Question

If A is non-singular matrix and if \(A^{-1}=\frac{1}{2}\begin{bmatrix} -10&-4\\2&1\end{bmatrix}\), then adj(A)=

• A. \(\begin{bmatrix} -1&-4\\2&10\end{bmatrix}\)
• B. \(\begin{bmatrix} 10&4\\-2&-1\end{bmatrix}\)
• C. \(\begin{bmatrix} 1&4\\-2&-10\end{bmatrix}\)
• D. \(\begin{bmatrix} -10&-4\\2&1\end{bmatrix}\)
• E. \(\begin{bmatrix} -1&-4\\10&2\end{bmatrix}\)

Answer 1

The Correct Option is B

We know that the adjugate of a matrix \( A \), denoted \( \text{adj}(A) \), is related to the inverse of \( A \) by the following formula: \[ A^{-1} = \frac{1}{\text{det}(A)} \cdot \text{adj}(A) \] Since \( A^{-1} = \frac{1}{2} \begin{bmatrix} -10 & -4 \\ 2 & 1 \end{bmatrix} \), we can find \( \text{adj}(A) \) as follows: \[ \text{adj}(A) = 2 \cdot A^{-1} = 2 \cdot \frac{1}{2} \begin{bmatrix} -10 & -4 \\ 2 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 4 \\ -2 & -1 \end{bmatrix} \]

The correct option is (B) : \(\begin{bmatrix} 10&4\\-2&-1\end{bmatrix}\)

Answer 2

We are given that \(A^{-1}=\frac{1}{2}\begin{bmatrix} -10& -4\\ 2& 1\end{bmatrix}\).

We know that \(A^{-1} = \frac{adj(A)}{|A|}\) which means \(adj(A) = |A| A^{-1}\).

First, let's find \(A^{-1}\):

\(A^{-1}=\frac{1}{2}\begin{bmatrix} -10& -4\\ 2& 1\end{bmatrix} = \begin{bmatrix} -5& -2\\ 1& \frac{1}{2}\end{bmatrix}\)

Now, let's find the determinant of \(A^{-1}\):

\(|A^{-1}| = (-5)(\frac{1}{2}) - (-2)(1) = -\frac{5}{2} + 2 = -\frac{1}{2}\)

Since \(|A^{-1}| = \frac{1}{|A|}\), we have \(|A| = \frac{1}{|A^{-1}|} = \frac{1}{-\frac{1}{2}} = -2\).

Now, we can find adj(A) using the formula \(adj(A) = |A| A^{-1}\):

\(adj(A) = -2 \cdot \frac{1}{2}\begin{bmatrix} -10& -4\\ 2& 1\end{bmatrix} = -1 \cdot \begin{bmatrix} -10& -4\\ 2& 1\end{bmatrix} = \begin{bmatrix} 10& 4\\ -2& -1\end{bmatrix}\)

Therefore, \(adj(A) = \begin{bmatrix} 10& 4\\ -2& -1\end{bmatrix}\).