Given that
The matrix given on the RHS of the equation is a 2 × 3matrix and the one given on the LHS of the equation is as a 2 × 3 matrix. Therefore, X has to be a 2 × 2 matrix.
Equating the corresponding elements of the two matrices, we have
a + 4c = − 7, 2a + 5c = −8, 3a + 6c = −9
b + 4d = 2, 2b + 5d = 4, 3b + 6d = 6
Now, a + 4c = − 7 ⇒ a = − 7 − 4c
2a + 5c = −8 ⇒ −14 − 8c + 5c = −8
⇒−3c = 6 ⇒ c = −2
∴ a = − 7 − 4(−2) = − 7 + 8 = 1
Now, b + 4d = 2 ⇒ b = 2 − 4d and 2b + 5d = 4 ⇒ 4 − 8d + 5d = 4
∴ −3d = 0 ⇒ d = 0
∴ b = 2 − 4(0) = 2
Thus, a = 1, b = 2, c = − 2, d = 0
Hence, the required matrix X is [(1, -2), (2, 0)]
