If X is a square matrix of order 2 such that \(X\begin{bmatrix}1&2\\0&1\end{bmatrix} = \begin{bmatrix}3&4\\1&5\end{bmatrix}\), then X is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\begin{bmatrix}3&-2\\1&3\end{bmatrix}\)
Let \(X = \begin{bmatrix}a&b\\c&d\end{bmatrix}\). Multiplying row by column:
\( \displaystyle \begin{bmatrix}a&b\\c&d\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix} = \begin{bmatrix}a&2a + b\\c&2c + d\end{bmatrix} \)
By equality of matrices with \(\begin{bmatrix}3&4\\1&5\end{bmatrix}\),
\(a = 3\), and \(6 + b = 4\), so \(b = -2\)
\(c = 1\), and \(2 + d = 5\), so \(d = 3\)
\( \displaystyle X = \begin{bmatrix}3&-2\\1&3\end{bmatrix} \)
Why the other options are incorrect
Option (2) \(\begin{bmatrix}1&-6\\1&5\end{bmatrix}\) — Incorrect. This solves the equation with X on the right of the given matrix, which is a different equation because matrix multiplication is not commutative; here XB has first row (1, −4), not (3, 4).
Option (3) \(\begin{bmatrix}3&10\\1&7\end{bmatrix}\) — Incorrect. This is \(\begin{bmatrix}3&4\\1&5\end{bmatrix}\begin{bmatrix}1&2\\0&1\end{bmatrix}\), a multiplication instead of solving for X.
Option (4) \(\begin{bmatrix}3&2\\1&-3\end{bmatrix}\) — Incorrect. Its product with the given matrix has first row (3, 8), not (3, 4).