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BITSAT 2022
Exam
BITSAT
Year
2022
Subject
Mathematics
Chapter
Matrices

BITSAT 2022 · Mathematics · Matrices

Let A=[ array ccc 1 & -1 & 1 2 & 1 & -3 1 & 1 & 1 array ] and 10...

Let A=\left[1−1121−3111\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right] and 10 B=\left[422−50α1−23\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & \alpha \\ 1 & -2 & 3\end{array}\right] If BB is the inverse of AA, then the value of α\alpha is

Options

  1. A4
  2. B3
  3. C−4-4
  4. D5

Correct answer

D. 5

Step-by-step solution

A=\left[1−1121−3111\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]\therefore|A|=\left[1−1121−3111\begin{array}{ccc}1 & -1 & 1 \\ 2 & 1 & -3 \\ 1 & 1 & 1\end{array}\right]=1(1+3)+1(2+3)+1(2-1)=4+5+1=10\Rightarrow \operatorname{Adj} A=\left[4−5120−2253\begin{array}{ccc}4 & -5 & 1 \\ 2 & 0 & -2 \\ 2 & 5 & 3\end{array}\right]^T=\left[422−5051−23\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\Rightarrow \quad B=A^{-1}=\frac{1}{10}\left[422−5051−23\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]\left[\because A^{-1}=\frac{1}{|A|} \operatorname{Adj} A\right]\Rightarrow 10 B=\left[422−5051−23\begin{array}{ccc}4 & 2 & 2 \\ -5 & 0 & 5 \\ 1 & -2 & 3\end{array}\right]Hence,Hence,\alpha=5
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