CBSE Class 12 Mathematics Determinants VBQs Set 02

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 \( = \begin{vmatrix} x+4 & 2x & 2x \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \)
\( = \begin{vmatrix} 5x+4 & 5x+4 & 5x+4 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = (5x+4) \begin{vmatrix} 1 & 1 & 1 \\ 2x & x+4 & 2x \\ 2x & 2x & x+4 \end{vmatrix} \) [Taking \( (5x+4) \) common from \( R_1 \)]
\( = (5x+4) \begin{vmatrix} 1 & 0 & 0 \\ 2x & 4-x & 0 \\ 2x & 0 & 4-x \end{vmatrix} \) [Applying \( C_2 \rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_1 \)]
\( = (5x+4) [1 \{(4-x)^2 - 0\} + 0 + 0] \) [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 \( = \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \)
\( = \begin{vmatrix} a+b & p+q & x+y \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} \) [Applying \( R_1 \leftrightarrow R_3 \) and \( R_3 \leftrightarrow R_2 \)]
Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \), we get
\( = \begin{vmatrix} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} = 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} \)
\( = 2 \begin{vmatrix} a & p & x \\ b+c & q+r & y+z \\ c+a & r+p & z+x \end{vmatrix} \) [Applying \( R_1 \rightarrow R_1 - R_2 \)]
\( = 2 \begin{vmatrix} a & p & x \\ b+c & q+r & y+z \\ c & r & z \end{vmatrix} \) [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 \( = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} \)
\( = abc \begin{vmatrix} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \) [Taking out \( a, b, c \) common from \( R_1, R_2 \) and \( R_3 \)]
\( = abc \begin{vmatrix} \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 & \frac{1}{a}+\frac{1}{b}+\frac{1}{c}+1 \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \) [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
\( \because \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = 0 \Rightarrow abc \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 \right) = 0 \Rightarrow \frac{1}{a} + \frac{1}{b} + \frac{1}{c} + 1 = 0 \) [\( a, b, c \) 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
\( = \frac{1}{abc} \begin{vmatrix} a(b+c)^2 & ba^2 & a^2c \\ ab^2 & b(a+c)^2 & b^2c \\ ac^2 & bc^2 & c(a+b)^2 \end{vmatrix} \)
\( = \frac{1}{abc} abc \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (a+c)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \) [Taking common \( a, b \) and \( c \) from \( C_1, C_2 \) and \( C_3 \) respectively]
Applying \( C_1 \rightarrow C_1 - C_3 \) and \( C_2 \rightarrow C_2 - C_3 \), we get
\( = \begin{vmatrix} (b+c)^2-a^2 & 0 & a^2 \\ 0 & (a+c)^2-b^2 & b^2 \\ c^2-(a+b)^2 & c^2-(a+b)^2 & (a+b)^2 \end{vmatrix} \)
\( = \begin{vmatrix} (b+c+a)(b+c-a) & 0 & a^2 \\ 0 & (a+c+b)(a+c-b) & b^2 \\ (c+a+b)(c-a-b) & (c+a+b)(c-a-b) & (a+b)^2 \end{vmatrix} \)
\( = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & a+c-b & b^2 \\ c-a-b & c-a-b & (a+b)^2 \end{vmatrix} \) [Taking common \( (a+b+c) \) from \( C_1 \) and \( C_2 \)]
\( = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & a+c-b & b^2 \\ -2b & -2a & 2ab \end{vmatrix} \) [\( R_3 \rightarrow R_3 - (R_1 + R_2) \)]
\( = \frac{(a+b+c)^2}{ab} \begin{vmatrix} ab+ac-a^2 & 0 & a^2 \\ 0 & bc+ba-b^2 & b^2 \\ -2ab & -2ab & 2ab \end{vmatrix} \) [Multiplying \( a \) in \( C_1 \) and \( b \) in \( C_2 \)]
\( = \frac{(a+b+c)^2}{ab} \begin{vmatrix} ab+ac & a^2 & a^2 \\ b^2 & bc+ba & b^2 \\ 0 & 0 & 2ab \end{vmatrix} \) [\( C_1 \rightarrow C_1 + C_3 \) and \( C_2 \rightarrow C_2 + C_3 \)]
\( = \frac{(a+b+c)^2}{ab} . ab . 2ab \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ 0 & 0 & 1 \end{vmatrix} \) [Taking \( a, b \) and \( 2ab \) common from \( R_1, R_2 \) and \( R_3 \) respectively]
\( = 2ab(a+b+c)^2 \begin{vmatrix} b+c & a \\ b & c+a \end{vmatrix} \)
\( = 2ab(a+b+c)^2 \{(b+c)(c+a) - ab\} \)
\( = 2abc(a+b+c)^3 = \text{RHS} \)

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 \( = \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \)
\( = \begin{vmatrix} 2(b+c) & 2(c+a) & 2(a+b) \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = 2 \begin{vmatrix} b+c & c+a & a+b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \) [Taking 2 common from \( R_1 \)]
\( = 2 \begin{vmatrix} c & 0 & a \\ -c & 0 & -a \\ a+b & b+c & c+a \end{vmatrix} \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = 2 \begin{vmatrix} 0 & c & b \\ -c & 0 & -a \\ -b & -a & 0 \end{vmatrix} \) [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 \( = \begin{vmatrix} a^3 & 2 & a \\ b^3 & 2 & b \\ c^3 & 2 & c \end{vmatrix} \)
\( = \begin{vmatrix} a^3 & 2 & a \\ b^3-a^3 & 0 & b-a \\ c^3-a^3 & 0 & c-a \end{vmatrix} \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = (b-a)(c-a) \begin{vmatrix} a^3 & 2 & a \\ b^2+a^2+ab & 0 & 1 \\ c^2+a^2+ac & 0 & 1 \end{vmatrix} \) [Taking common \( (b-a) \) from \( R_2 \) and \( (c-a) \) from \( R_3 \)]
\( = (b-a)(c-a) \begin{vmatrix} a^3 & 2 & a \\ a^2+b^2+ab & 0 & 1 \\ c^2-b^2+ac-ab & 0 & 0 \end{vmatrix} \) [Applying \( R_3 \rightarrow R_3 - R_2 \)]
Expanding along \( R_3 \), we get
\( = (b-a)(c-a)(c^2-b^2+ac-ab) \times 2 = 2 (b-a)(c-a)(c-b)(c+b+a) \)
\( = 2 (a-b)(b-c)(c-a)(a+b+c) = \text{RHS} \)

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 determinant, 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 \( = \begin{vmatrix} 1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \)
\( = \begin{vmatrix} 1+a+a^2 & a+1+a^2 & a^2+a+1 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \) [Applying \( R_1 \rightarrow R_1 + R_2 + R_3 \)]
\( = (1+a+a^2) \begin{vmatrix} 1 & 1 & 1 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix} \) [Taking out \( (1+a+a^2) \) from first row]
\( = (1+a+a^2) \begin{vmatrix} 0 & 1 & 1 \\ a^2-1 & 1 & a \\ a-a^2 & a^2 & 1 \end{vmatrix} \) [Applying \( C_1 \rightarrow C_1 - C_2 \)]
\( = (1+a+a^2) \begin{vmatrix} 0 & 0 & 1 \\ a^2-1 & 1-a & a \\ a-a^2 & a^2-1 & 1 \end{vmatrix} \) [Applying \( C_2 \rightarrow C_2 - C_3 \)]
Expanding along \( R_1 \), we have
\( = (1+a+a^2) [(a^2-1)^2 - a(1-a)^2] = (1+a+a^2) [(a+1)^2(a-1)^2 - a(a-1)^2] \)
\( = (1+a+a^2)(a-1)^2 [a^2+1+a] = (1+a+a^2)(a-1)^2 [a^2+1+a] \)
\( = (a-1)^2 (1+a+a^2)^2 = (1-a)^2 (1+a+a^2)^2 \)
\( = [(1-a)(1+a+a^2)]^2 = (1-a^3)^2 = \text{RHS} \)

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. 
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
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & xy-z^2-zx+y^2 & yz-x^2-xy+z^2 \\ 0 & yz-x^2-zx+y^2 & zx-y^2-xy+z^2 \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & x(y-z)+(y^2-z^2) & y(z-x)+(z^2-x^2) \\ 0 & z(y-x)+(y^2-x^2) & x(z-y)+(z^2-y^2) \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2) \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & (y-z)\cdot(x+y+z) & (z-x)\cdot(x+y+z) \\ 0 & (y-x)\cdot(x+y+z) & (z-y)\cdot(x+y+z) \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( R_2 \) and \( R_3 \), we get
\( = (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2 \begin{vmatrix} 1 & zx-y^2 & xy-z^2 \\ 0 & y-z & z-x \\ 0 & y-x & z-y \end{vmatrix} \)
\( = (xy+yz+zx-x^2-y^2-z^2)(x+y+z)^2 \{1 . (yz-y^2-z^2+zy-yz+xy+xz-x^2)\} \)
\( = (xy+yz+zx-x^2-y^2-z^2)^2 (x+y+z)^2 \)
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 \( (xy+yz+zx-x^2-y^2-z^2)^2 (x+y+z) \).

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 \( = \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix} \)
\( = \begin{vmatrix} a+b+c & b & c \\ a+b+c & c & a \\ a+b+c & a & b \end{vmatrix} \) [Applying \( C_1 \rightarrow C_1 + C_2 + C_3 \)]
\( = (a+b+c) \begin{vmatrix} 1 & b & c \\ 1 & c & a \\ 1 & a & b \end{vmatrix} \) [Taking out \( (a+b+c) \) from \( C_1 \)]
\( = (a+b+c) \begin{vmatrix} 1 & b & c \\ 0 & c-b & a-c \\ 0 & a-b & b-c \end{vmatrix} \) [Applying \( R_2 \rightarrow R_2 - R_1 \) and \( R_3 \rightarrow R_3 - R_1 \)]
\( = (a+b+c) \begin{vmatrix} c-b & a-c \\ a-b & b-c \end{vmatrix} \) [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 \( = -(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \)
\( = -(a+b+c)(a^2+ab\omega^2+ac\omega+ab\omega+b^2\omega^3+bc\omega^2+ac\omega^2+bc\omega^4+c^2\omega^3) \)
\( = -(a+b+c)[a^2+b^2+c^2+ab(\omega^2+\omega)+bc(\omega^2+\omega^4)+ca(\omega+\omega^2)] \) [\( \because \omega^3=1 \)]
\( = -(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = \text{LHS} \) [\( \because \omega^2+\omega+1=0 \) and \( \omega^4=\omega^3.\omega=\omega \)]

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

Question. 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: \( (3abc - a^3 - b^3 - c^3) \)

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: By applying \( C_1 \to C_1 + C_2 + C_3 \), we get \( (a + b + c) \) as a common factor in the first column. The remaining determinant is \( (a^2 + b^2 + c^2 - ab - bc - ca) \), which cannot be zero as \( a \neq b \neq c \). Thus, \( a + b + c = 0 \).

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 \( \triangle ABD \) is 3 sq units.
Answer: \( 3x - y = 0 \); \( k = \pm 2 \)

Question. Using properties of determinants, prove the following: \( \begin{vmatrix} 1 + x & 1 & 1 \\ 1 & 1 + y & 1 \\ 1 & 1 & 1 + z \end{vmatrix} = xyz + xy + yz + zx \) 
Answer: Divide \( R_1, R_2, R_3 \) by \( x, y, z \) respectively and take \( xyz \) common. The determinant then simplifies to \( 1 + \frac{1}{x} + \frac{1}{y} + \frac{1}{z} \), which equals \( xyz + xy + yz + zx \).

Question. Show that: \( \begin{vmatrix} 3x & -x + y & -x + z \\ x - y & 3y & z - y \\ x - z & y - z & 3z \end{vmatrix} = 3(x + y + z)(xy + yz + xz) \)
Answer: Use property \( C_1 \to C_1 + C_2 + C_3 \). Taking \( (x + y + z) \) common from \( C_1 \) and applying row operations leads to \( 3(x + y + z)(xy + yz + xz) \).

Question. Using properties of determinants, prove that : \( \begin{vmatrix} 1 & 1 & 1 + 3x \\ 1 + 3y & 1 & 1 \\ 1 & 1 + 3z & 1 \end{vmatrix} = 9(3xyz + xy + yz + zx) \) 
Answer: Applying \( C_1 \to C_1 - C_2 \) and \( C_3 \to C_3 - C_2 \), then expanding the determinant results in \( 9(3xyz + xy + yz + zx) \).

Question. Using properties of determinant, prove that: \( \begin{vmatrix} a^2 + 2a & 2a + 1 & 1 \\ 2a + 1 & a + 2 & 1 \\ 3 & 3 & 1 \end{vmatrix} = (a - 1)^3 \) 
Answer: Applying \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \) creates \( (a - 1) \) factors in the rows, leading to \( (a - 1)^3 \).

Question. If \( f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix} \), using properties of determinants, find the value of \( f(2x) - f(x) \). 
Answer: \( ax(2a + 3x) \)

Question. Using the properties of determinants, solve the following for \( x \) : \( \begin{vmatrix} x + 2 & x + 6 & x - 1 \\ x + 6 & x - 1 & x + 2 \\ x - 1 & x + 2 & x + 6 \end{vmatrix} = 0 \)  
Answer: \( x = -\frac{7}{3} \)

Question. Using the properties of determinants, prove the following: \( \begin{vmatrix} 1 & x & x + 1 \\ 2x & x(x - 1) & x(x + 1) \\ 3x(x - 1) & x(x - 1)(x - 2) & x(x + 1)(x - 1) \end{vmatrix} = 6x^2(1 - x^2) \) 
Answer: Taking common factors from columns and rows and applying property \( R_2 \to R_2 - 2xR_1 \) helps in evaluating to \( 6x^2(1 - x^2) \).

Question. If \( x, y, z \) are in GP, then using properties of determinants, show that \( \begin{vmatrix} px + y & x & y \\ py + z & y & z \\ 0 & px + y & py + z \end{vmatrix} = 0 \), where \( x \neq y \neq z \) and \( p \) is any real number. 
Answer: Since \( x, y, z \) are in GP, \( y^2 = xz \). Applying \( C_1 \to C_1 - (pC_2 + C_3) \), the first column becomes zero due to the GP property, making the determinant 0.

Question. Using properties of determinants, prove that \( \begin{vmatrix} 1 & 1 + p & 1 + p + q \\ 3 & 4 + 3p & 2 + 4p + 3q \\ 4 & 7 + 4p & 2 + 7p + 4q \end{vmatrix} = 1 \) 
Answer: Applying \( R_2 \to R_2 - 3R_1 \) and \( R_3 \to R_3 - 4R_1 \), then simplifying row 3, evaluates the determinant to 1.

Question. Without expanding the determinant at any stage, prove that \( \begin{vmatrix} 0 & 2 & -3 \\ -2 & 0 & 4 \\ 3 & -4 & 0 \end{vmatrix} = 0 \).
Answer: The determinant is of a skew-symmetric matrix of odd order (3), hence its value is 0.

Question. Using properties of determinants, prove that: \( \begin{vmatrix} (b + c)^2 & a^2 & bc \\ (c + a)^2 & b^2 & ca \\ (a + b)^2 & c^2 & ab \end{vmatrix} = (a - b)(b - c)(c - a)(a + b + c)(a^2 + b^2 + c^2) \)
Answer: Expanding \( (b+c)^2 = b^2 + c^2 + 2bc \) and applying properties like \( C_1 \to C_1 - C_2 - 2C_3 \) and row subtractions proves the identity.

Question. Using properties of determinants, prove the following: \( \begin{vmatrix} x & x + y & x + 2y \\ x + 2y & x & x + y \\ x + y & x + 2y & x \end{vmatrix} = 9y^2(x + y) \) 
Answer: Applying \( R_1 \to R_1 + R_2 + R_3 \) makes the first row \( 3(x + y) \). Then taking \( 3(x + y) \) common and applying row operations results in \( 9y^2(x + y) \).

Question. Using properties of determinants, prove that \( \begin{vmatrix} b + c & a & a \\ b & c + a & b \\ c & c & a + b \end{vmatrix} = 4abc \) 
Answer: Applying \( R_1 \to R_1 - (R_2 + R_3) \) results in a row with zeros and then expanding leads to \( 4abc \).

Choose and write the correct option in the following questions.

Question. If \( x, y \in R \), then the determinant \( \Delta = \begin{vmatrix} \cos x & -\sin x & 1 \\ \sin x & \cos x & 1 \\ \cos (x + y) & -\sin (x + y) & 0 \end{vmatrix} \) lies in the interval
(a) \( [-\sqrt{2}, \sqrt{2}] \)
(b) \( [-1, 1] \)
(c) \( [-\sqrt{2}, 1] \)
(d) \( [-1, -\sqrt{2}] \)
Answer: (a)

Question. The value of \( \begin{vmatrix} \sin 10^\circ & -\cos 10^\circ \\ \sin 80^\circ & \cos 80^\circ \end{vmatrix} \) is
(a) 0
(b) 1
(c) -1
(d) \( \frac{1}{2} \)
Answer: (b)

Question. The value(s) of \( k \) if area of triangle with vertices \( (-2, 0), (0, 4) \) and \( (0, k) \) is 4 sq. units is
(a) 0, 4
(b) -8
(c) 0, 8
(d) 0 only
Answer: (c)

Question. The value of \( x \) for which the matrix \( \begin{bmatrix} 5 - x & x + 1 \\ 2 & 4 \end{bmatrix} \) is singular, is
(a) 0
(b) 1
(c) 2
(d) 3
Answer: (d)