Q9 Q3.) Solve the matrix equation 4X + 5A = B for X.

ANSWER

[tex]X = \begin{bmatrix} 1 & 7/4 \\ 8&39/4\\ -4&23/2\end{bmatrix}[/tex]

EXPLANATION
Solve for x, noting that scalar multiplication on matrices is distributive:

[tex]\begin{aligned} 4X + 5A &= B \\ 4X &= B + (-5A) \\ X &= \tfrac{1}{4}( B + (-5A) ) \\ X &= \tfrac{1}{4} B + (-\tfrac{5}{4}A) \end{aligned}[/tex]

Note that

[tex]\begin{aligned} \tfrac{1}{4} B &= \frac{1}{4} \begin{bmatrix} 4 & -8 \\ 2 & -1 \\ 9 & 1 \end{bmatrix} \\ \\ &= \begin{bmatrix} (1/4) \cdot 4 & (1/4) \cdot-8 \\ (1/4) \cdot2 & (1/4)\cdot-1 \\ (1/4)\cdot9 & (1/4)\cdot1 \end{bmatrix} \\ \\ &= \begin{bmatrix} 1 & -2 \\ 1/2 & -1/4\\ 9/4 & 1/4\end{bmatrix} \end{aligned}[/tex]

and

[tex]\begin{aligned} (-\tfrac{5}{4}A) &= -\frac{5}{4} \begin{bmatrix} 0 & -3 \\ -6 & -8 \\ 5 & -9 \end{bmatrix} \\ \\ &= \begin{bmatrix} (-5/4)\cdot 0 &(-5/4)\cdot -3 \\ (-5/4)\cdot-6 &(-5/4)\cdot -8 \\ (-5/4)\cdot5 &(-5/4)\cdot -9 \end{bmatrix} \\ \\ &= \begin{bmatrix} 0 & 15/4 \\ 15/2 & 10 \\ -25/4 & 45/4 \end{bmatrix} \end{aligned} [/tex]

Therefore, since we add corresponding entries with matrix addition:

[tex]\begin{aligned} X &= \tfrac{1}{4} B + (-\tfrac{5}{4}A) \\ &= \begin{bmatrix} 1 & -2 \\ 1/2 & -1/4\\ 9/4 & 1/4 \end{bmatrix} + \begin{bmatrix} 0 & 15/4 \\ 15/2 & 10 \\ -25/4 & 45/4 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 7/4 \\ 8&39/4\\ -4&23/2\end{bmatrix} \end{aligned}[/tex]