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BITSAT 2024
Exam
BITSAT
Year
2024
Subject
Mathematics
Chapter
Matrices

BITSAT 2024 · Mathematics · Matrices

If A=(1)/(3)[ array ccc 1 & 2 & 2 2 & 1 & -2 a & 2 & b array ] is...

If A=\frac{1}{3}\left[12221−2a2b\begin{array}{ccc}1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b\end{array}$\right] is an orthogonal matrix, then

Options

  1. Aa=−2,b=−1a=-2, b=-1
  2. Ba=2,b=1a=2, b=1
  3. Ca=2,b=−1a=2, b=-1
  4. Da=−2,b=1a=-2, b=1

Correct answer

A. a=−2,b=−1a=-2, b=-1

Step-by-step solution

We know that, if Ais an orthogonal matrix, then \begin{gathered} \Rightarrow \quad \ AA^T=I \\ \Rightarrow \\ \Rightarrow \quad\left[12221−2a2b\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right] \cdot \frac{1}{3}\left[12a2122−2b\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[100010001\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ \Rightarrow\left[12221−2a2b\begin{array}{ccc} 1 & 2 & 2 \\ 2 & 1 & -2 \\ a & 2 & b \end{array}\right]\left[12a2122−2b\begin{array}{ccc} 1 & 2 & a \\ 2 & 1 & 2 \\ 2 & -2 & b \end{array}\right]=\left[900090009\begin{array}{lll} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \\ \left[90a+4+2b0+4+2b92a+2−2b2a+2−2ba2+4+b2\begin{array}{ccc} 9 & 0 & a+4+2 b \\ 0+4+2 b & 9 & 2 a+2-2 b \\ 2 a+2-2 b & a^2+4+b^2 \end{array}\right] \\ \quad=\left[900090009\begin{array}{ccc} 9 & 0 & 0 \\ 0 & 9 & 0 \\ 0 & 0 & 9 \end{array}\right] \\ \Rightarrow \quad a+4+2 b=0 \text{ and } 2 a+2-2 b=0 \\ a+2 b=-4 \ldots(i)\\ a-b=-1 \ldots(ii) \end{gathered} By solving Eqs. (i) and (ii), we get a=−2,b=−1a=-2, b=-1
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