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Let A and B be square matrices of 3rd order and A be an orthogonal matrix and B is a skew symmetric matrix. Then which of the following is not true.

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Numerical value of |A| is 1

|B| = 0

|AB| = 1

|AB| = 0

answer is C.

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Detailed Solution

$A A^{T} = I, B^{T} = - B$

1) $\begin{array}{rcl} |A A^{T}| & = & 1 \\ | A |^{2} & = & 1 \Rightarrow | A | = \pm 1 \end{array}$

(1) is true

2) $|B^{T}| = | - B |$

$\begin{array}{rcl} | B | & = & - | B | \\ | B | & = & 0 \end{array}$

(2) is true

3) $| AB | = 1$ is false

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