CBSE Class 12 Mathematics Determinants Worksheet Set 08

Very Short Answer Questions [1 mark]

Question. For what value of \( x \), the following matrix is singular?
\( \begin{bmatrix} 5-x & x+1 \\ 2 & 4 \end{bmatrix} \)
Answer: Let \( \begin{vmatrix} 5-x & x+1 \\ 2 & 4 \end{vmatrix} = 0 \)
For \( A \) to be singular, \( |A| = 0 \)
\( \implies \quad \begin{vmatrix} 5-x & x+1 \\ 2 & 4 \end{vmatrix} = 0 \)
\( \implies \quad 4(5-x) - 2(x+1) = 0 \)
\( \implies \quad 20 - 4x - 2x - 2 = 0 \)
\( \implies \quad 18 = 6x \)
\( \implies \quad x = 3 \).

Question. Let \( A \) be a square matrix of order \( 3 \times 3 \). Write the value of \( |2A| \), where \( |A| = 4 \).
Answer: \( \because \quad |2A| = 2^n |A| \), where \( n \) is order of matrix \( A \).
Here \( |A| = 4 \) and \( n = 3 \)
\( |2A| = 2^3 \times 4 = 32 \)

Question. If \( \begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} \), then write the value of \( x \).
Answer: Given \( \begin{vmatrix} x+1 & x-1 \\ x-3 & x+2 \end{vmatrix} = \begin{vmatrix} 4 & -1 \\ 1 & 3 \end{vmatrix} \)
\( \implies \quad (x+1)(x+2) - (x-1)(x-3) = 12 + 1 \)
\( \implies \quad x^2 + 2x + x + 2 - x^2 + 3x + x - 3 = 13 \)
\( \implies \quad 7x - 1 = 13 \)
\( \implies \quad 7x = 14 \)
\( \implies \quad x = 2 \)

Question. If \( A = [a_{ij}] \) is a matrix of order \( 2 \times 2 \), such that \( |A| = -15 \) and \( C_{ij} \) represents the cofactor of \( a_{ij} \), then find \( a_{21}C_{21} + a_{22}C_{22} \).
Answer: Given, \( A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix} \quad \therefore \quad |A| = \begin{vmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{vmatrix} \)
Expanding along \( R_2 \)
\( \implies \quad -15 = a_{21} \cdot C_{21} + a_{22} \cdot C_{22} \quad [C_{ij} = \text{Cofactor of } a_{ij}] \)
\( \implies \quad a_{21} \cdot C_{21} + a_{22} \cdot C_{22} = -15 \)

Question. Write the value of the following determinant:
\( \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix} \)
Answer: \( \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6x & 9x & 12x \end{vmatrix} = 3x \begin{vmatrix} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4 \end{vmatrix} \quad \) [Taking out \( 3x \) common from \( R_3 \)]
\( = 3x \times 0 = 0 \quad [\because R_1 = R_3] \)

Question. Write the value of the following determinant:
\( \begin{vmatrix} a-b & b-c & c-a \\ b-c & c-a & a-b \\ c-a & a-b & b-c \end{vmatrix} \)
Answer: Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( = \begin{vmatrix} 0 & b-c & c-a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{vmatrix} = 0 \quad [\because \text{ All elements of } C_1 \text{ are zero}] \)

Question. Show that the points \( (1, 0), (6, 0), (0, 0) \) are collinear.
Answer: Since \( \begin{vmatrix} 1 & 0 & 1 \\ 6 & 0 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0 \)
Hence, \( (1, 0), (6, 0) \) and \( (0, 0) \) are collinear.

Question. What positive value of \( x \) makes the following pair of determinants equal?
\( \begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} = \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix} \)
Answer: \( \because \begin{vmatrix} 2x & 3 \\ 5 & x \end{vmatrix} = \begin{vmatrix} 16 & 3 \\ 5 & 2 \end{vmatrix} \)
\( \implies \quad 2x^2 - 15 = 32 - 15 \)
\( \implies \quad 2x^2 = 32 \)
\( \implies \quad x^2 = 16 \)
\( \implies \quad x = \pm 4 \)
\( \implies \quad x = 4 \quad \text{(+ve value)}. \)

Question. Evaluate: \( \begin{vmatrix} \cos 15^{\circ} & \sin 15^{\circ} \\ \sin 75^{\circ} & \cos 75^{\circ} \end{vmatrix} \)
Answer: Expanding the determinant, we get
\( \cos 15^{\circ} \cdot \cos 75^{\circ} - \sin 15^{\circ} \cdot \sin 75^{\circ} = \cos (15^{\circ} + 75^{\circ}) = \cos 90^{\circ} = 0 \)
[Note : \( \cos(A+B) = \cos A \cdot \cos B - \sin A \cdot \sin B \)]

Question. Write the value of the following determinant:
\( \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \)
Answer: Let \( \Delta = \begin{vmatrix} 102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 - 6R_3 \), we get
\( \Delta = \begin{vmatrix} 0 & 0 & 0 \\ 1 & 3 & 4 \\ 17 & 3 & 6 \end{vmatrix} = 0 \quad [\because \text{ Each element of } R_1 \text{ is zero}] \)

Question. If \( A_{ij} \) is the cofactor of the element \( a_{ij} \) of the determinant \( \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \), then write the value of \( a_{32} \cdot A_{32} \).
Answer: \( a_{32} \cdot A_{32} = 5 \times (-1)^{3+2} \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = -5 (8 - 30) = -5 \times -22 = 110 \)

Question. If \( A \) is a square matrix and \( |A| = 2 \), then write the value of \( |AA'| \), where \( A' \) is the transpose of matrix \( A \).
Answer: \( |AA'| = |A| \cdot |A'| = |A| \cdot |A| = |A|^2 = 2^2 = 4 \)
[Note: \( |AB| = |A| \cdot |B| \) and \( |A| = |A^T| \), where \( A \) and \( B \) are square matrices.]

Question. If \( A \) is a \( 3 \times 3 \) invertible matrix, then what will be the value of \( k \) if \( \text{det} (A^{-1}) = (\text{det} A)^k \).
Answer: Given, \( \text{det} (A^{-1}) = (\text{det} A)^k \)
\( \implies \quad |A^{-1}| = |A|^k \)
\( \implies \quad \frac{1}{|A|} = |A|^k \)
\( \implies \quad |A|^{-1} = |A|^k \)
\( \implies \quad k = -1 \)

Short Answer Questions [2 marks]

Question. Evaluate the determinant: \( \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} \)
Answer: Let \( \Delta = \begin{vmatrix} x^2 - x + 1 & x - 1 \\ x + 1 & x + 1 \end{vmatrix} \)
\( = (x + 1) \begin{vmatrix} x^2 - x + 1 & x - 1 \\ 1 & 1 \end{vmatrix} \quad \) [Taking common \( (x + 1) \) from \( R_2 \)]
\( = (x + 1) \{ x^2 - x + 1 - x + 1 \} = (x + 1) (x^2 - 2x + 2) \)
\( = x^3 - 2x^2 + 2x + x^2 - 2x + 2 = x^3 - x^2 + 2 \)

Question. What is the value of the following determinant:
\( \begin{vmatrix} 4 & a & b+c \\ 4 & b & c+a \\ 4 & c & a+b \end{vmatrix} \)
Answer: Here, \( \Delta = \begin{vmatrix} 4 & a & b+c \\ 4 & b & c+a \\ 4 & c & a+b \end{vmatrix} = \begin{vmatrix} 4 & a & a+b+c \\ 4 & b & a+b+c \\ 4 & c & a+b+c \end{vmatrix} \quad \) [Applying \( C_3 \rightarrow C_3 + C_2 \)]
\( \Delta = 4(a+b+c) \begin{vmatrix} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{vmatrix} \quad \) [Taking out common 4 from \( C_1 \) and \( a+b+c \) from \( C_3 \)]
\( \Delta = 4(a+b+c) \cdot 0 = 0 \quad \begin{bmatrix} \because C_1 = C_3 \\ \therefore \Delta = 0 \end{bmatrix} \)

Question. Write the value of \( \Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Answer: Here, \( \Delta = \begin{vmatrix} x+y & y+z & z+x \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 + R_2 \), we get
\( = \begin{vmatrix} x+y+z & x+y+z & x+y+z \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_1 \), we get
\( = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ -3 & -3 & -3 \end{vmatrix} \)
Applying \( R_3 \rightarrow R_3 + 3R_1 \), we get
\( = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ z & x & y \\ 0 & 0 & 0 \end{vmatrix} = 0 \quad [\because \text{ Each element of } R_3 \text{ is zero}] \)

Question. Without expanding evaluate the determinant:
\( \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (a^y + a^{-y})^2 & (a^y - a^{-y})^2 & 1 \\ (a^z + a^{-z})^2 & (a^z + a^{-z})^2 & 1 \end{vmatrix} \), where \( a > 0 \) and \( x, y, z \in R \)
Answer: Here \( \Delta = \begin{vmatrix} (a^x + a^{-x})^2 & (a^x - a^{-x})^2 & 1 \\ (a^y + a^{-y})^2 & (a^y - a^{-y})^2 & 1 \\ (a^z + a^{-z})^2 & (a^z + a^{-z})^2 & 1 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 - C_2 \), we get
\( \Delta = \begin{vmatrix} 4 & (a^x - a^{-x})^2 & 1 \\ 4 & (a^y - a^{-y})^2 & 1 \\ 4 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \quad \) [Using \( (a+b)^2 - (a-b)^2 = 4ab \)]
Taking out 4 from \( C_1 \), we get
\( \Delta = 4 \begin{vmatrix} 1 & (a^x - a^{-x})^2 & 1 \\ 1 & (a^y - a^{-y})^2 & 1 \\ 1 & (a^z - a^{-z})^2 & 1 \end{vmatrix} \)
\( \implies \quad \Delta = 4 \times 0 = 0. \quad [\because C_1 \text{ and } C_3 \text{ are identical}] \)

Long Answer Questions-I [4 marks]

Question. Using properties of determinants, prove that
\( \begin{vmatrix} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix} = (x+y+z)^3. \)
Answer: LHS \( = \begin{vmatrix} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix} \)
Applying \( R_2 \leftrightarrow R_3 \), then \( R_1 \leftrightarrow R_2 \), we get
\( = \begin{vmatrix} x-y-z & 2x & 2x \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \)
Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get
\( = \begin{vmatrix} x+y+z & y+z+x & z+x+y \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \)
Taking out \( (x+y+z) \) from first row, we get
\( = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get
\( = (x+y+z) \begin{vmatrix} 0 & 0 & 1 \\ 0 & (y+z+x) & 2y \\ (x+y+z) & z+x+y & z-x-y \end{vmatrix} \)
Expanding along first row, we get
\( = (x+y+z) (x+y+z)^2 = (x+y+z)^3 = \text{RHS} \)

Question. Using properties of determinants, find the value of \( x \) for which \( \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} = 0 \).
Answer: We have, \( \begin{vmatrix} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} 12+x & 4+x & 4+x \\ 12+x & 4-x & 4+x \\ 12+x & 4+x & 4-x \end{vmatrix} = 0 \)
\( \implies \quad (12+x) \begin{vmatrix} 1 & 4+x & 4+x \\ 1 & 4-x & 4+x \\ 1 & 4+x & 4-x \end{vmatrix} = 0 \)
Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \), we get
\( (12+x) \begin{vmatrix} 1 & 4+x & 4+x \\ 0 & -2x & 0 \\ 0 & 0 & -2x \end{vmatrix} = 0 \)
\( \implies \quad (x+12)(4x^2) = 0 \)
\( \implies \quad x = 0, -12 \)

Question. If \( x, y, z \) are different and \( \Delta = \begin{vmatrix} x & x^2 & x^3-1 \\ y & y^2 & y^3-1 \\ z & z^2 & z^3-1 \end{vmatrix} = 0 \), then using properties of determinants, show that \( xyz = 1 \).
Answer: We have \( \begin{vmatrix} x & x^2 & x^3-1 \\ y & y^2 & y^3-1 \\ z & z^2 & z^3-1 \end{vmatrix} = 0 \)
\( \implies \quad \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} + \begin{vmatrix} x & x^2 & -1 \\ y & y^2 & -1 \\ z & z^2 & -1 \end{vmatrix} = 0 \)
\( \implies \quad xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + \begin{vmatrix} -1 & x^2 & x \\ -1 & y^2 & y \\ -1 & z^2 & z \end{vmatrix} = 0 \quad \left[ \begin{matrix} \text{In det 1 taking } x, y, z \text{ common} \\ \text{from each row and in det 2} \\ \text{using } C_1 \leftrightarrow C_3 \end{matrix} \right] \)
\( \implies \quad xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + \begin{vmatrix} -1 & x & x^2 \\ -1 & y & y^2 \\ -1 & z & z^2 \end{vmatrix} = 0 \quad [\text{Applying } C_2 \leftrightarrow C_3 \text{ in det 2}] \)
\( \implies \quad xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} + (-1) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \)
\( \implies \quad \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} (xyz - 1) = 0 \)
If \( \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \)
\( \implies \quad \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix} = 0 \quad \left[ \begin{matrix} \text{Applying } \\ R_2 \rightarrow R_2 - R_1 \\ R_3 \rightarrow R_3 - R_1 \end{matrix} \right] \)
\( \implies \quad \begin{vmatrix} y-x & y^2-x^2 \\ z-x & z^2-x^2 \end{vmatrix} = 0 \)
\( \implies \quad (y-x)(z-x) \begin{vmatrix} 1 & y+x \\ 1 & z+x \end{vmatrix} = 0 \)
\( \implies \quad (y-x)(z-x)(z+x-y-x) = 0 \)
\( \implies \quad (y-x)(z-x)(z-y) = 0 \)
\( \implies \quad x = y \text{ or } z = x \text{ or } y = z \text{, which is a contradiction.} \)
Hence, \( (xyz - 1) = 0 \)
\( \implies \quad xyz = 1 \).

Question. Prove that: \( \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3 \)
OR
Prove that: \( \begin{vmatrix} x+y+2z & x & y \\ z & y+z+2x & y \\ z & x & z+x+2y \end{vmatrix} = 2(x+y+z)^3 \)
Answer: LHS \( = \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( = \begin{vmatrix} 2(a+b+c) & a & b \\ 2(a+b+c) & b+c+2a & b \\ 2(a+b+c) & a & c+a+2b \end{vmatrix} \)
Taking \( 2(a+b+c) \) common from \( C_1 \), we get
\( = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get
\( = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{vmatrix} \)
Taking \( (a+b+c) \) common from \( R_2 \) and \( R_3 \), we get
\( = 2(a+b+c)^3 \begin{vmatrix} 1 & a & b \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = 2(a+b+c)^3 [1 - 0] = 2(a+b+c)^3 = \text{RHS} \)
OR
For solution replace \( a \rightarrow x, b \rightarrow y \) and \( c \rightarrow z \) in above solution.

Question. If \( \Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} = -4 \) then find the value of \( \begin{vmatrix} a^3-1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} \).
Answer: We have \( \Delta = \begin{vmatrix} 1 & a & a^2 \\ a & a^2 & 1 \\ a^2 & 1 & a \end{vmatrix} \)
\( C_{11} = a^3 - 1; \quad C_{12} = 0; \quad C_{13} = a - a^4 \)
\( C_{21} = 0; \quad C_{22} = a - a^4; \quad C_{23} = a^3 - 1 \)
\( C_{31} = a - a^4; \quad C_{32} = a^3 - 1; \quad C_{33} = 0 \)
Where \( C_{ij} = \text{Co-factor of } a_{ij} \, (i, j)^{\text{th}} \text{ element of determinant } \Delta \). Let \( \Delta_1 \) be the determinant made by corresponding co-factor of each element of determinant \( \Delta \).
i.e., \( \Delta_1 = \begin{vmatrix} C_{11} & C_{12} & C_{13} \\ C_{21} & C_{22} & C_{23} \\ C_{31} & C_{32} & C_{33} \end{vmatrix} \)
We know that \( \Delta_1 = \Delta^2 \quad [\because \Delta_1 = \Delta^{n-1} \text{ where each element of } \Delta_1 \text{ is cofactor of} \)
\( \therefore \quad \Delta_1 = (-4)^2 = 16 \quad \text{ corresponding element of } \Delta \text{ and } n \text{ is order of the determinant]} \)
\( \implies \quad \begin{vmatrix} a^3-1 & 0 & a-a^4 \\ 0 & a-a^4 & a^3-1 \\ a-a^4 & a^3-1 & 0 \end{vmatrix} = 16 \)

Question. Using properties of determinant, solve for \( x \):
\( \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0 \)
Answer: Let \( \Delta = \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} \)
\( = \begin{vmatrix} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \)]
\( = (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \end{vmatrix} \)
\( = (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
Expanding along \( C_1 \), we get
\( = (3a-x) (4x^2 - 0) = 4x^2 (3a-x) \)
Now, given that \( \Delta = 0 \)
Therefore, \( 4x^2 (3a-x) = 0 \)
\( \implies \quad x = 0, \text{ or } x = 3a \)
Hence, required values of \( x \) are \( x = 0, 3a \).

Question. Using property of determinant, prove the following:
\( \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9b^2 (a+b) \)
Answer: LHS \( = \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \)
\( = \begin{vmatrix} 3(a+b) & 3(a+b) & 3(a+b) \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \quad \) [Applying \( R_1 = R_1 + R_2 + R_3 \)]
\( = 3(a+b) \begin{vmatrix} 1 & 1 & 1 \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} \quad \) [Taking \( 3(a+b) \) common from \( R_1 \)]
\( = 3(a+b) \begin{vmatrix} 0 & 0 & 1 \\ b & -b & a+b \\ b & 2b & a \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 - C_3, C_2 \rightarrow C_2 - C_3 \)]
Expanding along \( R_1 \), we get
\( = 3(a+b) \{ 1(2b^2 + b^2) \} = 9b^2(a+b) = \text{RHS} \)

Question. Using properties of determinants, find the value of \( k \) if \( \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} = k(x^3 + y^3). \)
Answer: We have \( k(x^3 + y^3) = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix} \)
\( = \begin{vmatrix} 2x+2y & y & x+y \\ 2x+2y & x+y & x \\ 2x+2y & x & y \end{vmatrix} \quad \) [Using \( C_1 \rightarrow C_1 + C_2 + C_3 \)]
\( = (2x+2y) \begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix} \quad \) [Taking \( (2x+2y) \) common from \( C_1 \)]
\( = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - R_1, R_3 \rightarrow R_3 - R_1 \)]
\( = 2(x+y) \begin{vmatrix} x & -y \\ x-y & -x \end{vmatrix} \)
\( = 2(x+y) (-x^2 + xy - y^2) = -2(x+y)(x^2 - xy + y^2) \)
\( \implies \quad k(x^3 + y^3) = -2(x^3 + y^3) \)
Comparing the coefficient of \( (x^3 + y^3) \) on both the sides, we get
\( k = -2 \)

Question. Using properties of determinants show that
\( \begin{vmatrix} 1 & 1 & 1+x \\ 1 & 1+y & 1 \\ 1+z & 1 & 1 \end{vmatrix} = -(xyz + yz + zx + xy) \)
Answer: LHS \( = \begin{vmatrix} 1 & 1 & 1+x \\ 1 & 1+y & 1 \\ 1+z & 1 & 1 \end{vmatrix} \)
Apply \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get
\( = \begin{vmatrix} 1 & 1 & 1+x \\ 0 & y & -x \\ z & 0 & -x \end{vmatrix} \)
Expanding by \( R_1 \), we get
\( = 1 [-yx - 0] - 1 [+zx] + (1+x) (-zy) = -yx - zx - zy - xyz \)
\( = -(xy + xz + zy + xyz) = \text{RHS} \)

Question. Using properties of determinant, prove that:
\( \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix} = a^2(a+x+y+z) \)
Answer: LHS \( = \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix} \)
\( = \begin{vmatrix} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \)]
\( = (a+x+y+z) \begin{vmatrix} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} \quad \) [Taking out \( (a+x+y+z) \) common from \( C_1 \)]
\( = (a+x+y+z) \begin{vmatrix} 0 & -a & 0 \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} \quad \) [Apply \( R_1 \rightarrow R_1 - R_2 \)]
Expanding along \( R_1 \), we get
\( = (a+x+y+z) \{ 0 + a(a+z-z) \} \)
\( = a^2(a+x+y+z) = \text{RHS} \)

Question. Using properties of determinant, prove the following:
\( \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3 \)
Answer: LHS \( = \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} \)
\( = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 5x+4y & 4 & 2 \\ 10x+8y & 8 & 3 \end{vmatrix} \quad \) [Taking out \( x \) from \( C_2 \) and \( C_3 \)]
\( = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 3x+2y & 2 & 0 \\ 7x+5y & 5 & 0 \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \)]
Expanding along \( C_3 \), we get
\( = x^2 [ 1 \{ (3x+2y)5 - 2(7x+5y) \} - 0 + 0 ] = x^2 (15x+10y - 14x - 10y) \)
\( = x^2(x) = x^3 = \text{RHS} \)

Question. In a triangle ABC, if
\( \begin{vmatrix} 1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A + \sin^2 A & \sin B + \sin^2 B & \sin C + \sin^2 C \end{vmatrix} = 0 \)
then prove that \( \Delta ABC \) is an isosceles triangle.
Answer: Let \( \Delta = \begin{vmatrix} 1 & 1 & 1 \\ 1+\sin A & 1+\sin B & 1+\sin C \\ \sin A + \sin^2 A & \sin B + \sin^2 B & \sin C + \sin^2 C \end{vmatrix} \)
\( = \begin{vmatrix} 1 & 0 & 0 \\ 1+\sin A & \sin B - \sin A & \sin C - \sin A \\ \sin A + \sin^2 A & \sin^2 B - \sin^2 A + \sin B - \sin A & \sin^2 C - \sin^2 A + \sin C - \sin A \end{vmatrix} \)
Applying \( C_2 \rightarrow C_2 - C_1 \) and \( C_3 \rightarrow C_3 - C_1 \), we get
\( = \begin{vmatrix} 1 & 0 & 0 \\ 1+\sin A & \sin B - \sin A & \sin C - \sin A \\ \sin A + \sin^2 A & (\sin B - \sin A)(\sin B + \sin A + 1) & (\sin C - \sin A)(\sin C + \sin A + 1) \end{vmatrix} \)
\( = (\sin B - \sin A)(\sin C - \sin A) \begin{vmatrix} 1 & 0 & 0 \\ 1+\sin A & 1 & 1 \\ \sin A + \sin^2 A & \sin B + \sin A + 1 & \sin C + \sin A + 1 \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( (\sin B - \sin A) (\sin C - \sin A) [ \sin C + \sin A + 1 - \sin B - \sin A - 1 ] \)
\( = (\sin B - \sin A) (\sin C - \sin A) (\sin C - \sin B) \)
\( \because \quad \Delta = 0 \)
\( \implies \quad (\sin B - \sin A) (\sin C - \sin B) (\sin C - \sin A) = 0 \)
\( \implies \quad \sin B - \sin A = 0 \text{ or } \sin C - \sin B = 0 \text{ or } \sin C - \sin A = 0 \)
\( \implies \quad B = A \text{ or } C = B \text{ or } C = A \)
\( \implies \quad \Delta ABC \text{ is an isosceles triangle.} \)

Question. Without expanding, show that: \( \begin{vmatrix} \text{cosec}^2 \theta & \cot^2 \theta & 1 \\ \cot^2 \theta & \text{cosec}^2 \theta & -1 \\ 42 & 40 & 2 \end{vmatrix} = 0 \)
Answer: Given, \( \Delta = \begin{vmatrix} \text{cosec}^2 \theta & \cot^2 \theta & 1 \\ \cot^2 \theta & \text{cosec}^2 \theta & -1 \\ 42 & 40 & 2 \end{vmatrix} \)
\( = \begin{vmatrix} \text{cosec}^2 \theta - \cot^2 \theta - 1 & \cot^2 \theta & 1 \\ \cot^2 \theta - \text{cosec}^2 \theta + 1 & \text{cosec}^2 \theta & -1 \\ 0 & 40 & 2 \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 - C_2 - C_3 \)]
\( = \begin{vmatrix} 1 - 1 & \cot^2 \theta & 1 \\ -1 + 1 & \text{cosec}^2 \theta & -1 \\ 0 & 40 & 2 \end{vmatrix} \quad [\because \text{cosec}^2 \theta - \cot^2 \theta = 1] \)
\( = \begin{vmatrix} 0 & \cot^2 \theta & 1 \\ 0 & \text{cosec}^2 \theta & -1 \\ 0 & 40 & 2 \end{vmatrix} = 0 \quad [\because \text{All elements of } C_1 \text{ are } 0] \)

Question. Prove the following using properties of determinant:
\( \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2(3abc - a^3 - b^3 - c^3) \)
Answer: LHS \( = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \)
\( = \begin{vmatrix} 2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = 2(a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \quad \) [Taking \( 2(a+b+c) \) common from \( R_1 \)]
\( = 2(a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ c+a & b-c & b-a \\ a+b & c-a & c-b \end{vmatrix} \quad \) [Applying \( C_2 \rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_1 \)]
\( = 2(a+b+c) [1(bc - b^2 - c^2 + bc - bc + ac + ab - a^2)] \quad \) [Expanding along \( R_1 \)]
\( = 2(a+b+c) (bc + ac + ab - a^2 - b^2 - c^2) \)
\( = -2(a+b+c) (a^2 + b^2 + c^2 - ab - bc - ca) = -2(a^3 + b^3 + c^3 - 3abc) \)
\( = 2(3abc - a^3 - b^3 - c^3) = \text{RHS} \)

Question. If \( a+b+c \neq 0 \) and \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \), then using properties of determinants, prove that \( a = b = c \).
Answer: We have \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} (a+b+c) & b & c \\ (a+b+c) & c & a \\ (a+b+c) & a & b \end{vmatrix} = 0 \)
Taking \( (a+b+c) \) common from \( C_1 \), we get
\( (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \)
\( \implies \quad \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \quad [\because a+b+c \neq 0] \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get
\( \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} = 0 \)
Expanding along \( C_1 \), we get
\( 1 \{ (c-b)(b-c) - (a-c)(a-b) \} - 0 + 0 = 0 \)
\( \implies \quad - (b-c)^2 - (a-c)(a-b) = 0 \)
\( \implies \quad -b^2 - c^2 + 2bc - a^2 + ab + ac - bc = 0 \)
\( \implies \quad a^2 + b^2 + c^2 - bc - ab - ac = 0 \)
\( \implies \quad \frac{1}{2} [ 2a^2 + 2b^2 + 2c^2 - 2bc - 2ab - 2ac ] = 0 \)
\( \implies \quad (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \)
\( \implies \quad (a-b)^2 = 0 ; \, (b-c)^2 = 0 ; \, (c-a)^2 = 0 \)
\( \implies \quad a-b = 0 ; \, b-c = 0 ; \, c-a = 0 ; \)
\( \implies \quad a = b = c \)

Question. Using properties of determinants, show that \( \Delta ABC \) is an isosceles if:
\( \begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0 \)
Answer: We have \( \begin{vmatrix} 1 & 1 & 1 \\ 1+\cos A & 1+\cos B & 1+\cos C \\ \cos^2 A + \cos A & \cos^2 B + \cos B & \cos^2 C + \cos C \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get
\( \implies \quad \begin{vmatrix} 0 & 0 & 1 \\ \cos A - \cos C & \cos B - \cos C & 1+\cos C \\ \cos^2 A + \cos A - \cos^2 C - \cos C & \cos^2 B + \cos B - \cos^2 C - \cos C & \cos^2 C + \cos C \end{vmatrix} = 0 \)
\( \implies \quad \begin{vmatrix} 0 & 0 & 1 \\ \cos A - \cos C & \cos B - \cos C & 1+\cos C \\ (\cos A - \cos C)(\cos A + \cos C + 1) & (\cos B - \cos C)(\cos B + \cos C + 1) & \cos^2 C + \cos C \end{vmatrix} = 0 \)
Taking common \( (\cos A - \cos C) \) from \( C_1 \) and \( (\cos B - \cos C) \) from \( C_2 \), we get
\( \implies \quad (\cos A - \cos C) (\cos B - \cos C) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & 1+\cos C \\ \cos A + \cos C + 1 & \cos B + \cos C + 1 & \cos^2 C + \cos C \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 - C_2 \), we get
\( \implies \quad (\cos A - \cos C) (\cos B - \cos C) \begin{vmatrix} 0 & 0 & 1 \\ 0 & 1 & 1+\cos C \\ \cos A - \cos B & \cos B + \cos C + 1 & \cos^2 C + \cos C \end{vmatrix} = 0 \)
Expanding along \( R_1 \), we get
\( \implies \quad (\cos A - \cos C) (\cos B - \cos C) (\cos B - \cos A) = 0 \)
\( \implies \quad \cos A - \cos C = 0 \quad \text{i.e., } \cos A = \cos C \)
or, \quad \( \cos B - \cos C = 0 \quad \text{i.e., } \cos B = \cos C \)
or, \quad \( \cos B - \cos A = 0 \quad \text{i.e., } \cos B = \cos A \)
\( \implies \quad A = C \text{ or } B = C \text{ or } B = A \)
Hence, \( \Delta ABC \) is an isosceles triangle.

Question. Using properties of determinant, prove the following:
\( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz (x-y)(y-z)(z-x) \)
Answer: LHS \( = \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} \)
\( = xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} \quad \) [Taking \( x, y, z \) common from \( C_1, C_2, C_3 \) respectively]
\( = xyz \begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{vmatrix} \quad \) [Applying \( C_2 \rightarrow C_2 - C_1 ; C_3 \rightarrow C_3 - C_1 \)]
\( = xyz (y-x)(z-x) \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ x^2 & y+x & z+x \end{vmatrix} \quad \left[ \begin{matrix} \text{Taking common } (y-x) \text{ and } (z-x) \\ \text{from } C_2 \text{ and } C_3 \text{ respectively} \end{matrix} \right] \)
\( = xyz (y-x)(z-x) [ 1(z+x - y-x) ] \quad \) [Expanding along \( R_1 \)]
\( = xyz (y-x)(z-x)(z-y) = xyz (x-y)(y-z)(z-x) \)

Question. Using properties of determinants, prove the following:
\( \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2 \)
OR
\( \begin{vmatrix} x+\lambda & 2x & 2x \\ 2x & x+\lambda & 2x \\ 2x & 2x & x+\lambda \end{vmatrix} = (5x+\lambda)(\lambda-x)^2 \)
Answer: LHS \( = \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \)
\( = \begin{vmatrix} 5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = (5x+4) \begin{vmatrix} 1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \quad \) [Taking \( (5x+4) \) common from \( R_1 \)]
\( = (5x+4) \begin{vmatrix} 1 & 0 & 0 \\ 2x & 4-x & 0 \\ 2x & 0 & 4-x \end{vmatrix} \quad \) [Applying \( C_2 \rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_1 \)]
\( = (5x+4) [ 1 \{ (4-x)^2 - 0 \} + 0 + 0 ] \quad \) [Expanding along \( R_1 \)]
\( = (5x+4) (4-x)^2 = \text{RHS} \)
OR
Solve as above by putting \( \lambda \) instead of 4.

Question. Using properties of determinants, prove that
\( \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \)
OR
\( \begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \)
Answer: LHS \( = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \)
\( = \begin{vmatrix} b+c & q+r & y+z \\ a+b & p+q & x+y \\ c+a & r+p & z+x \end{vmatrix} \quad \) [Applying \( R_1 \leftrightarrow R_3 \) and \( R_3 \leftrightarrow R_2 \)]
Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get
\( = \begin{vmatrix} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} = 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} \)
\( = 2 \begin{vmatrix} a & p & x \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 - R_2 \)]
\( = 2 \begin{vmatrix} a & p & x \\ b+c & q+r & y+z \\ c & r & z \end{vmatrix} \quad \) [Applying \( R_3 \rightarrow R_3 - R_1 \)]
Again applying \( R_2 \rightarrow R_2 - R_3 \), we get
\( = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \text{RHS} \)

Question. Using properties of determinant, prove the following:
\( \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = ab + bc + ca + abc \)
OR
If \( a, b \) and \( c \) are all non-zero and \( \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0 \), then prove that \( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 = 0 \)
Answer: LHS \( = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} \)
\( = abc \begin{vmatrix} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \quad \) [Taking out \( a, b, c \) common from \( R_1, R_2 \) and \( R_3 \)]
\( = abc \begin{vmatrix} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = abc \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) \begin{vmatrix} 1 & 1 & 1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \)
Applying \( C_2 \rightarrow C_2 - C_1 ; C_3 \rightarrow C_3 - C_1 \), we get
\( = abc \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) \begin{vmatrix} 1 & 0 & 0 \\ \frac{1}{b} & 1 & 0 \\ \frac{1}{c} & 0 & 1 \end{vmatrix} = abc \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) \times \{ 1(1-0) - 0 + 0 \} \)
\( = abc \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) = ab + bc + ca + abc = \text{RHS} \)
OR
\( \because \quad \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0 \)
\( \implies \quad abc \left( \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \right) = 0 \)
\( \implies \quad \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 = 0 \quad [a, b, c \text{ are non-zero}] \)

Question. Using properties of determinants, show the following:
\( \begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3 \)
Answer: LHS \( = \begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix} \)
Multiplying \( R_1, R_2 \) and \( R_3 \) by \( a, b \) and \( c \) respectively, we get
\( = \frac{1}{abc} \begin{vmatrix} a(b+c)^2 & ba^2 & a^2c \\ ab^2 & b(a+c)^2 & b^2c \\ ac^2 & bc^2 & c(a+b)^2 \end{vmatrix} \)
\( = \frac{1}{abc} abc \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \quad \left[ \begin{matrix} \text{Taking common } a, b \text{ and } c \text{ from } C_1, C_2 \text{ and } C_3 \\ \text{respectively} \end{matrix} \right] \)
Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get
\( = \begin{vmatrix} (b+c)^2 - a^2 & 0 & a^2 \\ 0 & (c+a)^2 - b^2 & b^2 \\ c^2 - (a+b)^2 & c^2 - (a+b)^2 & (a+b)^2 \end{vmatrix} \)
\( = \begin{vmatrix} (b+c+a)(b+c-a) & 0 & a^2 \\ 0 & (c+a+b)(c+a-b) & b^2 \\ (c+a+b)(c-a-b) & (c+a+b)(c-a-b) & (a+b)^2 \end{vmatrix} \)
\( = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ c-a-b & c-a-b & (a+b)^2 \end{vmatrix} \quad \left[ \begin{matrix} \text{Taking common } (a+b+c) \\ \text{from } C_1 \text{ and } C_2 \end{matrix} \right] \)
\( = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ -2b & -2a & 2ab \end{vmatrix} \quad [R_3 \rightarrow R_3 - (R_1 + R_2)] \)
\( = \frac{(a+b+c)^2}{ab} \begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ 0 & bc+ba-b^2 & b^2 \\ -2ab & -2ab & 2ab \end{vmatrix} \quad [\text{Multiplying } a \text{ in } C_1 \text{ and } b \text{ in } C_2] \)
\( = \frac{(a+b+c)^2}{ab} \begin{vmatrix} ab+ac & a^2 & a^2 \\ b^2 & bc+ba & b^2 \\ 0 & 0 & 2ab \end{vmatrix} \quad [C_1 \rightarrow C_1 + C_3 \text{ and } C_2 \rightarrow C_2 + C_3] \)
\( = \frac{(a+b+c)^2}{ab} \cdot 2ab \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ 0 & 0 & 1 \end{vmatrix} \quad \left[ \begin{matrix} \text{Taking } a, b \text{ and } 2ab \text{ common} \\ \text{from } R_1, R_2 \text{ and } R_3 \text{ respectively} \end{matrix} \right] \)
\( = 2ab (a+b+c)^2 \begin{vmatrix} b+c & a \\ b & c+a \end{vmatrix} \)
\( = 2ab (a+b+c)^2 \{ (b+c)(c+a) - ab \} \)
\( = 2abc(a+b+c)^3 = \text{RHS} \)

Question. Using properties of determinant, show that:
\( \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc \)
Answer: LHS \( = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \)
\( = \begin{vmatrix} 2(b+c) & 2(c+a) & 2(a+b) \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = 2 \begin{vmatrix} (b+c) & (c+a) & (a+b) \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \quad \) [Taking 2 common from \( R_1 \)]
\( = 2 \begin{vmatrix} (b+c) & (c+a) & (a+b) \\ -c & 0 & -a \\ -b & -a & 0 \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = 2 \begin{vmatrix} 0 & c & b \\ -c & 0 & -a \\ -b & -a & 0 \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
Expanding along \( R_1 \), we get
\( = 2 [ 0 - c(0 - ab) + b(ac - 0) ] = 2[abc + abc] = 4abc = \text{RHS} \)

Question. Using properties of determinants, prove that
\( \begin{vmatrix} a^3 & 2 & a \\ b^3 & 2 & b \\ c^3 & 2 & c \end{vmatrix} = 2(a-b)(b-c)(c-a)(a+b+c). \)
Answer: LHS \( = \begin{vmatrix} a^3 & 2 & a \\ b^3 & 2 & b \\ c^3 & 2 & c \end{vmatrix} \)
\( = \begin{vmatrix} a^3 & 2 & a \\ b^3-a^3 & 0 & b-a \\ c^3-a^3 & 0 & c-a \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = (b-a)(c-a) \begin{vmatrix} a^3 & 2 & a \\ b^2+a^2+ab & 0 & 1 \\ c^2+a^2+ac & 0 & 1 \end{vmatrix} \quad \left[ \begin{matrix} \text{Taking common } (b-a) \\ \text{from } R_2 \text{ and } (c-a) \text{ from } R_3 \end{matrix} \right] \)
\( = (b-a)(c-a) \begin{vmatrix} a^3 & 2 & a \\ a^2+b^2+ab & 0 & 1 \\ c^2-b^2+ac-ab & 0 & 0 \end{vmatrix} \quad \) [Applying \( R_3 \rightarrow R_3 - R_2 \)]
Expanding along \( R_3 \), we get
\( = (b-a)(c-a)(c^2-b^2+ac-ab) 2 = 2(b-a)(c-a)(c-b)(c+b+a) \)
\( = 2(a-b)(b-c)(c-a)(a+b+c) = \text{RHS} \)

Question. Using properties of determinant, prove that:
\( \begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} = a^3 \)
Answer: LHS \( = \begin{vmatrix} a & a+b & a+b+c \\ 2a & 3a+2b & 4a+3b+2c \\ 3a & 6a+3b & 10a+6b+3c \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - 2R_1 \) and \( R_3 \rightarrow R_3 - 3R_1 \), we get
\( = \begin{vmatrix} a & a+b & a+b+c \\ 0 & a & 2a+b \\ 0 & 3a & 7a+3b \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = a[7a^2 + 3ab - 6a^2 - 3ab] \)
\( = a \times a^2 = a^3 = \text{RHS} \)

Question. Using properties of determinants, prove the following:
\( \begin{vmatrix} 1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} = (1-a^3)^2 \)
Answer: LHS \( = \begin{vmatrix} 1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \)
\( = \begin{vmatrix} 1+a^2+a & a+1+a^2 & a^2+a+1 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \quad \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = (1+a+a^2) \begin{vmatrix} 1 & 1 & 1 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \quad \) [Taking out \( (1+a+a^2) \) from first row]
\( = (1+a+a^2) \begin{vmatrix} 0 & 1 & 1 \\ a^2-1 & 1 & a \\ a-a^2 & a^2 & 1 \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 - C_2 \)]
\( = (1+a+a^2) \begin{vmatrix} 0 & 0 & 1 \\ a^2-1 & 1-a & a \\ a-a^2 & a^2-1 & 1 \end{vmatrix} \quad \) [Applying \( C_2 \rightarrow C_2 - C_3 \)]
Expanding along \( R_1 \), we have
\( = (1+a+a^2) [ (a^2-1)^2 - a(1-a)^2 ] = (1+a+a^2) [ (a+1)^2(a-1)^2 - a(a-1)^2 ] \)
\( = (a-1)^2 (1+a+a^2) [ (a+1)^2 - a ] = (1-a)^2 (1+a+a^2)^2 \)
\( = [ (1-a)(1+a+a^2) ]^2 = (1-a^3)^2 = \text{RHS} \)

Question. Prove that \( \begin{vmatrix} yz-x^2 & zx-y^2 & xy-z^2 \\ zx-y^2 & xy-z^2 & yz-x^2 \\ xy-z^2 & yz-x^2 & zx-y^2 \end{vmatrix} \) is divisible by \( (x+y+z) \), and hence find the quotient.
OR
Prove that \( \begin{vmatrix} bc-a^2 & ca-b^2 & ab-c^2 \\ ca-b^2 & ab-c^2 & bc-a^2 \\ ab-c^2 & bc-a^2 & ca-b^2 \end{vmatrix} \) is divisible by \( (a+b+c) \) and find the quotient.
Answer: We have
\( \Delta = \begin{vmatrix} yz-x^2 & zx-y^2 & xy-z^2 \\ zx-y^2 & xy-z^2 & yz-x^2 \\ xy-z^2 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( = \begin{vmatrix} xy+yz+zx-x^2-y^2-z^2 & zx-y^2 & xy-z^2 \\ xy+yz+zx-x^2-y^2-z^2 & xy-z^2 & yz-x^2 \\ xy+yz+zx-x^2-y^2-z^2 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Taking \( (xy+yz+zx-x^2-y^2-z^2) \) common from \( C_1 \), we get
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 1 & xy-z^2 & yz-x^2 \\ 1 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & xy-z^2-zx+y^2 & yz-x^2-xy+z^2 \\ 0 & yz-x^2-zx+y^2 & zx-y^2-xy+z^2 \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & x(y-z) + (y+z)(y-z) & y(z-x) + (z+x)(z-x) \\ 0 & z(y-x) + (y+x)(y-x) & x(z-y) + (z+y)(z-y) \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & (y-z)(x+y+z) & (z-x)(x+y+z) \\ 0 & (y-x)(x+y+z) & (z-y)(x+y+z) \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_2 \) and \( R_3 \), we get
\( = (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2 \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & y-z & z-x \\ 0 & y-x & z-y \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2 \{ 1 \cdot ((y-z)(z-y) - (y-x)(z-x)) \} \)
\( = (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2 \{ 1 \cdot (yz - y^2 - z^2 + zy - yz + xy + xz - x^2) \} \)
\( = (xy+yz+zx-x^2-y^2-z^2)^2 (x+y+z)^2 \)
Hence, \( \begin{vmatrix} yz-x^2 & zx-y^2 & xy-z^2 \\ zx-y^2 & xy-z^2 & yz-x^2 \\ xy-z^2 & yz-x^2 & zx-y^2 \end{vmatrix} \) is divisible by \( (x+y+z) \)
and quotient is \( (xy+yz+zx-x^2-y^2-z^2)^2 (x+y+z) \).
OR
Solve as above by putting \( a, b, c \) instead of \( x, y, z \).

Question. If \( a \neq b \neq c \) and \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \), then using properties of determinants, prove that \( a+b+c=0 \).
Answer: We have \( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = 0 \)
Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} (a+b+c) & b & c \\ (a+b+c) & c & a \\ (a+b+c) & a & b \end{vmatrix} = 0 \)
Taking \( (a+b+c) \) common from \( C_1 \), we get
\( (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} = 0 \)
Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \), we get
\( (a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} = 0 \)
Expanding along \( C_1 \), we get
\( (a+b+c) [ 1 \{ (c-b)(b-c) - (a-c)(a-b) \} - 0 + 0 ] = 0 \)
\( \implies \quad (a+b+c) [ (c-b)(b-c) - (a-c)(a-b) ] = 0 \)
\( \implies \quad (a+b+c) [ -(b-c)^2 - (a-c)(a-b) ] = 0 \)
\( \implies \quad (a+b+c) [ -(b^2+c^2-2bc) - (a^2-ab-ac+bc) ] = 0 \)
\( \implies \quad (a+b+c) [ -(b^2+c^2-2bc+a^2-ab-ac+bc) ] = 0 \)
\( \implies \quad (a+b+c) [ a^2+b^2+c^2-ab-bc-ca ] = 0 \)
\( \implies \quad (a+b+c) \frac{1}{2} [ 2a^2+2b^2+2c^2-2bc-2ab-2ac ] = 0 \)
\( \implies \quad (a+b+c) [ (a-b)^2+(b-c)^2+(c-a)^2 ] = 0 \)
\( \implies \quad (a+b+c) = 0 \quad [\because a \neq b \neq c \)
\( \implies \quad (a-b)^2+(b-c)^2+(c-a)^2 \neq 0] \)

Question. If \( a, b, c \) are real numbers, then prove that
\( \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = -(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \)
where \( \omega \) is a complex number and cube root of unity.
Answer: LHS \( = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \)
\( = \begin{vmatrix} a+b+c & b & c \\ b+c+a & c & a \\ c+a+b & a & b \end{vmatrix} \quad \) [Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \)]
\( = (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \quad \) [Taking out \( (a+b+c) \) from \( C_1 \)]
\( = (a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = (a+b+c) \begin{vmatrix} c-b & a-c \\ a-b & b-c \end{vmatrix} \quad \) [Expanding along \( C_1 \)]
\( = (a+b+c) \{ -(b-c)^2 - (a-c)(a-b) \} \)
\( = -(a+b+c) (a^2+b^2+c^2-ab-bc-ca) \) and
RHS \( = -(a+b+c) (a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \)
\( = -(a+b+c) (a^2 + ab\omega^2 + ac\omega + ab\omega + b^2\omega^3 + bc\omega^2 + ac\omega^2 + bc\omega^4 + c^2\omega^3) \)
\( = -(a+b+c) [ (a^2+b^2+c^2) + ab(\omega^2+\omega) + bc(\omega^2+\omega^4) + ca(\omega+\omega^2) ] \quad [\because \omega^3=1] \)
\( = -(a+b+c) (a^2+b^2+c^2-ab-bc-ca) = \text{LHS} \quad [\because \omega^2+\omega+1=0 \text{ and } \omega^4=\omega^3 \cdot \omega = \omega] \)

Question. Find the equation of the line joining \( A(1, 3) \) and \( B(0, 0) \) using determinants and find \( k \) if \( D(k, 0) \) is a point such that the area of \( \Delta ABD \) is 3 sq units.
Answer: Let \( P(x, y) \) be any point on the line \( AB \). Then,
\( ar(\Delta ABP) = 0 \)
\( \implies \quad \frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ x & y & 1 \end{vmatrix} = 0 \)
\( \implies \quad \frac{1}{2} \{ 1(0-y) - 3(0-x) + 1(0-0) \} = 0 \)
\( \implies \quad 3x - y = 0 \), which is the required equation of line \( AB \).
Now, area (\( \Delta ABD \)) = 3 sq units
\( \implies \quad \frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3 \)
\( \implies \quad \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 6 \)
\( \implies \quad 1(0-0) - 3(0-k) + 1(0-0) = \pm 6 \)
\( \implies \quad 3k = \pm 6 \)
\( \implies \quad k = \pm 2 \)

Question. Let \( f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\sin t & t & 2t \\ \sin t & t & t \end{vmatrix} \), then find \( \lim_{t \to 0} \frac{f(t)}{t^2} \).
Answer: Given, \( f(t) = \begin{vmatrix} \cos t & t & 1 \\ 2\sin t & t & 2t \\ \sin t & t & t \end{vmatrix} = \begin{vmatrix} \cos t & t & 1 \\ 0 & -t & 0 \\ \sin t & t & t \end{vmatrix} \quad \) [Applying \( R_2 \rightarrow R_2 - 2R_3 \)]
\( = t \begin{vmatrix} \cos t & 1 & 1 \\ 0 & -1 & 0 \\ \sin t & 1 & t \end{vmatrix} \)
Expanding along \( R_2 \), we get
\( t [ (-1)(t\cos t - \sin t) ] = -t^2\cos t + t\sin t \)
\( \therefore \quad \lim_{t \to 0} \frac{f(t)}{t^2} = \lim_{t \to 0} \frac{-t^2\cos t + t\sin t}{t^2} = \lim_{t \to 0} \left( \frac{-t^2\cos t}{t^2} + \frac{t\sin t}{t^2} \right) \)
\( = \lim_{t \to 0} \left( -\cos t + \frac{\sin t}{t} \right) = -1 + \lim_{t \to 0} \frac{\sin t}{t} = -1 + 1 = 0 \)