First slide

Applications of determinant

Question

The number of matrices $A = [\begin{array}{ll} a b \\ c d \end{array}]$ (where $a, b, c, d \in R$ ) such that $A^{- 1} = - A$ is:

Moderate

A

0
B

1
C

2
D

infinite

Solution

As $A^{- 1}$ exists, $\det (A) = a d - b c \neq 0$ .
Also, $\begin{array}{l} \det (- A) = \det [\begin{array}{rr} - a & - b \\ - c & - d \end{array}] = a d - b c \\ = \det (A) \end{array}$
Since $\det (A^{- 1}) = \frac{1}{\det (A)}, A^{- 1} = - A$ implies
$\begin{array}{l} \frac{1}{a d - b c} = a d - b c \\ \Rightarrow (a d - b c)^{2} = 1 \Rightarrow a d - b c = \pm 1 \end{array}$
If $a d - b c = 1$ , then $A^{- 1} = - A$
gives
$\begin{array}{l} A^{- 1} = [\begin{array}{lc} d & - b \\ - c & a \end{array}] = [\begin{array}{cc} - a & - b \\ - c & - d \end{array}] \\ \Rightarrow a + d = 0 \\ ∴ a (- a) - b c = 1 \end{array}$
$\Rightarrow b c = - (1 + a^{2}) \Rightarrow b c \neq 0$ and
$b = - \frac{1 + a^{2}}{c}$
$∴ A = [\begin{array}{cc} c & (1 + a^{2}) \\ a & c \\ c & - a \end{array}],$ where $c \neq 0$
Thus, there are infinite number of such matrices.

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