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CBSE Sample Paper 2022-23 · Section D · 5 marks

Q35: Find A−1A^{-1} and solve 2x−3y+5z=112x - 3y + 5z = 11⁠, 3x+2y−4z=−53x + 2y - 4z = -5⁠, x+y−2z=−3x + y - 2z = -3

Sample 2022-23Ch 4 · DeterminantsCh 3 · Matrices
The question

If

A=[2−3532−411−2],A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix},

find A−1A^{-1}⁠. Use A−1A^{-1} to solve the following system of equations

2x−3y+5z=112x - 3y + 5z = 11⁠, 3x+2y−4z=−53x + 2y - 4z = -5⁠, x+y−2z=−3x + y - 2z = -3

Exam answer
A=[2−3532−411−2]A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix} ∣A∣=2(−4+4)+3(−6+4)+5(3−2)=2(0)+3(−2)+5(1)=−1≠0\begin{aligned} |A| &= 2(-4 + 4) + 3(-6 + 4) \\ &\quad + 5(3 - 2) \\ &= 2(0) + 3(-2) + 5(1) \\ &= -1 \ne 0 \end{aligned}
Why?

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

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

The three 2×22 \times 2 determinants, each as (top left)(bottom right) minus (top right)(bottom left):

∣2−41−2∣=(2)(−2)−(−4)(1)=−4+4=0∣3−41−2∣=(3)(−2)−(−4)(1)=−6+4=−2∣3211∣=(3)(1)−(2)(1)=3−2=1\begin{aligned} \begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} &= (2)(-2) - (-4)(1) \\ &= -4 + 4 = 0 \\[4pt] \begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} &= (3)(-2) - (-4)(1) \\ &= -6 + 4 = -2 \\[4pt] \begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} &= (3)(1) - (2)(1) \\ &= 3 - 2 = 1 \end{aligned}

Watch out: the middle entry is −3-3⁠, and with its minus sign it gives −(−3)=+3-(-3) = +3⁠.

Multiply by the entries and their signs, then add:

2(0)=0+3(−2)=−65(1)=50−6=−6−6+5=−1\begin{aligned} 2(0) &= 0 \\ +3(-2) &= -6 \\ 5(1) &= 5 \\[4pt] 0 - 6 &= -6 \\ -6 + 5 &= -1 \end{aligned}

So A−1A^{-1} exists, and A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{|A|}\operatorname{adj}A⁠.

Why?

A square matrix has an inverse exactly when its determinant is not zero. Then the inverse is the adjoint divided by the determinant.

The cofactors of the elements of AA are

A11=0, A12=2, A13=1A21=−1, A22=−9, A23=−5A31=2, A32=23, A33=13\begin{aligned} A_{11} &= 0,\ A_{12} = 2,\ A_{13} = 1 \\ A_{21} &= -1,\ A_{22} = -9,\ A_{23} = -5 \\ A_{31} &= 2,\ A_{32} = 23,\ A_{33} = 13 \end{aligned}
Why?

The cofactor of the entry in row ii⁠, column jj 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. The sign is −- when i+ji + j is odd.

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

[+−+−+−+−+]  on  A=[2−3532−411−2]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{bmatrix}

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

A11=+∣2−41−2∣=+((2)(−2)−(−4)(1))=+(−4+4)=0A12=−∣3−41−2∣=−((3)(−2)−(−4)(1))=−(−6+4)=−(−2)=2A13=+∣3211∣=+((3)(1)−(2)(1))=+(3−2)=1\begin{aligned} A_{11} &= +\begin{vmatrix} 2 & -4 \\ 1 & -2 \end{vmatrix} \\ &= +\big((2)(-2) - (-4)(1)\big) \\ &= +(-4 + 4) = 0 \\[4pt] A_{12} &= -\begin{vmatrix} 3 & -4 \\ 1 & -2 \end{vmatrix} \\ &= -\big((3)(-2) - (-4)(1)\big) \\ &= -(-6 + 4) = -(-2) = 2 \\[4pt] A_{13} &= +\begin{vmatrix} 3 & 2 \\ 1 & 1 \end{vmatrix} \\ &= +\big((3)(1) - (2)(1)\big) \\ &= +(3 - 2) = 1 \end{aligned}

Row 2 (cover row 2):

A21=−∣−351−2∣=−((−3)(−2)−(5)(1))=−(6−5)=−1A22=+∣251−2∣=+((2)(−2)−(5)(1))=+(−4−5)=−9A23=−∣2−311∣=−((2)(1)−(−3)(1))=−(2+3)=−5\begin{aligned} A_{21} &= -\begin{vmatrix} -3 & 5 \\ 1 & -2 \end{vmatrix} \\ &= -\big((-3)(-2) - (5)(1)\big) \\ &= -(6 - 5) = -1 \\[4pt] A_{22} &= +\begin{vmatrix} 2 & 5 \\ 1 & -2 \end{vmatrix} \\ &= +\big((2)(-2) - (5)(1)\big) \\ &= +(-4 - 5) = -9 \\[4pt] A_{23} &= -\begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} \\ &= -\big((2)(1) - (-3)(1)\big) \\ &= -(2 + 3) = -5 \end{aligned}

Row 3 (cover row 3):

A31=+∣−352−4∣=+((−3)(−4)−(5)(2))=+(12−10)=2A32=−∣253−4∣=−((2)(−4)−(5)(3))=−(−8−15)=−(−23)=23A33=+∣2−332∣=+((2)(2)−(−3)(3))=+(4+9)=13\begin{aligned} A_{31} &= +\begin{vmatrix} -3 & 5 \\ 2 & -4 \end{vmatrix} \\ &= +\big((-3)(-4) - (5)(2)\big) \\ &= +(12 - 10) = 2 \\[4pt] A_{32} &= -\begin{vmatrix} 2 & 5 \\ 3 & -4 \end{vmatrix} \\ &= -\big((2)(-4) - (5)(3)\big) \\ &= -(-8 - 15) = -(-23) = 23 \\[4pt] A_{33} &= +\begin{vmatrix} 2 & -3 \\ 3 & 2 \end{vmatrix} \\ &= +\big((2)(2) - (-3)(3)\big) \\ &= +(4 + 9) = 13 \end{aligned}
adj⁡A=[0−122−9231−513]\operatorname{adj}A = \begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}
Why?

The adjoint is the transpose of the matrix of cofactors: the cofactors of row 1 (0,2,1)(0, 2, 1) become column 1, and so on.

A−1=1(−1)[0−122−9231−513]A^{-1} = \frac{1}{(-1)}\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}

The given system is AX=BAX = B⁠, where

X=[xyz],B=[11−5−3]X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\quad B = \begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix}
Why?

The rows of AA hold the coefficients of xx⁠, yy⁠, zz in the three equations, in order. BB holds the right-hand sides.

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

Why?

Multiply AX=BAX = B on the left by A−1A^{-1}⁠. Since A−1A=IA^{-1}A = I and IX=XIX = X⁠, this gives X=A−1BX = A^{-1}B⁠.

[xyz]=1(−1)[0−122−9231−513][11−5−3]=1(−1)[0+5−622+45−6911+25−39]=1(−1)[−1−2−3]=[123]\begin{aligned} \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac{1}{(-1)}\begin{bmatrix} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{bmatrix}\begin{bmatrix} 11 \\ -5 \\ -3 \end{bmatrix} \\ &= \frac{1}{(-1)}\begin{bmatrix} 0 + 5 - 6 \\ 22 + 45 - 69 \\ 11 + 25 - 39 \end{bmatrix} \\ &= \frac{1}{(-1)}\begin{bmatrix} -1 \\ -2 \\ -3 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix} \end{aligned}
Why?

Each entry of the product is a row of adj⁡A\operatorname{adj}A times the column BB⁠. For the first row:

0(11)+(−1)(−5)+2(−3)=0+5−6=−1\begin{aligned} &0(11) + (-1)(-5) + 2(-3) \\ &= 0 + 5 - 6 = -1 \end{aligned}

The other two rows the same way:

Row 2: (2)(11)+(−9)(−5)+(23)(−3)=22+45−69=67−69=−2Row 3: (1)(11)+(−5)(−5)+(13)(−3)=11+25−39=36−39=−3\begin{aligned} &\text{Row 2: } (2)(11) + (-9)(-5) \\ &\quad + (23)(-3) \\ &= 22 + 45 - 69 \\ &= 67 - 69 = -2 \\[4pt] &\text{Row 3: } (1)(11) + (-5)(-5) \\ &\quad + (13)(-3) \\ &= 11 + 25 - 39 \\ &= 36 - 39 = -3 \end{aligned}

Then multiply every entry by 1(−1)\frac{1}{(-1)}⁠, which changes every sign.

−1−1=1−2−1=2−3−1=3\begin{aligned} \frac{-1}{-1} &= 1 \\ \frac{-2}{-1} &= 2 \\ \frac{-3}{-1} &= 3 \end{aligned}

Check in the first equation: 2(1)−3(2)+5(3)=112(1) - 3(2) + 5(3) = 11⁠.

2(1)−3(2)+5(3)=2−6+15=−4+15=11 ✓\begin{aligned} &2(1) - 3(2) + 5(3) \\ &= 2 - 6 + 15 \\ &= -4 + 15 = 11\ \checkmark \end{aligned}

∴ x=1, y=2, z=3\therefore\ x = 1,\ y = 2,\ z = 3

Where this comes from

Chapter 4, Determinants: Inverse by the adjoint; solving a system by A⁻¹.

Question type: Solve equations by the matrix method

How to spot it: click to see

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, nine cofactors give adj⁡A\operatorname{adj}A⁠, and the equations are AX=BAX = B with B=(11,−5,−3)B = (11, -5, -3)⁠, so X=A−1BX = A^{-1}B⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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