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CBSE Sample Paper 2023-24 · Section D · 5 marks

Q34: Matrix method: 2x+3y+10z=4\frac2x + \frac3y + \frac{10}{z} = 4⁠, 4x−6y+5z=1\frac4x - \frac6y + \frac5z = 1⁠, 6x+9y−20z=2\frac6x + \frac9y - \frac{20}{z} = 2

Sample 2023-24Ch 4 · DeterminantsCh 3 · Matrices
The question

Using the matrix method, solve the following system of linear equations :

2x+3y+10z=4\dfrac2x + \dfrac3y + \dfrac{10}{z} = 4⁠, 4x−6y+5z=1\dfrac4x - \dfrac6y + \dfrac5z = 1⁠, 6x+9y−20z=2\dfrac6x + \dfrac9y - \dfrac{20}{z} = 2⁠.

Exam answer

The given system of equations can be written in the form AX=BAX = B⁠, where

A=[23104−6569−20]A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix} X=[1x1y1z],B=[412]X = \begin{bmatrix} \frac1x \\ \frac1y \\ \frac1z \end{bmatrix},\quad B = \begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix}
Why?

The equations are not linear in xx⁠, yy⁠, zz⁠, but they are linear in 1x\frac1x⁠, 1y\frac1y⁠, 1z\frac1z⁠. Writing u=1xu = \frac1x⁠, v=1yv = \frac1y⁠, w=1zw = \frac1z⁠, the first equation is 2u+3v+10w=42u + 3v + 10w = 4⁠.

AA holds the coefficients row by row, XX the unknowns and BB the right sides.

∣A∣=2(120−45)−3(−80−30)+10(36+36)=150+330+720=1200≠0\begin{aligned} |A| &= 2(120 - 45) - 3(-80 - 30) \\ &\quad + 10(36 + 36) \\ &= 150 + 330 + 720 \\ &= 1200 \ne 0 \end{aligned}
Why?

Expand along the first row:

∣A∣=a11M11−a12M12+a13M13\begin{aligned} |A| &= a_{11}M_{11} - a_{12}M_{12} \\ &\quad + a_{13}M_{13} \end{aligned}

Each entry multiplies the 2×22 \times 2 determinant left when its row and column are covered:

∣A∣=2∣−659−20∣−3∣456−20∣+10∣4−669∣\begin{aligned} |A| &= 2\begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} - 3\begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} \\ &\quad + 10\begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} \end{aligned}

The three 2 by 2 minors are

M11=(−6)(−20)−(5)(9)M12=(4)(−20)−(5)(6)M13=(4)(9)−(−6)(6)\begin{aligned} M_{11} &= (-6)(-20) - (5)(9) \\ M_{12} &= (4)(-20) - (5)(6) \\ M_{13} &= (4)(9) - (-6)(6) \end{aligned}

Work each one out, (top left)(bottom right) minus (top right)(bottom left):

M11=(−6)(−20)−(5)(9)=120−45=75M12=(4)(−20)−(5)(6)=−80−30=−110M13=(4)(9)−(−6)(6)=36−(−36)=36+36=72\begin{aligned} M_{11} &= (-6)(-20) - (5)(9) \\ &= 120 - 45 = 75 \\[4pt] M_{12} &= (4)(-20) - (5)(6) \\ &= -80 - 30 = -110 \\[4pt] M_{13} &= (4)(9) - (-6)(6) \\ &= 36 - (-36) \\ &= 36 + 36 = 72 \end{aligned}

Then 2(75)=1502(75) = 150⁠, −3(−110)=330-3(-110) = 330 and 10(72)=72010(72) = 720⁠.

Multiply by the entries and their signs, then add:

2(75)=150−3(−110)=+33010(72)=720150+330=480480+720=1200\begin{aligned} 2(75) &= 150 \\ -3(-110) &= +330 \\ 10(72) &= 720 \\[4pt] 150 + 330 &= 480 \\ 480 + 720 &= 1200 \end{aligned}

Watch out: the middle term carries a minus sign.

∴ A−1\therefore\ A^{-1} exists.

Why?

A matrix has an inverse exactly when its determinant is not 0. Then the system has a unique solution, X=A−1BX = A^{-1}B⁠. Always write ∣A∣≠0|A| \ne 0⁠: it is the reason the method works.

The cofactors of the elements of AA are

A11=120−45=75A12=−(−80−30)=110A13=36+36=72\begin{aligned} A_{11} &= 120 - 45 = 75 \\ A_{12} &= -(-80 - 30) = 110 \\ A_{13} &= 36 + 36 = 72 \end{aligned}
Why?

Each cofactor is a signed minor:

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

where MijM_{ij} is the determinant left after deleting row ii and column jj⁠. For example, deleting row 1 and column 1 leaves

M11=∣−659−20∣=120−45\begin{aligned} M_{11} &= \begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} \\ &= 120 - 45 \end{aligned}

The signs (−1)i+j(-1)^{i + j} follow the pattern + − ++\ -\ + in rows 1 and 3, and − + −-\ +\ - in row 2. That is why some cofactors have a minus sign in front of the bracket.

[+−+−+−+−+]  on  A=[23104−6569−20]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 2 & 3 & 10 \\ 4 & -6 & 5 \\ 6 & 9 & -20 \end{bmatrix}

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

A11=+∣−659−20∣=+((−6)(−20)−(5)(9))=+(120−45)=75A12=−∣456−20∣=−((4)(−20)−(5)(6))=−(−80−30)=−(−110)=110A13=+∣4−669∣=+((4)(9)−(−6)(6))=+(36+36)=72\begin{aligned} A_{11} &= +\begin{vmatrix} -6 & 5 \\ 9 & -20 \end{vmatrix} \\ &= +\big((-6)(-20) - (5)(9)\big) \\ &= +(120 - 45) = 75 \\[4pt] A_{12} &= -\begin{vmatrix} 4 & 5 \\ 6 & -20 \end{vmatrix} \\ &= -\big((4)(-20) - (5)(6)\big) \\ &= -(-80 - 30) \\ &= -(-110) = 110 \\[4pt] A_{13} &= +\begin{vmatrix} 4 & -6 \\ 6 & 9 \end{vmatrix} \\ &= +\big((4)(9) - (-6)(6)\big) \\ &= +(36 + 36) = 72 \end{aligned}
A21=−(−60−90)=150A22=−40−60=−100A23=−(18−18)=0\begin{aligned} A_{21} &= -(-60 - 90) = 150 \\ A_{22} &= -40 - 60 = -100 \\ A_{23} &= -(18 - 18) = 0 \end{aligned}
Why?

Row 2 (cover row 2; the signs are − + −-\ +\ -⁠):

A21=−∣3109−20∣=−((3)(−20)−(10)(9))=−(−60−90)=−(−150)=150A22=+∣2106−20∣=+((2)(−20)−(10)(6))=+(−40−60)=−100A23=−∣2369∣=−((2)(9)−(3)(6))=−(18−18)=−(0)=0\begin{aligned} A_{21} &= -\begin{vmatrix} 3 & 10 \\ 9 & -20 \end{vmatrix} \\ &= -\big((3)(-20) - (10)(9)\big) \\ &= -(-60 - 90) \\ &= -(-150) = 150 \\[4pt] A_{22} &= +\begin{vmatrix} 2 & 10 \\ 6 & -20 \end{vmatrix} \\ &= +\big((2)(-20) - (10)(6)\big) \\ &= +(-40 - 60) = -100 \\[4pt] A_{23} &= -\begin{vmatrix} 2 & 3 \\ 6 & 9 \end{vmatrix} \\ &= -\big((2)(9) - (3)(6)\big) \\ &= -(18 - 18) = -(0) = 0 \end{aligned}
A31=15+60=75A32=−(10−40)=30A33=−12−12=−24\begin{aligned} A_{31} &= 15 + 60 = 75 \\ A_{32} &= -(10 - 40) = 30 \\ A_{33} &= -12 - 12 = -24 \end{aligned}
Why?

Row 3 (cover row 3; the signs are + − ++\ -\ +⁠):

A31=+∣310−65∣=+((3)(5)−(10)(−6))=+(15−(−60))=15+60=75A32=−∣21045∣=−((2)(5)−(10)(4))=−(10−40)=−(−30)=30A33=+∣234−6∣=+((2)(−6)−(3)(4))=+(−12−12)=−24\begin{aligned} A_{31} &= +\begin{vmatrix} 3 & 10 \\ -6 & 5 \end{vmatrix} \\ &= +\big((3)(5) - (10)(-6)\big) \\ &= +(15 - (-60)) \\ &= 15 + 60 = 75 \\[4pt] A_{32} &= -\begin{vmatrix} 2 & 10 \\ 4 & 5 \end{vmatrix} \\ &= -\big((2)(5) - (10)(4)\big) \\ &= -(10 - 40) \\ &= -(-30) = 30 \\[4pt] A_{33} &= +\begin{vmatrix} 2 & 3 \\ 4 & -6 \end{vmatrix} \\ &= +\big((2)(-6) - (3)(4)\big) \\ &= +(-12 - 12) = -24 \end{aligned}

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

adj A=[7515075110−10030720−24]\text{adj } A = \begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}
Why?

adj AA is the transpose of the matrix of cofactors: the cofactors of row 1 are written down column 1, and so on.

So the first column is A11,A12,A13=75,110,72A_{11}, A_{12}, A_{13} = 75, 110, 72⁠.

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

A−1=1∣A∣(adj A)A^{-1} = \frac{1}{|A|}(\text{adj } A) =11200[7515075110−10030720−24]= \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}

As AX=B⇒X=A−1BAX = B \Rightarrow X = A^{-1}B⁠,

Why?

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

The order matters: A−1BA^{-1}B⁠, not BA−1BA^{-1}⁠.

[1x1y1z]=11200[7515075110−10030720−24][412]\begin{aligned} &\begin{bmatrix} \frac1x \\ \frac1y \\ \frac1z \end{bmatrix} \\ &= \frac{1}{1200}\begin{bmatrix} 75 & 150 & 75 \\ 110 & -100 & 30 \\ 72 & 0 & -24 \end{bmatrix}\begin{bmatrix} 4 \\ 1 \\ 2 \end{bmatrix} \end{aligned} =11200[300+150+150440−100+60288+0−48]= \frac{1}{1200}\begin{bmatrix} 300 + 150 + 150 \\ 440 - 100 + 60 \\ 288 + 0 - 48 \end{bmatrix}
Why?

Multiply each row of adj AA by the column BB⁠, entry by entry, and add. For row 1:

75(4)+150(1)+75(2)=300+150+150\begin{aligned} &75(4) + 150(1) + 75(2) \\ &= 300 + 150 + 150 \end{aligned}

All three rows, products first, then the sum:

Row 1: (75)(4)+(150)(1)+(75)(2)=300+150+150=600Row 2: (110)(4)+(−100)(1)+(30)(2)=440−100+60=400Row 3: (72)(4)+(0)(1)+(−24)(2)=288+0−48=240\begin{aligned} &\text{Row 1: } (75)(4) + (150)(1) + (75)(2) \\ &= 300 + 150 + 150 = 600 \\[4pt] &\text{Row 2: } (110)(4) + (-100)(1) + (30)(2) \\ &= 440 - 100 + 60 = 400 \\[4pt] &\text{Row 3: } (72)(4) + (0)(1) + (-24)(2) \\ &= 288 + 0 - 48 = 240 \end{aligned}
=11200[600400240]=[121315]= \frac{1}{1200}\begin{bmatrix} 600 \\ 400 \\ 240 \end{bmatrix} = \begin{bmatrix} \frac12 \\ \frac13 \\ \frac15 \end{bmatrix}
Why?

Divide each entry by 1200:

6001200=12,4001200=13,2401200=15\begin{aligned} \frac{600}{1200} &= \frac12,\quad \frac{400}{1200} = \frac13, \\ \frac{240}{1200} &= \frac15 \end{aligned}

Each time, divide the top and the bottom by the top:

6001200=600÷6001200÷600=124001200=400÷4001200÷400=132401200=240÷2401200÷240=15\begin{aligned} \frac{600}{1200} &= \frac{600 \div 600}{1200 \div 600} = \frac12 \\ \frac{400}{1200} &= \frac{400 \div 400}{1200 \div 400} = \frac13 \\ \frac{240}{1200} &= \frac{240 \div 240}{1200 \div 240} = \frac15 \end{aligned}

Thus, 1x=12\dfrac1x = \dfrac12⁠, 1y=13\dfrac1y = \dfrac13⁠, 1z=15\dfrac1z = \dfrac15⁠.

Why?

The matrix gave 1x\frac1x⁠, 1y\frac1y⁠, 1z\frac1z⁠, so take reciprocals to get xx⁠, yy⁠, zz⁠.

1x=12⇒x=21y=13⇒y=31z=15⇒z=5\begin{aligned} \frac1x = \frac12 &\Rightarrow x = 2 \\ \frac1y = \frac13 &\Rightarrow y = 3 \\ \frac1z = \frac15 &\Rightarrow z = 5 \end{aligned}

Check in the first equation: 22+33+105=1+1+2=4\frac22 + \frac33 + \frac{10}{5} = 1 + 1 + 2 = 4⁠.

The other two equations check the same way:

42−63+55=2−2+1=1 ✓62+93−205=3+3−4=2 ✓\begin{aligned} &\frac42 - \frac63 + \frac55 \\ &= 2 - 2 + 1 = 1\ \checkmark \\[4pt] &\frac62 + \frac93 - \frac{20}{5} \\ &= 3 + 3 - 4 = 2\ \checkmark \end{aligned}

Watch out: do not stop at 12\frac12⁠, 13\frac13⁠, 15\frac15⁠. Those are not xx⁠, yy⁠, zz⁠.

Hence, x=2x = 2⁠, y=3y = 3⁠, z=5z = 5⁠.

Where this comes from

Chapter 4, Determinants: Matrix method: equations in 1/x, 1/y, 1/z.

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: The equations are linear in 1x\frac1x⁠, 1y\frac1y⁠, 1z\frac1z⁠, so take those as the unknowns in XX⁠. Then ∣A∣=1200≠0|A| = 1200 \ne 0⁠, and X=A−1BX = A^{-1}B gives 12,13,15\frac12, \frac13, \frac15⁠, so x=2x = 2⁠, y=3y = 3⁠, z=5z = 5⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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