Let A be a square matrix of order 2 such that \(A\begin{bmatrix}1\\1\end{bmatrix} = \begin{bmatrix}3\\4\end{bmatrix}\) and \(A\begin{bmatrix}1\\2\end{bmatrix} = \begin{bmatrix}5\\7\end{bmatrix}\). Then \(A^{-1}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\)
Let \(A = \begin{bmatrix}a & b\\c & d\end{bmatrix}\). Multiplying out the two given products:
\(a + b = 3, \quad c + d = 4\)
\(a + 2b = 5, \quad c + 2d = 7\)
Subtracting the first line from the second gives b = 2 and d = 3, and then a = 1 and c = 1:
\(A = \begin{bmatrix}1 & 2\\1 & 3\end{bmatrix}\)
\(|A| = 3 - 2 = 1\)
For a matrix of order 2, the adjoint is found by interchanging the diagonal elements and changing the signs of the other two:
\(\mathrm{adj}\,A = \begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix}\)
\( \displaystyle A^{-1} = \frac{1}{|A|}\,\mathrm{adj}\,A = \begin{bmatrix}3 & -2\\-1 & 1\end{bmatrix} \)
Why the other options are incorrect
Option (1) \(\begin{bmatrix}3 & -1\\-2 & 1\end{bmatrix}\) — Incorrect. This is the matrix of cofactors of A, not transposed; the adjoint is its transpose.
Option (2) \(\begin{bmatrix}1 & -2\\-1 & 3\end{bmatrix}\) — Incorrect. Here the signs are changed but the diagonal elements 1 and 3 are not interchanged.
Option (3) \(\begin{bmatrix}-3 & 2\\1 & -1\end{bmatrix}\) — Incorrect. This is \(-\mathrm{adj}\,A\), which would be the inverse only if |A| were −1; here |A| = 1.