If A=[102−11−2021] and adj A=[5a−2110−2−2b] then (b, a)=

A⋅adj A=|A| I [102−11−2021][5a−2110−2−26]=[100010001] ⇒[1a−42b−205−a−2−2b00b]=[100010001]⇒ a−4=0 a=4, b=1 (b,a)=(1,4)

If A=[102−11−2021] and adj A=[5a−2110−2−2b] then (b, a)=

A⋅adj A=|A| I [102−11−2021][5a−2110−2−26]=[100010001] ⇒[1a−42b−205−a−2−2b00b]=[100010001]⇒ a−4=0 a=4, b=1 (b,a)=(1,4)

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If $A = [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & - 2 \\ 0 & 2 & 1 \end{matrix}]$ and $a d j A = [\begin{matrix} 5 & a & - 2 \\ 1 & 1 & 0 \\ - 2 & - 2 & b \end{matrix}]$ then $(b, a) =$

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$(- 1, - 4)$

$(- 4, 1)$

$(4, 1)$

$(1, 4)$

answer is B.

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Detailed Solution

$A \cdot a d j A = | A | I [\begin{matrix} 1 & 0 & 2 \\ - 1 & 1 & - 2 \\ 0 & 2 & 1 \end{matrix}] [\begin{matrix} 5 & a & - 2 \\ 1 & 1 & 0 \\ - 2 & - 2 & 6 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow [\begin{matrix} 1 & a - 4 & 2 b - 2 \\ 0 & 5 - a & - 2 - 2 b \\ 0 & 0 & b \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] \Rightarrow a - 4 = 0 a = 4, b = 1 (b, a) = (1, 4)$