Determinant | Business Statistics MGT 202

This unit covers the meaning and order of a determinant, the sign scheme for expansion, evaluation of determinants up to order three, the properties of determinants (with proofs of standard results) and the solution of simultaneous linear equations by Cramer's rule. Board exam questions are listed at the end by exam year, with marks in brackets.

Meaning of Determinant

The arrangement of numbers in a square form, enclosed between two vertical lines, is known as a determinant. A matrix is enclosed in brackets; a determinant is enclosed in vertical bars and has a single numerical value.

$$\begin{vmatrix} a & b \\ c & d \end{vmatrix} \ \text{(determinant)} \qquad \begin{bmatrix} a & b \\ c & d \end{bmatrix} \ \text{(matrix)}$$

In a determinant, the elements in horizontal lines are said to be in rows and those in vertical lines in columns. The number of rows and columns together forms the order of a determinant (rows $\times$ columns), for example $2 \times 2$ or $3 \times 3$.

Notation and Sign Scheme

A third-order determinant is written with element $a_{ij}$ in row $i$ and column $j$. When expanding along any row or column, each element takes the sign shown in the sign scheme.

$$|A| = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \qquad \begin{vmatrix} + & - \\ - & + \end{vmatrix}_{2\times2} \qquad \begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix}_{3\times3}$$

Evaluation of Determinants

Second Order

$$\begin{aligned} \begin{vmatrix} a & b \\ c & d \end{vmatrix} &= ad - bc \\ \begin{vmatrix} 2 & 3 \\ 4 & 5 \end{vmatrix} &= (2 \times 5) - (4 \times 3) = 10 - 12 = -2 \\ \begin{vmatrix} 2 & -3 \\ 4 & 5 \end{vmatrix} &= (2 \times 5) - (4 \times (-3)) = 10 + 12 = 22 \end{aligned}$$

Third Order

A third-order determinant is expanded along any row or column: each element is multiplied by the $2 \times 2$ determinant left after deleting its row and column, with the sign from the sign scheme. The value is the same whichever row or column is used.

Example: evaluate $\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \\ 8 & 9 & 0 \end{vmatrix}$.

$$\begin{aligned} \text{Along } R_1: &\ 2\begin{vmatrix} 6 & 7 \\ 9 & 0 \end{vmatrix} - 3\begin{vmatrix} 5 & 7 \\ 8 & 0 \end{vmatrix} + 4\begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} = 2(0-63) - 3(0-56) + 4(45-48) = -126 + 168 - 12 = 30 \\ \text{Along } C_1: &\ 2\begin{vmatrix} 6 & 7 \\ 9 & 0 \end{vmatrix} - 5\begin{vmatrix} 3 & 4 \\ 9 & 0 \end{vmatrix} + 8\begin{vmatrix} 3 & 4 \\ 6 & 7 \end{vmatrix} = 2(0-63) - 5(0-36) + 8(21-24) = 30 \\ \text{Along } R_3: &\ 8\begin{vmatrix} 3 & 4 \\ 6 & 7 \end{vmatrix} - 9\begin{vmatrix} 2 & 4 \\ 5 & 7 \end{vmatrix} + 0 = 8(21-24) - 9(14-20) + 0 = 30 \end{aligned}$$

Example (2065): evaluate $\begin{vmatrix} 42 & 1 & 6 \\ 28 & 7 & 4 \\ 14 & 3 & 2 \end{vmatrix}$.

$$\begin{aligned} \text{Along } C_1: &\ 42\begin{vmatrix} 7 & 4 \\ 3 & 2 \end{vmatrix} - 28\begin{vmatrix} 1 & 6 \\ 3 & 2 \end{vmatrix} + 14\begin{vmatrix} 1 & 6 \\ 7 & 4 \end{vmatrix} = 42(14-12) - 28(2-18) + 14(4-42) = 84 + 448 - 532 = 0 \\ \text{Alternatively: } &\ 7\begin{vmatrix} 6 & 1 & 6 \\ 4 & 7 & 4 \\ 2 & 3 & 2 \end{vmatrix} = 7 \times 0 = 0 \quad [\because C_1 = C_3 \text{ after taking 7 common from } C_1] \end{aligned}$$

Properties of Determinants

If each element of a row (or column) of a determinant is zero, then its value is zero.

$$\begin{vmatrix} 0 & 1 & 5 \\ 0 & 8 & 7 \\ 0 & -6 & -2 \end{vmatrix} = 0 \qquad \begin{vmatrix} 8 & 2 & 3 \\ 5 & 1 & 6 \\ 0 & 0 & 0 \end{vmatrix} = 0$$
If two rows (or columns) of a determinant are the same, the value of the determinant is zero.

$$\begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 7 \\ 2 & 3 & 4 \end{vmatrix} = 0 \quad [\because R_1 = R_3]$$
The value of a determinant remains unaltered if all rows and columns are interchanged.

$$\begin{vmatrix} 7 & 8 & 6 \\ -1 & 4 & -5 \\ 0 & 1 & 4 \end{vmatrix} = \begin{vmatrix} 7 & -1 & 0 \\ 8 & 4 & 1 \\ 6 & -5 & 4 \end{vmatrix}$$
If two rows (or columns) are interchanged, the sign of the determinant changes without changing its numerical value.

$$\begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 8 \\ 6 & 7 & 9 \end{vmatrix} = -\begin{vmatrix} 4 & 5 & 8 \\ 1 & 2 & 3 \\ 6 & 7 & 9 \end{vmatrix}$$
Multiplication of a determinant by a number $k$ is equal to the original determinant with each element of one row (or column) multiplied by $k$.

$$k\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = \begin{vmatrix} ka_1 & b_1 & c_1 \\ ka_2 & b_2 & c_2 \\ ka_3 & b_3 & c_3 \end{vmatrix}$$
If each element of a row (or column) is equal to some scalar times the corresponding element of another row (or column), the value of the determinant is zero.

$$\begin{vmatrix} 4 & 1 & 5 \\ 8 & 2 & 8 \\ 12 & 3 & 7 \end{vmatrix} = 0 \quad [\because C_1 = 4 \times C_2]$$
If every element of a row (or column) is expressed as a sum of two terms, the determinant can be expressed as a sum of two determinants.

$$\begin{vmatrix} x+y & z & y \\ a+b & c & d \\ p+q & r & s \end{vmatrix} = \begin{vmatrix} x & z & y \\ a & c & d \\ p & r & s \end{vmatrix} + \begin{vmatrix} y & z & y \\ b & c & d \\ q & r & s \end{vmatrix}$$

Operations such as $C_1 \to C_1 + C_2 + C_3$ or $R_1 \to R_1 - R_2$ do not change the value of a determinant; they are used to create zeros or a common factor before expanding.

Cramer's Rule

For a system of linear equations, let $D$ be the determinant of the coefficients, and $D_1, D_2, D_3$ the determinants obtained by replacing the coefficients of $x$, $y$, $z$ respectively by the constants. If $D \neq 0$:

$$x = \frac{D_1}{D}, \qquad y = \frac{D_2}{D}, \qquad z = \frac{D_3}{D}$$

Example: Two Variables (2075)

The prices $P$ and $Q$ satisfy:

	Coefficient of $P$	Coefficient of $Q$	Constant
Hari	7	−9	70
Gopal	9	−14	−80

$$\begin{aligned} D &= \begin{vmatrix} 7 & -9 \\ 9 & -14 \end{vmatrix} = -98 + 81 = -17 \\ D_1 &= \begin{vmatrix} 70 & -9 \\ -80 & -14 \end{vmatrix} = -980 - 720 = -1700 \\ D_2 &= \begin{vmatrix} 7 & 70 \\ 9 & -80 \end{vmatrix} = -560 - 630 = -1190 \\ P &= \frac{D_1}{D} = \frac{-1700}{-17} = 100, \qquad Q = \frac{D_2}{D} = \frac{-1190}{-17} = 70 \end{aligned}$$

The prices of $P$ and $Q$ are Rs. 100 and Rs. 70 respectively.

Example: Three Variables (2072)

	Coefficient of $x$	Coefficient of $y$	Coefficient of $z$	Constant
A	3	5	−4	6,000
B	2	−3	1	5,000
C	−1	4	6	13,000

$$\begin{aligned} D &= \begin{vmatrix} 3 & 5 & -4 \\ 2 & -3 & 1 \\ -1 & 4 & 6 \end{vmatrix} = 3(-18-4) - 5(12+1) - 4(8-3) = -66 - 65 - 20 = -151 \\ D_1 &= \begin{vmatrix} 6000 & 5 & -4 \\ 5000 & -3 & 1 \\ 13000 & 4 & 6 \end{vmatrix} = 6000(-18-4) - 5000(30+16) + 13000(5-12) = -453000 \\ D_2 &= \begin{vmatrix} 3 & 6000 & -4 \\ 2 & 5000 & 1 \\ -1 & 13000 & 6 \end{vmatrix} = -6000(12+1) + 5000(18-4) - 13000(3+8) = -151000 \\ D_3 &= \begin{vmatrix} 3 & 5 & 6000 \\ 2 & -3 & 5000 \\ -1 & 4 & 13000 \end{vmatrix} = 6000(8-3) - 5000(12+5) + 13000(-9-10) = -302000 \\ x &= \frac{-453000}{-151} = 3000, \quad y = \frac{-151000}{-151} = 1000, \quad z = \frac{-302000}{-151} = 2000 \end{aligned}$$

The prices per unit of commodities $x$, $y$ and $z$ are Rs. 3,000, Rs. 1,000 and Rs. 2,000 respectively.

Worked Examples Using Properties

Solve for $x$ (2013)

Solve $\begin{vmatrix} 7+x & 7-x & 7-x \\ 7-x & 7+x & 7-x \\ 7-x & 7-x & 7+x \end{vmatrix} = 0$.

$$\begin{aligned} C_1 \to C_1 + C_2 + C_3: &\quad (21-x)\begin{vmatrix} 1 & 7-x & 7-x \\ 1 & 7+x & 7-x \\ 1 & 7-x & 7+x \end{vmatrix} = 0 \\ R_1 \to R_1 - R_2: &\quad (21-x)\begin{vmatrix} 0 & -2x & 0 \\ 1 & 7+x & 7-x \\ 1 & 7-x & 7+x \end{vmatrix} = 0 \\ \text{Expanding } R_1: &\quad (21-x) \cdot 2x\,[(7+x) - (7-x)] = (21-x) \cdot 4x^2 = 0 \end{aligned}$$

Either $21 - x = 0$ or $4x^2 = 0$, so $x = 21$ or $x = 0$.

Prove: $\begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \end{vmatrix} = abc(a-b)(b-c)(c-a)$ (2074)

$$\begin{aligned} \text{L.H.S.} &= abc\begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix} && \text{(taking } a, b, c \text{ common from } C_1, C_2, C_3) \\ &= abc\begin{vmatrix} 0 & 0 & 1 \\ a-b & b-c & c \\ a^2-b^2 & b^2-c^2 & c^2 \end{vmatrix} && (C_1 \to C_1 - C_2,\ C_2 \to C_2 - C_3) \\ &= abc(a-b)(b-c)\begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & c \\ a+b & b+c & c^2 \end{vmatrix} \\ &= abc(a-b)(b-c)\,[(b+c) - (a+b)] && (\text{expanding } R_1) \\ &= abc(a-b)(b-c)(c-a) = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} a+b+c & -c & -b \\ -c & a+b+c & -a \\ -b & -a & a+b+c \end{vmatrix} = 2(a+b)(b+c)(c+a)$ (2072)

$$\begin{aligned} \text{L.H.S.} &= \begin{vmatrix} a+b & -(b+c) & -b \\ a+b & b+c & -a \\ -(a+b) & b+c & a+b+c \end{vmatrix} && (C_1 \to C_1 + C_2,\ C_2 \to C_2 + C_3) \\ &= (a+b)(b+c)\begin{vmatrix} 1 & -1 & -b \\ 1 & 1 & -a \\ -1 & 1 & a+b+c \end{vmatrix} \\ &= (a+b)(b+c)\begin{vmatrix} 1 & -1 & -b \\ 1 & 1 & -a \\ 0 & 0 & a+c \end{vmatrix} && (R_3 \to R_3 + R_1) \\ &= (a+b)(b+c)(c+a)(1+1) = 2(a+b)(b+c)(c+a) = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{vmatrix} = 1 + x^2 + y^2 + z^2$ (2074)

$$\begin{aligned} \text{L.H.S.} &= \frac{1}{xyz}\begin{vmatrix} x(x^2+1) & xy^2 & xz^2 \\ x^2y & y(y^2+1) & yz^2 \\ x^2z & y^2z & z(z^2+1) \end{vmatrix} && (\text{multiplying } C_1, C_2, C_3 \text{ by } x, y, z) \\ &= \begin{vmatrix} x^2+1 & y^2 & z^2 \\ x^2 & y^2+1 & z^2 \\ x^2 & y^2 & z^2+1 \end{vmatrix} && (\text{taking } x, y, z \text{ common from } R_1, R_2, R_3) \\ &= (1+x^2+y^2+z^2)\begin{vmatrix} 1 & y^2 & z^2 \\ 1 & y^2+1 & z^2 \\ 1 & y^2 & z^2+1 \end{vmatrix} && (C_1 \to C_1 + C_2 + C_3) \\ &= (1+x^2+y^2+z^2)\begin{vmatrix} 0 & -1 & 0 \\ 1 & y^2+1 & z^2 \\ 1 & y^2 & z^2+1 \end{vmatrix} && (R_1 \to R_1 - R_2) \\ &= (1+x^2+y^2+z^2)(z^2 + 1 - z^2) = 1 + x^2 + y^2 + z^2 = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{vmatrix} = (x+y+z)(x-z)^2$ (2071)

$$\begin{aligned} \text{L.H.S.} &= (x+y+z)\begin{vmatrix} 2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z \end{vmatrix} && (R_1 \to R_1 + R_2 + R_3) \\ &= (x+y+z)\begin{vmatrix} 0 & 1 & 1 \\ 0 & z & x \\ x-z & y & z \end{vmatrix} && (C_1 \to C_1 - C_2 - C_3) \\ &= (x+y+z)(x-z)\begin{vmatrix} 1 & 1 \\ z & x \end{vmatrix} = (x+y+z)(x-z)^2 = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ bc & ca & ab \end{vmatrix} = (a-b)(b-c)(c-a)(ab+bc+ca)$ (2068)

$$\begin{aligned} \text{L.H.S.} &= \frac{1}{abc}\begin{vmatrix} a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ abc & abc & abc \end{vmatrix} = \begin{vmatrix} a^2 & b^2 & c^2 \\ a^3 & b^3 & c^3 \\ 1 & 1 & 1 \end{vmatrix} && (\text{multiplying } C_1, C_2, C_3 \text{ by } a, b, c) \\ &= (a-b)(b-c)\begin{vmatrix} a+b & b+c & c^2 \\ a^2+ab+b^2 & b^2+bc+c^2 & c^3 \\ 0 & 0 & 1 \end{vmatrix} && (C_1 \to C_1 - C_2,\ C_2 \to C_2 - C_3) \\ &= (a-b)(b-c)\,[(a+b)(b^2+bc+c^2) - (b+c)(a^2+ab+b^2)] && (\text{expanding } R_3) \\ &= (a-b)(b-c)\,(ac^2 + bc^2 - a^2b - a^2c) \\ &= (a-b)(b-c)\,[b(c^2-a^2) + ac(c-a)] \\ &= (a-b)(b-c)(c-a)(ab+bc+ca) = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc\left(1 + \frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)$ (2063)

$$\begin{aligned} \text{L.H.S.} &= abc\begin{vmatrix} 1+\frac{1}{a} & \frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & 1+\frac{1}{b} & \frac{1}{c} \\ \frac{1}{a} & \frac{1}{b} & 1+\frac{1}{c} \end{vmatrix} && (\text{dividing } C_1, C_2, C_3 \text{ by } a, b, c) \\ &= abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\begin{vmatrix} 1 & \frac{1}{b} & \frac{1}{c} \\ 1 & 1+\frac{1}{b} & \frac{1}{c} \\ 1 & \frac{1}{b} & 1+\frac{1}{c} \end{vmatrix} && (C_1 \to C_1 + C_2 + C_3) \\ &= abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\begin{vmatrix} 0 & -1 & 0 \\ 1 & 1+\frac{1}{b} & \frac{1}{c} \\ 1 & \frac{1}{b} & 1+\frac{1}{c} \end{vmatrix} && (R_1 \to R_1 - R_2) \\ &= abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\left(1+\frac{1}{c}-\frac{1}{c}\right) = abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = \text{R.H.S.} \end{aligned}$$

Prove: $\begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix} = 2(x+y+z)^3$

$$\begin{aligned} \text{L.H.S.} &= 2(x+y+z)\begin{vmatrix} 1 & x & y \\ 1 & y+z+2x & y \\ 1 & x & z+x+2y \end{vmatrix} && (C_1 \to C_1 + C_2 + C_3) \\ &= 2(x+y+z)\begin{vmatrix} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{vmatrix} && (R_2 \to R_2 - R_1,\ R_3 \to R_3 - R_1) \\ &= 2(x+y+z)(x+y+z)^2 = 2(x+y+z)^3 = \text{R.H.S.} \end{aligned}$$

Important Questions

2078 Exam

Find the value of the determinant: [2 marks]

$$\begin{vmatrix} 1 & 5 & 6 \\ 2 & 3 & 7 \\ 3 & 5 & 6 \end{vmatrix}$$
Solve the following equations by using the determinant or matrix method: [10 marks]

$$\begin{aligned} 2x + 2y + z &= 5 \\ 4x + 2y + 3z &= 10 \\ x + 3y + 4z &= 7 \end{aligned}$$

2079 Exam

Find the value of the determinant of the following matrix: [2 marks]

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 1 \\ 9 & 3 & 2 \end{bmatrix}$$
The marks distribution of 100 students of a college is as follows: [10 marks]

Marks 10–20 20–30 30–40 40–50 50–60 60–70 70–80

No. of students 8 16 25 22 15 9 5

Determine the limits of marks of the middle 80% of students.
Solve the following system of linear equations by using the determinant or matrix method: [10 marks]

$$\begin{aligned} 2x - y + 2z &= -1 \\ x - 2y + 3z &= 4 \\ 4x + y + z &= 4 \end{aligned}$$

Marks	10–20	20–30	30–40	40–50	50–60	60–70	70–80
No. of students	8	16	25	22	15	9	5

2080 Exam

Evaluate: [2 marks]

$$|A| = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$$
Solve the following equations by using the matrix or determinant method: [10 marks]

$$\begin{aligned} 3x + 2y + 5z &= 10 \\ 2x - 3y + 7z &= 9 \\ x + y + z &= 5 \end{aligned}$$

2081 Exam

Find the value of the determinant: [2 marks]

$$\begin{vmatrix} 3 & 2 & 1 \\ 2 & 3 & 5 \\ 3 & 5 & 1 \end{vmatrix}$$
Solve the following equations by using the determinant or matrix method: [10 marks]

$$\begin{aligned} 5x + 3y + z &= 16 \\ 2x + y + 3z &= 19 \\ x + 2y + 4z &= 25 \end{aligned}$$

2082 Exam

Find the value of the following determinant: [2 marks]

$$\begin{vmatrix} 1 & a & b+c \\ 1 & b & c+a \\ 1 & c & a+b \end{vmatrix}$$
Solve the following equations by using the determinant or matrix method: [10 marks]

$$\begin{aligned} 2x - y + z &= -2 \\ x + y - 2z &= -9 \\ x + 2y + z &= 9 \end{aligned}$$

2083 Exam

Find the value of the determinant of the following matrix: [2 marks]

$$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 5 & 1 \\ 9 & 3 & 2 \end{bmatrix}$$

Business Statistics

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