First slide

Applications of determinant

Question

If A is singular matrix, then adj A is

Moderate

A

singular
B

non-singular
C

symmetric
D

not defined

Solution

$Since A is \underline{\underline{singular}} | A | = 0$
$\begin{aligned} | a d j A | = | A |^{n - 1} \\ | a d j A | = 0^{n - 1} \\ ∣ adj A ∣= 0 \\ \Rightarrow A d j A is singular \end{aligned}$

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