COMEDK · Maths · 21. Matrices
If \(A=\left[\begin{array}{ccc}-1 & 1 & 2 \\ 1 & 2 & 3 \\ 3 & 1 & 1\end{array}\right]\) then the inverse of \((A I)^t\) (where \(\mathrm{I}\) is an identity matrix) is
- A \(\left[\begin{array}{ccc}
1 & 8 & -5 \\
-1 & 7 & -4 \\
0 & 5 & 3
\end{array}\right]\) - B \(\left[\begin{array}{ccc}
1 & -1 & 1 \\
-8 & 7 & -5 \\
5 & -4 & 3
\end{array}\right]\) - C \(\left[\begin{array}{ccc}
1 & -8 & 5 \\
-1 & 7 & -4 \\
1 & -5 & 3
\end{array}\right]\) - D \(\left[\begin{array}{ccc}
-1 & 1 & -1 \\
8 & -7 & 5 \\
-5 & 4 & -3
\end{array}\right]\)
Answer & Solution
Correct Answer
(C) \(\left[\begin{array}{ccc}
1 & -8 & 5 \\
-1 & 7 & -4 \\
1 & -5 & 3
\end{array}\right]\)
Step-by-step Solution
Detailed explanation
Since \(AI = A\), we need \((A^t)^{-1}\). \(A^t = \begin{bmatrix} -1 & 1 & 3 \\ 1 & 2 & 1 \\ 2 & 3 & 1 \end{bmatrix}\) \(|A^t| = -1(2-3) - 1(1-2) + 3(3-4) = 1 + 1 - 3 = -1\) Cofactors of \(A^t\): \(C_{11} = -1,\ C_{12} = 1,\ C_{13} = -1\)…
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