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NCERT Exemplar · Q19

Q.If XX and YY are 2×22 \times 2 matrices, then solve the following matrix equations for XX and YY: 2X+3Y=[2340]2X + 3Y = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}, 3X+2Y=[−221−5]3X + 2Y = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}.

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We treat the two matrix equations as a system of linear equations in the unknown matrices XX and YY. By eliminating one variable (using the same elimination method as for numbers), we solve for XX and YY individually. The final answers are X=[−20−1−3]X = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix} and Y=[2122]Y = \begin{bmatrix} 2 & 1 \\ 2 & 2 \end{bmatrix}.

The key insight is that matrix equations of this form behave exactly like ordinary linear equations — as long as we remember that matrix multiplication is not involved here. Each equation is a linear combination of the unknown matrices XX and YY with scalar coefficients (2, 3, etc.). So we can solve them using the same elimination or substitution method we use for numbers.

Think of it this way: if we had 2x+3y=a2x + 3y = a and 3x+2y=b3x + 2y = b, we’d solve for xx and yy by eliminating one variable. The same logic applies here — except xx and yy are 2×22 \times 2 matrices, and aa and bb are the given 2×22 \times 2 matrices on the right-hand sides.

Let’s label the equations for clarity:

2X+3Y=A,where A=[2340]2X + 3Y = A, \quad \text{where } A = \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix}

3X+2Y=B,where B=[−221−5]3X + 2Y = B, \quad \text{where } B = \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix}

We want to isolate XX and YY.


Step-by-step solution

Step 1: Eliminate YY to find XX.

Multiply equation (1) by 2 and equation (2) by 3, so that the coefficients of YY become 6 and 6 — then subtract.

From (1):

2×(2X+3Y)=2A⇒4X+6Y=2A2 \times (2X + 3Y) = 2A \quad \Rightarrow \quad 4X + 6Y = 2A

From (2):

3×(3X+2Y)=3B⇒9X+6Y=3B3 \times (3X + 2Y) = 3B \quad \Rightarrow \quad 9X + 6Y = 3B

Now subtract the first result from the second:

(9X+6Y)−(4X+6Y)=3B−2A(9X + 6Y) - (4X + 6Y) = 3B - 2A

The 6Y6Y terms cancel, leaving:

5X=3B−2A5X = 3B - 2A

Step 2: Compute 3B−2A3B - 2A.

First, 3B3B:

3×[−221−5]=[−663−15]3 \times \begin{bmatrix} -2 & 2 \\ 1 & -5 \end{bmatrix} = \begin{bmatrix} -6 & 6 \\ 3 & -15 \end{bmatrix}

Next, 2A2A:

2×[2340]=[4680]2 \times \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 4 & 6 \\ 8 & 0 \end{bmatrix}

Now subtract:

3B−2A=[−6−46−63−8−15−0]=[−100−5−15]3B - 2A = \begin{bmatrix} -6 - 4 & 6 - 6 \\ 3 - 8 & -15 - 0 \end{bmatrix} = \begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix}

So we have:

5X=[−100−5−15]5X = \begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix}

Step 3: Solve for XX.

Divide both sides by 5 (i.e., multiply by 15\frac{1}{5}):

X=15[−100−5−15]=[−20−1−3]X = \frac{1}{5} \begin{bmatrix} -10 & 0 \\ -5 & -15 \end{bmatrix} = \begin{bmatrix} -2 & 0 \\ -1 & -3 \end{bmatrix}

Tip

Dividing a matrix by a scalar means dividing every entry — no matrix inversion needed here.

Step 4: Eliminate XX to find YY.

We can use a similar trick. Multiply equation (1) by 3 and equation (2) by 2, so the coefficients of XX become 6 and 6.

From (1):

3×(2X+3Y)=3A⇒6X+9Y=3A3 \times (2X + 3Y) = 3A \quad \Rightarrow \quad 6X + 9Y = 3A

From (2):

2×(3X+2Y)=2B⇒6X+4Y=2B2 \times (3X + 2Y) = 2B \quad \Rightarrow \quad 6X + 4Y = 2B

Now subtract the second from the first:

(6X+9Y)−(6X+4Y)=3A−2B(6X + 9Y) - (6X + 4Y) = 3A - 2B

The 6X6X terms cancel, giving:

5Y=3A−2B5Y = 3A - 2B

Step 5: Compute 3A−2B3A - 2B.

First, 3A3A:

3×[2340]=[69120]3 \times \begin{bmatrix} 2 & 3 \\ 4 & 0 \end{bmatrix} = \begin{bmatrix} 6 & 9 \\ 12 & 0 \end{bmatrix}

Next, 2B2B: …

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