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Worked Examples · Example 9

Q.Find XX and YY, if X+Y=[5209]X + Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} and X−Y=[360−1]X - Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}.

Puducherry CbseNCERTSubjective· 3mImportance★★★★★
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✓ Free question

We solve a system of two matrix equations by adding and subtracting them, just like solving for two numbers. Adding eliminates YY to find XX, and subtracting eliminates XX to find YY. The result is X=[4404]X = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix} and Y=[1−205]Y = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}.

The problem gives you two matrix equations:

X+Y=[5209],X−Y=[360−1].X + Y = \begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix}, \quad X - Y = \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix}.

You need to find the matrices XX and YY. This looks exactly like a system of linear equations — except the “unknowns” are matrices, not numbers. But the same logic works: if you had x+y=ax+y = a and x−y=bx-y = b, you’d add to get 2x=a+b2x = a+b, then subtract to get 2y=a−b2y = a-b. Matrix addition works entry-wise, so the same trick applies perfectly.

Let’s do it step by step.

  1. Add the two equations Adding the left sides: (X+Y)+(X−Y)=2X(X+Y) + (X-Y) = 2X. Adding the right sides:

[5209]+[360−1]=[5+32+60+09+(−1)]=[8808].\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} + \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 5+3 & 2+6 \\ 0+0 & 9+(-1) \end{bmatrix} = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}.

So 2X=[8808]2X = \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix}.

  1. Solve for XX Divide every entry by 2:

X=12[8808]=[4404].X = \frac{1}{2} \begin{bmatrix} 8 & 8 \\ 0 & 8 \end{bmatrix} = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix}.

  1. Subtract the second equation from the first Left sides: (X+Y)−(X−Y)=2Y(X+Y) - (X-Y) = 2Y. Right sides:

[5209]−[360−1]=[5−32−60−09−(−1)]=[2−4010].\begin{bmatrix} 5 & 2 \\ 0 & 9 \end{bmatrix} - \begin{bmatrix} 3 & 6 \\ 0 & -1 \end{bmatrix} = \begin{bmatrix} 5-3 & 2-6 \\ 0-0 & 9-(-1) \end{bmatrix} = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}.

So 2Y=[2−4010]2Y = \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix}.

  1. Solve for YY Divide by 2:

Y=12[2−4010]=[1−205].Y = \frac{1}{2} \begin{bmatrix} 2 & -4 \\ 0 & 10 \end{bmatrix} = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}.

Tip

You can always check your answer by plugging back:

X+Y=[4+14+(−2)0+04+5]=[5209]X+Y = \begin{bmatrix}4+1 & 4+(-2) \\ 0+0 & 4+5\end{bmatrix} = \begin{bmatrix}5 & 2 \\ 0 & 9\end{bmatrix} — matches.

X−Y=[4−14−(−2)0−04−5]=[360−1]X-Y = \begin{bmatrix}4-1 & 4-(-2) \\ 0-0 & 4-5\end{bmatrix} = \begin{bmatrix}3 & 6 \\ 0 & -1\end{bmatrix} — matches.

Watch out

A common mistake is to forget that matrix subtraction is entry-wise, just like addition. Also, when subtracting, be careful with signs: 9−(−1)=109 - (-1) = 10, not 88.

✓Final answer

The matrices are X=[4404]X = \begin{bmatrix} 4 & 4 \\ 0 & 4 \end{bmatrix} and Y=[1−205]Y = \begin{bmatrix} 1 & -2 \\ 0 & 5 \end{bmatrix}.

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