrow a_{4} + a_{6} = 4 \\ \Rightarrow a_{7} + a_{9} = 4 \end{matrix}\}$ (i)

Also, $[\begin{matrix} a_{1} & a_{2} & a_{3} \\ a_{4} & a_{5} & a_{6} \\ a_{7} & a_{8} & a_{9} \end{matrix}] [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 1 \\ 3 \\ 1 \end{matrix}]$

$\begin{matrix} \Rightarrow a_{3} = 1 \Rightarrow a_{1} = 2 \\ \Rightarrow a_{6} = 3 \Rightarrow a_{4} = 1 \\ \Rightarrow a_{9} = 1 \Rightarrow a_{7} = 3 \end{matrix}\}$ [From (i)]

Now, $[\begin{matrix} a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix}] [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 3 \\ 1 \\ 1 \end{matrix}]$

$\begin{matrix} \Rightarrow a_{7} = 3 \\ \Rightarrow a_{8} = 1 \\ \Rightarrow a_{9} = 1 \end{matrix}\}$ (ii)

Also, $[\begin{matrix} a_{1} & a_{4} & a_{7} \\ a_{2} & a_{5} & a_{8} \\ a_{3} & a_{6} & a_{9} \end{matrix}] [\begin{matrix} 1 \\ 0 \\ 1 \end{matrix}] = [\begin{matrix} 5 \\ 2 \\ 2 \end{matrix}]$

$\begin{matrix} \Rightarrow a_{1} + a_{7} = 5 \Rightarrow a_{1} = 2 \\ \Rightarrow a_{2} + a_{8} = 2 \Rightarrow a_{2} = 1 \\ \Rightarrow a_{3} + a_{9} = 2 \Rightarrow a_{3} = 1 \end{matrix}\}$ [From (ii)]

$| A | = | \begin{matrix} 2 & 1 & 1 \\ 1 & a_{5} & 3 \\ 3 & 1 & 1 \end{matrix} | \Rightarrow - a_{5} + 3 = 1 \Rightarrow a_{5} = 2$

$∴ A = | \begin{matrix} 2 & 1 & 1 \\ 1 & 2 & 3 \\ 3 & 1 & 1 \end{matrix} |$

Now, $| adj A \cdot (A + I) | = | adj (A + I) | | adj A |$

$| A + I | = | \begin{matrix} 3 & 1 & 1 \\ 1 & 3 & 3 \\ 3 & 1 & 2 \end{matrix} | = 8$

$∴ | adj (A^{2} + A) | = 64 \times 1 = 64$

Solution

1 16

2 25

3 49

4 64

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