CBSE Class 12 Mathematics Determinants Notes Set 04

Important Notes and Formulas

Definition

(1) Consider two equations, \( a_1x + b_1y = 0 \) .....(i) and \( a_2x + b_2y = 0 \) .....(ii)
Multiplying (i) by \( b_2 \) and (ii) by \( b_1 \) and subtracting, dividing by x, we get, \( a_1b_2 - a_2b_1 = 0 \)
The result \( a_1b_2 - a_2b_1 \) is represented by \( \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} \)
which is known as a determinant of order two and \( a_1b_2 - a_2b_1 \) is the expansion of this determinant. The horizontal lines are called rows and the vertical lines are called columns.
Now let us consider three homogeneous linear equations
\( a_1x + b_1y + c_1z = 0,\ a_2x + b_2y + c_2z = 0 \) and \( a_3x + b_3y + c_3z = 0 \)
Eliminating x, y, z from the above three equations we obtain
\( a_1(b_2c_3 - b_3c_2) - b_1(a_2c_3 - a_3c_2) + c_1(a_2b_3 - a_3b_2) = 0 \) .....(iii)
The L.H.S. of (iii) is represented by \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \)
It contains three rows and three columns, so it is called a determinant of third order.
Note: The number of elements in a second order determinant is \( 2^2 = 4 \) and the number of elements in a third order determinant is \( 3^2 = 9 \).

(2) Rows and columns of a determinant: In a determinant the horizontal lines counting from the top 1st, 2nd, 3rd, ..... are respectively known as rows and denoted by \( R_1, R_2, R_3, ...... \) and the vertical lines counting from left to right, 1st, 2nd, 3rd, ..... are respectively known as columns and denoted by \( C_1, C_2, C_3, ..... \)

(3) Shape and constituents of a determinant: The shape of every determinant is square. If a determinant is of order n, then it contains n rows and n columns.
i.e., Number of constituents in a determinant = \( n^2 \)

(4) Sign system for expansion of determinant: The sign systems for order 2, order 3, order 4, ..... are given by
\( \begin{vmatrix} + & - \\ - & + \end{vmatrix},\ \begin{vmatrix} + & - & + \\ - & + & - \\ + & - & + \end{vmatrix},\ \begin{vmatrix} + & - & + & - \\ - & + & - & + \\ + & - & + & - \\ - & + & - & + \end{vmatrix},\ ...... \)

Expansion of Determinants

Unlike a matrix, a determinant is not just a table of numerical data but (quite differently) a short hand way of writing an algebraic expression, whose value can be computed when the values of the terms or elements are known.

(1) The 4 numbers \( a_1, b_1, a_2, b_2 \) arranged as \( \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} \) form a determinant of second order. These numbers are called the elements of the determinant. The value of the determinant is defined as
\( \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} = a_1b_2 - a_2b_1 \)
The expanded form of the determinant has 2! terms.

(2) The 9 numbers \( a_r, b_r, c_r\ (r = 1, 2, 3) \) arranged as \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) form a determinant of third order. Take any row (or column); the value of the determinant is the sum of the products of the elements of the row (or column) and the corresponding determinant obtained by omitting the row and the column of the element, with a proper sign given by the rule \( (-1)^{i+j} \), where i and j are the numbers of the row and the column respectively of the element of the row (or the column) chosen. Thus
\( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1\begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1\begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1\begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \)
The diagonal through the left-hand top corner which contains the elements \( a_1, b_2, c_3 \) is called the leading diagonal or principal diagonal and the terms are called the leading terms. The expanded form of the determinant has 3! terms.

Short cut method or Sarrus diagram method: To find the value of a third order determinant, the following method is also useful.

Taking the products of the R.H.S. diagonal elements as positive and the L.H.S. diagonal elements as negative and adding them, we get the value of the determinant as
\( = a_1b_2c_3 + b_1c_2a_3 + c_1a_2b_3 - c_1b_2a_3 - a_1c_2b_3 - b_1a_2c_3 \)
Note: This method does not work for determinants of order greater than three.

Minors and Cofactors

(1) Minor of an element: If we take an element of the determinant and delete (remove) the row and column containing that element, the determinant left is called the minor of that element. It is denoted by \( M_{ij} \).
Consider the determinant \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \), then the determinant of minors is \( M = \begin{vmatrix} M_{11} & M_{12} & M_{13} \\ M_{21} & M_{22} & M_{23} \\ M_{31} & M_{32} & M_{33} \end{vmatrix} \),
where \( M_{11} = \) minor of \( a_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} \), \( M_{12} = \) minor of \( a_{12} = \begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} \),
\( M_{13} = \) minor of \( a_{13} = \begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \)
Similarly, we can find the minors of the other elements. Using this concept the value of the determinant can be
\( \Delta = a_{11}M_{11} - a_{12}M_{12} + a_{13}M_{13} \)
or, \( \Delta = -a_{21}M_{21} + a_{22}M_{22} - a_{23}M_{23} \) or, \( \Delta = a_{31}M_{31} - a_{32}M_{32} + a_{33}M_{33} \).

(2) Cofactor of an element: The cofactor of an element \( a_{ij} \) (i.e. the element in the \( i^{th} \) row and \( j^{th} \) column) is defined as \( (-1)^{i+j} \) times the minor of that element. It is denoted by \( C_{ij} \) or \( A_{ij} \) or \( F_{ij} \).
\( C_{ij} = (-1)^{i+j}M_{ij} \)
If \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \), then the determinant of cofactors is \( C = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix} \), where
\( C_{11} = (-1)^{1+1}M_{11} = +M_{11} \), \( C_{12} = (-1)^{1+2}M_{12} = -M_{12} \) and \( C_{13} = (-1)^{1+3}M_{13} = +M_{13} \)
Similarly, we can find the cofactors of the other elements.

Note:
(i) The sum of the products of the elements of any row with their corresponding cofactors is equal to the value of the determinant, i.e. \( \Delta = a_{11}C_{11} + a_{12}C_{12} + a_{13}C_{13} = a_{11}C_{11} + a_{21}C_{21} + a_{31}C_{31} \), where the capital letters \( C_{11}, C_{12}, C_{13} \) etc. denote the cofactors of \( a_{11}, a_{12}, a_{13} \) etc.
(ii) In general, it should be noted that \( a_{i1}C_{j1} + a_{i2}C_{j2} + a_{i3}C_{j3} = 0 \), if i ≠ j or \( a_{1i}C_{1j} + a_{2i}C_{2j} + a_{3i}C_{3j} = 0 \), if i ≠ j
(iii) If \( \Delta' \) is the determinant formed by replacing the elements of a determinant Δ by their corresponding cofactors, then if Δ = 0, then \( \Delta' = 0 \), and \( \Delta' = \Delta^{n-1} \), where n is the order of the determinant.

Application of Determinants in Co-ordinate Geometry

(1) The area of a triangle whose vertices are \( (x_r, y_r);\ r = 1, 2, 3 \) is
\( \Delta = \frac{1}{2}[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)] = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \)
(2) The equation of a straight line passing through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is \( \begin{vmatrix} x & y & 1 \\ x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \end{vmatrix} = 0 \)

Solved Examples

Example 1. If a ≠ b ≠ c, the value of x which satisfies the equation \( \begin{vmatrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{vmatrix} = 0 \) is [EAMCET 1988; Karnataka CET 1991; MNR 1980; MP PET 1988, 99, 2001; DCE 2001]
(a) \( x = 0 \)
(b) \( x = a \)
(c) \( x = b \)
(d) \( x = c \)
Answer:
Ans. (a) \( x = 0 \)
Solution: Expanding the determinant along \( R_1 \), we get
\( \Delta = -(x - a)[-(x + b)(x - c)] + (x - b)[(x + a)(x + c)] = 0 \) ⇒ \( 2x^3 - 2(ab + bc - ca)x = 0 \)
⇒ Either x = 0 or \( x^2 = ab + bc - ca \). Since x = 0 satisfies the given equation, (a) is correct.
Trick: On putting x = 0, we observe that the determinant becomes \( \Delta_{x=0} = \begin{vmatrix} 0 & -a & -b \\ a & 0 & -c \\ b & c & 0 \end{vmatrix} = 0 \)
∴ x = 0 is a root of the given equation.

Example 2. If \( p\lambda^4 + q\lambda^3 + r\lambda^2 + s\lambda + t = \begin{vmatrix} \lambda^2 + 3\lambda & \lambda - 1 & \lambda + 3 \\ \lambda + 1 & 2 - \lambda & \lambda - 4 \\ \lambda - 3 & \lambda + 4 & 3\lambda \end{vmatrix} \), then the value of t is [IIT 1981]
(a) 16
(b) 18
(c) 17
(d) 19
Answer:
Ans. (b) 18
Solution: Since it is an identity in λ, it is satisfied by every value of λ. Now put λ = 0 in the given equation, we have
\( t = \begin{vmatrix} 0 & -1 & 3 \\ 1 & 2 & -4 \\ -3 & 4 & 0 \end{vmatrix} = -12 + 30 = 18 \)

Example 3. The cofactor of the element 4 in the determinant \( \begin{vmatrix} 1 & 3 & 5 & 1 \\ 2 & 3 & 4 & 2 \\ 8 & 0 & 1 & 1 \\ 0 & 2 & 1 & 1 \end{vmatrix} \) is [MP PET 1987]
(a) 4
(b) 10
(c) −10
(d) −4
Answer:
Ans. (b) 10
Solution: The cofactor of the element 4, in the 2nd row and 3rd column, is
\( (-1)^{2+3}\begin{vmatrix} 1 & 3 & 1 \\ 8 & 0 & 1 \\ 0 & 2 & 1 \end{vmatrix} = -\{1(-2) - 3(8 - 0) + 1 \cdot 16\} = 10 \)

Example 4. If \( \Delta = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) and \( A_1, B_1, C_1 \) denote the cofactors of \( a_1, b_1, c_1 \) respectively, then the value of the determinant \( \begin{vmatrix} A_1 & B_1 & C_1 \\ A_2 & B_2 & C_2 \\ A_3 & B_3 & C_3 \end{vmatrix} \) is
(a) Δ
(b) \( \Delta^2 \)
(c) \( \Delta^3 \)
(d) 0
Answer:
Ans. (b) \( \Delta^2 \)
Solution: The given determinant \( \Delta' \) is formed by replacing every element of Δ by its cofactor.
According to the property of cofactors, \( \Delta' = \Delta^{n-1} = \Delta^{3-1} = \Delta^2 \) (∵ here n = 3)

Example 5. If the value of a third order determinant is 11, then the value of the square of the determinant formed by the cofactors will be
(a) 11
(b) 121
(c) 1331
(d) 14641
Answer:
Ans. (d) 14641
Solution: \( \Delta' = \Delta^{n-1} = \Delta^{3-1} = \Delta^2 = (11)^2 = 121 \). But we have to find the value of the square of the determinant, so the required value is \( (121)^2 = 14641 \).

Example 6. If the system of equations x + 2y − 3z = 1, (k + 3)z = 3, (2k + 1)x + z = 0 is inconsistent, then the value of k is [Roorkee 2000]
(a) −3
(b) \( \frac{1}{2} \)
(c) 0
(d) 2
Answer:
Ans. (a) −3
Solution: For the equations to be inconsistent, the determinant of the coefficient matrix must be zero, i.e. |A| = 0.
∴ \( |A| = \begin{vmatrix} 1 & 2 & -3 \\ 0 & 0 & k + 3 \\ 2k + 1 & 0 & 1 \end{vmatrix} = 1(0 - 0) - 2[0 - (k + 3)(2k + 1)] - 3(0 - 0) = 2(k + 3)(2k + 1) \)
|A| = 0 ⇒ k = −3 or \( k = -\frac{1}{2} \)
For k = −3, the second equation becomes 0 · z = 3, which is not true for any z. Hence the system is inconsistent for k = −3.

Example 7. The equations x + y + z = 6, x + 2y + 3z = 10, x + 2y + mz = n give an infinite number of values of the triplet (x, y, z) if
(a) m = 3, n ∈ R
(b) m = 3, n ≠ 10
(c) m = 3, n = 10
(d) None of these
Answer:
Ans. (c) m = 3, n = 10
Solution: Each of the first three options contains m = 3. When m = 3, the last two equations become x + 2y + 3z = 10 and x + 2y + 3z = n.
Obviously, when n = 10 these equations become the same. So we are left with only two independent equations to find the values of the three unknowns.
Consequently, there will be infinite solutions.

Example 8. The value of λ for which the system of equations 2x − y − z = 12, x − 2y + z = −4, x + y + λz = 4 has no solution is [IIT Screening 2004]
(a) 3
(b) −3
(c) 2
(d) −2
Answer:
Ans. (d) −2
Solution: Here \( A = \begin{bmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{bmatrix} \) and \( B = \begin{bmatrix} 12 \\ -4 \\ 4 \end{bmatrix} \).
\( |A| = \begin{vmatrix} 2 & -1 & -1 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda \end{vmatrix} = -3\lambda - 6 \). For no solution the necessary condition is −3λ − 6 = 0 ⇒ λ = −2.
For λ = −2, \( \text{adj } A = \begin{bmatrix} 3 & -3 & -3 \\ 3 & -3 & -3 \\ 3 & -3 & -3 \end{bmatrix} \) and \( (\text{adj } A)B = \begin{bmatrix} 36 + 12 - 12 \\ 36 + 12 - 12 \\ 36 + 12 - 12 \end{bmatrix} = \begin{bmatrix} 36 \\ 36 \\ 36 \end{bmatrix} \neq O \).
Hence for λ = −2 there is no solution for the given system of equations.