Q.Find X and Y, if

Concept: Matrix Addition — adding or subtracting matrices is done element-wise, so we can treat each entry independently.

(i) Add the two equations:

$(X+Y)+(X-Y)=2X = \begin{bmatrix} 7+3 & 0+0 \\ 2+0 & 5+3 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix}$

So $X = \frac12\begin{bmatrix} 10 & 0 \\ 2 & 8 \end{bmatrix} = \begin{bmatrix} 5 & 0 \\ 1 & 4 \end{bmatrix}$.

Subtract the second from the first:

$(X+Y)-(X-Y)=2Y = \begin{bmatrix} 7-3 & 0-0 \\ 2-0 & 5-3 \end{bmatrix} = \begin{bmatrix} 4 & 0 \\ 2 & 2 \end{bmatrix}$ …

The core idea

Matrices of the same order add and subtract entry-by-entry, so two matrix equations in $X$ and $Y$ behave exactly like a pair of scalar equations. Eliminate one unknown, solve for the other, then back-substitute.

Part (i): $X + Y = A$, $X - Y = B$

with $A = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix}$, $B = \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix}$.

Add the equations to eliminate $Y$:

Subtract the equations to eliminate $X$:

Check: $X + Y = \begin{bmatrix} 7 & 0 \\ 2 & 5 \end{bmatrix} = A$. ✓

Part (ii): $2X + 3Y = C$, $3X + 2Y = D$

with $C = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}$, $D = \begin{bmatrix} 2 & -2 \\ -1 & 5 \end{bmatrix}$.

Eliminate $Y$: multiply the first equation by $2$ and the second by $3$, then subtract:

Eliminate $X$: multiply the first equation by $3$ and the second by $2$, then subtract: …

Method: Eliminating one unknown from a pair of matrix equations

When two matrix equations both involve the same unknown matrices $X$ and $Y$, treat them exactly like a pair of simultaneous scalar equations and combine them so one unknown cancels.

Steps

Step 1: Use that the operations are entry-wise

Matrices of the same order add and subtract entry-by-entry, so any linear combination of the two equations is legitimate — every elimination move from ordinary algebra carries over.

Step 2: Eliminate one unknown …

Common Mistakes

Mistake 1: Scaling only one equation instead of both when the coefficients differ

Why it's wrong: to eliminate $Y$ from $2X+3Y$ and $3X+2Y$ the $Y$-coefficients must be made equal, which needs BOTH equations scaled (here $\times2$ and $\times3$). Correct approach: multiply each full equation by the right constant, then subtract.

Mistake 2: Dividing only some entries by the scalar …