The question
A school wants to allocate students into three clubs : Sports, Music and Drama, under following conditions :
The number of students in Sports club should be equal to the sum of the number of students in Music and Drama club.
The number of students in Music club should be 20 more than half the number of students in Sports club.
The total number of students to be allocated in all three clubs are 180.
Find the number of students allocated to different clubs, using matrix method.
Exam answer Show all explanations
Let x x x y y y z z z x = y + z x = y + z x = y + z y = x 2 + 20 y = \frac{x}{2} + 20 y = 2 x + 20 x + y + z = 180 x + y + z = 180 x + y + z = 180
x − y − z = 0 x − 2 y = − 40 x + y + z = 180 \begin{aligned} x - y - z &= 0 \\ x - 2y &= -40 \\ x + y + z &= 180 \end{aligned} x − y − z x − 2 y x + y + z = 0 = − 40 = 180 Why?
Turn each condition into an equation. "Sports equals Music plus Drama" gives x = y + z x = y + z x = y + z y = x 2 + 20 y = \frac{x}{2} + 20 y = 2 x + 20
Then bring the unknowns to the left side. Multiplying y = x 2 + 20 y = \frac{x}{2} + 20 y = 2 x + 20 2 2 2 2 y = x + 40 2y = x + 40 2 y = x + 40 x − 2 y = − 40 x - 2y = -40 x − 2 y = − 40
x = y + z x − y − z = 0 y = x 2 + 20 2 y = x + 40 2 y − x = 40 x − 2 y = − 40 ( × − 1 ) \begin{aligned} x &= y + z \\ x - y - z &= 0 \\[4pt] y &= \frac{x}{2} + 20 \\ 2y &= x + 40 \\ 2y - x &= 40 \\ x - 2y &= -40 \quad (\times\, -1) \end{aligned} x x − y − z y 2 y 2 y − x x − 2 y = y + z = 0 = 2 x + 20 = x + 40 = 40 = − 40 ( × − 1 ) These can be written as A X = B AX = B A X = B
A = [ 1 − 1 − 1 1 − 2 0 1 1 1 ] A = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -2 & 0 \\ 1 & 1 & 1 \end{bmatrix} A = 1 1 1 − 1 − 2 1 − 1 0 1 X = [ x y z ] , B = [ 0 − 40 180 ] X = \begin{bmatrix} x \\ y \\ z \end{bmatrix},\quad B = \begin{bmatrix} 0 \\ -40 \\ 180 \end{bmatrix} X = x y z , B = 0 − 40 180 Why?
Row i i i A A A x , y , z x, y, z x , y , z i i i z z z 0 0 0 B B B
∣ A ∣ = 1 ( − 2 − 0 ) + 1 ( 1 − 0 ) − 1 ( 1 + 2 ) = − 2 + 1 − 3 = − 4 ≠ 0 \begin{aligned} |A| &= 1(-2 - 0) + 1(1 - 0) \\ &\quad - 1(1 + 2) \\ &= -2 + 1 - 3 = -4 \ne 0 \end{aligned} ∣ A ∣ = 1 ( − 2 − 0 ) + 1 ( 1 − 0 ) − 1 ( 1 + 2 ) = − 2 + 1 − 3 = − 4 = 0 Why?
Expand along the first row, with the signs + , − , + +, -, + + , − , +
1 [ ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) ] − ( − 1 ) [ ( 1 ) ( 1 ) − ( 0 ) ( 1 ) ] + ( − 1 ) [ ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) ] \begin{aligned} &1\big[(-2)(1) - (0)(1)\big] \\ &- (-1)\big[(1)(1) - (0)(1)\big] \\ &+ (-1)\big[(1)(1) - (-2)(1)\big] \end{aligned} 1 [ ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) ] − ( − 1 ) [ ( 1 ) ( 1 ) − ( 0 ) ( 1 ) ] + ( − 1 ) [ ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) ] Each 2 × 2 2 \times 2 2 × 2
∣ − 2 0 1 1 ∣ = ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) = − 2 − 0 = − 2 ∣ 1 0 1 1 ∣ = ( 1 ) ( 1 ) − ( 0 ) ( 1 ) = 1 − 0 = 1 ∣ 1 − 2 1 1 ∣ = ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) = 1 + 2 = 3 \begin{aligned} \begin{vmatrix} -2 & 0 \\ 1 & 1 \end{vmatrix} &= (-2)(1) - (0)(1) \\ &= -2 - 0 = -2 \\[4pt] \begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} &= (1)(1) - (0)(1) \\ &= 1 - 0 = 1 \\[4pt] \begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} &= (1)(1) - (-2)(1) \\ &= 1 + 2 = 3 \end{aligned} − 2 1 0 1 1 1 0 1 1 1 − 2 1 = ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) = − 2 − 0 = − 2 = ( 1 ) ( 1 ) − ( 0 ) ( 1 ) = 1 − 0 = 1 = ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) = 1 + 2 = 3 Multiply by the entries and their signs, then add:
1 ( − 2 ) = − 2 − ( − 1 ) ( 1 ) = + 1 ( − 1 ) ( 3 ) = − 3 − 2 + 1 = − 1 − 1 − 3 = − 4 \begin{aligned} 1(-2) &= -2 \\ -(-1)(1) &= +1 \\ (-1)(3) &= -3 \\[4pt] -2 + 1 &= -1 \\ -1 - 3 &= -4 \end{aligned} 1 ( − 2 ) − ( − 1 ) ( 1 ) ( − 1 ) ( 3 ) − 2 + 1 − 1 − 3 = − 2 = + 1 = − 3 = − 1 = − 4 Watch out: the middle entry is − 1 -1 − 1 − ( − 1 ) [ … ] -(-1)[\dots] − ( − 1 ) [ … ] + 1 ( 1 − 0 ) +1(1 - 0) + 1 ( 1 − 0 )
So A − 1 A^{-1} A − 1 X = A − 1 B X = A^{-1}B X = A − 1 B
Why?
A square matrix has an inverse exactly when its determinant is not zero. Multiplying A X = B AX = B A X = B A − 1 A^{-1} A − 1 X = A − 1 B X = A^{-1}B X = A − 1 B
Cofactors: A 11 = − 2 A_{11} = -2 A 11 = − 2 A 12 = − 1 A_{12} = -1 A 12 = − 1 A 13 = 3 A_{13} = 3 A 13 = 3 A 21 = 0 A_{21} = 0 A 21 = 0 A 22 = 2 A_{22} = 2 A 22 = 2 A 23 = − 2 A_{23} = -2 A 23 = − 2 A 31 = − 2 A_{31} = -2 A 31 = − 2 A 32 = − 1 A_{32} = -1 A 32 = − 1 A 33 = − 1 A_{33} = -1 A 33 = − 1
Why?
The cofactor is A i j = ( − 1 ) i + j M i j A_{ij} = (-1)^{i + j}M_{ij} A ij = ( − 1 ) i + j M ij M i j M_{ij} M ij i i i j j j ( − 1 ) i + j (-1)^{i + j} ( − 1 ) i + j
[ + − + − + − + − + ] on A = [ 1 − 1 − 1 1 − 2 0 1 1 1 ] \begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}\ \text{ on }\ A = \begin{bmatrix} 1 & -1 & -1 \\ 1 & -2 & 0 \\ 1 & 1 & 1 \end{bmatrix} + − + − + − + − + on A = 1 1 1 − 1 − 2 1 − 1 0 1 Row 1 (cover row 1 and one column each time):
A 11 = + ∣ − 2 0 1 1 ∣ = + ( ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) ) = + ( − 2 − 0 ) = − 2 A 12 = − ∣ 1 0 1 1 ∣ = − ( ( 1 ) ( 1 ) − ( 0 ) ( 1 ) ) = − ( 1 − 0 ) = − 1 A 13 = + ∣ 1 − 2 1 1 ∣ = + ( ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) ) = + ( 1 + 2 ) = 3 \begin{aligned} A_{11} &= +\begin{vmatrix} -2 & 0 \\ 1 & 1 \end{vmatrix} \\ &= +\big((-2)(1) - (0)(1)\big) \\ &= +(-2 - 0) = -2 \\[4pt] A_{12} &= -\begin{vmatrix} 1 & 0 \\ 1 & 1 \end{vmatrix} \\ &= -\big((1)(1) - (0)(1)\big) \\ &= -(1 - 0) = -1 \\[4pt] A_{13} &= +\begin{vmatrix} 1 & -2 \\ 1 & 1 \end{vmatrix} \\ &= +\big((1)(1) - (-2)(1)\big) \\ &= +(1 + 2) = 3 \end{aligned} A 11 A 12 A 13 = + − 2 1 0 1 = + ( ( − 2 ) ( 1 ) − ( 0 ) ( 1 ) ) = + ( − 2 − 0 ) = − 2 = − 1 1 0 1 = − ( ( 1 ) ( 1 ) − ( 0 ) ( 1 ) ) = − ( 1 − 0 ) = − 1 = + 1 1 − 2 1 = + ( ( 1 ) ( 1 ) − ( − 2 ) ( 1 ) ) = + ( 1 + 2 ) = 3 Row 2 (cover row 2):
A 21 = − ∣ − 1 − 1 1 1 ∣ = − ( ( − 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = − ( − 1 + 1 ) = − ( 0 ) = 0 A 22 = + ∣ 1 − 1 1 1 ∣ = + ( ( 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = + ( 1 + 1 ) = 2 A 23 = − ∣ 1 − 1 1 1 ∣ = − ( ( 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = − ( 1 + 1 ) = − 2 \begin{aligned} A_{21} &= -\begin{vmatrix} -1 & -1 \\ 1 & 1 \end{vmatrix} \\ &= -\big((-1)(1) - (-1)(1)\big) \\ &= -(-1 + 1) = -(0) = 0 \\[4pt] A_{22} &= +\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\ &= +\big((1)(1) - (-1)(1)\big) \\ &= +(1 + 1) = 2 \\[4pt] A_{23} &= -\begin{vmatrix} 1 & -1 \\ 1 & 1 \end{vmatrix} \\ &= -\big((1)(1) - (-1)(1)\big) \\ &= -(1 + 1) = -2 \end{aligned} A 21 A 22 A 23 = − − 1 1 − 1 1 = − ( ( − 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = − ( − 1 + 1 ) = − ( 0 ) = 0 = + 1 1 − 1 1 = + ( ( 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = + ( 1 + 1 ) = 2 = − 1 1 − 1 1 = − ( ( 1 ) ( 1 ) − ( − 1 ) ( 1 ) ) = − ( 1 + 1 ) = − 2 Row 3 (cover row 3):
A 31 = + ∣ − 1 − 1 − 2 0 ∣ = + ( ( − 1 ) ( 0 ) − ( − 1 ) ( − 2 ) ) = + ( 0 − 2 ) = − 2 A 32 = − ∣ 1 − 1 1 0 ∣ = − ( ( 1 ) ( 0 ) − ( − 1 ) ( 1 ) ) = − ( 0 + 1 ) = − 1 A 33 = + ∣ 1 − 1 1 − 2 ∣ = + ( ( 1 ) ( − 2 ) − ( − 1 ) ( 1 ) ) = + ( − 2 + 1 ) = − 1 \begin{aligned} A_{31} &= +\begin{vmatrix} -1 & -1 \\ -2 & 0 \end{vmatrix} \\ &= +\big((-1)(0) - (-1)(-2)\big) \\ &= +(0 - 2) = -2 \\[4pt] A_{32} &= -\begin{vmatrix} 1 & -1 \\ 1 & 0 \end{vmatrix} \\ &= -\big((1)(0) - (-1)(1)\big) \\ &= -(0 + 1) = -1 \\[4pt] A_{33} &= +\begin{vmatrix} 1 & -1 \\ 1 & -2 \end{vmatrix} \\ &= +\big((1)(-2) - (-1)(1)\big) \\ &= +(-2 + 1) = -1 \end{aligned} A 31 A 32 A 33 = + − 1 − 2 − 1 0 = + ( ( − 1 ) ( 0 ) − ( − 1 ) ( − 2 ) ) = + ( 0 − 2 ) = − 2 = − 1 1 − 1 0 = − ( ( 1 ) ( 0 ) − ( − 1 ) ( 1 ) ) = − ( 0 + 1 ) = − 1 = + 1 1 − 1 − 2 = + ( ( 1 ) ( − 2 ) − ( − 1 ) ( 1 ) ) = + ( − 2 + 1 ) = − 1 Watch out for the sign pattern: + − + +\ -\ + + − + − + − -\ +\ - − + −
adj A = [ − 2 0 − 2 − 1 2 − 1 3 − 2 − 1 ] \text{adj}\,A = \begin{bmatrix} -2 & 0 & -2 \\ -1 & 2 & -1 \\ 3 & -2 & -1 \end{bmatrix} adj A = − 2 − 1 3 0 2 − 2 − 2 − 1 − 1 Why?
adj A \text{adj}\,A adj A 1 1 1 1 1 1
A 11 , A 12 , A 13 = − 2 , − 1 , 3 → col 1 A 21 , A 22 , A 23 = 0 , 2 , − 2 → col 2 A 31 , A 32 , A 33 = − 2 , − 1 , − 1 → col 3 \begin{aligned} A_{11}, A_{12}, A_{13} &= -2,\ -1,\ 3 \to \text{col } 1 \\ A_{21}, A_{22}, A_{23} &= 0,\ 2,\ -2 \to \text{col } 2 \\ A_{31}, A_{32}, A_{33} &= -2,\ -1,\ -1 \to \text{col } 3 \end{aligned} A 11 , A 12 , A 13 A 21 , A 22 , A 23 A 31 , A 32 , A 33 = − 2 , − 1 , 3 → col 1 = 0 , 2 , − 2 → col 2 = − 2 , − 1 , − 1 → col 3 A − 1 = 1 ∣ A ∣ adj A = 1 4 [ 2 0 2 1 − 2 1 − 3 2 1 ] \begin{aligned} A^{-1} &= \frac{1}{|A|}\,\text{adj}\,A \\ &= \frac14\begin{bmatrix} 2 & 0 & 2 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{bmatrix} \end{aligned} A − 1 = ∣ A ∣ 1 adj A = 4 1 2 1 − 3 0 − 2 2 2 1 1 Why?
1 ∣ A ∣ = − 1 4 \frac{1}{|A|} = -\frac14 ∣ A ∣ 1 = − 4 1 1 4 \frac14 4 1 − 1 -1 − 1 − 1 -1 − 1 adj A \text{adj}\,A adj A
A − 1 = − 1 4 [ − 2 0 − 2 − 1 2 − 1 3 − 2 − 1 ] = 1 4 [ 2 0 2 1 − 2 1 − 3 2 1 ] \begin{aligned} A^{-1} &= -\frac14\begin{bmatrix} -2 & 0 & -2 \\ -1 & 2 & -1 \\ 3 & -2 & -1 \end{bmatrix} \\ &= \frac14\begin{bmatrix} 2 & 0 & 2 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{bmatrix} \end{aligned} A − 1 = − 4 1 − 2 − 1 3 0 2 − 2 − 2 − 1 − 1 = 4 1 2 1 − 3 0 − 2 2 2 1 1 X = A − 1 B = 1 4 [ 2 0 2 1 − 2 1 − 3 2 1 ] [ 0 − 40 180 ] = 1 4 [ 360 260 100 ] = [ 90 65 25 ] \begin{aligned} X &= A^{-1}B \\ &= \frac14\begin{bmatrix} 2 & 0 & 2 \\ 1 & -2 & 1 \\ -3 & 2 & 1 \end{bmatrix}\begin{bmatrix} 0 \\ -40 \\ 180 \end{bmatrix} \\ &= \frac14\begin{bmatrix} 360 \\ 260 \\ 100 \end{bmatrix} = \begin{bmatrix} 90 \\ 65 \\ 25 \end{bmatrix} \end{aligned} X = A − 1 B = 4 1 2 1 − 3 0 − 2 2 2 1 1 0 − 40 180 = 4 1 360 260 100 = 90 65 25 Why?
Multiply each row by the column B B B
0 + 0 + 360 = 360 0 + 80 + 180 = 260 0 − 80 + 180 = 100 \begin{aligned} 0 + 0 + 360 &= 360 \\ 0 + 80 + 180 &= 260 \\ 0 - 80 + 180 &= 100 \end{aligned} 0 + 0 + 360 0 + 80 + 180 0 − 80 + 180 = 360 = 260 = 100 Each row in full, products first:
Row 1: ( 2 ) ( 0 ) + ( 0 ) ( − 40 ) + ( 2 ) ( 180 ) = 0 + 0 + 360 = 360 Row 2: ( 1 ) ( 0 ) + ( − 2 ) ( − 40 ) + ( 1 ) ( 180 ) = 0 + 80 + 180 = 260 Row 3: ( − 3 ) ( 0 ) + ( 2 ) ( − 40 ) + ( 1 ) ( 180 ) = 0 − 80 + 180 = 100 \begin{aligned} &\text{Row 1: } (2)(0) + (0)(-40) + (2)(180) \\ &= 0 + 0 + 360 = 360 \\[4pt] &\text{Row 2: } (1)(0) + (-2)(-40) + (1)(180) \\ &= 0 + 80 + 180 = 260 \\[4pt] &\text{Row 3: } (-3)(0) + (2)(-40) + (1)(180) \\ &= 0 - 80 + 180 = 100 \end{aligned} Row 1: ( 2 ) ( 0 ) + ( 0 ) ( − 40 ) + ( 2 ) ( 180 ) = 0 + 0 + 360 = 360 Row 2: ( 1 ) ( 0 ) + ( − 2 ) ( − 40 ) + ( 1 ) ( 180 ) = 0 + 80 + 180 = 260 Row 3: ( − 3 ) ( 0 ) + ( 2 ) ( − 40 ) + ( 1 ) ( 180 ) = 0 − 80 + 180 = 100 Then divide each by 4 4 4
360 4 = 90 260 4 = 65 100 4 = 25 \begin{aligned} \frac{360}{4} &= 90 \\ \frac{260}{4} &= 65 \\ \frac{100}{4} &= 25 \end{aligned} 4 360 4 260 4 100 = 90 = 65 = 25 So x = 90 x = 90 x = 90 y = 65 y = 65 y = 65 z = 25 z = 25 z = 25
Why?
Check with the conditions: 65 + 25 = 90 65 + 25 = 90 65 + 25 = 90 90 90 90 45 45 45 45 + 20 = 65 45 + 20 = 65 45 + 20 = 65 90 + 65 + 25 = 180 90 + 65 + 25 = 180 90 + 65 + 25 = 180
x − y − z = 90 − 65 − 25 = 25 − 25 = 0 ✓ x − 2 y = 90 − 2 ( 65 ) = 90 − 130 = − 40 ✓ x + y + z = 90 + 65 + 25 = 155 + 25 = 180 ✓ \begin{aligned} x - y - z &= 90 - 65 - 25 \\ &= 25 - 25 = 0\ \checkmark \\[4pt] x - 2y &= 90 - 2(65) \\ &= 90 - 130 = -40\ \checkmark \\[4pt] x + y + z &= 90 + 65 + 25 \\ &= 155 + 25 = 180\ \checkmark \end{aligned} x − y − z x − 2 y x + y + z = 90 − 65 − 25 = 25 − 25 = 0 ✓ = 90 − 2 ( 65 ) = 90 − 130 = − 40 ✓ = 90 + 65 + 25 = 155 + 25 = 180 ✓ ∴ \therefore ∴
Where this comes from Chapter 4, Determinants : System of linear 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: Sports x x x y y y z z z x − y − z = 0 x - y - z = 0 x − y − z = 0 x − 2 y = − 40 x - 2y = -40 x − 2 y = − 40 x + y + z = 180 x + y + z = 180 x + y + z = 180 ∣ A ∣ = − 4 ≠ 0 |A| = -4 \ne 0 ∣ A ∣ = − 4 = 0 adj A \operatorname{adj}A adj A X = A − 1 B X = A^{-1}B X = A − 1 B
Quick lesson · Worked NCERT solutions · Interactive lesson
Also uses Chapter 3, Matrices .