Problem Statement
Given $A = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$. Find a $2 \times 2$ matrix $X$ such that $AX = B$.
Verified Solution & Marking Scheme
Set up matrix X with unknown elements
Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$.
$AX = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix} \begin{bmatrix} a & b \\ c & d \end{bmatrix} = \begin{bmatrix} a + 3c & b + 3d \\ 2a + 4c & 2b + 4d \end{bmatrix}$
Given $AX = \begin{bmatrix} 1 & 2 \\ 2 & 4 \end{bmatrix}$.
Equate elements and solve for a and c
1. $a + 3c = 1 \implies a = 1 - 3c$
2. $2a + 4c = 2 \implies a + 2c = 1$
Substitute $a = 1 - 3c$:
$(1 - 3c) + 2c = 1 \implies 1 - c = 1 \implies c = 0$
Then $a = 1 - 3(0) = 1$.
Equate elements and solve for b and d
3. $b + 3d = 2 \implies b = 2 - 3d$
4. $2b + 4d = 4 \implies b + 2d = 2$
Substitute $b = 2 - 3d$:
$(2 - 3d) + 2d = 2 \implies 2 - d = 2 \implies d = 0$
Then $b = 2 - 3(0) = 2$.
Therefore:
$X = \begin{bmatrix} 1 & 2 \\ 0 & 0 \end{bmatrix}$