If $A = ∣ ∣ ∣ ∣ \begin{matrix} x & 5 & 2 2 & y & 3 1 & 1 & z \end{matrix} ∣ ∣ ∣ ∣, x y z = 80, 3 x + 2 y + 10 z = 20,$ then prove that A adj A $∣ ∣ ∣ ∣ \begin{matrix} 79 & 0 & 0 0 & 79 & 0 0 & 0 & 79 \end{matrix} ∣ ∣ ∣ ∣$

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Solution

A = $∣ ∣ ∣ ∣ \begin{matrix} x & 5 & 2 2 & y & 3 1 & 1 & z \end{matrix} ∣ ∣ ∣ ∣$ xyz=80
3x+2y+10z=80
$A a d j A = ∣ ∣ ∣ ∣ \begin{matrix} 79 & 0 & 0 0 & 79 & 0 0 & 0 & 79 \end{matrix} ∣ ∣ ∣ ∣$
A=x(yz-3)-5(2z-3)+2(2-y)
=xyz-3x-10z+15+4-2y
=80+15+4-(3x+2y+10z)
=80+19-20
=79
$A (a d j A) = | A | I = ∣ ∣ ∣ ∣ ∣ \begin{matrix} | A | & 0 & 0 0 & | A | & 0 0 & 0 & | A | \end{matrix} ∣ ∣ ∣ ∣ ∣$
$| A | = 79$
$∴ A (a d j A) = ∣ ∣ ∣ ∣ \begin{matrix} 79 & 0 & 0 0 & 79 & 0 0 & 0 & 79 \end{matrix} ∣ ∣ ∣ ∣$

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