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CBSE Sample Paper 2020-21 · Part B: Section V · 5 marks

Q36: A−1A^{-1} and a system with matrix A′A'⁠, or AB=6IAB = 6I and a system

Sample 2020-21Ch 4 · DeterminantsCh 3 · Matrices
The question

(a) If A=[120−2−1−20−11]A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}⁠, find A−1A^{-1}⁠. Hence

Solve the system of equations;

x−2y=102x−y−z=8−2y+z=7\begin{aligned} x - 2y &= 10 \\ 2x - y - z &= 8 \\ -2y + z &= 7 \end{aligned}

OR

(b) Evaluate the product AB, where

A=[1−10234012] and B=[22−4−42−42−15]\begin{aligned} A &= \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \text{ and } \\ B &= \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \end{aligned}

Hence solve the system of linear equations

x−y=32x+3y+4z=17y+2z=7\begin{aligned} x - y &= 3 \\ 2x + 3y + 4z &= 17 \\ y + 2z &= 7 \end{aligned}
Exam answer

(a)

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

Expand along the first row:

∣A∣=a11A11+a12A12+a13A13\begin{aligned} |A| &= a_{11}A_{11} + a_{12}A_{12} \\ &\quad + a_{13}A_{13} \end{aligned}

For a11=1a_{11} = 1⁠, delete row 1 and column 1: what is left gives (−1)(1)−(−2)(−1)=−1−2(-1)(1) - (-2)(-1) = -1 - 2⁠. For a12=2a_{12} = 2⁠, the minor is (−2)(1)−(−2)(0)(-2)(1) - (-2)(0)⁠, with a minus sign in front. The third term is 0, because a13=0a_{13} = 0⁠.

With the signs ++⁠, −-⁠, ++⁠:

∣A∣=1∣−1−2−11∣−2∣−2−201∣+0∣−2−10−1∣\begin{aligned} |A| &= 1\begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} - 2\begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} \\ &\quad + 0\begin{vmatrix} -2 & -1 \\ 0 & -1 \end{vmatrix} \end{aligned}

Each 2×22 \times 2 determinant is (top left)(bottom right) minus (top right)(bottom left):

∣−1−2−11∣=(−1)(1)−(−2)(−1)=−1−2=−3∣−2−201∣=(−2)(1)−(−2)(0)=−2−0=−2∣−2−10−1∣=(−2)(−1)−(−1)(0)=2−0=2\begin{aligned} \begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} &= (-1)(1) - (-2)(-1) \\ &= -1 - 2 = -3 \\[4pt] \begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} &= (-2)(1) - (-2)(0) \\ &= -2 - 0 = -2 \\[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(2)=0−3+4+0=1\begin{aligned} 1(-3) &= -3 \\ -2(-2) &= +4 \\ 0(2) &= 0 \\[4pt] -3 + 4 + 0 &= 1 \end{aligned}

A is non-singular, therefore A−1A^{-1} exists.

Why?

A square matrix has an inverse exactly when its determinant is not zero. So always state ∣A∣≠0|A| \ne 0 before finding A−1A^{-1}⁠.

The cofactors of A are

A11=−1−2=−3A12=−(−2−0)=2A13=2−0=2\begin{aligned} A_{11} &= -1 - 2 = -3 \\ A_{12} &= -(-2 - 0) = 2 \\ A_{13} &= 2 - 0 = 2 \end{aligned} A21=−(2−0)=−2A22=1−0=1A23=−(−1−0)=1\begin{aligned} A_{21} &= -(2 - 0) = -2 \\ A_{22} &= 1 - 0 = 1 \\ A_{23} &= -(-1 - 0) = 1 \end{aligned} A31=−4−0=−4A32=−(−2−0)=2A33=−1+4=3\begin{aligned} A_{31} &= -4 - 0 = -4 \\ A_{32} &= -(-2 - 0) = 2 \\ A_{33} &= -1 + 4 = 3 \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 2×22 \times 2 determinant left after deleting row ii and column jj⁠. The sign is ++ when i+ji + j is even and −- when it is odd, which is why A12A_{12}⁠, A21A_{21}⁠, A23A_{23} and A32A_{32} have a minus in front.

For example, A33A_{33}⁠: deleting row 3 and column 3 leaves (1)(−1)−(2)(−2)=−1+4(1)(-1) - (2)(-2) = -1 + 4⁠.

All nine, with the sign pattern

[+−+−+−+−+]  on  A=[120−2−1−20−11]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{bmatrix}

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

A11=+∣−1−2−11∣=+((−1)(1)−(−2)(−1))=+(−1−2)=−3A12=−∣−2−201∣=−((−2)(1)−(−2)(0))=−(−2−0)=−(−2)=2A13=+∣−2−10−1∣=+((−2)(−1)−(−1)(0))=+(2−0)=2\begin{aligned} A_{11} &= +\begin{vmatrix} -1 & -2 \\ -1 & 1 \end{vmatrix} \\ &= +\big((-1)(1) - (-2)(-1)\big) \\ &= +(-1 - 2) = -3 \\[4pt] A_{12} &= -\begin{vmatrix} -2 & -2 \\ 0 & 1 \end{vmatrix} \\ &= -\big((-2)(1) - (-2)(0)\big) \\ &= -(-2 - 0) = -(-2) = 2 \\[4pt] A_{13} &= +\begin{vmatrix} -2 & -1 \\ 0 & -1 \end{vmatrix} \\ &= +\big((-2)(-1) - (-1)(0)\big) \\ &= +(2 - 0) = 2 \end{aligned}

Row 2 (cover row 2):

A21=−∣20−11∣=−((2)(1)−(0)(−1))=−(2−0)=−2A22=+∣1001∣=+((1)(1)−(0)(0))=+(1−0)=1A23=−∣120−1∣=−((1)(−1)−(2)(0))=−(−1−0)=−(−1)=1\begin{aligned} A_{21} &= -\begin{vmatrix} 2 & 0 \\ -1 & 1 \end{vmatrix} \\ &= -\big((2)(1) - (0)(-1)\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 & -1 \end{vmatrix} \\ &= -\big((1)(-1) - (2)(0)\big) \\ &= -(-1 - 0) = -(-1) = 1 \end{aligned}

Row 3 (cover row 3):

A31=+∣20−1−2∣=+((2)(−2)−(0)(−1))=+(−4−0)=−4A32=−∣10−2−2∣=−((1)(−2)−(0)(−2))=−(−2−0)=−(−2)=2A33=+∣12−2−1∣=+((1)(−1)−(2)(−2))=+(−1+4)=3\begin{aligned} A_{31} &= +\begin{vmatrix} 2 & 0 \\ -1 & -2 \end{vmatrix} \\ &= +\big((2)(-2) - (0)(-1)\big) \\ &= +(-4 - 0) = -4 \\[4pt] A_{32} &= -\begin{vmatrix} 1 & 0 \\ -2 & -2 \end{vmatrix} \\ &= -\big((1)(-2) - (0)(-2)\big) \\ &= -(-2 - 0) = -(-2) = 2 \\[4pt] A_{33} &= +\begin{vmatrix} 1 & 2 \\ -2 & -1 \end{vmatrix} \\ &= +\big((1)(-1) - (2)(-2)\big) \\ &= +(-1 + 4) = 3 \end{aligned}
adj⁡A=[−3−2−4212213]\operatorname{adj}A = \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix}
Why?

The adjoint is the transpose of the matrix of cofactors. So A11,A12,A13A_{11}, A_{12}, A_{13} go down the first column, and so on.

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

Watch out: writing the cofactors in rows, without transposing, gives the wrong matrix.

A−1=1∣A∣adj⁡A=[−3−2−4212213]\begin{aligned} A^{-1} &= \frac{1}{|A|}\operatorname{adj}A \\ &= \begin{bmatrix} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{bmatrix} \end{aligned}
Why?

Since ∣A∣=1|A| = 1⁠, dividing by ∣A∣|A| changes nothing, so A−1=adj⁡AA^{-1} = \operatorname{adj}A⁠. Multiplying out AA−1AA^{-1} gives II⁠, which checks the work.

For example, row 1 of AA times each column of A−1A^{-1}⁠:

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

That is row 1 of II⁠. The other rows work the same way.

The given equations can be written as

[1−202−1−10−21][xyz]=[1087]\begin{bmatrix} 1 & -2 & 0 \\ 2 & -1 & -1 \\ 0 & -2 & 1 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}
Why?

Each row holds the coefficients of xx⁠, yy⁠, zz in one equation; a missing variable has coefficient 0. Compare with A: the rows of this matrix are the columns of A, so it is the transpose A′A'⁠. That is why the question says "hence".

which is of the form A′X=BA'X = B⁠, where A′A' is the transpose of A and B=[1087]B = \begin{bmatrix} 10 \\ 8 \\ 7 \end{bmatrix}⁠.

⇒X=(A′)−1B=(A−1)′B\Rightarrow X = (A')^{-1}B = (A^{-1})'B
Why?

Multiply both sides of A′X=BA'X = B on the left by (A′)−1(A')^{-1}⁠. Then use the property

(A′)−1=(A−1)′(A')^{-1} = (A^{-1})'

so we only need to transpose the A−1A^{-1} already found: its rows become columns.

[xyz]=[−322−211−423][1087]=[−30+16+14−20+8+7−40+16+21]=[0−5−3]\begin{aligned} \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} \\ &= \begin{bmatrix} -30 + 16 + 14 \\ -20 + 8 + 7 \\ -40 + 16 + 21 \end{bmatrix} = \begin{bmatrix} 0 \\ -5 \\ -3 \end{bmatrix} \end{aligned}
Why?

Each entry is a row of the matrix times the column BB⁠. For the first row:

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

The other two rows:

y=(−2)(10)+1(8)+1(7)=−20+8+7=−12+7=−5z=(−4)(10)+2(8)+3(7)=−40+16+21=−24+21=−3\begin{aligned} y &= (-2)(10) + 1(8) + 1(7) \\ &= -20 + 8 + 7 \\ &= -12 + 7 = -5 \\[4pt] z &= (-4)(10) + 2(8) + 3(7) \\ &= -40 + 16 + 21 \\ &= -24 + 21 = -3 \end{aligned}

In the first row, −30+16=−14-30 + 16 = -14⁠, then −14+14=0-14 + 14 = 0⁠.

Check in the equations: 0−2(−5)=100 - 2(-5) = 10⁠, 0+5+3=80 + 5 + 3 = 8 and 10−3=710 - 3 = 7⁠.

x−2y=0−2(−5)=0+10=102x−y−z=2(0)−(−5)−(−3)=0+5+3=8−2y+z=−2(−5)+(−3)=10−3=7\begin{aligned} x - 2y &= 0 - 2(-5) = 0 + 10 = 10 \\ 2x - y - z &= 2(0) - (-5) - (-3) \\ &= 0 + 5 + 3 = 8 \\ -2y + z &= -2(-5) + (-3) \\ &= 10 - 3 = 7 \end{aligned}

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

(b)

AB=[1−10234012]×[22−4−42−42−15]=[600060006]=6I\begin{aligned} AB &= \begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix} \\ &\qquad \times \begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix} \\ &= \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix} = 6I \end{aligned}
Why?

The entry in row ii⁠, column jj of ABAB is row ii of A times column jj of B. For example, row 1 times column 1:

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

and row 1 times column 2 is 2−2+0=02 - 2 + 0 = 0⁠. The 6s on the diagonal and 0s elsewhere make 6I6I⁠.

All nine entries, products first:

(1,1):1(2)+(−1)(−4)+0(2)=2+4+0=6(1,2):1(2)+(−1)(2)+0(−1)=2−2+0=0(1,3):1(−4)+(−1)(−4)+0(5)=−4+4+0=0(2,1):2(2)+3(−4)+4(2)=4−12+8=0(2,2):2(2)+3(2)+4(−1)=4+6−4=6(2,3):2(−4)+3(−4)+4(5)=−8−12+20=0(3,1):0(2)+1(−4)+2(2)=0−4+4=0(3,2):0(2)+1(2)+2(−1)=0+2−2=0(3,3):0(−4)+1(−4)+2(5)=0−4+10=6\begin{aligned} (1,1)&: 1(2) + (-1)(-4) + 0(2) \\ &= 2 + 4 + 0 = 6 \\ (1,2)&: 1(2) + (-1)(2) + 0(-1) \\ &= 2 - 2 + 0 = 0 \\ (1,3)&: 1(-4) + (-1)(-4) + 0(5) \\ &= -4 + 4 + 0 = 0 \\[4pt] (2,1)&: 2(2) + 3(-4) + 4(2) \\ &= 4 - 12 + 8 = 0 \\ (2,2)&: 2(2) + 3(2) + 4(-1) \\ &= 4 + 6 - 4 = 6 \\ (2,3)&: 2(-4) + 3(-4) + 4(5) \\ &= -8 - 12 + 20 = 0 \\[4pt] (3,1)&: 0(2) + 1(-4) + 2(2) \\ &= 0 - 4 + 4 = 0 \\ (3,2)&: 0(2) + 1(2) + 2(-1) \\ &= 0 + 2 - 2 = 0 \\ (3,3)&: 0(-4) + 1(-4) + 2(5) \\ &= 0 - 4 + 10 = 6 \end{aligned}
⇒A(16B)=I⇒A−1=16B\Rightarrow A\left(\frac16B\right) = I \Rightarrow A^{-1} = \frac16B
Why?

Divide both sides of AB=6IAB = 6I by 6. If AC=IAC = I for a square matrix CC⁠, then CC is the inverse of A. Here C=16BC = \frac16B⁠.

The given equations can be written as

[1−10234012][xyz]=[3177]\begin{bmatrix} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{bmatrix}\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}

i.e. AX=DAX = D⁠, where D=[3177]D = \begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix}⁠. So X=A−1DX = A^{-1}D⁠.

Why?

This time the coefficient matrix is A itself, so no transpose is needed. Multiplying AX=DAX = D on the left by A−1A^{-1} gives X=A−1DX = A^{-1}D⁠, and A−1=16BA^{-1} = \frac16B from above.

[xyz]=16[22−4−42−42−15][3177]=16[12−624]=[2−14]\begin{aligned} \begin{bmatrix} x \\ y \\ z \end{bmatrix} &= \frac16\begin{bmatrix} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{bmatrix}\begin{bmatrix} 3 \\ 17 \\ 7 \end{bmatrix} \\ &= \frac16\begin{bmatrix} 12 \\ -6 \\ 24 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ 4 \end{bmatrix} \end{aligned}
Why?

Row by row, times the column (3,17,7)(3, 17, 7)⁠:

Row 1: (2)(3)+(2)(17)+(−4)(7)=6+34−28=12Row 2: (−4)(3)+(2)(17)+(−4)(7)=−12+34−28=−6Row 3: (2)(3)+(−1)(17)+(5)(7)=6−17+35=24\begin{aligned} &\text{Row 1: } (2)(3) + (2)(17) + (-4)(7) \\ &= 6 + 34 - 28 = 12 \\[4pt] &\text{Row 2: } (-4)(3) + (2)(17) + (-4)(7) \\ &= -12 + 34 - 28 = -6 \\[4pt] &\text{Row 3: } (2)(3) + (-1)(17) + (5)(7) \\ &= 6 - 17 + 35 = 24 \end{aligned}

Then divide each by 6. Check: 2−(−1)=32 - (-1) = 3⁠, 4−3+16=174 - 3 + 16 = 17 and −1+8=7-1 + 8 = 7⁠.

126=2,−66=−1,246=4\begin{aligned} \frac{12}{6} = 2,\quad \frac{-6}{6} = -1,\quad \frac{24}{6} = 4 \end{aligned} x−y=2−(−1)=32x+3y+4z=2(2)+3(−1)+4(4)=4−3+16=17y+2z=−1+2(4)=−1+8=7\begin{aligned} x - y &= 2 - (-1) = 3 \\ 2x + 3y + 4z &= 2(2) + 3(-1) + 4(4) \\ &= 4 - 3 + 16 = 17 \\ y + 2z &= -1 + 2(4) = -1 + 8 = 7 \end{aligned}

∴ x=2\therefore\ x = 2⁠, y=−1y = -1⁠, z=4z = 4

Where this comes from

Chapter 4, Determinants: Inverse of a matrix; solving a system of equations.

Question type: (a) Solve equations by the matrix method; (b) Multiply two matrices + Solve equations by the matrix method (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: Find A−1A^{-1}⁠: ∣A∣=1≠0|A| = 1 \ne 0⁠, so A−1=adj⁡AA^{-1} = \operatorname{adj}A⁠. The coefficient matrix of the equations is A′A'⁠, not AA⁠, so X=(A−1)′BX = (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, multiply AA and BB row by column: AB=6IAB = 6I⁠, so A−1=16BA^{-1} = \frac16B⁠.

(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 coefficient matrix of the equations is AA itself, so X=A−1D=16BDX = A^{-1}D = \frac16BD with D=(3,17,7)D = (3, 17, 7)⁠, giving x=2x = 2⁠, y=−1y = -1⁠, z=4z = 4⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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