(i)
On equating the corresponding elements, we get
⇒ 2a - c = - 1 ........... (i)
2b - d = - 8 .......... (ii)
a = 1 ......... (iii)
b = - 2 ......... (iv)
- 3a + 4c = 9 .......... (v)
- 3b + 4d = 22 .......... (vi)
On solving these equations, we get
a = 1, b = - 2, c = 3, d = 4
Thus, A = \(\begin{bmatrix} 1 & -2 \\ 3 & 4 \\ \end{bmatrix}\)
(ii) It is given that:
A \(\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ \end{bmatrix}\) = \(\begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \\ \end{bmatrix}\)
The matrix given on the R.H.S. of the equation is a 2 × 3 matrix and the one given on the L.H.S. of the equation is a 2 × 3 matrix.
Therefore, A is a 2 × 2 matrix.
Equating the corresponding elements of the two matrices, we get
a + 4c = - 7
2a + 5c = 8 ........ (ii)
3a + 6c = - 9 ......... (iii)
b + 4d = 2 ....... (iv)
2b + 5d = 4 ........ (v)
3b + 6d = 6 ........ (vi)
Now, from (i), we have
a + 4c = - 7
⇒ a - 7 - 4c .......... (vii)
From (ii), we have
∴ 2a + 5c = - 8 ⇒ - 14 - 8c + 5c = - 8
⇒ - 3c = 6 ⇒ c = - 2
Substituting c = - 2 in Eq. (vii), we get
∴ a = - 7 - 4(- 2)
= - 7 + 8 = 1
Now, from (iv), we get
b + 4d = 2
⇒ b = 2 - 4d ...... (viii)
From eqs. (y) and (viii), we get
⇒ 4 - 8d + 5d = 4
⇒ - 3d = 0
⇒ d = 0
Substituting d = 0 in Eq. (viii), we get
∴ b = 2 - 4(0) = 2
Thus, a = 1, b = 2, c = - 2, d = 0
Hence, the required matrix A is \(\begin{bmatrix} 1 & -2 \\ 2 & 0 \\ \end{bmatrix}\)
