MHT CET · Maths · Matrices
The inverse matrix of \(\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\) is
- A \(\left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]\)
- B \(\left[\begin{array}{rrr}\frac{1}{2} & -4 & \frac{5}{2} \\ 1 & -6 & 3 \\ 1 & 2 & -1\end{array}\right]\)
- C \(\frac{1}{2}\left[\begin{array}{lll}1 & 2 & 3 \\ 3 & 2 & 1 \\ 4 & 2 & 3\end{array}\right]\)
- D \(\frac{1}{2}\left[\begin{array}{rrr}1 & -1 & -1 \\ -8 & 6 & -2 \\ 5 & -3 & 1\end{array}\right]\)
Answer & Solution
Correct Answer
(A) \(\left[\begin{array}{rrr}\frac{1}{2} & -\frac{1}{2} & \frac{1}{2} \\ -4 & 3 & -1 \\ \frac{5}{2} & -\frac{3}{2} & \frac{1}{2}\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Let
\(A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\)
\(|A|=0-1(1-9)+2(1-6)\)
\(=8-10=-2 \neq 0\)
\(\operatorname{adj}(A)=\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
\(\therefore A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)
\(=\frac{1}{-2}\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
\(=\left[\begin{array}{rrr}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]\)
\(A=\left[\begin{array}{lll}0 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\)
\(|A|=0-1(1-9)+2(1-6)\)
\(=8-10=-2 \neq 0\)
\(\operatorname{adj}(A)=\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
\(\therefore A^{-1}=\frac{\operatorname{adj}(A)}{|A|}\)
\(=\frac{1}{-2}\left[\begin{array}{rrr}-1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1\end{array}\right]\)
\(=\left[\begin{array}{rrr}1 / 2 & -1 / 2 & 1 / 2 \\ -4 & 3 & -1 \\ 5 / 2 & -3 / 2 & 1 / 2\end{array}\right]\)
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