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CBSE Board Examination 2024 · Section D · 5 marks

Q32: Find A−1A^{-1} and solve a system of three equations, or find unknown entries from A−1A^{-1}

Board 2024Ch 4 · DeterminantsCh 3 · Matrices
The question

(a) If A=[1−202−1−10−21]A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}⁠, find A−1A^{-1} and use it to solve the following system of equations :

x−2y=10x - 2y = 10⁠, 2x−y−z=82x - y - z = 8⁠, −2y+z=7-2y + z = 7

OR

(b) If A=[−1a212x311]A = \begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix} and A−1=[1−11−87−5by3]A^{-1} = \begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix}⁠, find the value of (a+x)−(b+y)(a + x) - (b + y)⁠.

Exam answer

(a)

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

Expand along the first row, whose entries are 1,−2,01, -2, 0⁠. The signs go +,−,++, -, +⁠:

∣A∣=1∣−1−1−21∣−(−2)∣2−101∣+0\begin{aligned} |A| &= 1\begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} \\ &\quad - (-2)\begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} + 0 \end{aligned}

The minors are (−1)(1)−(−1)(−2)=−1−2(-1)(1) - (-1)(-2) = -1 - 2 and 2(1)−(−1)(0)=2−02(1) - (-1)(0) = 2 - 0⁠.

∣−1−1−21∣=(−1)(1)−(−1)(−2)=−1−2=−3∣2−101∣=(2)(1)−(−1)(0)=2−0=2\begin{aligned} \begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} &= (-1)(1) - (-1)(-2) \\ &= -1 - 2 = -3 \\[4pt] \begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} &= (2)(1) - (-1)(0) \\ &= 2 - 0 = 2 \end{aligned}

Multiply by the entries and their signs, then add:

1(−3)=−3−(−2)(2)=+40(…)=0−3+4+0=1\begin{aligned} 1(-3) &= -3 \\ -(-2)(2) &= +4 \\ 0(\ldots) &= 0 \\[4pt] -3 + 4 + 0 &= 1 \end{aligned}

Watch out: −(−2)=+2-(-2) = +2⁠.

Hence A−1A^{-1} exists. The cofactors of A are

Why?

A square matrix has an inverse exactly when its determinant is not zero.

A11=−3, A12=−2, A13=−4A21=2, A22=1, A23=2A31=2, A32=1, A33=3\begin{aligned} A_{11} &= -3,\ A_{12} = -2,\ A_{13} = -4 \\ A_{21} &= 2,\ A_{22} = 1,\ A_{23} = 2 \\ A_{31} &= 2,\ A_{32} = 1,\ A_{33} = 3 \end{aligned}
Why?

The cofactor of the entry in row i, column j is the minor with a sign:

Aij=(−1)i+jMijA_{ij} = (-1)^{i + j}M_{ij}

where MijM_{ij} is the determinant left after deleting row i and column j. For example,

A12=−∣2−101∣=−2A33=∣1−22−1∣=3\begin{aligned} A_{12} &= -\begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} = -2 \\ A_{33} &= \begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix} = 3 \end{aligned}

The signs (−1)i+j(-1)^{i + j} follow the pattern

[+−+−+−+−+]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}

placed on

A=[1−202−1−10−21]A = \begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}

Row 1 (cover row 1 and one column each time):

A11=+∣−1−1−21∣=+((−1)(1)−(−1)(−2))=+(−1−2)=−3A12=−∣2−101∣=−((2)(1)−(−1)(0))=−(2−0)=−2A13=+∣2−10−2∣=+((2)(−2)−(−1)(0))=+(−4−0)=−4\begin{aligned} A_{11} &= +\begin{vmatrix} -1 & -1 \\ -2 & 1 \end{vmatrix} \\ &= +\big((-1)(1) - (-1)(-2)\big) \\ &= +(-1 - 2) = -3 \\[4pt] A_{12} &= -\begin{vmatrix} 2 & -1 \\ 0 & 1 \end{vmatrix} \\ &= -\big((2)(1) - (-1)(0)\big) \\ &= -(2 - 0) = -2 \\[4pt] A_{13} &= +\begin{vmatrix} 2 & -1 \\ 0 & -2 \end{vmatrix} \\ &= +\big((2)(-2) - (-1)(0)\big) \\ &= +(-4 - 0) = -4 \end{aligned}

Row 2 (cover row 2):

A21=−∣−20−21∣=−((−2)(1)−(0)(−2))=−(−2−0)=2A22=+∣1001∣=+((1)(1)−(0)(0))=+(1−0)=1A23=−∣1−20−2∣=−((1)(−2)−(−2)(0))=−(−2−0)=2\begin{aligned} A_{21} &= -\begin{vmatrix} -2 & 0 \\ -2 & 1 \end{vmatrix} \\ &= -\big((-2)(1) - (0)(-2)\big) \\ &= -(-2 - 0) = 2 \\[4pt] A_{22} &= +\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} \\ &= +\big((1)(1) - (0)(0)\big) \\ &= +(1 - 0) = 1 \\[4pt] A_{23} &= -\begin{vmatrix} 1 & -2 \\ 0 & -2 \end{vmatrix} \\ &= -\big((1)(-2) - (-2)(0)\big) \\ &= -(-2 - 0) = 2 \end{aligned}

Row 3 (cover row 3):

A31=+∣−20−1−1∣=+((−2)(−1)−(0)(−1))=+(2−0)=2A32=−∣102−1∣=−((1)(−1)−(0)(2))=−(−1−0)=1A33=+∣1−22−1∣=+((1)(−1)−(−2)(2))=+(−1+4)=3\begin{aligned} A_{31} &= +\begin{vmatrix} -2 & 0 \\ -1 & -1 \end{vmatrix} \\ &= +\big((-2)(-1) - (0)(-1)\big) \\ &= +(2 - 0) = 2 \\[4pt] A_{32} &= -\begin{vmatrix} 1 & 0 \\ 2 & -1 \end{vmatrix} \\ &= -\big((1)(-1) - (0)(2)\big) \\ &= -(-1 - 0) = 1 \\[4pt] A_{33} &= +\begin{vmatrix} 1 & -2 \\ 2 & -1 \end{vmatrix} \\ &= +\big((1)(-1) - (-2)(2)\big) \\ &= +(-1 + 4) = 3 \end{aligned}
adj⁡A=[−322−211−423]\operatorname{adj}A = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}
Why?

adj A is the transpose of the matrix of cofactors. The cofactors of row 1, −3,−2,−4-3, -2, -4⁠, become column 1 of adj A, and so on.

Watch out: forgetting to transpose is the most common slip here.

A−1=1∣A∣adj⁡A=[−322−211−423]\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix} \end{aligned}
Why?

Here ∣A∣=1|A| = 1⁠, so A−1A^{-1} is adj A itself.

The given system can be written as AX=BAX = B⁠, where

X=[xyz], B=[1087]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}
Why?

The coefficients of x,y,zx, y, z in each equation form a row of A. For example, x−2y=10x - 2y = 10 is 1x−2y+0z=101x - 2y + 0z = 10⁠, which is row 1 of A, (1,−2,0)(1, -2, 0)⁠. The right-hand sides form B.

∣A∣≠0|A| \ne 0⁠, so the system has a unique solution X=A−1BX = A^{-1}B⁠.

Why?

Multiplying AX=BAX = B on the left by A−1A^{-1} gives A−1AX=A−1BA^{-1}AX = A^{-1}B⁠, and A−1A=IA^{-1}A = I⁠, so X=A−1BX = A^{-1}B⁠.

[xyz]=[−322−211−423][1087]\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{bmatrix}\begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix} =[−30+16+14−20+8+7−40+16+21]=[0−5−3]= \begin{bmatrix} -30 + 16 + 14 \\ -20 + 8 + 7 \\ -40 + 16 + 21 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix}
Why?

Each entry is a row of A−1A^{-1} times the column B. For the first entry:

(−3)(10)+2(8)+2(7)=−30+16+14=0\begin{aligned} &(-3)(10) + 2(8) + 2(7) \\ &= -30 + 16 + 14 = 0 \end{aligned}

All three rows, products first, then added two at a time:

Row 1: (−3)(10)+(2)(8)+(2)(7)=−30+16+14=−14+14=0Row 2: (−2)(10)+(1)(8)+(1)(7)=−20+8+7=−12+7=−5Row 3: (−4)(10)+(2)(8)+(3)(7)=−40+16+21=−24+21=−3\begin{aligned} &\text{Row 1: } (-3)(10) + (2)(8) + (2)(7) \\ &= -30 + 16 + 14 \\ &= -14 + 14 = 0 \\[4pt] &\text{Row 2: } (-2)(10) + (1)(8) + (1)(7) \\ &= -20 + 8 + 7 \\ &= -12 + 7 = -5 \\[4pt] &\text{Row 3: } (-4)(10) + (2)(8) + (3)(7) \\ &= -40 + 16 + 21 \\ &= -24 + 21 = -3 \end{aligned}

A quick check in the first equation: x−2y=0−2(−5)=10x - 2y = 0 - 2(-5) = 10⁠.

x−2y=0−2(−5)=0+10=10 ✓\begin{aligned} x - 2y &= 0 - 2(-5) \\ &= 0 + 10 = 10\ \checkmark \end{aligned}

∴ x=0\therefore\ x = 0⁠, y=−5y = -5⁠, z=−3z = -3⁠.

(b)

AA−1=IAA^{-1} = I⁠, so

[−1a212x311][1−11−87−5by3]=I\begin{bmatrix} -1 & a & 2 \\ 1 & 2 & x \\ 3 & 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & -1 & 1 \\ -8 & 7 & -5 \\ b & y & 3 \end{bmatrix} = I
Why?

A matrix times its inverse is the identity matrix. Multiplying out the product gives equations for the unknowns a, x, b, y.

Multiplying, the rows of AA−1AA^{-1} are

Row 1: −1−8a+2b-1 - 8a + 2b⁠, 1+7a+2y1 + 7a + 2y⁠, 5−5a5 - 5a⁠;

Row 2: −15+bx-15 + bx⁠, 13+xy13 + xy⁠, 3x−93x - 9⁠;

Row 3: −5+b-5 + b⁠, 4+y4 + y⁠, 11⁠.

Why?

The entry in row i, column j is row i of A times column j of A−1A^{-1}⁠, entry by entry, added. For example, row 1 times column 1:

(−1)(1)+a(−8)+2(b)(-1)(1) + a(-8) + 2(b)

and row 3 times column 3: 3(1)+1(−5)+1(3)=13(1) + 1(-5) + 1(3) = 1⁠.

All nine entries, products first:

(1,1):(−1)(1)+a(−8)+2(b)=−1−8a+2b(1,2):(−1)(−1)+a(7)+2(y)=1+7a+2y(1,3):(−1)(1)+a(−5)+2(3)=−1−5a+6=5−5a\begin{aligned} (1,1)&: (-1)(1) + a(-8) + 2(b) \\ &= -1 - 8a + 2b \\[4pt] (1,2)&: (-1)(-1) + a(7) + 2(y) \\ &= 1 + 7a + 2y \\[4pt] (1,3)&: (-1)(1) + a(-5) + 2(3) \\ &= -1 - 5a + 6 = 5 - 5a \end{aligned} (2,1):(1)(1)+2(−8)+x(b)=1−16+bx=−15+bx(2,2):(1)(−1)+2(7)+x(y)=−1+14+xy=13+xy(2,3):(1)(1)+2(−5)+x(3)=1−10+3x=3x−9\begin{aligned} (2,1)&: (1)(1) + 2(-8) + x(b) \\ &= 1 - 16 + bx = -15 + bx \\[4pt] (2,2)&: (1)(-1) + 2(7) + x(y) \\ &= -1 + 14 + xy = 13 + xy \\[4pt] (2,3)&: (1)(1) + 2(-5) + x(3) \\ &= 1 - 10 + 3x = 3x - 9 \end{aligned} (3,1):(3)(1)+(1)(−8)+(1)(b)=3−8+b=−5+b(3,2):(3)(−1)+(1)(7)+(1)(y)=−3+7+y=4+y(3,3):(3)(1)+(1)(−5)+(1)(3)=3−5+3=1\begin{aligned} (3,1)&: (3)(1) + (1)(-8) + (1)(b) \\ &= 3 - 8 + b = -5 + b \\[4pt] (3,2)&: (3)(-1) + (1)(7) + (1)(y) \\ &= -3 + 7 + y = 4 + y \\[4pt] (3,3)&: (3)(1) + (1)(-5) + (1)(3) \\ &= 3 - 5 + 3 = 1 \end{aligned}

Comparing them with the corresponding elements of the identity matrix II (1 on the diagonal, 0 elsewhere),

Why?

Two matrices are equal only when every entry of one equals the entry in the same place of the other. So each of the nine entries above gives an equation. We need only four of them, one for each unknown.

−5+b=0⇒b=55−5a=0⇒a=14+y=0⇒y=−43x−9=0⇒x=3\begin{aligned} -5 + b = 0 &\Rightarrow b = 5 \\ 5 - 5a = 0 &\Rightarrow a = 1 \\ 4 + y = 0 &\Rightarrow y = -4 \\ 3x - 9 = 0 &\Rightarrow x = 3 \end{aligned}
Why?

Choose the entries that contain only one unknown: (3, 1), (1, 3), (3, 2) and (2, 3). All four are off the diagonal, so each equals 0.

−5+b=0⇒b=55−5a=0⇒5a=5⇒a=14+y=0⇒y=−43x−9=0⇒3x=9⇒x=3\begin{aligned} -5 + b = 0 &\Rightarrow b = 5 \\ 5 - 5a = 0 &\Rightarrow 5a = 5 \\ &\Rightarrow a = 1 \\ 4 + y = 0 &\Rightarrow y = -4 \\ 3x - 9 = 0 &\Rightarrow 3x = 9 \\ &\Rightarrow x = 3 \end{aligned}

The other entries then check out. For example, entry (1, 1) is −1−8+10=1-1 - 8 + 10 = 1⁠.

(1,1):−1−8(1)+2(5)=−1−8+10=1 ✓(1,2):1+7(1)+2(−4)=1+7−8=0 ✓\begin{aligned} (1,1)&: -1 - 8(1) + 2(5) \\ &= -1 - 8 + 10 = 1\ \checkmark \\[4pt] (1,2)&: 1 + 7(1) + 2(-4) \\ &= 1 + 7 - 8 = 0\ \checkmark \end{aligned}
(a+x)−(b+y)=(1+3)−(5−4)=4−1\begin{aligned} &(a + x) - (b + y) \\ &= (1 + 3) - (5 - 4) \\ &= 4 - 1 \end{aligned}
Why?

Substitute a=1a = 1⁠, x=3x = 3⁠, b=5b = 5⁠, y=−4y = -4⁠. Watch out: b+y=5+(−4)=1b + y = 5 + (-4) = 1⁠.

a+x=1+3=4b+y=5+(−4)=5−4=14−1=3\begin{aligned} a + x &= 1 + 3 = 4 \\ b + y &= 5 + (-4) = 5 - 4 = 1 \\ 4 - 1 &= 3 \end{aligned}

∴ (a+x)−(b+y)=3\therefore\ (a + x) - (b + y) = 3⁠.

Where this comes from

Chapter 4, Determinants: Inverse by adjoint; solving a linear system.

Question type: (a) Solve equations by the matrix method; (b) Multiply two matrices + Equal matrices (combined)

How to spot it: click to see

(a) 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: A−1A^{-1} is found and then used: ∣A∣=1≠0|A| = 1 \ne 0⁠, so A−1=adj⁡AA^{-1} = \operatorname{adj}A⁠. The coefficient rows of the equations are the rows of AA⁠, so X=A−1BX = A^{-1}B with B=(10,8,7)B = (10, 8, 7)⁠.

(b) 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, AA−1=IAA^{-1} = I⁠: multiply AA by the given A−1A^{-1} row by column, keeping a,x,b,ya, x, b, y as letters.

(b) Ch 3, type 2. Equal matrices: match entry with entry

Spot it: two matrices joined by an equals sign, with unknowns such as xx or a−ba - b inside.

Here: The product equals II⁠, so match entries: the four off-diagonal entries with one letter each give b=5b = 5⁠, a=1a = 1⁠, y=−4y = -4⁠, x=3x = 3⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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