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CBSE Sample Paper 2024-25 · Section D · 5 marks

Q33: Path y=ax2+bx+cy = ax^2 + bx + c of a headed football by the matrix method

Sample 2024-25Ch 4 · DeterminantsCh 3 · Matrices
The question

The equation of the path traversed by the ball headed by the footballer is y=ax2+bx+cy = ax^2 + bx + c⁠; (where 0≤x≤140 \le x \le 14 and a,b,c∈Ra, b, c \in R and a≠0a \ne 0⁠) with respect to a XY-coordinate system in the vertical plane. The ball passes through the points (2,15)(2, 15)⁠, (4,25)(4, 25) and (14,15)(14, 15)⁠. Determine the values of a, b and c by solving the system of linear equations in a, b and c, using matrix method. Also find the equation of the path traversed by the ball.

Exam answer

y=ax2+bx+cy = ax^2 + bx + c passes through (2,15)(2, 15)⁠, (4,25)(4, 25) and (14,15)(14, 15)⁠:

4a+2b+c=1516a+4b+c=25196a+14b+c=15\begin{aligned} 4a + 2b + c &= 15 \\ 16a + 4b + c &= 25 \\ 196a + 14b + c &= 15 \end{aligned}
Why?

A point lies on the curve when its coordinates satisfy the equation. Put each (x,y)(x, y) into y=ax2+bx+cy = ax^2 + bx + c⁠:

(2,15):15=4a+2b+c(4,25):25=16a+4b+c(14,15):15=196a+14b+c\begin{aligned} (2, 15)&: 15 = 4a + 2b + c \\ (4, 25)&: 25 = 16a + 4b + c \\ (14, 15)&: 15 = 196a + 14b + c \end{aligned}

(142=19614^2 = 196⁠.) This gives three linear equations in aa⁠, bb⁠, cc⁠.

The set of equations can be represented in the matrix form as AX=BAX = B⁠, where

A=[4211641196141],A = \begin{bmatrix} 4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1 \end{bmatrix}, X=[abc], B=[152515]X = \begin{bmatrix} a \\ b \\ c \end{bmatrix},\ B = \begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix}
Why?

Row ii of AA holds the coefficients of aa⁠, bb⁠, cc in equation ii⁠. Then row ii of AXAX is the left side of equation ii⁠, and BB holds the right sides.

∣A∣=4(4−14)−2(16−196)+1(224−784)=−40+360−560=−240≠0\begin{aligned} \lvert A\rvert &= 4(4 - 14) - 2(16 - 196) \\ &\quad + 1(224 - 784) \\ &= -40 + 360 - 560 \\ &= -240 \ne 0 \end{aligned}
Why?

Expand along the first row with the signs ++⁠, −-⁠, ++⁠. Each entry multiplies the 2×22 \times 2 determinant left when its row and column are covered:

∣A∣=4∣41141∣−2∣1611961∣+1∣16419614∣\begin{aligned} \lvert A\rvert &= 4\begin{vmatrix} 4 & 1 \\ 14 & 1 \end{vmatrix} - 2\begin{vmatrix} 16 & 1 \\ 196 & 1 \end{vmatrix} \\ &\quad + 1\begin{vmatrix} 16 & 4 \\ 196 & 14 \end{vmatrix} \end{aligned}

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

∣41141∣=(4)(1)−(1)(14)=4−14=−10∣1611961∣=(16)(1)−(1)(196)=16−196=−180∣16419614∣=(16)(14)−(4)(196)=224−784=−560\begin{aligned} \begin{vmatrix} 4 & 1 \\ 14 & 1 \end{vmatrix} &= (4)(1) - (1)(14) \\ &= 4 - 14 = -10 \\[4pt] \begin{vmatrix} 16 & 1 \\ 196 & 1 \end{vmatrix} &= (16)(1) - (1)(196) \\ &= 16 - 196 = -180 \\[4pt] \begin{vmatrix} 16 & 4 \\ 196 & 14 \end{vmatrix} &= (16)(14) - (4)(196) \\ &= 224 - 784 = -560 \end{aligned}

Multiply by the entries and their signs, then add:

4(−10)=−40−2(−180)=+3601(−560)=−560−40+360=320320−560=−240\begin{aligned} 4(-10) &= -40 \\ -2(-180) &= +360 \\ 1(-560) &= -560 \\[4pt] -40 + 360 &= 320 \\ 320 - 560 &= -240 \end{aligned}

Hence A−1A^{-1} exists, and the system has a unique solution X=A−1BX = A^{-1}B⁠.

Why?

A square matrix has an inverse exactly when its determinant is not 0. Multiplying AX=BAX = B on the left by A−1A^{-1} gives X=A−1BX = A^{-1}B⁠, and this is the only solution.

Cofactors of AA⁠:

A11=−10, A12=180,A13=−560, A21=12,A22=−192, A23=336,A31=−2, A32=12,A33=−16\begin{aligned} &A_{11} = -10,\ A_{12} = 180, \\ &A_{13} = -560,\ A_{21} = 12, \\ &A_{22} = -192,\ A_{23} = 336, \\ &A_{31} = -2,\ A_{32} = 12, \\ &A_{33} = -16 \end{aligned}
Why?

The cofactor is Aij=(−1)i+jMijA_{ij} = (-1)^{i + j}M_{ij}⁠, where the minor MijM_{ij} is the determinant left after covering row ii and column jj⁠. The signs (−1)i+j(-1)^{i + j} follow the pattern

[+−+−+−+−+]  on  A=[4211641196141]\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 4 & 2 & 1 \\ 16 & 4 & 1 \\ 196 & 14 & 1 \end{bmatrix}

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

A11=+∣41141∣=+((4)(1)−(1)(14))=+(4−14)=−10A12=−∣1611961∣=−((16)(1)−(1)(196))=−(16−196)=−(−180)=180A13=+∣16419614∣=+((16)(14)−(4)(196))=+(224−784)=−560\begin{aligned} A_{11} &= +\begin{vmatrix} 4 & 1 \\ 14 & 1 \end{vmatrix} \\ &=+\big((4)(1) - (1)(14)\big) \\ &= +(4 - 14) = -10 \\[4pt] A_{12} &= -\begin{vmatrix} 16 & 1 \\ 196 & 1 \end{vmatrix} \\ &=-\big((16)(1) - (1)(196)\big) \\ &= -(16 - 196) = -(-180) = 180 \\[4pt] A_{13} &= +\begin{vmatrix} 16 & 4 \\ 196 & 14 \end{vmatrix} \\ &=+\big((16)(14) - (4)(196)\big) \\ &= +(224 - 784) = -560 \end{aligned}

Row 2 (cover row 2):

A21=−∣21141∣=−((2)(1)−(1)(14))=−(2−14)=−(−12)=12A22=+∣411961∣=+((4)(1)−(1)(196))=+(4−196)=−192A23=−∣4219614∣=−((4)(14)−(2)(196))=−(56−392)=−(−336)=336\begin{aligned} A_{21} &= -\begin{vmatrix} 2 & 1 \\ 14 & 1 \end{vmatrix} \\ &=-\big((2)(1) - (1)(14)\big) \\ &= -(2 - 14) = -(-12) = 12 \\[4pt] A_{22} &= +\begin{vmatrix} 4 & 1 \\ 196 & 1 \end{vmatrix} \\ &=+\big((4)(1) - (1)(196)\big) \\ &= +(4 - 196) = -192 \\[4pt] A_{23} &= -\begin{vmatrix} 4 & 2 \\ 196 & 14 \end{vmatrix} \\ &=-\big((4)(14) - (2)(196)\big) \\ &= -(56 - 392) = -(-336) = 336 \end{aligned}

Row 3 (cover row 3):

A31=+∣2141∣=+((2)(1)−(1)(4))=+(2−4)=−2A32=−∣41161∣=−((4)(1)−(1)(16))=−(4−16)=−(−12)=12A33=+∣42164∣=+((4)(4)−(2)(16))=+(16−32)=−16\begin{aligned} A_{31} &= +\begin{vmatrix} 2 & 1 \\ 4 & 1 \end{vmatrix} \\ &=+\big((2)(1) - (1)(4)\big) \\ &= +(2 - 4) = -2 \\[4pt] A_{32} &= -\begin{vmatrix} 4 & 1 \\ 16 & 1 \end{vmatrix} \\ &=-\big((4)(1) - (1)(16)\big) \\ &= -(4 - 16) = -(-12) = 12 \\[4pt] A_{33} &= +\begin{vmatrix} 4 & 2 \\ 16 & 4 \end{vmatrix} \\ &=+\big((4)(4) - (2)(16)\big) \\ &= +(16 - 32) = -16 \end{aligned}

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

adj⁡A=[−1012−2180−19212−560336−16]\operatorname{adj}A = \begin{bmatrix} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{bmatrix}
Why?

adj⁡A\operatorname{adj}A is the transpose of the matrix of cofactors. The cofactors of row 1 of AA (−10-10⁠, 180, −560-560⁠) form column 1 of adj⁡A\operatorname{adj}A⁠, and so on.

X=A−1B=1∣A∣(adj⁡A)B\begin{aligned} X &= A^{-1}B \\ &= \frac{1}{\lvert A\rvert}(\operatorname{adj}A)B \end{aligned}
Why?

The inverse is

A−1=1∣A∣adj⁡AA^{-1} = \frac{1}{\lvert A\rvert}\operatorname{adj}A

Here 1∣A∣=−1240\frac{1}{\lvert A\rvert} = -\frac{1}{240}⁠.

(adj⁡A)B=[−1012−2180−19212−560336−16][152515]=5[−1012−2180−19212−560336−16][353]=5[24−384−48]\begin{aligned} &(\operatorname{adj}A)B \\ &= \begin{bmatrix} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{bmatrix}\begin{bmatrix} 15 \\ 25 \\ 15 \end{bmatrix} \\ &= 5\begin{bmatrix} -10 & 12 & -2 \\ 180 & -192 & 12 \\ -560 & 336 & -16 \end{bmatrix}\begin{bmatrix} 3 \\ 5 \\ 3 \end{bmatrix} \\ &= 5\begin{bmatrix} 24 \\ -384 \\ -48 \end{bmatrix} \end{aligned}
Why?

Every entry of BB is a multiple of 5, so take 5 out first: the numbers stay small.

Multiply each row by the column (3,5,3)(3, 5, 3)⁠:

Row 1: (−10)(3)+(12)(5)+(−2)(3)=−30+60−6=24Row 2: (180)(3)+(−192)(5)+(12)(3)=540−960+36=−384Row 3: (−560)(3)+(336)(5)+(−16)(3)=−1680+1680−48=−48\begin{aligned} &\text{Row 1: } (-10)(3) + (12)(5) + (-2)(3) \\ &= -30 + 60 - 6 = 24 \\[4pt] &\text{Row 2: } (180)(3) + (-192)(5) + (12)(3) \\ &= 540 - 960 + 36 = -384 \\[4pt] &\text{Row 3: } (-560)(3) + (336)(5) + (-16)(3) \\ &= -1680 + 1680 - 48 = -48 \end{aligned}
X=−5240[24−384−48]=[−1281]\begin{aligned} X &= -\frac{5}{240}\begin{bmatrix} 24 \\ -384 \\ -48 \end{bmatrix} \\ &= \begin{bmatrix} -\frac12 \\ 8 \\ 1 \end{bmatrix} \end{aligned}
Why?

Divide the top and bottom by 5: −5240=−5÷5240÷5=−148-\frac{5}{240} = -\frac{5 \div 5}{240 \div 5} = -\frac{1}{48}⁠. Then

−148(24)=−2448=−12−148(−384)=38448=8−148(−48)=4848=1\begin{aligned} -\frac{1}{48}(24) &= -\frac{24}{48} = -\frac12 \\ -\frac{1}{48}(-384) &= \frac{384}{48} = 8 \\ -\frac{1}{48}(-48) &= \frac{48}{48} = 1 \end{aligned}

∴ a=−12\therefore\ a = -\dfrac12⁠, b=8b = 8⁠, c=1c = 1

Why?

a=−12≠0a = -\frac12 \ne 0⁠, as the question requires. Check with (4,25)(4, 25)⁠:

−12(4)2+8(4)+1=−12(16)+32+1=−8+32+1=25 ✓\begin{aligned} &-\frac12(4)^2 + 8(4) + 1 \\ &= -\frac12(16) + 32 + 1 \\ &= -8 + 32 + 1 = 25\ \checkmark \end{aligned}

So, the equation of the path is y=−12x2+8x+1y = -\dfrac12x^2 + 8x + 1⁠.

Where this comes from

Chapter 4, Determinants: System of three equations by the matrix method.

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 word problem: put each point (2,15)(2, 15)⁠, (4,25)(4, 25)⁠, (14,15)(14, 15) into y=ax2+bx+cy = ax^2 + bx + c to get one equation in a,b,ca, b, c⁠. Then ∣A∣=−240≠0|A| = -240 \ne 0⁠, and X=A−1BX = A^{-1}B gives a=−12a = -\frac12⁠, b=8b = 8⁠, c=1c = 1⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 3, Matrices.

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