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BITSAT 2024 Shift 1 Paper

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Question : 67 of 130

Marks: +1, -0

If

A=\frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}

is an orthogonal matrix, then

$a=-2,\; b=-1$
$a=2,\; b=1$
$a=2,\; b=-1$
$a=-2,\; b=1$

Solution:

We know that, if

A

is an orthogonal matrix, then

A A^T = I

\Rightarrow \frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix} \cdot \frac{1}{3}\begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}

\Rightarrow \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{bmatrix}\begin{bmatrix} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{bmatrix} = \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}

\Rightarrow \begin{bmatrix} 9 & 0 & a+4+2b \\ 0 & 9 & 2a+2-2b \\ a+2b & 2a+2-2b & a^2+4+b^2 \end{bmatrix}

= \begin{bmatrix} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{bmatrix}

\Rightarrow \;\; a+4+2b=0

and

2a+2-2b=0

\Rightarrow \;\; a+2b=-4 \;\; \dots

(i)

and

a-b=-1 \;\; \dots

(ii)

By solving Eqs. (i) and (ii), we get

a=-2,\; b=-1

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