CBSE Class 12 Mathematics HOTs Determinants Set 05

Question. If \( A = \begin{bmatrix} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{bmatrix} \), then write the cofactor of the element \( a_{21} \) of its \( 2^{\text{nd}} \) row.
Answer: We have, \( A = \begin{bmatrix} 5 & 6 & -3 \\ -4 & 3 & 2 \\ -4 & -7 & 3 \end{bmatrix} \)
\( \therefore \) Cofactor of \( a_{21} = (-1)^{2+1} \begin{vmatrix} 6 & -3 \\ -7 & 3 \end{vmatrix} = -1 (18 - 21) = 3 \).

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: Let \( \Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \)
Now, \( a_{32} = 5 \)
\( A_{32} = \) cofactor of \( a_{32} \) in \( \Delta = (-1)^{3+2} \begin{vmatrix} 2 & 5 \\ 6 & 4 \end{vmatrix} = - (8 - 30) = 22 \)
\( \therefore a_{32} \cdot A_{32} = 5 \cdot 22 = 110. \)

Question. If \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \), write the minor of the element \( a_{23} \).
Answer: Here, \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \)
\( \therefore \) Minor of \( a_{23} = \begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = 10 - 3 = 7 \).

Question. If \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \), write the cofactor of the element \( a_{32} \).
Answer: Here, \( \Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix} \)
\( \therefore \) Cofactor of \( a_{32} = (-1)^{3+2} \begin{vmatrix} 5 & 8 \\ 2 & 1 \end{vmatrix} = -(5 - 16) = 11 \).

Question. If \( \Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 5 & 3 & 8 \end{vmatrix} \), write the minor of element \( a_{22} \).
Answer: We have, \( \Delta = \begin{vmatrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 5 & 3 & 8 \end{vmatrix} \)
\( \dots \) Minor of \( a_{22} = \begin{vmatrix} 1 & 3 \\ 5 & 8 \end{vmatrix} = 8 - 15 = -7 \).

Question. Find the minor of the element of second row and third column (\( a_{23} \)) in the determinant \( \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \).
Answer: Let \( \Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \)
\( \therefore \) Minor of \( a_{23} = \begin{vmatrix} 2 & -3 \\ 1 & 5 \end{vmatrix} = 10 + 3 = 13 \).

Question. Find the co-factor of \( a_{12} \) in the following: \( \begin{bmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{bmatrix} \).
Answer: Let \( \Delta = \begin{vmatrix} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{vmatrix} \)
\( \therefore \) Cofactor of \( a_{12} = (-1)^{1+2} \begin{vmatrix} 6 & 4 \\ 1 & -7 \end{vmatrix} = -(-42 - 4) = 46 \).

Question. In the interval \( \pi/2 < x < \pi \), find the value of \( x \) for which the matrix \( \begin{bmatrix} 2 \sin x & 3 \\ 1 & 2 \sin x \end{bmatrix} \) is singular.
Answer: For the matrix \( \begin{bmatrix} 2 \sin x & 3 \\ 1 & 2 \sin x \end{bmatrix} \) to be singular, its determinant = 0
\( \therefore \begin{vmatrix} 2 \sin x & 3 \\ 1 & 2 \sin x \end{vmatrix} = 0 \)
\( \Rightarrow 4 \sin^2 x - 3 = 0 \Rightarrow \sin^2 x = \frac{3}{4} \Rightarrow \sin x = \pm \frac{\sqrt{3}}{2} \)
\( \therefore x = \frac{2\pi}{3} \quad \left[ \because \frac{\pi}{2} < x < \pi \right] \).

Question. Using properties of determinants, prove that \[ \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} = (5x+4)(4-x)^2 \]
Answer: L.H.S. = \( \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} 5x+4 & 2x & 2x \\ 5x+4 & x+4 & 2x \\ 5x+4 & 2x & x+4 \end{vmatrix} \)
Taking \( (5x+4) \) common from \( C_1 \), we get
\( (5x+4) \begin{vmatrix} 1 & 2x & 2x \\ 1 & x+4 & 2x \\ 1 & 2x & x+4 \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get
\( (5x+4) \begin{vmatrix} 1 & 2x & 2x \\ 0 & 4-x & 0 \\ 0 & 0 & 4-x \end{vmatrix} \)
Taking \( (4-x) \) common from \( R_2 \) and \( R_3 \) both, we get
\( (5x+4)(4-x)^2 \begin{vmatrix} 1 & 2x & 2x \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix} \)
\( = (5x+4)(4-x)^2 = \text{R.H.S.} \)

Question. Using properties of determinants, solve the following for \( x \): \[ \begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0 \]
Answer: Given, \( \begin{vmatrix} x-2 & 2x-3 & 3x-4 \\ x-4 & 2x-9 & 3x-16 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0 \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get
\( \begin{vmatrix} 2 & 6 & 12 \\ 4 & 18 & 48 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0 \)
Taking 2 common from both \( R_1 \) and \( R_2 \), we get
\( 4 \begin{vmatrix} 1 & 3 & 6 \\ 2 & 9 & 24 \\ x-8 & 2x-27 & 3x-64 \end{vmatrix} = 0 \)
Applying \( C_2 \to C_2 - 2C_1 \) and \( C_3 \to C_3 - 3C_1 \), we get
\( 4 \begin{vmatrix} 1 & 1 & 3 \\ 2 & 5 & 18 \\ x-8 & -11 & -40 \end{vmatrix} = 0 \)
Applying \( R_2 \to R_2 - 2R_1 \), we get
\( 4 \begin{vmatrix} 1 & 1 & 3 \\ 0 & 3 & 12 \\ x-8 & -11 & -40 \end{vmatrix} = 0 \)
Expanding along \( C_1 \), we get
\( \implies 4 [ 1 \cdot (3(-40) - 12(-11)) + (x-8)(12 - 9) ] = 0 \)
\( \implies 4 [ (-120 + 132) + 3(x-8) ] = 0 \)
\( \implies 4 [ 12 + 3x - 24 ] = 0 \)
\( \implies 3x - 12 = 0 \implies x = 4. \)

Question. Using properties of determinants, solve the following for \( x \): \[ \begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix} = 0, \quad a \neq 0 \]
Answer: Given, \( \begin{vmatrix} x+a & x & x \\ x & x+a & x \\ x & x & x+a \end{vmatrix} = 0 \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} 3x+a & x & x \\ 3x+a & x+a & x \\ 3x+a & x & x+a \end{vmatrix} = 0 \)
Taking \( (3x+a) \) common from \( C_1 \), we get
\( (3x+a) \begin{vmatrix} 1 & x & x \\ 1 & x+a & x \\ 1 & x & x+a \end{vmatrix} = 0 \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get
\( (3x+a) \begin{vmatrix} 1 & x & x \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} = 0 \)
\( \implies a^2(3x+a) = 0 \)
Since \( a \neq 0 \), we have:
\( 3x+a = 0 \implies x = -\frac{a}{3}. \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} = 1+a^2+b^2+c^2 \]
Answer: L.H.S. = \( \begin{vmatrix} a^2+1 & ab & ac \\ ab & b^2+1 & bc \\ ca & cb & c^2+1 \end{vmatrix} \)
Multiplying \( R_1, R_2, R_3 \) by \( a, b, c \) respectively, and dividing by \( abc \), we get:
\( \frac{1}{abc} \begin{vmatrix} a(a^2+1) & a^2b & a^2c \\ ab^2 & b(b^2+1) & b^2c \\ c^2a & c^2b & c(c^2+1) \end{vmatrix} \)
Taking out common factors \( a, b, c \) from \( C_1, C_2, C_3 \) respectively, we get:
\( = \begin{vmatrix} a^2+1 & a^2 & a^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Applying \( R_1 \to R_1 + R_2 + R_3 \), we get:
\( \begin{vmatrix} 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 & 1+a^2+b^2+c^2 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Taking \( (1+a^2+b^2+c^2) \) common from \( R_1 \), we get:
\( (1+a^2+b^2+c^2) \begin{vmatrix} 1 & 1 & 1 \\ b^2 & b^2+1 & b^2 \\ c^2 & c^2 & c^2+1 \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( (1+a^2+b^2+c^2) \begin{vmatrix} 1 & 0 & 0 \\ b^2 & 1 & 0 \\ c^2 & 0 & 1 \end{vmatrix} \)
Expanding along \( R_1 \), we get:
\( = (1+a^2+b^2+c^2)[1(1-0)] = 1+a^2+b^2+c^2 = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3 \]
Answer: L.H.S. = \( \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( \begin{vmatrix} (b+c)^2 & (a-b-c)(a+b+c) & (a-b-c)(a+b+c) \\ b^2 & (c+a-b)(a+b+c) & 0 \\ c^2 & 0 & (a+b-c)(a+b+c) \end{vmatrix} \)
Taking \( (a+b+c) \) common from \( C_2 \) and \( C_3 \), we get:
\( (a+b+c)^2 \begin{vmatrix} (b+c)^2 & a-b-c & a-b-c \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \)
Applying \( R_1 \to R_1 - (R_2 + R_3) \), we get:
\( (a+b+c)^2 \begin{vmatrix} 2bc & -2c & -2b \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \)
Applying \( C_2 \to C_2 + \frac{1}{b} C_1, C_3 \to C_3 + \frac{1}{c} C_1 \), we get:
\( (a+b+c)^2 \begin{vmatrix} 2bc & 0 & 0 \\ b^2 & a+c & b^2/c \\ c^2 & c^2/b & a+b \end{vmatrix} \)
Expanding along \( R_1 \), we get:
\( = (a+b+c)^2 \cdot 2bc \left[ (a+c)(a+b) - \frac{b^2}{c} \cdot \frac{c^2}{b} \right] \)
\( = 2bc(a+b+c)^2 [ a^2+ab+ac+bc - bc ] \)
\( = 2abc(a+b+c)^3 = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3 \]
Answer: L.H.S. = \( \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} \)
Applying \( C_1 \to C_1 - bC_3, C_2 \to C_2 + aC_3 \), 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 \( (1+a^2+b^2) \) common from \( C_1 \) and \( C_2 \) both, 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 \( R_3 \to R_3 - bR_1 \), we get:
\( (1+a^2+b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & -a & 1-a^2+b^2 \end{vmatrix} \)
Applying \( R_3 \to R_3 + aR_2 \), we get:
\( (1+a^2+b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ 0 & 0 & 1+a^2+b^2 \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( = (1+a^2+b^2)^2 \cdot 1[ 1(1+a^2+b^2) ] = (1+a^2+b^2)^3 = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix} = 1 \]
Answer: L.H.S. = \( \begin{vmatrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 1+3p+2q \\ 3 & 6+3p & 1+6p+3q \end{vmatrix} \)
Applying \( R_2 \to R_2 - 2R_1, R_3 \to R_3 - 3R_1 \), we get:
\( \begin{vmatrix} 1 & 1+p & 1+p+q \\ 0 & 1 & -1+p \\ 0 & 3 & -2+3p \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( = 1 [ 1(-2+3p) - 3(-1+p) ] \)
\( = -2+3p + 3 - 3p = 1 = \text{R.H.S.} \)

Question. If \( x \neq y \neq z \) and \[ \begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0, \] then show that \( xyz = -1 \).
Answer: Given determinant is equal to \( 0 \). Expressing it as sum of two determinants:
\[ \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = 0 \]
Taking \( x, y, z \) common from \( R_1, R_2, R_3 \) respectively in the second determinant, we get:
\[ \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \]
Interchanging \( C_2 \leftrightarrow C_3 \) then \( C_1 \leftrightarrow C_2 \) in the first determinant, we get:
\[ \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} (1 + xyz) = 0 \]
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get:
\[ (1 + xyz) \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix} = 0 \]
Taking \( (y-x) \) and \( (z-x) \) common, we get:
\[ (1+xyz)(y-x)(z-x)(z-y) = 0 \]
Since \( x \neq y \neq z \), we have \( (x-y) \neq 0, (y-z) \neq 0, (z-x) \neq 0 \).
Thus, \( 1 + xyz = 0 \implies xyz = -1 \).

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3 \]
Answer: L.H.S. = \( \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \)
Applying \( R_1 \to R_1 + R_2 + R_3 \), 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 \( R_1 \), we get:
\( (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} \)
Expanding along \( R_1 \), we get:
\( = (a+b+c) [ 1 \cdot (-(a+b+c))(-(a+b+c)) ] = (a+b+c)^3 = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} = (x-y)(y-z)(z-x)(xy+yz+zx) \]
Answer: By replacing \( a, b, c \) with \( x, y, z \) respectively in the proof of Question 45, we get:
L.H.S. = \( \begin{vmatrix} x & x^2 & yz \\ y & y^2 & zx \\ z & z^2 & xy \end{vmatrix} \)
Applying \( R_1 \to xR_1, R_2 \to yR_2, R_3 \to zR_3 \), we obtain:
\( = \frac{1}{xyz} \begin{vmatrix} x^2 & x^3 & xyz \\ y^2 & y^3 & xyz \\ z^2 & z^3 & xyz \end{vmatrix} \)
Taking \( xyz \) common from \( C_3 \), we get:
\( = \begin{vmatrix} x^2 & x^3 & 1 \\ y^2 & y^3 & 1 \\ z^2 & z^3 & 1 \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get:
\( = \begin{vmatrix} x^2-y^2 & x^3-y^3 & 0 \\ y^2-z^2 & y^3-z^3 & 0 \\ z^2 & z^3 & 1 \end{vmatrix} \)
Taking \( (x-y) \) common from \( R_1 \) and \( (y-z) \) common from \( R_2 \), we get:
\( = (x-y)(y-z) \begin{vmatrix} x+y & x^2+xy+y^2 & 0 \\ y+z & y^2+yz+z^2 & 0 \\ z^2 & z^3 & 1 \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1 \), we get:
\( = (x-y)(y-z) \begin{vmatrix} x+y & x^2+xy+y^2 & 0 \\ z-x & (z-x)(x+y+z) & 0 \\ z^2 & z^3 & 1 \end{vmatrix} \)
Taking \( (z-x) \) common from \( R_2 \), we get:
\( = (x-y)(y-z)(z-x) \begin{vmatrix} x+y & x^2+xy+y^2 & 0 \\ 1 & x+y+z & 0 \\ z^2 & z^3 & 1 \end{vmatrix} \)
Expanding along \( C_3 \), we get:
\( = (x-y)(y-z)(z-x) [ (x+y)(x+y+z) - (x^2+xy+y^2) ] \)
\( = (x-y)(y-z)(z-x) (xy+yz+zx) = \text{R.H.S.} \)

Question. Using properties of determinants, show that \[ \begin{vmatrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{vmatrix} = (x+y+z)(x-z)^2 \]
Answer: L.H.S. = \( \begin{vmatrix} y+z & x & y \\ z+x & z & x \\ x+y & y & z \end{vmatrix} \)
Applying \( R_1 \to R_1 + R_2 + R_3 \), we get:
\( \begin{vmatrix} 2(x+y+z) & x+y+z & x+y+z \\ z+x & z & x \\ x+y & y & z \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_1 \), we get:
\( (x+y+z) \begin{vmatrix} 2 & 1 & 1 \\ z+x & z & x \\ x+y & y & z \end{vmatrix} \)
Applying \( C_1 \to C_1 - 2C_3, C_2 \to C_2 - C_3 \), we get:
\( (x+y+z) \begin{vmatrix} 0 & 0 & 1 \\ z-x & z-x & x \\ x+y-2z & y-z & z \end{vmatrix} \)
Expanding along \( R_1 \), we get:
\( = (x+y+z) \cdot 1[ (z-x)(y-z) - (z-x)(x+y-2z) ] \)
\( = (x+y+z)(z-x) [ y-z - (x+y-2z) ] \)
\( = (x+y+z)(z-x)(z-x) = (x+y+z)(x-z)^2 = \text{R.H.S.} \)

Question. Prove, using the properties of determinants \[ \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Answer: L.H.S. = \( \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:
\( \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{vmatrix} \)
Taking \( 2 \) common from \( C_1 \), we get:
\( 2 \begin{vmatrix} a+b+c & c+a & a+b \\ a+b+c & a+b & b+c \\ a+b+c & b+c & c+a \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( 2 \begin{vmatrix} a+b+c & -b & -c \\ a+b+c & -c & -a \\ a+b+c & -a & -b \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:
\( 2 \begin{vmatrix} a & -b & -c \\ b & -c & -a \\ c & -a & -b \end{vmatrix} \)
Taking \( -1 \) common from \( C_2 \) and \( C_3 \), we get:
\( 2 \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} = \text{R.H.S.} \)

Question. Prove the following, using properties of determinants. \[ \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{vmatrix} = (x-y)(y-z)(z-x)(x+y+z) \]
Answer: By replacing \( \alpha, \beta, \gamma \) with \( x, y, z \) respectively in the proof of Question 54, we get:
L.H.S. = \( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ y+z & z+x & x+y \end{vmatrix} \)
Applying \( R_3 \to R_3 + R_1 \), we get:
\( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x+y+z & x+y+z & x+y+z \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_3 \), we get:
\( (x+y+z) \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} \)
Applying \( C_1 \to C_1 - C_2, C_2 \to C_2 - C_3 \), we get:
\( (x+y+z) \begin{vmatrix} x-y & y-z & z \\ x^2-y^2 & y^2-z^2 & z^2 \\ 0 & 0 & 1 \end{vmatrix} \)
Taking \( (x-y) \) common from \( C_1 \) and \( (y-z) \) common from \( C_2 \), we get:
\( (x+y+z)(x-y)(y-z) \begin{vmatrix} 1 & 1 & z \\ x+y & y+z & z^2 \\ 0 & 0 & 1 \end{vmatrix} \)
Expanding along \( R_3 \), we get:
\( = (x-y)(y-z)(z-x)(x+y+z) = \text{R.H.S.} \)

Question. Using properties of determinants, prove that \[ \begin{vmatrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a-1)^3 \]
Answer: L.H.S. = \( \begin{vmatrix} a^2+2a & 2a+1 & 1 \\ 2a+1 & a+2 & 1 \\ 3 & 3 & 1 \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get:
\( \begin{vmatrix} a^2-1 & a-1 & 0 \\ 2(a-1) & a-1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \)
Taking \( (a-1) \) common from \( R_1 \) and \( R_2 \), we get:
\( (a-1)^2 \begin{vmatrix} a+1 & 1 & 0 \\ 2 & 1 & 0 \\ 3 & 3 & 1 \end{vmatrix} \)
Expanding along \( C_3 \), we get:
\( = (a-1)^2 [ 1(a+1 - 2) ] = (a-1)^3 = \text{R.H.S.} \)

Question. Prove that \[ \begin{vmatrix} 1 & x & x^3 \\ 1 & y & y^3 \\ 1 & z & z^3 \end{vmatrix} = (x-y)(y-z)(z-x)(x+y+z) \]
Answer: Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get:
\( \begin{vmatrix} 0 & x-y & x^3-y^3 \\ 0 & y-z & y^3-z^3 \\ 1 & z & z^3 \end{vmatrix} \)
Taking \( (x-y) \) and \( (y-z) \) common from \( R_1 \) and \( R_2 \) respectively, we get:
\( (x-y)(y-z) \begin{vmatrix} 0 & 1 & x^2+xy+y^2 \\ 0 & 1 & y^2+yz+z^2 \\ 1 & z & z^3 \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( = (x-y)(y-z) [ 1 \cdot \{ (y^2+yz+z^2) - (x^2+xy+y^2) \} ] \)
\( = (x-y)(y-z) [ (z^2-x^2) + y(z-x) ] \)
\( = (x-y)(y-z) [ (z-x)(z+x) + y(z-x) ] \)
\( = (x-y)(y-z)(z-x)(x+y+z) = \text{R.H.S.} \)

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.
Answer: Let \( \Delta = \begin{vmatrix} yz-x^2 & zx-y^2 & xy-z^2 \\ zx-y^2 & xy-z^2 & yz-x^2 \\ xy-z^2 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:
\( \Delta = \begin{vmatrix} -(x^2+y^2+z^2-xy-yz-zx) & zx-y^2 & xy-z^2 \\ -(x^2+y^2+z^2-xy-yz-zx) & xy-z^2 & yz-x^2 \\ -(x^2+y^2+z^2-xy-yz-zx) & yz-x^2 & zx-y^2 \end{vmatrix} \)
Taking \( -(x^2+y^2+z^2-xy-yz-zx) \) common from \( C_1 \), we get:
\( \Delta = -(x^2+y^2+z^2-xy-yz-zx) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 1 & xy-z^2 & yz-x^2 \\ 1 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_3, R_2 \to R_2 - R_3 \), we get:
\( \Delta = -(x^2+y^2+z^2-xy-yz-zx) \begin{vmatrix} 0 & (x-y)(x+y+z) & (y-z)(x+y+z) \\ 0 & (x-z)(x+y+z) & (y-x)(x+y+z) \\ 1 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_1 \) and \( R_2 \), we get:
\( \Delta = -(x^2+y^2+z^2-xy-yz-zx)(x+y+z)^2 \begin{vmatrix} 0 & x-y & y-z \\ 0 & x-z & y-x \\ 1 & yz-x^2 & zx-y^2 \end{vmatrix} \)
\( = -(x+y+z)(x^3+y^3+z^3-3xyz) \begin{vmatrix} 0 & x-y & y-z \\ 0 & x-z & y-x \\ 1 & yz-x^2 & zx-y^2 \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( \Delta = -(x+y+z)(x^3+y^3+z^3-3xyz) [ (x-y)(y-x) - (x-z)(y-z) ] \)
\( = -(x+y+z)(x^3+y^3+z^3-3xyz) [ -(x^2+y^2+z^2-xy-yz-zx) ] \)
\( = (x+y+z)(x^3+y^3+z^3-3xyz)(x^2+y^2+z^2-xy-yz-zx) \)
Hence, \( \Delta \) is divisible by \( (x+y+z) \) and the quotient is:
\( (x^3+y^3+z^3-3xyz)(x^2+y^2+z^2-xy-yz-zx) \).

Question. Using properties of determinants, show that \( \Delta ABC \) is 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: Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( \begin{vmatrix} 1 & 0 & 0 \\ 1+\cos A & \cos B - \cos A & \cos C - \cos A \\ \cos^2 A + \cos A & (\cos B - \cos A)(\cos B + \cos A + 1) & (\cos C - \cos A)(\cos C + \cos A + 1) \end{vmatrix} = 0 \)
Taking \( (\cos B - \cos A) \) common from \( C_2 \) and \( (\cos C - \cos A) \) common from \( C_3 \), we get:
\( (\cos B - \cos A)(\cos C - \cos A) \begin{vmatrix} 1 & 0 & 0 \\ 1+\cos A & 1 & 1 \\ \cos^2 A + \cos A & \cos B + \cos A + 1 & \cos C + \cos A + 1 \end{vmatrix} = 0 \)
Expanding along \( R_1 \), we get:
\( \implies (\cos B - \cos A)(\cos C - \cos A) [ (\cos C + \cos A + 1) - (\cos B + \cos A + 1) ] = 0 \)
\( \implies (\cos B - \cos A)(\cos C - \cos A)(\cos C - \cos B) = 0 \)
\( \implies \cos B = \cos A \), or \( \cos C = \cos A \), or \( \cos C = \cos B \)
\( \implies B = A \), or \( C = A \), or \( C = B \).
Thus, \( \Delta ABC \) is an isosceles triangle.

Question. 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: Given, \( \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0 \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get:
\( \begin{vmatrix} a & -b & 0 \\ 0 & b & -c \\ 1 & 1 & 1+c \end{vmatrix} = 0 \)
Expanding along \( R_1 \), we get:
\( \implies a[ b(1+c) - (-c)(1) ] - (-b)[ 0(1+c) - (-c)(1) ] = 0 \)
\( \implies a(b + bc + c) + bc = 0 \)
\( \implies ab + abc + ac + bc = 0 \)
Dividing both sides by \( abc \), we get:
\( \frac{1}{c} + 1 + \frac{1}{b} + \frac{1}{a} = 0 \implies \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 = 0 \).

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: L.H.S. = \( \begin{vmatrix} (b+c)^2 & ab & ca \\ ab & (a+c)^2 & bc \\ ac & bc & (a+b)^2 \end{vmatrix} \)
Applying \( R_1 \to aR_1, R_2 \to bR_2, R_3 \to cR_3 \), we get:
\( \frac{1}{abc} \begin{vmatrix} a(b+c)^2 & a^2b & a^2c \\ ab^2 & b(a+c)^2 & b^2c \\ ac^2 & bc^2 & c(a+b)^2 \end{vmatrix} \)
Taking \( a, b, c \) common from \( C_1, C_2, C_3 \) respectively, we get:
\( = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get:
\( \begin{vmatrix} (b+c)^2 & (a-b-c)(a+b+c) & (a-b-c)(a+b+c) \\ b^2 & (c+a-b)(a+b+c) & 0 \\ c^2 & 0 & (a+b-c)(a+b+c) \end{vmatrix} \)
Taking \( (a+b+c) \) common from \( C_2 \) and \( C_3 \), we get:
\( (a+b+c)^2 \begin{vmatrix} (b+c)^2 & a-b-c & a-b-c \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \)
Applying \( R_1 \to R_1 - (R_2 + R_3) \), we get:
\( (a+b+c)^2 \begin{vmatrix} 2bc & -2c & -2b \\ b^2 & c+a-b & 0 \\ c^2 & 0 & a+b-c \end{vmatrix} \)
Applying \( C_2 \to C_2 + \frac{1}{b} C_1, C_3 \to C_3 + \frac{1}{c} C_1 \), we get:
\( (a+b+c)^2 \begin{vmatrix} 2bc & 0 & 0 \\ b^2 & a+c & b^2/c \\ c^2 & c^2/b & a+b \end{vmatrix} \)
Expanding along \( R_1 \), we get:
\( = (a+b+c)^2 \cdot 2bc \left[ (a+c)(a+b) - \frac{b^2}{c} \cdot \frac{c^2}{b} \right] \)
\( = 2bc(a+b+c)^2 [ a^2+ab+ac+bc - bc ] \)
\( = 2abc(a+b+c)^3 = \text{R.H.S.} \)

Question. If \( a, b, c \) are positive and unequal, show that the following determinant is negative. \[ \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \]
Answer: We have \( \Delta = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get:
\( \Delta = \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} = (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get:
\( = (a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( = (a+b+c) [ (c-b)(b-c) - (a-b)(a-c) ] \)
\( = -(a+b+c) [ a^2+b^2+c^2 - ab - bc - ca ] \)
\( = -\frac{1}{2}(a+b+c)[ (a-b)^2 + (b-c)^2 + (c-a)^2 ] \)
Since \( a, b, c \) are positive and unequal, the term \( (a-b)^2 + (b-c)^2 + (c-a)^2 \) is strictly positive, and \( (a+b+c) \) is positive. Thus, \( \Delta \) is negative.

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} x & x^2 & 1+px^3 \\ y & y^2 & 1+py^3 \\ z & z^2 & 1+pz^3 \end{vmatrix} = (1+pxyz)(x-y)(y-z)(z-x) \]
Answer: L.H.S. = \( \begin{vmatrix} x & x^2 & 1+px^3 \\ y & y^2 & 1+py^3 \\ z & z^2 & 1+pz^3 \end{vmatrix} \)
Expressing the determinant as the sum of two determinants:
\( = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & px^3 \\ y & y^2 & py^3 \\ z & z^2 & pz^3 \end{vmatrix} \)
Taking \( p \) common from \( C_3 \), \( x \) from \( R_1 \), \( y \) from \( R_2 \), and \( z \) from \( R_3 \) in the second determinant, we get:
\( = \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + pxyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
Applying \( C_2 \leftrightarrow C_3 \) then \( C_1 \leftrightarrow C_2 \) in the first determinant, we get:
\( = (1+pxyz) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get:
\( = (1+pxyz) \begin{vmatrix} 1 & x & x^2 \\ 0 & y-x & y^2-x^2 \\ 0 & z-x & z^2-x^2 \end{vmatrix} \)
Taking \( (y-x) \) and \( (z-x) \) common, we get:
\( = (1+pxyz)(y-x)(z-x) \begin{vmatrix} 1 & x & x^2 \\ 0 & 1 & y+x \\ 0 & 1 & z+x \end{vmatrix} \)
Expanding along \( C_1 \), we get:
\( = (1+pxyz)(y-x)(z-x)(z-y) \)
\( = (1+pxyz)(x-y)(y-z)(z-x) = \text{R.H.S.} \)

Question. Find the equation of the line joining \( A(1, 3) \) and \( B(0, 0) \) using determinants and find the value of \( k \) if \( D(k, 0) \) is a point such that area of \( \Delta ABD \) is 3 square units.
Answer: The equation of the line joining \( A(1, 3) \) and \( B(0, 0) \) is:
\( \begin{vmatrix} x & y & 1 \\ 1 & 3 & 1 \\ 0 & 0 & 1 \end{vmatrix} = 0 \)
\( \implies 1(3x - y) = 0 \implies y = 3x \)
Now, the area of \( \Delta ABD \) is 3 square units with vertices \( A(1, 3), B(0, 0), D(k, 0) \):
\( \frac{1}{2} \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 3 \)
\( \implies \begin{vmatrix} 1 & 3 & 1 \\ 0 & 0 & 1 \\ k & 0 & 1 \end{vmatrix} = \pm 6 \)
Expanding along \( C_2 \), we get:
\( -3(0 - k) = \pm 6 \implies 3k = \pm 6 \implies k = \pm 2. \)