CBSE Class 12 Mathematics Determinants Notes Set 02

This page gives the complete notes for Chapter 4 Determinants of Class 12 Mathematics (CBSE), covering expansion, area of a triangle, minors, cofactors, adjoint and inverse of a matrix, and solving linear equations by the matrix method. It also includes solved examples with step marks, Do Yourself practice questions and their answers.

Important Notes and Formulas

Definition

When an algebraic or numerical expression is expressed in a square form containing some rows and columns, this square form is named as a determinant of that expression. For example, when the expression \( a_1b_2 - a_2b_1 \) is expressed in the form \( \begin{vmatrix} a_1 & b_1 \\ a_2 & b_2 \end{vmatrix} \), then it is called a determinant of order 2. Clearly a determinant of order 2 contains 2 rows and 2 columns. Similarly \( \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} \) is a determinant of order 3.
Obviously in every determinant, the number of rows and columns are equal and this number is called the order of that determinant.

Note:
(i) For matrix A, |A| is read as determinant of A and not modulus of A.
(ii) Only square matrices have determinant.

Representation of a Determinant

Generally we use Δ or |A| symbols to express a determinant, and a determinant of order 3 is represented by
\[ \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \]
It should be noted that the (i, j)^th element (the element of the i^th row and j^th column) of the determinant has been expressed by a_ij, i = 1, 2, 3; j = 1, 2, 3. The elements for which i = j are called diagonal elements and the diagonal containing them is called principal diagonal or simply diagonal of the determinant. For the above determinant a₁₁, a₂₂, a₃₃ are diagonal elements.

Expansion of Determinant or Value of a Determinant

(i) Expansion along first row (R₁):
The expression which has been expressed in a determinant form is called the value of that determinant.
Let \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \) be a third order determinant.
To find its value we expand it by any row or column as the sum of three determinants of order 2. If we expand it by first row then
\[ \Delta = (-1)^{1+1} a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + (-1)^{1+2} a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + (-1)^{1+3} a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \]
\[ = a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} - a_{12}\begin{vmatrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{vmatrix} + a_{13}\begin{vmatrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{vmatrix} \]
= a₁₁a₂₂a₃₃ − a₁₁a₃₂a₂₃ − a₁₂a₂₁a₃₃ + a₁₂a₃₁a₂₃ + a₁₃a₂₁a₃₂ − a₁₃a₃₁a₂₂   ...(i)

(ii) Expansion along first column (C₁):
\[ |A| = (-1)^{1+1} a_{11}\begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} + (-1)^{2+1} a_{21}\begin{vmatrix} a_{12} & a_{13} \\ a_{32} & a_{33} \end{vmatrix} + (-1)^{3+1} a_{31}\begin{vmatrix} a_{12} & a_{13} \\ a_{22} & a_{23} \end{vmatrix} \]
= a₁₁a₂₂a₃₃ − a₁₁a₂₃a₃₂ − a₂₁a₁₂a₃₃ + a₂₁a₃₂a₁₃ + a₃₁a₁₂a₂₃ − a₃₁a₁₃a₂₂   ...(ii)
Values of |A| in (i) and (ii) are same and we can find the value of |A| by expanding along any row (R₁, R₂, R₃) or column (C₁, C₂, C₃).

Note: For easier calculation we shall expand the determinant along that row or column which contains the maximum number of zeros.

Area of Triangle

The area of a triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by \( \frac{1}{2}\left[x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)\right] \). Now this expression can be written in the form of a determinant as
\[ \Delta = \frac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \quad \text{or} \quad \frac{1}{2}\begin{vmatrix} x_1 & x_2 & x_3 \\ y_1 & y_2 & y_3 \\ 1 & 1 & 1 \end{vmatrix} \]

Note:
(i) Since area is a positive quantity, we always take the absolute value of the determinant.
(ii) The area of the triangle formed by three collinear points is zero.
(iii) If area is given, then for calculation take both positive and negative values.

Minors

The determinant that is left by cancelling the row and column intersecting at a particular element is called the minor of that element. Minor of an element a_ij is denoted by M_ij.

Note: Minor of an element of a determinant of order n (n ≥ 2) is a determinant of order (n − 1).

Example:
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 minor of a₁₁ is \( M_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} \), similarly \( M_{32} = \begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix} \).

Cofactors

The cofactor of an element a_ij is denoted by C_ij and is equal to (−1)^{i + j} · M_ij, where M_ij is the minor of element a_ij.

Example: \( C_{11} = (-1)^{1+1} M_{11} = M_{11} = \begin{vmatrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{vmatrix} \), similarly \( C_{32} = (-1)^{3+2} M_{32} = -M_{32} = -\begin{vmatrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{vmatrix} \).

Note:
(i) The sum of products of the elements of any row with their corresponding cofactors is equal to the value of the determinant, i.e. Δ = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃
(ii) The sum of the products of elements of any row with the corresponding cofactors of another row is equal to zero, i.e. a₁₁C₂₁ + a₁₂C₂₂ + a₁₃C₂₃ = 0
(iii) If the order of a determinant (Δ) is n, then the value of the determinant formed by replacing every element by its cofactor is Δⁿ⁻¹.

Adjoint of a Matrix

If every element of a square matrix A is replaced by its cofactor in |A|, then the transpose of the matrix so obtained is called the adjoint of A and it is denoted by adj A.
Thus if A = [a_ij] is a square matrix and F_ij is the cofactor of a_ij in |A|, then adj A = [F_ij]^T ⇒ (adj A)_ij = F_ji.
Hence if
\[ A = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{bmatrix}, \text{ then adj } A = \begin{bmatrix} F_{11} & F_{12} & \cdots & F_{1n} \\ F_{21} & F_{22} & \cdots & F_{2n} \\ \vdots & \vdots & & \vdots \\ F_{n1} & F_{n2} & \cdots & F_{nn} \end{bmatrix}^T = \begin{bmatrix} F_{11} & F_{21} & \cdots & F_{n1} \\ F_{12} & F_{22} & \cdots & F_{n2} \\ \vdots & \vdots & & \vdots \\ F_{1n} & F_{2n} & \cdots & F_{nn} \end{bmatrix} \]

Properties of Adjoint Matrix:
If A, B are square matrices of order n and I_n is the corresponding unit matrix, then
(i) A (adj A) = |A| I_n = (adj A) A. (Thus A (adj A) is always a scalar matrix.)
(ii) |adj A| = |A|ⁿ⁻¹
(iii) A is singular ⇒ |A| = 0
(iv) A is non-singular ⇒ |A| ≠ 0

Trick: For a square matrix of order 2, given by \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \), adj A can also be obtained by interchanging a₁₁ and a₂₂ and by changing the sign of a₁₂ and a₂₁.
Example:

Inverse Matrix (For Non-singular Matrix)

If A is a square matrix of order m and if there exists another square matrix B of the same order m such that AB = BA = I, then B is called the inverse matrix of A and it is denoted by A⁻¹, and A is called an invertible matrix.
Also A (adj A) = |A| I = (adj A) A
⇒ \( A\,\frac{\text{adj}(A)}{|A|} = I = \frac{\text{adj}(A)}{|A|}\,A \)
⇒ \( A^{-1} = \frac{\text{adj}(A)}{|A|} \quad [\,|A| \neq 0\,] \)
Thus A⁻¹ exists ⇔ |A| ≠ 0.

Note:
(1) Matrix A is called invertible if A⁻¹ exists.
(2) A rectangular matrix does not have an inverse matrix because BA and AB should be defined and be equal. (So A and B should be square matrices of the same order.)
(3) If B is the inverse of A, then A is also the inverse of B.

Properties of Inverse Matrix:
(i) (A⁻¹)⁻¹ = A    (ii) (AB)⁻¹ = B⁻¹A⁻¹    (iii) \( |A^{-1}| = \frac{1}{|A|} = |A|^{-1} \)

Solved Examples

Illustration 1. Find the value of \( A = \begin{vmatrix} 6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4 \end{vmatrix} \)   [1 Mark]
Answer:
\[ A = 6(-1)^{1+1}\begin{vmatrix} -3 & 1 \\ 1 & -4 \end{vmatrix} + (-7)(-1)^{1+2}\begin{vmatrix} 1 & 1 \\ 2 & -4 \end{vmatrix} + 8(-1)^{1+3}\begin{vmatrix} 1 & -3 \\ 2 & 1 \end{vmatrix} \]
= 6(12 − 1) + 7(−4 − 2) + 8(1 + 6) = 66 − 42 + 56 = 80   [1]

Illustration 2. Find the area of the triangle whose vertices are \( A(at_1^2, 2at_1) \), \( B(at_2^2, 2at_2) \) and \( C(at_3^2, 2at_3) \).   [4 Marks]
Answer:
\[ \text{Area of } \triangle ABC = \frac{1}{2}\left|\,\Delta\,\right|, \text{ where } \Delta = \begin{vmatrix} at_1^2 & 2at_1 & 1 \\ at_2^2 & 2at_2 & 1 \\ at_3^2 & 2at_3 & 1 \end{vmatrix} \]
Expanding Δ along the third column (C₃):
\[ \Delta = 1\begin{vmatrix} at_2^2 & 2at_2 \\ at_3^2 & 2at_3 \end{vmatrix} - 1\begin{vmatrix} at_1^2 & 2at_1 \\ at_3^2 & 2at_3 \end{vmatrix} + 1\begin{vmatrix} at_1^2 & 2at_1 \\ at_2^2 & 2at_2 \end{vmatrix} \]
\[ = 2a^2\left[(t_2^2t_3 - t_2t_3^2) - (t_1^2t_3 - t_1t_3^2) + (t_1^2t_2 - t_1t_2^2)\right] \][2]
\[ = 2a^2\left[t_1^2(t_2 - t_3) - t_1(t_2^2 - t_3^2) + t_2t_3(t_2 - t_3)\right] \]
\[ = 2a^2(t_2 - t_3)\left[t_1^2 - t_1(t_2 + t_3) + t_2t_3\right] = 2a^2(t_2 - t_3)(t_1 - t_2)(t_1 - t_3) \]
\[ \therefore\; \text{Area} = \frac{1}{2}\left|\,2a^2(t_1 - t_2)(t_2 - t_3)(t_1 - t_3)\,\right| \]
= a² |(t₁ − t₂)(t₂ − t₃)(t₃ − t₁)| square units.   [2]

Illustration 3. Find the minors and the cofactors of each entry of the third row of the determinant A and hence evaluate det A.
\[ A = \begin{vmatrix} 6 & -7 & 8 \\ 1 & -3 & 1 \\ 2 & 1 & -4 \end{vmatrix} \][1 Mark]
Answer:
Now \( M_{31} = \begin{vmatrix} -7 & 8 \\ -3 & 1 \end{vmatrix} \) = −7 − (−24) = 17, \( M_{32} = \begin{vmatrix} 6 & 8 \\ 1 & 1 \end{vmatrix} \) = 6 − 8 = −2 and
\( M_{33} = \begin{vmatrix} 6 & -7 \\ 1 & -3 \end{vmatrix} \) = 6 × (−3) − (−7) × 1 = −11.
Also, C₃₁ = (−1)³⁺¹ M₃₁ = M₃₁ = 17, C₃₂ = (−1)³⁺² M₃₂ = −M₃₂ = 2
and C₃₃ = (−1)³⁺³ M₃₃ = M₃₃ = −11.
Further, det A = a₃₁C₃₁ + a₃₂C₃₂ + a₃₃C₃₃ (Expansion with the help of third row)
= 2 × 17 + 1 × 2 + (−4) × (−11) = 80

Illustration 4. Find the adjoint of the matrix \( A = \begin{bmatrix} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{bmatrix} \) and verify that A (adj A) = (adj A) A = |A| I₃.   [4 Marks]
Answer:
Here, \( |A| = \begin{vmatrix} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{vmatrix} \)
= 4(1 − 11) − (−6)(1 + 4) + 1(−11 − 4) = −40 + 30 − 15 = −25   [1]
Cofactors of the elements of first row are:
\( C_{11} = (-1)^{1+1}\begin{vmatrix} -1 & 1 \\ 11 & -1 \end{vmatrix} = -10,\; C_{12} = (-1)^{1+2}\begin{vmatrix} -1 & 1 \\ -4 & -1 \end{vmatrix} = -5,\; C_{13} = (-1)^{1+3}\begin{vmatrix} -1 & -1 \\ -4 & 11 \end{vmatrix} = -15 \)
Cofactors of the elements of second row are:
\( C_{21} = (-1)^{2+1}\begin{vmatrix} -6 & 1 \\ 11 & -1 \end{vmatrix} = 5,\; C_{22} = (-1)^{2+2}\begin{vmatrix} 4 & 1 \\ -4 & -1 \end{vmatrix} = 0,\; C_{23} = (-1)^{2+3}\begin{vmatrix} 4 & -6 \\ -4 & 11 \end{vmatrix} = -20 \)
Cofactors of the elements of third row are:
\( C_{31} = (-1)^{3+1}\begin{vmatrix} -6 & 1 \\ -1 & 1 \end{vmatrix} = -5,\; C_{32} = (-1)^{3+2}\begin{vmatrix} 4 & 1 \\ -1 & 1 \end{vmatrix} = -5,\; C_{33} = (-1)^{3+3}\begin{vmatrix} 4 & -6 \\ -1 & -1 \end{vmatrix} = -10 \)
\[ \therefore\; \text{adj } A = \begin{bmatrix} -10 & -5 & -15 \\ 5 & 0 & -20 \\ -5 & -5 & -10 \end{bmatrix}^T = \begin{bmatrix} -10 & 5 & -5 \\ -5 & 0 & -5 \\ -15 & -20 & -10 \end{bmatrix} \][1½]
Verification: \( A(\text{adj } A) = \begin{bmatrix} 4 & -6 & 1 \\ -1 & -1 & 1 \\ -4 & 11 & -1 \end{bmatrix}\begin{bmatrix} -10 & 5 & -5 \\ -5 & 0 & -5 \\ -15 & -20 & -10 \end{bmatrix} \)
\[ = \begin{bmatrix} -40+30-15 & 20-0-20 & -20+30-10 \\ 10+5-15 & -5-0-20 & 5+5-10 \\ 40-55+15 & -20+0+20 & 20-55+10 \end{bmatrix} \]
\[ = \begin{bmatrix} -25 & 0 & 0 \\ 0 & -25 & 0 \\ 0 & 0 & -25 \end{bmatrix} = -25\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = (-25)\, I_3 = |A|\, I_3 \][1]
Similarly, we can show that (adj A) A = |A| I₃.
Hence A (adj A) = |A| I₃ = (adj A) A.   [½]

Illustration 5. If \( A = \begin{bmatrix} 2 & 1 \\ 5 & 3 \end{bmatrix} \) and \( B^{-1} = \begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix} \), compute (AB)⁻¹.   [4 Marks]
Answer:
By reversal law, we have
\( (AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}A^{-1} \)   [1]
In this case, we have to compute A⁻¹ only.
Here, \( \det A = \begin{vmatrix} 2 & 1 \\ 5 & 3 \end{vmatrix} \) = 6 − 5 = 1   [1]
Now C₁₁ = 3, C₁₂ = −5, C₂₁ = −1, C₂₂ = 2
⇒ \( A^{-1} = \frac{1}{|A|}(\text{adj } A) = \frac{1}{1}\begin{bmatrix} 3 & -5 \\ -1 & 2 \end{bmatrix}^T = \begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} \)   [1]
Hence, \( (AB)^{-1} = \begin{bmatrix} 4 & 5 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 3 & -1 \\ -5 & 2 \end{bmatrix} = \begin{bmatrix} 12-25 & -4+10 \\ 9-20 & -3+8 \end{bmatrix} = \begin{bmatrix} -13 & 6 \\ -11 & 5 \end{bmatrix} \)   [1]

Illustration 6. Solve using matrices, for x, y and z:
2x − y − z = 7
3x + y − z = 7
x + y − z = 3.   [4 Marks]
Answer:
The given system of equations can be written as AX = B   ...(i)
where \( A = \begin{bmatrix} 2 & -1 & -1 \\ 3 & 1 & -1 \\ 1 & 1 & -1 \end{bmatrix},\; X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}\text{ and } B = \begin{bmatrix} 7 \\ 7 \\ 3 \end{bmatrix} \).
Here, \( \det A = \begin{vmatrix} 2 & -1 & -1 \\ 3 & 1 & -1 \\ 1 & 1 & -1 \end{vmatrix} \) = 2(−1 + 1) − (−1)(−3 + 1) + (−1)(3 − 1) = −2 − 2 = −4 ≠ 0   [1]
Therefore, the given system is consistent and has a unique solution given by X = A⁻¹B (from (i)).
To find A⁻¹, we have to find the cofactors of each element.
A₁₁ = 0, A₁₂ = 2, A₁₃ = 2, A₂₁ = −2, A₂₂ = −1, A₂₃ = −3, A₃₁ = 2, A₃₂ = −1 and A₃₃ = 5
\[ \therefore\; A^{-1} = \frac{1}{\det A}(\text{adj } A) = \frac{1}{-4}\begin{bmatrix} 0 & 2 & 2 \\ -2 & -1 & -3 \\ 2 & -1 & 5 \end{bmatrix}^T = -\frac{1}{4}\begin{bmatrix} 0 & -2 & 2 \\ 2 & -1 & -1 \\ 2 & -3 & 5 \end{bmatrix} \][2]
Then from (i), \( X = A^{-1}B = -\frac{1}{4}\begin{bmatrix} 0 & -2 & 2 \\ 2 & -1 & -1 \\ 2 & -3 & 5 \end{bmatrix}\begin{bmatrix} 7 \\ 7 \\ 3 \end{bmatrix} \)
\[ \Rightarrow \begin{bmatrix} x \\ y \\ z \end{bmatrix} = -\frac{1}{4}\begin{bmatrix} 0-14+6 \\ 14-7-3 \\ 14-21+15 \end{bmatrix} = -\frac{1}{4}\begin{bmatrix} -8 \\ 4 \\ 8 \end{bmatrix} = \begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix} \]
⇒ x = 2, y = −1, z = −2   [1]

Do Yourself

Do Yourself 1
(i) Prove that the points (a, b), (a′, b′), (a − a′, b − b′) are collinear if ab′ = a′b.
(ii) Find the value of x if the area of triangle is 35 square units with vertices (x, 4), (2, −6) and (5, 4).

Do Yourself 2
Write the minors and cofactors of each element of the first column of the following determinants:
(i) \( A = \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} \)     (ii) \( A = \begin{vmatrix} a & h & g \\ h & b & f \\ g & f & c \end{vmatrix} \)

Do Yourself 3
If \( A = \begin{bmatrix} 2 & 3 \\ 1 & 4 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \), verify that adj (AB) = (adj B)(adj A).

Do Yourself 4
(i) Given \( A = \begin{bmatrix} 2 & -3 \\ -4 & 7 \end{bmatrix} \), compute A⁻¹ and show that 2A⁻¹ = 9I − A.
(ii) Using matrix method, solve the following system of equations:
3x + 2y − 2z = 3
x + 2y + 3z = 6
2x − y + z = 2.

Answers for Do Yourself

Do Yourself 1: (ii) x = −2, 12

Do Yourself 2:
(i) M₁₁ = a(b² − c²), M₂₁ = b(a² − c²) and M₃₁ = c(a² − b²)
C₁₁ = a(b² − c²), C₂₁ = −b(a² − c²) and C₃₁ = c(a² − b²)
(ii) M₁₁ = bc − f², M₂₁ = hc − fg and M₃₁ = hf − bg
C₁₁ = bc − f², C₂₁ = fg − ch and C₃₁ = hf − bg

Do Yourself 4: (i) \( A^{-1} = \frac{1}{2}\begin{bmatrix} 7 & 3 \\ 4 & 2 \end{bmatrix} \)    (ii) x = 1, y = 1, z = 1