Given,
AC = B
⇒ [ 2 − 1 − 4 5 ] C = [ − 3 2 ] ⇒ [ 2 − 1 − 4 5 ] C is a 2 × 1 matrix, but [ 2 − 1 − 4 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 1 − 4 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ] \Rightarrow \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix } C = \begin{bmatrix}[r] -3 \ 2 \end{bmatrix } \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix } C \text{ is a } 2 \times 1 \text{ matrix, but} \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix } \text{ is a } 2 \times 2 \text{ matrix}. \\[0.5em] \therefore \text{C is a } 1 \times 2 \text{ matrix}. \\[0.5em] \text{Let matrix C } = \begin{bmatrix}[r] x \ y \end{bmatrix } \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2 & -1 \ -4 & 5 \end{bmatrix } \begin{bmatrix}[r] x \ y \end{bmatrix } = \begin{bmatrix}[r] -3 \ 2 \end{bmatrix } \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2 \times x + (-1) \times y \ -4 \times x + 5 \times y \end{bmatrix } = \begin{bmatrix}[r] -3 \ 2 \end{bmatrix } \\[0.5em] \Rightarrow \begin{bmatrix}[r] 2x - y \ -4x + 5y \end{bmatrix } = \begin{bmatrix}[r] -3 \ 2 \end{bmatrix } \\[0.5em] ⇒ [ 2 − 4 − 1 5 ] C = [ − 3 2 ] ⇒ [ 2 − 4 − 1 5 ] C is a 2 × 1 matrix, but [ 2 − 4 − 1 5 ] is a 2 × 2 matrix . ∴ C is a 1 × 2 matrix . Let matrix C = [ x y ] ⇒ [ 2 − 4 − 1 5 ] [ x y ] = [ − 3 2 ] ⇒ [ 2 × x + ( − 1 ) × y − 4 × x + 5 × y ] = [ − 3 2 ] ⇒ [ 2 x − y − 4 x + 5 y ] = [ − 3 2 ]
By definition of equality of matrices we get,
⇒ 2x - y = -3 or y = 2x + 3 (…Eq 1)
⇒ -4x + 5y = 2 (…Eq 2)
Putting value of y from Eq 1 in Eq 2,
⇒ -4x + 5y = 2− 13 6 -\dfrac{13}{6} − 6 13
Now finding value of y,
y = 2 x + 3 = 2 ( − 13 6 ) + 3 = − 26 6 + 3 = − 26 + 18 6 = − 8 6 = − 4 3 . y = 2x + 3 \\[1em] = 2\Big(-\dfrac{13}{6}\Big) + 3 \\[1em] = -\dfrac{26}{6} + 3 \\[1em] = \dfrac{-26 + 18}{6} \\[1em] = -\dfrac{8}{6} \\[1em] = -\dfrac{4}{3}. y = 2 x + 3 = 2 ( − 6 13 ) + 3 = − 6 26 + 3 = 6 − 26 + 18 = − 6 8 = − 3 4 .
∴ x = − 13 6 and y = − 4 3 . -\dfrac{13}{6} \text{ and y } = -\dfrac{4}{3}. − 6 13 and y = − 3 4 .
Since,
C = [ x y ] ∴ C = [ − 13 6 − 4 3 ] \text{C }= \begin{bmatrix}[r] x \ y \end{bmatrix } \\[1em] \therefore C = \begin{bmatrix}[r] -\dfrac{13}{6} \ -\dfrac{4}{3} \end{bmatrix } C = [ x y ] ∴ C = ⎣ ⎡ − 6 13 − 3 4 ⎦ ⎤
Hence, the matrix C = [ − 13 6 − 4 3 ] . \begin{bmatrix}[r] -\dfrac{13}{6} \ -\dfrac{4}{3} \end{bmatrix }. ⎣ ⎡ − 6 13 − 3 4 ⎦ ⎤ .
