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CBSE Sample Paper 2025-26 · Section D · 5 marks

Q32: Find ABAB and hence solve a system of three equations

Sample 2025-26Ch 3 · MatricesCh 4 · Determinants
The question

For two matrices

A=[3−6−12−5−1−241]A = \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ -2 & 4 & 1 \end{bmatrix}

and

B=[1−2−10−1−1203],B = \begin{bmatrix} 1 & -2 & -1 \\ 0 & -1 & -1 \\ 2 & 0 & 3 \end{bmatrix},

find the product ABAB and hence solve the system of equations:

3x−6y−z=33x - 6y - z = 3

2x−5y−z+2=02x - 5y - z + 2 = 0

−2x+4y+z=5-2x + 4y + z = 5

Exam answer
AB=[3−6−12−5−1−241]×[1−2−10−1−1203]=[100010001]=I\begin{aligned} AB &= \begin{bmatrix} 3 & -6 & -1 \\ 2 & -5 & -1 \\ -2 & 4 & 1 \end{bmatrix} \\ &\quad\times\begin{bmatrix} 1 & -2 & -1 \\ 0 & -1 & -1 \\ 2 & 0 & 3 \end{bmatrix} \\ &= \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = I \end{aligned}
Why?

Each entry of ABAB is a row of AA times a column of BB⁠: multiply matching entries and add. For example, row 1 of AA times column 1 of BB⁠:

3(1)+(−6)(0)+(−1)(2)=3+0−2=1\begin{aligned} &3(1) + (-6)(0) + (-1)(2) \\ &= 3 + 0 - 2 = 1 \end{aligned}

Watch the signs: two negatives multiply to a positive. Row 1 times column 2 is 3(−2)+(−6)(−1)+(−1)(0)3(-2) + (-6)(-1) + (-1)(0)⁠, which is −6+6+0=0-6 + 6 + 0 = 0⁠.

All nine entries, row by row:

Row 1: 3+0−2=13 + 0 - 2 = 1⁠, −6+6+0=0-6 + 6 + 0 = 0⁠, −3+6−3=0-3 + 6 - 3 = 0⁠.

Row 2: 2+0−2=02 + 0 - 2 = 0⁠, −4+5+0=1-4 + 5 + 0 = 1⁠, −2+5−3=0-2 + 5 - 3 = 0⁠.

Row 3: −2+0+2=0-2 + 0 + 2 = 0⁠, 4−4+0=04 - 4 + 0 = 0⁠, 2−4+3=12 - 4 + 3 = 1⁠.

In full, with each product written out. The columns of BB are (1,0,2)(1, 0, 2)⁠, (−2,−1,0)(-2, -1, 0) and (−1,−1,3)(-1, -1, 3)⁠.

Row 1 of AA is (3,−6,−1)(3, -6, -1)⁠:

(1,1):3(1)+(−6)(0)+(−1)(2)=3+0−2=1(1,2):3(−2)+(−6)(−1)+(−1)(0)=−6+6+0=0(1,3):3(−1)+(−6)(−1)+(−1)(3)=−3+6−3=0\begin{aligned} (1, 1)&: 3(1) + (-6)(0) + (-1)(2) \\ &= 3 + 0 - 2 = 1 \\[4pt] (1, 2)&: 3(-2) + (-6)(-1) + (-1)(0) \\ &= -6 + 6 + 0 = 0 \\[4pt] (1, 3)&: 3(-1) + (-6)(-1) + (-1)(3) \\ &= -3 + 6 - 3 = 0 \end{aligned}

Row 2 of AA is (2,−5,−1)(2, -5, -1)⁠:

(2,1):2(1)+(−5)(0)+(−1)(2)=2+0−2=0(2,2):2(−2)+(−5)(−1)+(−1)(0)=−4+5+0=1(2,3):2(−1)+(−5)(−1)+(−1)(3)=−2+5−3=0\begin{aligned} (2, 1)&: 2(1) + (-5)(0) + (-1)(2) \\ &= 2 + 0 - 2 = 0 \\[4pt] (2, 2)&: 2(-2) + (-5)(-1) + (-1)(0) \\ &= -4 + 5 + 0 = 1 \\[4pt] (2, 3)&: 2(-1) + (-5)(-1) + (-1)(3) \\ &= -2 + 5 - 3 = 0 \end{aligned}

Row 3 of AA is (−2,4,1)(-2, 4, 1)⁠:

(3,1):(−2)(1)+4(0)+1(2)=−2+0+2=0(3,2):(−2)(−2)+4(−1)+1(0)=4−4+0=0(3,3):(−2)(−1)+4(−1)+1(3)=2−4+3=1\begin{aligned} (3, 1)&: (-2)(1) + 4(0) + 1(2) \\ &= -2 + 0 + 2 = 0 \\[4pt] (3, 2)&: (-2)(-2) + 4(-1) + 1(0) \\ &= 4 - 4 + 0 = 0 \\[4pt] (3, 3)&: (-2)(-1) + 4(-1) + 1(3) \\ &= 2 - 4 + 3 = 1 \end{aligned}

So, A−1=BA^{-1} = B (and B−1=AB^{-1} = A⁠).

Why?

The inverse of a square matrix AA is the matrix that multiplies with it to give the identity:

AA−1=A−1A=IAA^{-1} = A^{-1}A = I

For square matrices, AB=IAB = I is enough to conclude B=A−1B = A^{-1}⁠.

The given system of equations is

3x−6y−z=33x - 6y - z = 3⁠, 2x−5y−z=−22x - 5y - z = -2⁠, −2x+4y+z=5-2x + 4y + z = 5⁠.

Why?

The second equation 2x−5y−z+2=02x - 5y - z + 2 = 0 is first written with the constant on the right: subtract 2 from both sides.

In matrix form it can be written as AX=CAX = C⁠, where

X=[xyz],C=[3−25]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\quad C = \begin{bmatrix} 3 \\ -2 \\ 5 \end{bmatrix}
Why?

The rows of coefficients of xx⁠, yy and zz are exactly the rows of AA⁠: (3, −6-6⁠, −1-1⁠), (2, −5-5⁠, −1-1⁠) and (−2-2⁠, 4, 1). That is why the product ABAB was asked for first.

∣A∣=3(−5+4)+6(2−2)−1(8−10)=−3+0+2=−1≠0\begin{aligned} |A| &= 3(-5 + 4) + 6(2 - 2) \\ &\quad - 1(8 - 10) \\ &= -3 + 0 + 2 = -1 \ne 0 \end{aligned}
Why?

Expand along the first row with the signs +,−,++, -, +⁠. Each entry multiplies the 2 by 2 determinant left when its row and column are covered:

3:(−5)(1)−(−1)(4)−(−6):(2)(1)−(−1)(−2)−1:(2)(4)−(−5)(−2)\begin{aligned} 3&: (-5)(1) - (-1)(4) \\ -(-6)&: (2)(1) - (-1)(-2) \\ -1&: (2)(4) - (-5)(-2) \end{aligned}

These give −5+4-5 + 4⁠, 2−22 - 2 and 8−108 - 10⁠.

In full:

∣A∣=3∣−5−141∣−(−6)∣2−1−21∣+(−1)∣2−5−24∣\begin{aligned} \lvert A\rvert &= 3\begin{vmatrix} -5 & -1 \\ 4 & 1 \end{vmatrix} - (-6)\begin{vmatrix} 2 & -1 \\ -2 & 1 \end{vmatrix} \\ &\quad + (-1)\begin{vmatrix} 2 & -5 \\ -2 & 4 \end{vmatrix} \end{aligned} ∣−5−141∣=(−5)(1)−(−1)(4)=−5+4=−1∣2−1−21∣=(2)(1)−(−1)(−2)=2−2=0∣2−5−24∣=(2)(4)−(−5)(−2)=8−10=−2\begin{aligned} \begin{vmatrix} -5 & -1 \\ 4 & 1 \end{vmatrix} &= (-5)(1) - (-1)(4) \\ &= -5 + 4 = -1 \\[4pt] \begin{vmatrix} 2 & -1 \\ -2 & 1 \end{vmatrix} &= (2)(1) - (-1)(-2) \\ &= 2 - 2 = 0 \\[4pt] \begin{vmatrix} 2 & -5 \\ -2 & 4 \end{vmatrix} &= (2)(4) - (-5)(-2) \\ &= 8 - 10 = -2 \end{aligned}

Multiply by the entries and their signs, then add:

3(−1)=−3−(−6)(0)=6(0)=0(−1)(−2)=+2−3+0=−3−3+2=−1\begin{aligned} 3(-1) &= -3 \\ -(-6)(0) = 6(0) &= 0 \\ (-1)(-2) &= +2 \\[4pt] -3 + 0 &= -3 \\ -3 + 2 &= -1 \end{aligned}

So, the system is consistent and has a unique solution given by X=A−1C=BCX = A^{-1}C = BC⁠.

Why?

∣A∣≠0|A| \ne 0⁠, so A−1A^{-1} exists. Multiply both sides of AX=CAX = C on the left by A−1A^{-1}⁠:

A−1AX=A−1C⇒X=A−1CA^{-1}AX = A^{-1}C \Rightarrow X = A^{-1}C

since A−1A=IA^{-1}A = I and IX=XIX = X⁠. We already know A−1=BA^{-1} = B⁠.

[xyz]=[1−2−10−1−1203][3−25]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 1 & -2 & -1 \\ 0 & -1 & -1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} 3 \\ -2 \\ 5 \end{bmatrix} =[3+4−50+2−56+0+15]=[2−321]= \begin{bmatrix} 3 + 4 - 5 \\ 0 + 2 - 5 \\ 6 + 0 + 15 \end{bmatrix} = \begin{bmatrix} 2 \\ -3 \\ 21 \end{bmatrix}
Why?

Each row of BB times the column CC⁠. For the first row:

1(3)+(−2)(−2)+(−1)(5)=3+4−5=2\begin{aligned} &1(3) + (-2)(-2) + (-1)(5) \\ &= 3 + 4 - 5 = 2 \end{aligned}

The second and third rows in the same way:

Row 2: 0(3)+(−1)(−2)+(−1)(5)=0+2−5=−3Row 3: 2(3)+0(−2)+3(5)=6+0+15=21\begin{aligned} &\text{Row 2: } 0(3) + (-1)(-2) + (-1)(5) \\ &= 0 + 2 - 5 = -3 \\[4pt] &\text{Row 3: } 2(3) + 0(-2) + 3(5) \\ &= 6 + 0 + 15 = 21 \end{aligned}

Check in the first equation: 3(2)−6(−3)−21=6+18−21=33(2) - 6(-3) - 21 = 6 + 18 - 21 = 3⁠.

The other two equations check as well:

2(2)−5(−3)−21=4+15−21=−2 ✓−2(2)+4(−3)+21=−4−12+21=5 ✓\begin{aligned} &2(2) - 5(-3) - 21 \\ &= 4 + 15 - 21 = -2\ \checkmark \\[4pt] &-2(2) + 4(-3) + 21 \\ &= -4 - 12 + 21 = 5\ \checkmark \end{aligned}

∴ x=2\therefore\ x = 2⁠, y=−3y = -3⁠, z=21z = 21⁠.

Where this comes from

Chapter 3, Matrices: Inverse from AB = I; solving a system of equations.

Question type: Multiply two matrices + Solve equations by the matrix method (combines 2 types)

How to spot it: click to see

Ch 3, type 5. Multiply two matrices: row times column

Spot it: two matrices written side by side, or ABAB⁠, BABA or A2A^2⁠.

Here: First, multiply AA and BB row by column: every entry comes out as in II⁠, so AB=IAB = I and A−1=BA^{-1} = B⁠.

Ch 4, type 9. Solve equations by the matrix method

Spot it: "solve, using the matrix method", with two or three equations. A word problem (prices, three numbers) is the same: write the equations first.

Here: The coefficients of the three equations are the rows of AA⁠, so the system is AX=CAX = C with C=(3,−2,5)C = (3, -2, 5)⁠. ∣A∣=−1≠0|A| = -1 \ne 0⁠, and X=A−1C=BCX = A^{-1}C = BC gives x=2x = 2⁠, y=−3y = -3⁠, z=21z = 21⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 4, Determinants.

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