First slide

Applications of determinant

Question

$If A = [\begin{array}{ccc} - 1 & - 2 & - 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{array}], Adj A = x A^{T}, then x =$

Moderate

A

0
B

1
C

2
D

3

Solution

$Cofactor of - 1_{11} is A_{11} = (- 1)^{1 + 1} |\begin{array}{cc} 1 & - 2 \\ - 2 & 1 \end{array}|$
$\begin{aligned} A_{11} = 1 - 4 \\ A_{11} = - 3 \end{aligned}$
$Given that Adj A = 3 A^{T}$
$\begin{aligned} - 3 = x (- 1) \\ x = 3 \end{aligned}$

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