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CBSE Board Examination 2023 · Section D · 5 marks

Q33: Find (AB)−1(AB)^{-1} from AA and B−1B^{-1}⁠, or solve a system by the matrix method

Board 2023Ch 4 · DeterminantsCh 3 · Matrices
The question

(a) If

A=[12−2−1300−21]A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}

and

B−1=[3−11−156−55−22],B^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix},

find (AB)−1(AB)^{-1}⁠.

OR

(b) Solve the following system of equations by matrix method :

x+2y+3z=6x + 2y + 3z = 6

2x−y+z=22x - y + z = 2

3x+2y−2z=33x + 2y - 2z = 3

Exam answer

(a)

(AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1}
Why?

The inverse of a product is the product of the inverses in reverse order. B−1B^{-1} is given, so only A−1A^{-1} has to be found.

Watch out: A−1B−1A^{-1}B^{-1} is wrong in general, because matrix multiplication is not commutative.

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

Expand along the first row, with signs +,−,++, -, +⁠:

∣A∣=a11M11−a12M12+a13M13|A| = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} ∣A∣=1∣30−21∣−2∣−1001∣+(−2)∣−130−2∣\begin{aligned} |A| &= 1\begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} - 2\begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} \\ &\quad + (-2)\begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} \end{aligned}

The minors are 3(1)−0(−2)=33(1) - 0(-2) = 3⁠, (−1)(1)−0(0)=−1(-1)(1) - 0(0) = -1 and (−1)(−2)−3(0)=2(-1)(-2) - 3(0) = 2⁠.

∣30−21∣=(3)(1)−(0)(−2)=3−0=3∣−1001∣=(−1)(1)−(0)(0)=−1−0=−1∣−130−2∣=(−1)(−2)−(3)(0)=2−0=2\begin{aligned} \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} &= (3)(1) - (0)(-2) \\ &= 3 - 0 = 3 \\[4pt] \begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} &= (-1)(1) - (0)(0) \\ &= -1 - 0 = -1 \\[4pt] \begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} &= (-1)(-2) - (3)(0) \\ &= 2 - 0 = 2 \end{aligned}

Multiply by the entries and their signs, then add:

1(3)=3−2(−1)=+2(−2)(2)=−43+2=55−4=1\begin{aligned} 1(3) &= 3 \\ -2(-1) &= +2 \\ (-2)(2) &= -4 \\[4pt] 3 + 2 &= 5 \\ 5 - 4 &= 1 \end{aligned}

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

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

A11=3, A12=1, A13=2A21=2, A22=1, A23=2A31=6, A32=2, A33=5\begin{aligned} A_{11} &= 3,\ A_{12} = 1,\ A_{13} = 2 \\ A_{21} &= 2,\ A_{22} = 1,\ A_{23} = 2 \\ A_{31} &= 6,\ A_{32} = 2,\ A_{33} = 5 \end{aligned}
Why?

Each cofactor is Aij=(−1)i+jMijA_{ij} = (-1)^{i + j}M_{ij}⁠, where the minor MijM_{ij} is the determinant left after deleting row ii and column jj⁠. For example,

A12=−[(−1)(1)−0(0)]=1A21=−[2(1)−(−2)(−2)]=2A31=2(0)−(−2)(3)=6\begin{aligned} A_{12} &= -[(-1)(1) - 0(0)] = 1 \\ A_{21} &= -[2(1) - (-2)(-2)] = 2 \\ A_{31} &= 2(0) - (-2)(3) = 6 \end{aligned}

All nine, one at a time. The signs (−1)i+j(-1)^{i + j} follow the pattern

[+−+−+−+−+]  on  A=[12−2−1300−21]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}

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

A11=+∣30−21∣=+((3)(1)−(0)(−2))=+(3−0)=3A12=−∣−1001∣=−((−1)(1)−(0)(0))=−(−1−0)=−(−1)=1A13=+∣−130−2∣=+((−1)(−2)−(3)(0))=+(2−0)=2\begin{aligned} A_{11} &= +\begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} \\ &= +\big((3)(1) - (0)(-2)\big) \\ &= +(3 - 0) = 3 \\[4pt] A_{12} &= -\begin{vmatrix} -1 & 0 \\ 0 & 1 \end{vmatrix} \\ &= -\big((-1)(1) - (0)(0)\big) \\ &= -(-1 - 0) = -(-1) = 1 \\[4pt] A_{13} &= +\begin{vmatrix} -1 & 3 \\ 0 & -2 \end{vmatrix} \\ &= +\big((-1)(-2) - (3)(0)\big) \\ &= +(2 - 0) = 2 \end{aligned}

Row 2 (cover row 2):

A21=−∣2−2−21∣=−((2)(1)−(−2)(−2))=−(2−4)=−(−2)=2A22=+∣1−201∣=+((1)(1)−(−2)(0))=+(1−0)=1A23=−∣120−2∣=−((1)(−2)−(2)(0))=−(−2−0)=−(−2)=2\begin{aligned} A_{21} &= -\begin{vmatrix} 2 & -2 \\ -2 & 1 \end{vmatrix} \\ &= -\big((2)(1) - (-2)(-2)\big) \\ &= -(2 - 4) = -(-2) = 2 \\[4pt] A_{22} &= +\begin{vmatrix} 1 & -2 \\ 0 & 1 \end{vmatrix} \\ &= +\big((1)(1) - (-2)(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) = 2 \end{aligned}

Row 3 (cover row 3):

A31=+∣2−230∣=+((2)(0)−(−2)(3))=+(0+6)=6A32=−∣1−2−10∣=−((1)(0)−(−2)(−1))=−(0−2)=−(−2)=2A33=+∣12−13∣=+((1)(3)−(2)(−1))=+(3+2)=5\begin{aligned} A_{31} &= +\begin{vmatrix} 2 & -2 \\ 3 & 0 \end{vmatrix} \\ &= +\big((2)(0) - (-2)(3)\big) \\ &= +(0 + 6) = 6 \\[4pt] A_{32} &= -\begin{vmatrix} 1 & -2 \\ -1 & 0 \end{vmatrix} \\ &= -\big((1)(0) - (-2)(-1)\big) \\ &= -(0 - 2) = -(-2) = 2 \\[4pt] A_{33} &= +\begin{vmatrix} 1 & 2 \\ -1 & 3 \end{vmatrix} \\ &= +\big((1)(3) - (2)(-1)\big) \\ &= +(3 + 2) = 5 \end{aligned}

Watch out: the sign (−1)i+j(-1)^{i + j} is minus when i+ji + j is odd.

adj⁡A=[326112225]\operatorname{adj}A = \begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}
Why?

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

A−1=1∣A∣adj⁡A=11[326112225]A^{-1} = \frac{1}{|A|}\operatorname{adj}A = \frac11\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix}
Why?

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

B−1A−1=[3−11−156−55−22][326112225]\begin{aligned} &B^{-1}A^{-1} \\ &= \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}\begin{bmatrix} 3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5 \end{bmatrix} \end{aligned} ∴ (AB)−1=[10721−49−34−103171236]\begin{aligned} &\therefore\ (AB)^{-1} \\ &= \begin{bmatrix} 10 & 7 & 21 \\ -49 & -34 & -103 \\ 17 & 12 & 36 \end{bmatrix} \end{aligned}
Why?

Each entry is a row of B−1B^{-1} times a column of A−1A^{-1}⁠. For example, row 1 times column 1, and row 2 times column 3:

3(3)−1(1)+1(2)=9−1+2=10\begin{aligned} &3(3) - 1(1) + 1(2) \\ &= 9 - 1 + 2 = 10 \end{aligned} −15(6)+6(2)−5(5)=−90+12−25=−103\begin{aligned} &-15(6) + 6(2) - 5(5) \\ &= -90 + 12 - 25 = -103 \end{aligned}

All nine entries. Row 1 of B−1B^{-1} is (3,−1,1)(3, -1, 1)⁠; the columns of A−1A^{-1} are (3,1,2)(3, 1, 2)⁠, (2,1,2)(2, 1, 2)⁠, (6,2,5)(6, 2, 5)⁠:

R1C1: (3)(3)+(−1)(1)+(1)(2)=9−1+2=10R1C2: (3)(2)+(−1)(1)+(1)(2)=6−1+2=7R1C3: (3)(6)+(−1)(2)+(1)(5)=18−2+5=21\begin{aligned} &\text{R1C1: } (3)(3) + (-1)(1) + (1)(2) \\ &= 9 - 1 + 2 = 10 \\[4pt] &\text{R1C2: } (3)(2) + (-1)(1) + (1)(2) \\ &= 6 - 1 + 2 = 7 \\[4pt] &\text{R1C3: } (3)(6) + (-1)(2) + (1)(5) \\ &= 18 - 2 + 5 = 21 \end{aligned}

Row 2 of B−1B^{-1} is (−15,6,−5)(-15, 6, -5)⁠:

R2C1: (−15)(3)+(6)(1)+(−5)(2)=−45+6−10=−49R2C2: (−15)(2)+(6)(1)+(−5)(2)=−30+6−10=−34R2C3: (−15)(6)+(6)(2)+(−5)(5)=−90+12−25=−103\begin{aligned} &\text{R2C1: } (-15)(3) + (6)(1) \\ &\quad + (-5)(2) \\ &= -45 + 6 - 10 = -49 \\[4pt] &\text{R2C2: } (-15)(2) + (6)(1) \\ &\quad + (-5)(2) \\ &= -30 + 6 - 10 = -34 \\[4pt] &\text{R2C3: } (-15)(6) + (6)(2) \\ &\quad + (-5)(5) \\ &= -90 + 12 - 25 = -103 \end{aligned}

Row 3 of B−1B^{-1} is (5,−2,2)(5, -2, 2)⁠:

R3C1: (5)(3)+(−2)(1)+(2)(2)=15−2+4=17R3C2: (5)(2)+(−2)(1)+(2)(2)=10−2+4=12R3C3: (5)(6)+(−2)(2)+(2)(5)=30−4+10=36\begin{aligned} &\text{R3C1: } (5)(3) + (-2)(1) + (2)(2) \\ &= 15 - 2 + 4 = 17 \\[4pt] &\text{R3C2: } (5)(2) + (-2)(1) + (2)(2) \\ &= 10 - 2 + 4 = 12 \\[4pt] &\text{R3C3: } (5)(6) + (-2)(2) + (2)(5) \\ &= 30 - 4 + 10 = 36 \end{aligned}

(b) The given system is AX=BAX = B⁠, where

A=[1232−1132−2],A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \end{bmatrix}, X=[xyz], B=[623]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\ B = \begin{bmatrix} 6 \\ 2 \\ 3 \end{bmatrix}
Why?

Each equation gives one row of A: the coefficients of xx⁠, yy and zz⁠. The unknowns form the column X, and the right sides form the column B.

∣A∣=1(2−2)−2(−4−3)+3(4+3)=0+14+21=35≠0\begin{aligned} |A| &= 1(2 - 2) - 2(-4 - 3) \\ &\quad + 3(4 + 3) \\ &= 0 + 14 + 21 = 35 \ne 0 \end{aligned}
Why?

Expand along the first row. The minors are

M11=(−1)(−2)−1(2)=0M12=2(−2)−1(3)=−7M13=2(2)−(−1)(3)=7\begin{aligned} M_{11} &= (-1)(-2) - 1(2) = 0 \\ M_{12} &= 2(-2) - 1(3) = -7 \\ M_{13} &= 2(2) - (-1)(3) = 7 \end{aligned}

so ∣A∣=1(0)−2(−7)+3(7)|A| = 1(0) - 2(-7) + 3(7)⁠.

Written out as 2×22 \times 2 determinants (top left)(bottom right) minus (top right)(bottom left):

∣−112−2∣=(−1)(−2)−(1)(2)=2−2=0∣213−2∣=(2)(−2)−(1)(3)=−4−3=−7∣2−132∣=(2)(2)−(−1)(3)=4+3=7\begin{aligned} \begin{vmatrix} -1 & 1 \\ 2 & -2 \end{vmatrix} &= (-1)(-2) - (1)(2) \\ &= 2 - 2 = 0 \\[4pt] \begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} &= (2)(-2) - (1)(3) \\ &= -4 - 3 = -7 \\[4pt] \begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} &= (2)(2) - (-1)(3) \\ &= 4 + 3 = 7 \end{aligned}

Multiply by the entries and their signs, then add:

1(0)=0−2(−7)=+143(7)=210+14=1414+21=35\begin{aligned} 1(0) &= 0 \\ -2(-7) &= +14 \\ 3(7) &= 21 \\[4pt] 0 + 14 &= 14 \\ 14 + 21 &= 35 \end{aligned}

So A−1A^{-1} exists and the system has a unique solution X=A−1BX = A^{-1}B⁠. The cofactors of A are

Why?

Since ∣A∣≠0|A| \ne 0⁠, A−1A^{-1} exists. Multiplying AX=BAX = B on the left by A−1A^{-1} gives X=A−1BX = A^{-1}B⁠, so there is exactly one solution.

A11=0, A12=7, A13=7A21=10, A22=−11, A23=4A31=5, A32=5, A33=−5\begin{aligned} A_{11} &= 0,\ A_{12} = 7,\ A_{13} = 7 \\ A_{21} &= 10,\ A_{22} = -11,\ A_{23} = 4 \\ A_{31} &= 5,\ A_{32} = 5,\ A_{33} = -5 \end{aligned}
Why?

Use Aij=(−1)i+jMijA_{ij} = (-1)^{i + j}M_{ij} again. For example,

A21=−[2(−2)−3(2)]=10A22=1(−2)−3(3)=−11A33=1(−1)−2(2)=−5\begin{aligned} A_{21} &= -[2(-2) - 3(2)] = 10 \\ A_{22} &= 1(-2) - 3(3) = -11 \\ A_{33} &= 1(-1) - 2(2) = -5 \end{aligned}

All nine, one at a time, on

A=[1232−1132−2]A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 1 \\ 3 & 2 & -2 \end{bmatrix}

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

A11=+∣−112−2∣=+((−1)(−2)−(1)(2))=+(2−2)=0A12=−∣213−2∣=−((2)(−2)−(1)(3))=−(−4−3)=−(−7)=7A13=+∣2−132∣=+((2)(2)−(−1)(3))=+(4+3)=7\begin{aligned} A_{11} &= +\begin{vmatrix} -1 & 1 \\ 2 & -2 \end{vmatrix} \\ &= +\big((-1)(-2) - (1)(2)\big) \\ &= +(2 - 2) = 0 \\[4pt] A_{12} &= -\begin{vmatrix} 2 & 1 \\ 3 & -2 \end{vmatrix} \\ &= -\big((2)(-2) - (1)(3)\big) \\ &= -(-4 - 3) = -(-7) = 7 \\[4pt] A_{13} &= +\begin{vmatrix} 2 & -1 \\ 3 & 2 \end{vmatrix} \\ &= +\big((2)(2) - (-1)(3)\big) \\ &= +(4 + 3) = 7 \end{aligned}

Row 2 (cover row 2):

A21=−∣232−2∣=−((2)(−2)−(3)(2))=−(−4−6)=−(−10)=10A22=+∣133−2∣=+((1)(−2)−(3)(3))=+(−2−9)=−11A23=−∣1232∣=−((1)(2)−(2)(3))=−(2−6)=−(−4)=4\begin{aligned} A_{21} &= -\begin{vmatrix} 2 & 3 \\ 2 & -2 \end{vmatrix} \\ &= -\big((2)(-2) - (3)(2)\big) \\ &= -(-4 - 6) = -(-10) = 10 \\[4pt] A_{22} &= +\begin{vmatrix} 1 & 3 \\ 3 & -2 \end{vmatrix} \\ &= +\big((1)(-2) - (3)(3)\big) \\ &= +(-2 - 9) = -11 \\[4pt] A_{23} &= -\begin{vmatrix} 1 & 2 \\ 3 & 2 \end{vmatrix} \\ &= -\big((1)(2) - (2)(3)\big) \\ &= -(2 - 6) = -(-4) = 4 \end{aligned}

Row 3 (cover row 3):

A31=+∣23−11∣=+((2)(1)−(3)(−1))=+(2+3)=5A32=−∣1321∣=−((1)(1)−(3)(2))=−(1−6)=−(−5)=5A33=+∣122−1∣=+((1)(−1)−(2)(2))=+(−1−4)=−5\begin{aligned} A_{31} &= +\begin{vmatrix} 2 & 3 \\ -1 & 1 \end{vmatrix} \\ &= +\big((2)(1) - (3)(-1)\big) \\ &= +(2 + 3) = 5 \\[4pt] A_{32} &= -\begin{vmatrix} 1 & 3 \\ 2 & 1 \end{vmatrix} \\ &= -\big((1)(1) - (3)(2)\big) \\ &= -(1 - 6) = -(-5) = 5 \\[4pt] A_{33} &= +\begin{vmatrix} 1 & 2 \\ 2 & -1 \end{vmatrix} \\ &= +\big((1)(-1) - (2)(2)\big) \\ &= +(-1 - 4) = -5 \end{aligned}
A−1=135[01057−11574−5]A^{-1} = \frac{1}{35}\begin{bmatrix} 0 & 10 & 5 \\ 7 & -11 & 5 \\ 7 & 4 & -5 \end{bmatrix}
Why?

A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A⁠, where adj A is the transpose of the cofactor matrix. The cofactors of row 1, namely 0, 7, 7, form column 1.

X=A−1B=135[01057−11574−5][623]=135[0+20+1542−22+1542+8−15]=135[353535]=[111]\begin{aligned} X &= A^{-1}B \\ &= \frac{1}{35}\begin{bmatrix} 0 & 10 & 5 \\ 7 & -11 & 5 \\ 7 & 4 & -5 \end{bmatrix}\begin{bmatrix} 6 \\ 2 \\ 3 \end{bmatrix} \\ &= \frac{1}{35}\begin{bmatrix} 0 + 20 + 15 \\ 42 - 22 + 15 \\ 42 + 8 - 15 \end{bmatrix} \\ &= \frac{1}{35}\begin{bmatrix} 35 \\ 35 \\ 35 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \end{aligned}
Why?

Multiply each row by the column B. For the first row, 0(6)+10(2)+5(3)=350(6) + 10(2) + 5(3) = 35⁠.

Row 1: (0)(6)+(10)(2)+(5)(3)=0+20+15=35Row 2: (7)(6)+(−11)(2)+(5)(3)=42−22+15=35Row 3: (7)(6)+(4)(2)+(−5)(3)=42+8−15=35\begin{aligned} &\text{Row 1: } (0)(6) + (10)(2) + (5)(3) \\ &= 0 + 20 + 15 = 35 \\[4pt] &\text{Row 2: } (7)(6) + (-11)(2) + (5)(3) \\ &= 42 - 22 + 15 = 35 \\[4pt] &\text{Row 3: } (7)(6) + (4)(2) + (-5)(3) \\ &= 42 + 8 - 15 = 35 \end{aligned}

Then 135(35)=3535=1\frac{1}{35}(35) = \frac{35}{35} = 1 in each place.

Check in the first equation: 1+2(1)+3(1)=61 + 2(1) + 3(1) = 6⁠.

2x−y+z=2(1)−1+1=2−1+1=2 ✓3x+2y−2z=3(1)+2(1)−2(1)=3+2−2=3 ✓\begin{aligned} 2x - y + z &= 2(1) - 1 + 1 \\ &= 2 - 1 + 1 = 2\ \checkmark \\[4pt] 3x + 2y - 2z &= 3(1) + 2(1) - 2(1) \\ &= 3 + 2 - 2 = 3\ \checkmark \end{aligned}

∴ x=1\therefore\ x = 1⁠, y=1y = 1⁠, z=1z = 1⁠.

Where this comes from

Chapter 4, Determinants: Inverse of a product; system by matrix method.

Question type: (a) Inverse of a matrix; (b) Solve equations by the matrix method

How to spot it: click to see

(a) Type 6. Inverse of a matrix: ∣A∣|A|⁠, then the adjoint, then divide

Spot it: "find the inverse" or "find A−1A^{-1}".

Here: (AB)−1=B−1A−1(AB)^{-1} = B^{-1}A^{-1} and B−1B^{-1} is given, so only A−1A^{-1} is needed: ∣A∣=1≠0|A| = 1 \ne 0⁠, nine cofactors, then A−1=adj⁡AA^{-1} = \operatorname{adj}A⁠. Finish by multiplying B−1A−1B^{-1}A^{-1}⁠.

(b) 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: Write AX=BAX = B with B=(6,2,3)B = (6, 2, 3)⁠. ∣A∣=35≠0|A| = 35 \ne 0⁠, so find adj⁡A\operatorname{adj}A⁠, and X=A−1BX = A^{-1}B gives x=y=z=1x = y = z = 1⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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