CBSE Class 12 Mathematics Determinants VBQs Set 01

BASIC CONCEPTS

1. Determinant: Every square matrix can be associated to an expression or a number which is known as its determinant.
Determinant of square matrix \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \) is given by
\( |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} = a_{11}a_{22} - a_{12}a_{21} \)
and determinant of a matrix \( A = \begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{bmatrix} \) is given by
\( |A| = \begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \end{vmatrix} = a_1 \begin{vmatrix} b_2 & c_2 \\ b_3 & c_3 \end{vmatrix} - b_1 \begin{vmatrix} a_2 & c_2 \\ a_3 & c_3 \end{vmatrix} + c_1 \begin{vmatrix} a_2 & b_2 \\ a_3 & b_3 \end{vmatrix} \)
This is known as the expansion of \( |A| \) along first row.
In fact, \( |A| \) can be expanded along any of its rows or columns.

2. Singular and Non-singular Matrix: A square matrix is a singular matrix if its determinant is zero. Otherwise, it is a non-singular matrix.

3. (i) Minor: Let \( A = [a_{ij}] \) be a square matrix of order \( n \). Then the minor \( M_{ij} \) of \( a_{ij} \) in \( A \) is the determinant of the sub-matrix of order \( (n - 1) \) obtained by leaving \( i^{th} \) row and \( j^{th} \) column of \( A \).
For example, if \( A = \begin{bmatrix} 1 & 2 & 3 \\ -3 & 2 & -1 \\ 2 & -4 & 3 \end{bmatrix} \), then
\( M_{11} = \begin{vmatrix} 2 & -1 \\ -4 & 3 \end{vmatrix} = 2, M_{12} = \begin{vmatrix} -3 & -1 \\ 2 & 3 \end{vmatrix} = -7 \) and so on.
(ii) Cofactor: The cofactor \( C_{ij} \) of \( a_{ij} \) in \( A = [a_{ij}]_{n \times n} \) is equal to \( (-1)^{i+j} \) times \( M_{ij} \).
For example, if \( A = \begin{bmatrix} 1 & 2 & 3 \\ -3 & 2 & -1 \\ 2 & -4 & 3 \end{bmatrix} \), then
\( C_{11} = (-1)^{1+1} M_{11} = M_{11} = 2 \) and \( C_{12} = (-1)^{1+2} M_{12} = -M_{12} = 7 \) and so on.

Some Important Properties of Determinants:

(i) Let \( A = [a_{ij}] \) be a square matrix of order \( n \), then the sum of the product of elements of any row (column) with their cofactors is always equal to \( |A| \) or, \( \text{det}(A) \), i.e.,
\( \sum_{j=1}^n a_{ij}C_{ij} = |A| \) and \( \sum_{i=1}^n a_{ij}C_{ij} = |A| \)

(ii) Let \( A = [a_{ij}] \) be a square matrix of order \( n \), then the sum of the product of elements of any row (column) with cofactors of the corresponding elements of some other row (column) is zero, i.e.,
\( \sum_{j=1}^n a_{ij}C_{kj} = 0 \) and \( \sum_{i=1}^n a_{ij}C_{ik} = 0 \), \( i \neq k \) or \( j \neq k \)

(iii) Let \( A = [a_{ij}] \) be a square matrix of order \( n \), then \( |A| = |A^T| \).
In other words, we say that the value of a determinant remains unchanged, if its rows and columns are interchanged.

(iv) Let \( A = [a_{ij}] \) be a square matrix of order \( n(n \geq 2) \) and \( B \) be a matrix obtained from \( A \) by interchanging any two rows (columns) of \( A \), then \( |B| = -|A| \).

(v) If any two rows (columns) of a square matrix \( A = [a_{ij}] \) of order \( n(n \geq 2) \) are identical, then value of its determinant is zero i.e., \( |A| = 0 \).

(vi) Let \( A = [a_{ij}] \) be a square matrix of order \( n \), and let \( B \) be the matrix obtained from \( A \) by multiplying each element of a row (column) of \( A \) by a scalar \( k \), then \( |B| = k|A| \).

(vii) Let \( A \) be a square matrix such that each element of a row (column) of \( A \) is expressed as the sum of two or more terms. Then the determinant of \( A \) can be expressed as the sum of the determinants of two or more matrices of the same order.

(viii) Let \( A \) be a square matrix and \( B \) be a matrix obtained from \( A \) by adding to a row (column) of \( A \) a scalar multiple of another row (column) of \( A \), then \( |B| = |A| \).

(ix) Let \( A \) be a square matrix of order \( n(n \geq 2) \) such that each element in a row (column) of \( A \) is zero, then \( |A| = 0 \).

(x) If \( A = [a_{ij}] \) is a diagonal matrix of order \( n(n \geq 2) \), then
\( |A| = a_{11} \cdot a_{22} \cdot a_{33} \dots a_{nn} \) i.e., \( |A| \) is the product of its diagonal elements.

(xi) If \( A \) and \( B \) are square matrices of the same order, then
\( |AB| = |A||B| \)

(xii) If \( A = [a_{ij}] \) is a triangular matrix of order \( n \), then
\( |A| = a_{11} \cdot a_{22} \cdot a_{33} \dots a_{nn} \) i.e., \( |A| \) is the product of its diagonal elements.

(xiii) If \( A = [a_{ij}] \) is a square matrix of order \( n \), then \( |kA| = k^n|A| \), because \( k \) is common from each row (or column) of \( kA \).

(xiv) We can take out any common factor from any one row or any one column of a given determinant.

5. Area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \) and \( (x_3, y_3) \) is given by
\( \Delta = \text{Numerical value of } \frac{1}{2} \begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \)
Note: Since area is positive quantity therefore we take absolute value of \( \Delta \).

6. (i) If \( A \) is a skew-symmetric matrix of odd order, then \( |A| = 0 \).
(ii) The determinant of a skew-symmetric matrix of even order is a perfect square.

7. Some Important Facts:
(i) Only square matrices have determinants.
(ii) We cannot equate the corresponding elements of equal determinants like matrices
i.e., \( \begin{vmatrix} x & y \\ z & w \end{vmatrix} = \begin{vmatrix} l & m \\ n & p \end{vmatrix} \not\Rightarrow x = l, y = m, z = n, w = p \)
(iii) In the case of matrices, we take out any common factor from each elements of matrix, while in the case of determinants we can take out common factor from any one row or any one column of the determinant.
(iv) If the value of determinant '\( \Delta \)' becomes zero by substituting \( x = a \) then \( (x - a) \) is factor of the determinant '\( \Delta \)'.
(v) If area is given then both positive and negative values of the determinant is taken for calculation.
(vi) To prove three points collinear, we show area of the triangle formed by these three points is zero.

Selected NCERT Questions

Question. If \( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \), then show that \( |2A| = 4|A| \).
Answer: We have,
\( A = \begin{bmatrix} 1 & 2 \\ 4 & 2 \end{bmatrix} \Rightarrow 2A = \begin{bmatrix} 2 & 4 \\ 8 & 4 \end{bmatrix} \)
\( \therefore \text{LHS} = |2A| = \begin{vmatrix} 2 & 4 \\ 8 & 4 \end{vmatrix} = 8 - 32 = -24 \)
\( \text{RHS} = 4|A| = 4 \begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix} = 4(2 - 8) = 4 \times (-6) = -24 \)
\( \therefore \text{LHS} = \text{RHS} \)
Hence Proved

Question. By using properties of determinant in problems 2 to 5 prove that:
\( \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} = 4a^2b^2c^2 \). 
Answer: \( \text{LHS} = \Delta = \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} \)
Taking \( a, b \) and \( c \) common from \( R_1, R_2 \) and \( R_3 \) respectively, we get
\( \Delta = abc \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} \)
Taking \( a, b \) and \( c \) common from \( C_1, C_2 \) and \( C_3 \) respectively, we get
\( \Delta = a^2b^2c^2 \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \)
Operating \( R_2 \rightarrow R_2 + R_1 \) and \( R_3 \rightarrow R_3 + R_1 \), we get
\( \Delta = a^2b^2c^2 \begin{vmatrix} -1 & 1 & 1 \\ 0 & 0 & 2 \\ 0 & 2 & 0 \end{vmatrix} \)
Interchanging \( C_2 \) and \( C_3 \), we get
\( \Delta = (-1) a^2b^2c^2 \begin{vmatrix} -1 & 1 & 1 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{vmatrix} \)
Since the determinant of a triangular matrix is product of its diagonal elements.
\( = (-1) a^2b^2c^2 (-1) \times (2) \times (2) = 4a^2b^2c^2 = \text{RHS} \)

Question. Prove that \( \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} = (a - b)(b - c)(c - a)(a + b + c) \). 
Answer: \( \text{LHS} = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^3 & b^3 & c^3 \end{vmatrix} \)
Operating \( C_2 \rightarrow C_2 - C_1 \) and \( C_3 \rightarrow C_3 - C_1 \), we get
\( = \begin{vmatrix} 1 & 0 & 0 \\ a & b - a & c - a \\ a^3 & b^3 - a^3 & c^3 - a^3 \end{vmatrix} \)
Taking \( (b - a) \) and \( (c - a) \) common from \( C_2 \) and \( C_3 \), we get
\( = (b - a)(c - a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 1 \\ a^3 & b^2 + ba + a^2 & c^2 + ca + a^2 \end{vmatrix} \)
Operating \( C_3 \rightarrow C_3 - C_2 \), we get
\( = (b - a)(c - a) \begin{vmatrix} 1 & 0 & 0 \\ a & 1 & 0 \\ a^3 & b^2 + ba + a^2 & c^2 + ca - b^2 - ba \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( = (b - a)(c - a)(c^2 + ca - b^2 - ba) = (b - a)(c - a)[c^2 - b^2 + a(c - b)] \)
\( = (b - a)(c - a)[(c - b)(c + b) + a(c - b)] = (b - a)(c - a)(c - b)[c + b + a] \)
\( = (a - b)(b - c)(c - a)(a + b + c) = \text{RHS} \).

Question. Prove that \( \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} = (x - y)(y - z)(z - x)(xy + yz + zx) \).
Answer: \( \text{LHS} = \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} \)
Operating \( R_1 \rightarrow R_1 - R_3 \) and \( R_2 \rightarrow R_2 - R_3 \), we get
\( = \begin{vmatrix} x - z & x^2 - z^2 & yz - xy \\ y - z & y^2 - z^2 & zx - xy \\ z & z^2 & xy \end{vmatrix} = \begin{vmatrix} (x - z) & (x - z)(x + z) & -y(x - z) \\ (y - z) & (y - z)(y + z) & -x(y - z) \\ z & z^2 & xy \end{vmatrix} \)
Taking \( (x - z) \) and \( (y - z) \) common from \( R_1 \) and \( R_2 \), we get
\( = (x - z)(y - z) \begin{vmatrix} 1 & x + z & -y \\ 1 & y + z & -x \\ z & z^2 & xy \end{vmatrix} \)
Operating \( R_2 \rightarrow R_2 - R_1 \), and \( R_3 \rightarrow R_3 - zR_1 \), we get
\( = (x - z)(y - z) \begin{vmatrix} 1 & x + z & -y \\ 0 & y - x & y - x \\ 0 & -xz & xy + yz \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( = (x - z)(y - z)[(y - x)(xy + yz) + xz(y - x)] \)
\( = (x - y)(y - z)(z - x)(xy + yz + zx) = \text{RHS} \)

Question. Prove that \( \begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix} = (a + b + c)^3 \)
Answer: \( \text{LHS} = \begin{vmatrix} a - b - c & 2a & 2a \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix} \)
Operating \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get
\( = \begin{vmatrix} a + b + c & a + b + c & a + b + c \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix} \)
Taking \( (a + b + c) \) common from first row, we get
\( = (a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b - c - a & 2b \\ 2c & 2c & c - a - b \end{vmatrix} \)
Operating \( C_2 \rightarrow C_2 - C_1 \) and \( C_3 \rightarrow C_3 - C_1 \), we get
\( = (a + b + c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -c - a - b & 0 \\ 2c & 0 & -a - b - c \end{vmatrix} \)
Since determinant of a triangular matrix is equal to product of its diagonal elements
\( \therefore = (a + b + c)(a + b + c)(a + b + c) = (a + b + c)^3 = \text{RHS} \)

Question. By using properties of determinant, show that:
\( \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} = (1 + a^2 + b^2)^3 \) 
Answer: \( \text{LHS} = \begin{vmatrix} 1 + a^2 - b^2 & 2ab & -2b \\ 2ab & 1 - a^2 + b^2 & 2a \\ 2b & -2a & 1 - a^2 - b^2 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 - bC_3 \), and \( C_2 \rightarrow C_2 + aC_3 \), we get
\( = \begin{vmatrix} (1 + a^2 + b^2) & 0 & -2b \\ 0 & (1 + a^2 + b^2) & 2a \\ b(1 + a^2 + b^2) & -a(1 + a^2 + b^2) & 1 - a^2 - b^2 \end{vmatrix} \)
Taking out \( (1 + a^2 + b^2) \) from \( C_1 \) and \( C_2 \) column, we get
\( = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1 - a^2 - b^2 \end{vmatrix} \)
Applying \( R_3 \rightarrow R_3 - bR_1 \), we get
\( = (1 + a^2 + b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & -a & 1 - a^2 + b^2 \end{vmatrix} \)
Expanding along first column, we get
\( = (1 + a^2 + b^2)^2 [1 - a^2 + b^2 + 2a^2] \)
\( = (1 + a^2 + b^2)^2 (1 + a^2 + b^2) = (1 + a^2 + b^2)^3 = \text{RHS} \)

Question. By using properties of determinant, show that:
\( \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} = 1 + a^2 + b^2 + c^2 \) 
Answer: \( \text{LHS} = \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} \)
\( = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & ab & ac \\ ab & b^2 + 1 & bc \\ ca & cb & c^2 + 1 \end{vmatrix} \) [Multiplying and dividing by \( abc \)]
Multiplying \( a \) in \( C_1 \), \( b \) in \( C_2 \) and \( c \) in \( C_3 \), we get
\( = \frac{1}{abc} \begin{vmatrix} a^3 + a & ab^2 & ac^2 \\ a^2b & b^3 + b & bc^2 \\ a^2c & b^2c & c^3 + c \end{vmatrix} \)
Taking \( a, b \) and \( c \) common from \( R_1, R_2 \) and \( R_3 \) respectively, we get
\( = \frac{abc}{abc} \begin{vmatrix} a^2 + 1 & b^2 & c^2 \\ a^2 & b^2 + 1 & c^2 \\ a^2 & b^2 & c^2 + 1 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( = \begin{vmatrix} 1 + a^2 + b^2 + c^2 & b^2 & c^2 \\ 1 + a^2 + b^2 + c^2 & b^2 + 1 & c^2 \\ 1 + a^2 + b^2 + c^2 & b^2 & c^2 + 1 \end{vmatrix} \)
Taking \( (1 + a^2 + b^2 + c^2) \) common from \( C_1 \), we get
\( = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 1 & b^2 & c^2 \\ 1 & b^2 + 1 & c^2 \\ 1 & b^2 & c^2 + 1 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 - R_3 \) and \( R_2 \rightarrow R_2 - R_3 \), we get
\( = (1 + a^2 + b^2 + c^2) \begin{vmatrix} 0 & 0 & -1 \\ 0 & 1 & -1 \\ 1 & b^2 & c^2 + 1 \end{vmatrix} \)

Expanding along first column, we get
\( = (1 + a^2 + b^2 + c^2) [1(0 \times (-1) - 1 \times (-1))] = (1 + a^2 + b^2 + c^2) \times 1 = \text{RHS} \)

Hence Proved

Question. Prove that: \[\begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2 \end{vmatrix} = 4a^2b^2c^2\] 
Answer: LHS \( = \begin{vmatrix} a^2 & bc & ac + c^2 \\ a^2 + ab & b^2 & ac \\ ab & b^2 + bc & c^2 \end{vmatrix} \)
\( = abc \begin{vmatrix} a & c & a + c \\ a + b & b & a \\ b & b + c & c \end{vmatrix} \) [Taking out \( a, b, c \) from \( C_1, C_2 \) and \( C_3 \)]
\( = abc \begin{vmatrix} 0 & c & a + c \\ 2b & b & a \\ 2b & b + c & c \end{vmatrix} \) [Applying \( C_1 \rightarrow C_1 + C_2 - C_3 \)]
\( = 2ab^2c \begin{vmatrix} 0 & c & a + c \\ 1 & b & a \\ 1 & b + c & c \end{vmatrix} \) [Taking out \( 2b \) from \( C_1 \)]
\( = 2ab^2c \begin{vmatrix} 0 & c & a + c \\ 0 & -c & a - c \\ 1 & b + c & c \end{vmatrix} \) [Applying \( R_2 \rightarrow R_2 - R_3 \)]
\( = 2ab^2c \cdot 1 \cdot \begin{vmatrix} c & a + c \\ -c & a - c \end{vmatrix} = 2ab^2c (ac - c^2 + ac + c^2) \) [Expanding by \( I \) column]
\( = 2ab^2c(2ac) = 4a^2b^2c^2 = \text{RHS} \)

Question. Prove: \[\begin{vmatrix} x & x^2 & 1 + px^3 \\ y & y^2 & 1 + py^3 \\ z & z^2 & 1 + pz^3 \end{vmatrix} = (1 + pxyz)(x - y)(y - z)(z - x)\] 
Answer: LHS \( \Delta = \begin{vmatrix} x & x^2 & 1 + px^3 \\ y & y^2 & 1 + py^3 \\ z & z^2 & 1 + pz^3 \end{vmatrix} \)
\( = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & px^3 \\ y & y^2 & py^3 \\ z & z^2 & pz^3 \end{vmatrix} = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} 1 & x & px^2 \\ 1 & y & py^2 \\ 1 & z & pz^2 \end{vmatrix} \) [Taking common \( x, y, z \) from \( R_1, R_2, R_3 \) respectively]
\( = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + (xyz)p \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \) [Taking \( p \) common from \( C_3 \)]
By changing (transforming) column to column in first determinant, we get
\( = \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + pxyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = (1 + pxyz) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 - R_3 \) and \( R_2 \rightarrow R_2 - R_3 \), we get
\( = (1 + pxyz) \begin{vmatrix} 0 & x - z & x^2 - z^2 \\ 0 & y - z & y^2 - z^2 \\ 1 & z & z^2 \end{vmatrix} \)
Taking out \( (x - z), (y - z) \) from \( R_1 \) and \( R_2 \) respectively, we get
\( = (1 + pxyz)(x - z)(y - z) \begin{vmatrix} 0 & 1 & x + z \\ 0 & 1 & y + z \\ 1 & z & z^2 \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = (1 + pxyz)(x - z)(y - z) [y + z - x - z] \)
\( = (1 + pxyz)(x - y)(y - z)(z - x) = \text{RHS} \).

Question. If \( a, b \) and \( c \) are real numbers and \( \Delta = \begin{vmatrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} = 0 \), then show that either \( a + b + c = 0 \) or \( a = b = c \). 
Answer: Given \( \Delta = \begin{vmatrix} b + c & c + a & a + b \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} \)
\( = \begin{vmatrix} 2(a + b + c) & 2(a + b + c) & 2(a + b + c) \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = 2(a + b + c) \begin{vmatrix} 1 & 1 & 1 \\ c + a & a + b & b + c \\ a + b & b + c & c + a \end{vmatrix} \) [Taking common \( 2(a + b + c) \) from \( R_1 \)]
Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get
\( = 2(a + b + c) \begin{vmatrix} 0 & 0 & 1 \\ a - b & a - c & b + c \\ a - c & b - a & c + a \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( = 2(a + b + c) [(a - b)(b - a) - (a - c)(a - c)] \)
\( = 2(a + b + c) [ab - a^2 - b^2 + ab - (a^2 + c^2 - 2ac)] \)
\( = 2(a + b + c) [ab - a^2 - b^2 + ab - a^2 - c^2 + 2ac] \)
\( = 2(a + b + c) [-2a^2 - b^2 - c^2 + 2ab + 2ac] \)
\( = -2(a + b + c) [a^2 + b^2 + c^2 - ab - bc - ca] \)
\( = -(a + b + c) [2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca] \)
\( = -(a + b + c) [(a - b)^2 + (b - c)^2 + (c - a)^2] \)
Now, given that \( \Delta = 0 \)
\( \Rightarrow \Delta = (a + b + c) [(a - b)^2 + (b - c)^2 + (c - a)^2] = 0 \)
So, either \( (a + b + c) = 0 \) or \( (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \) i.e., \( a = b = c \).

Question. Show that points \( A(a, b + c), B(b, c + a), C(c, a + b) \) are collinear.
Answer: We have,
Area of \( \Delta ABC = \frac{1}{2} \begin{vmatrix} a & b + c & 1 \\ b & c + a & 1 \\ c & a + b & 1 \end{vmatrix} \)
\( = \frac{1}{2} \begin{vmatrix} a & a + b + c & 1 \\ b & b + c + a & 1 \\ c & c + a + b & 1 \end{vmatrix} \) (Applying \( C_2 \rightarrow C_2 + C_1 \))
\( = \frac{1}{2} (a + b + c) \begin{vmatrix} a & 1 & 1 \\ b & 1 & 1 \\ c & 1 & 1 \end{vmatrix} \) (Taking \( (a + b + c) \) common from \( C_2 \))
\( = \frac{1}{2} \times (a + b + c) \times 0 \) (\(\because C_2 = C_3\))
\( \Rightarrow \text{ar}(\Delta ABC) = 0 \)
Since area of \( \Delta ABC \) is zero, therefore points \( A, B \) and \( C \) are collinear.
Hence proved.

Multiple Choice Questions

Question. If \(\begin{vmatrix} x & 2 \\ 18 & x \end{vmatrix} = \begin{vmatrix} 6 & 2 \\ 18 & 6 \end{vmatrix}\), then \( x \) is equal to
(a) 6
(b) \(\pm 6\)
(c) \(- 6\)
(d) 0
Answer: (b)

Question. The value of determinant \(\begin{vmatrix} a - b & b + c & a \\ b - c & c + a & b \\ c - a & a + b & c \end{vmatrix}\) 
(a) \( a^3 + b^3 + c^3 \)
(b) \( 3abc \)
(c) \( a^3 + b^3 + c^3 - 3abc \)
(d) None of these
Answer: (c)

Question. The area of a triangle with vertices \((-3, 0), (3, 0)\) and \((0, k)\) is 9 sq. units. The value of \( k \) will be
(a) 9
(b) 3
(c) \(-9\)
(d) 6
Answer: (b)

Question. If \( A, B \) and \( C \) are angles of a triangle, then the determinant \(\begin{vmatrix} -1 & \cos C & \cos B \\ \cos C & -1 & \cos A \\ \cos B & \cos A & -1 \end{vmatrix}\) is equal to
(a) 0
(b) \(-1\)
(c) 1
(d) None of these
Answer: (a)

Question. If \( f(x) = \begin{vmatrix} 0 & x - a & x - b \\ x + a & 0 & x - c \\ x + b & x + c & 0 \end{vmatrix}\), then 
(a) \( f(a) = 0 \)
(b) \( f(b) = 0 \)
(c) \( f(0) = 0 \)
(d) \( f(1) = 0 \)
Answer: (c)

Question. If \( x, y, z \) are all different from zero and \(\begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} = 0\), then value of \( x^{-1} + y^{-1} + z^{-1} \) is
(a) \( xyz \)
(b) \( x^{-1} y^{-1} z^{-1} \)
(c) \(-x -y -z\)
(d) \(-1\)
Answer: (d)

Question. There are two values of \( a \) which makes determinant \( \Delta = \begin{vmatrix} 1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2a \end{vmatrix} = 86 \), then sum of these numbers is
(a) 4
(b) 5
(c) \(- 4\)
(d) 9
Answer: (c)

Question. If \( A \) is a non-singular square matrix of order 3 such that \( A^2 = 3A \), then value of \( |A| \) is 
(a) \(-3\)
(b) 3
(c) 9
(d) 27
Answer: (d)

Question. If \(\begin{vmatrix} 2 & 3 & 2 \\ x & x & x \\ 4 & 9 & 1 \end{vmatrix} + 3 = 0\), then the value of \( x \) is 
(a) 3
(b) 0
(c) \(-1\)
(d) 1
Answer: (c)

Question. The value of the determinant \(\begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix}\) is 
(a) \( 9x^2(x + y) \)
(b) \( 9y^2(x + y) \)
(c) \( 3y^2(x + y) \)
(d) \( 7x^2(x + y) \)
Answer: (b)

Question. If \( a, b, c \) are in AP, then the value of determinant \( \Delta = \begin{vmatrix} x + 2 & x + 3 & x + 2a \\ x + 3 & x + 4 & x + 2b \\ x + 4 & x + 5 & x + 2c \end{vmatrix}\) is
(a) 0
(b) 1
(c) \( x \)
(d) \( 2x \)
Answer: (a)

Question. The value of \(\begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ a & b & c \\ a^2 & b^2 & c^2 \end{vmatrix}\) is
(a) \((a - b)(b - c)(c - a)\)
(b) \((b - a)(c - b)(c - a)\)
(c) \( a(b - c)(c - a) \)
(d) None of these
Answer: (a)

Question. The value of \(\begin{vmatrix} 0 & a - b & a - c \\ b - a & 0 & b - c \\ c - a & c - b & 0 \end{vmatrix}\) is
(a) \( a \)
(b) \( b \)
(c) 0
(d) None of these
Answer: (c)

Question. The value of \(\begin{vmatrix} 1 & \omega & \omega^2 \\ \omega & \omega^2 & 1 \\ \omega^2 & 1 & \omega \end{vmatrix}\) is
(a) 1
(b) \(-1\)
(c) 0
(d) \( \omega \)
Answer: (c)

Question. If area of triangle is 35 sq units with vertices \((2, -6), (5, 4)\) and \((k, 4)\), then \( k \) is
(a) 12
(b) \(-2\)
(c) \(-12, -2\)
(d) \( 12, -2 \)
Answer: (d)

Question. Let \( A \) be a square matrix of order \( 3 \times 3 \), then \( |KA| \) is equal to
(a) \( K|A| \)
(b) \( K^2|A| \)
(c) \( K^3|A| \)
(d) \( 3K|A| \)
Answer: (c)

Question. The value of \(\begin{vmatrix} 265 & 240 & 219 \\ 240 & 225 & 198 \\ 219 & 198 & 181 \end{vmatrix}\) is
(a) 0
(b) 1
(c) \(-1\)
(d) None
Answer: (a)

Question. The value of \(\begin{vmatrix} 1 & a & b + c \\ 1 & b & c + a \\ 1 & c & a + b \end{vmatrix}\) is
(a) 1
(b) 0
(c) \( a + b \)
(d) \( a - b \)
Answer: (b)

Question. If \( \Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix} \) and \( A_{ij} \) is cofactors of \( a_{ij} \), then value of \( \Delta \) is given by
(a) \( a_{11} A_{31} + a_{12} A_{32} + a_{13} A_{33} \)
(b) \( a_{11} A_{11} + a_{12} A_{21} + a_{13} A_{31} \)
(c) \( a_{21} A_{11} + a_{22} A_{12} + a_{23} A_{13} \)
(d) \( a_{11} A_{11} + a_{21} A_{21} + a_{31} A_{31} \)
Answer: (d)

Question. If \( A \) is a \( 3 \times 3 \) matrix such that \( |A| = 8 \), then \( |3A| \) equals 
(a) 8
(b) 24
(c) 72
(d) 216
Answer: (d)