First slide

Applications of determinant

Question

$A is an involutary matrix given by A = [\begin{array}{ccc} 0 & 1 & - 1 \\ 4 & - 3 & 4 \\ 3 & - 3 & 4 \end{array}], then the inverse of A / 2 will$

Moderate

A

$2 A$
B

$\frac{A^{- 1}}{2}$
C

$\frac{A}{2}$
D

$A^{2}$

Solution

A is involutary . Hence,
$A^{2} = I \Rightarrow A = A^{- 1}$
Also,
$\begin{array}{l} (k A)^{- 1} = \frac{1}{k} (A)^{- 1} \\ \Rightarrow {(\frac{1}{2} A)}^{- 1} \\ = 2 (A)^{- 1} \\ = 2 A \end{array}$

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