If A=a1 b1 c1a2 b2 c2a3 b3 c3 is a non-singular matrix then prove that A−1=Adj⁡Adet⁡A

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If $A = [\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}]$ is a non-singular matrix then prove that $A^{- 1} = \frac{Adj A}{\det A}$

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Detailed Solution

Given $A = [\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}]$ is a non singular

$Matrix \Rightarrow \det A \neq 0$

Let cofactors of $a_{1}, b_{1}, c_{1} \dots \dots$ are $A_{1}, B_{1}, C_{1} \dots \dots$ respectively

$∴ Adj A = {[\begin{array}{lll} A_{1} B_{1} C_{1} \\ A_{2} B_{2} C_{2} \\ A_{3} B_{3} C_{3} \end{array}]}^{T} = [\begin{array}{lll} A_{1} A_{2} A_{3} \\ B_{1} B_{2} B_{3} \\ C_{1} C_{2} C_{3} \end{array}]$

$A. (AdjA) = [\begin{array}{lll} a_{1} b_{1} c_{1} \\ a_{2} b_{2} c_{2} \\ a_{3} b_{3} c_{3} \end{array}] [\begin{array}{lll} A_{1} A_{2} A_{3} \\ B_{1} B_{2} B_{3} \\ C_{1} C_{2} C_{3} \end{array}]$

$= [\begin{array}{lll} a_{1} A_{1} + b_{1} B_{1} + c_{1} C_{1} a_{1} A_{2} + b_{1} B_{2} + c_{1} C_{2} a_{1} A_{3} + b_{1} B_{3} + c_{1} C_{3} \\ a_{2} A_{1} + b_{2} B_{1} + c_{2} C_{1} a_{2} A_{2} + b_{2} B_{2} + c_{2} C_{2} a_{2} A_{3} + b_{2} B_{3} + c_{2} C_{3} \\ a_{3} A_{1} + b_{3} B_{1} + c_{3} C_{1} a_{3} A_{2} + b_{3} B_{2} + c_{2} C_{2} a_{3} A_{3} + b_{3} B_{3} + c_{3} C_{3} \end{array}]$

$= [\begin{array}{ccc} \det A & 0 & 0 \\ 0 & \det A & 0 \\ 0 & 0 & \det A \end{array}] = \det A [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$