CBSE Class 12 Mathematics HOTs Determinants Set 04

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: Given, \(f(x) = \begin{vmatrix} a & -1 & 0 \\ ax & a & -1 \\ ax^2 & ax & a \end{vmatrix}\)
Taking \(a\) common from \(C_1\), we get:
\(f(x) = a \begin{vmatrix} 1 & -1 & 0 \\ x & a & -1 \\ x^2 & ax & a \end{vmatrix}\)
Applying \(C_2 \rightarrow C_2 + C_1\), we get:
\(f(x) = a \begin{vmatrix} 1 & 0 & 0 \\ x & x+a & -1 \\ x^2 & x^2+ax & a \end{vmatrix}\)
\(\Rightarrow f(x) = a[a(x+a) + (x^2+ax)]\)
\(\Rightarrow f(x) = a(a^2 + ax + x^2 + ax) = a(a^2 + 2ax + x^2)\)
Now, \(f(2x) = a\{a^2 + 2a(2x) + (2x)^2\} = a(a^2 + 4ax + 4x^2)\)
\(\therefore f(2x) - f(x) = a(a^2 + 4ax + 4x^2 - a^2 - 2ax - x^2) = ax(3x + 2a)\).

Question. Using properties of determinants, prove the following: \(\begin{vmatrix} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix} = 4a^2b^2c^2\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} a^2 & bc & ac+c^2 \\ a^2+ab & b^2 & ac \\ ab & b^2+bc & c^2 \end{vmatrix}\)
Taking \(a, b, c\) common from \(C_1, C_2, C_3\) respectively, we get:
\(abc \begin{vmatrix} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{vmatrix}\)
Applying \(C_3 \rightarrow C_3 - C_1 - C_2\), we get:
\(abc \begin{vmatrix} a & c & 0 \\ a+b & b & -2b \\ b & b+c & -2b \end{vmatrix}\)
Applying \(R_2 \rightarrow R_2 - R_3\), we get:
\(abc \begin{vmatrix} a & c & 0 \\ a & -c & 0 \\ b & b+c & -2b \end{vmatrix}\)
\(= abc(-2b)(-ac - ac) = -2abc^2(-2ac) = 4a^2b^2c^2 = \text{R.H.S.}\)

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: \(\text{L.H.S.} = \begin{vmatrix} 1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1 \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\), we get:
\(\begin{vmatrix} 1+a+a^2 & a & a^2 \\ 1+a+a^2 & 1 & a \\ 1+a+a^2 & a^2 & 1 \end{vmatrix}\)
Taking \((1+a+a^2)\) common from \(C_1\), we get:
\((1+a+a^2) \begin{vmatrix} 1 & a & a^2 \\ 1 & 1 & a \\ 1 & a^2 & 1 \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 - C_2\), we get:
\((1+a+a^2) \begin{vmatrix} 1-a & a & a^2 \\ 0 & 1 & a \\ 1-a^2 & a^2 & 1 \end{vmatrix}\)
Taking \((1-a)\) common from \(C_1\), we get:
\((1+a+a^2)(1-a) \begin{vmatrix} 1 & a & a^2 \\ 0 & 1 & a \\ 1+a & a^2 & 1 \end{vmatrix}\)
\(= (1-a^3) [ 1(1 - a^3) + (1+a)(a^2 - a^2) ]\)
\(= (1-a^3)(1-a^3) = (1-a^3)^2 = \text{R.H.S.}\)

Question. Using properties of determinants, prove that \(\begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} = -2\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} (a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1 \end{vmatrix} = \begin{vmatrix} a^2+3a+2 & a+2 & 1 \\ a^2+5a+6 & a+3 & 1 \\ a^2+7a+12 & a+4 & 1 \end{vmatrix}\)
Applying \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_2\), we get:
\(\begin{vmatrix} a^2+3a+2 & a+2 & 1 \\ 2a+4 & 1 & 0 \\ 2a+6 & 1 & 0 \end{vmatrix}\)
Applying \(R_3 \rightarrow R_3 - R_2\), we get:
\(\begin{vmatrix} a^2+3a+2 & a+2 & 1 \\ 2a+4 & 1 & 0 \\ 2 & 0 & 0 \end{vmatrix}\)
Expanding along \(C_3\), we get:
\(1(0 - 2) = -2 = \text{R.H.S.}\)

Question. Using properties of determinants, 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: Given, \(\begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\), we get:
\(\begin{vmatrix} 3a-x & a-x & a-x \\ 3a-x & a+x & a-x \\ 3a-x & a-x & a+x \end{vmatrix} = 0\)
Applying \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\), we get:
\(\begin{vmatrix} 3a-x & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0\)
\(\Rightarrow (3a-x)(2x)^2 = 0\)
\(\Rightarrow x = 0, 3a\).

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: \(\text{L.H.S.} = \begin{vmatrix} 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix}\)
Applying \(R_1 \rightarrow R_1 + R_2 + R_3\), we get:
\(\begin{vmatrix} x+y+z & x+y+z & x+y+z \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix}\)
Taking \((x+y+z)\) common from \(R_1\), we get:
\((x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ 2z & 2z & z-x-y \\ x-y-z & 2x & 2x \end{vmatrix}\)
Applying \(C_2 \rightarrow C_2 - C_1\) and \(C_3 \rightarrow C_3 - C_2\), we get:
\((x+y+z) \begin{vmatrix} 1 & 0 & 0 \\ 2z & 0 & -(x+y+z) \\ x-y-z & x+y+z & 0 \end{vmatrix}\)
\(= (x+y+z) \cdot 1 \cdot [0 - (-(x+y+z))(x+y+z)] = (x+y+z)^3 = \text{R.H.S.}\)

Question. Prove the following using properties of determinants: \(\begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\) and 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:
\(2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & b+c+a & 0 \\ 0 & 0 & c+a+b \end{vmatrix}\)
\(= 2(a+b+c) \cdot 1 \cdot (b+c+a) \cdot (c+a+b) = 2(a+b+c)^3 = \text{R.H.S.}\)

Question. Using properties of determinants, prove the following: \(\begin{vmatrix} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{vmatrix} = 1+x^2+y^2+z^2\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} x^2+1 & xy & xz \\ xy & y^2+1 & yz \\ xz & yz & z^2+1 \end{vmatrix}\)
Multiplying and dividing the determinant by \(xyz\), and applying \(R_1 \rightarrow \frac{1}{x}R_1\), \(R_2 \rightarrow \frac{1}{y}R_2\), \(R_3 \rightarrow \frac{1}{z}R_3\), we get:
\(\frac{xyz}{xyz} \begin{vmatrix} x+\frac{1}{x} & y & z \\ x & y+\frac{1}{y} & z \\ x & y & z+\frac{1}{z} \end{vmatrix}\)
Now, applying \(C_1 \rightarrow xC_1\), \(C_2 \rightarrow yC_2\), \(C_3 \rightarrow zC_3\), we get:
\(\begin{vmatrix} x^2+1 & y^2 & z^2 \\ x^2 & y^2+1 & z^2 \\ x^2 & y^2 & z^2+1 \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\), we get:
\(\begin{vmatrix} x^2+y^2+z^2+1 & y^2 & z^2 \\ x^2+y^2+z^2+1 & y^2+1 & z^2 \\ x^2+y^2+z^2+1 & y^2 & z^2+1 \end{vmatrix}\)
Taking \((x^2+y^2+z^2+1)\) common from \(C_1\), we get:
\((x^2+y^2+z^2+1) \begin{vmatrix} 1 & y^2 & z^2 \\ 1 & y^2+1 & z^2 \\ 1 & y^2 & z^2+1 \end{vmatrix}\)
Applying \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\), we get:
\((1+x^2+y^2+z^2) \begin{vmatrix} 1 & y^2 & z^2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}\)
\(= (1+x^2+y^2+z^2)(1)(1-0) = 1+x^2+y^2+z^2 = \text{R.H.S.}\)

Question. Using properties of determinants, prove the following: \(\begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = ab+bc+ca+abc\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix}\)
Applying \(R_2 \rightarrow R_2 - R_1\) and \(R_3 \rightarrow R_3 - R_1\), we get:
\(\begin{vmatrix} 1+a & 1 & 1 \\ -a & b & 0 \\ -a & 0 & c \end{vmatrix}\)
Expanding along \(R_3\), we get:
\(-a(0 - b) + c[b(1+a) + a] = ab + bc + abc + ca = ab + bc + ca + abc = \text{R.H.S.}\)

Question. Using properties of determinants, prove that \(\begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix} = 2\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix}\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 + C_2 + C_3\) and taking \(2\) common from \(C_1\), we get:
\(2 \begin{vmatrix} a+b+c & c+a & a+b \\ p+q+r & r+p & p+q \\ x+y+z & z+x & x+y \end{vmatrix}\)
Applying \(C_2 \rightarrow C_2 - C_1\) and \(C_3 \rightarrow C_3 - C_1\), we get:
\(2 \begin{vmatrix} a+b+c & -b & -c \\ p+q+r & -q & -r \\ x+y+z & -y & -z \end{vmatrix} = 2(-1)(-1) \begin{vmatrix} a+b+c & b & c \\ p+q+r & q & r \\ x+y+z & y & z \end{vmatrix}\)
Applying \(C_1 \rightarrow C_1 - C_2 - C_3\), we get:
\(2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} = \text{R.H.S.}\)

Question. Using properties of determinants, show that \(\begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3\).
Answer: \(\text{L.H.S.} = \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix}\)
Since each element in the first column of the determinant is the sum of two elements, the determinant can be expressed as the sum of two determinants:
\(\begin{vmatrix} x & x & x \\ 5x & 4x & 2x \\ 10x & 8x & 3x \end{vmatrix} + \begin{vmatrix} y & x & x \\ 4y & 4x & 2x \\ 8y & 8x & 3x \end{vmatrix}\)
Taking \(x\) common from \(R_1, R_2, R_3\) in the first determinant, and \(x\) common from \(C_2, C_3\) and \(y\) common from \(C_1\) in the second determinant, we get:
\(x^3 \begin{vmatrix} 1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3 \end{vmatrix} + yx^2 \begin{vmatrix} 1 & 1 & 1 \\ 4 & 4 & 2 \\ 8 & 8 & 3 \end{vmatrix}\)
\(= x^3 \begin{vmatrix} 1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3 \end{vmatrix} + yx^2 \cdot 0\) [since \(C_1\) and \(C_2\) are identical in the second determinant]
Applying \(C_1 \rightarrow C_1 - C_3\) and \(C_2 \rightarrow C_2 - C_3\) in the remaining determinant, we get:
\(x^3 \begin{vmatrix} 0 & 0 & 1 \\ 3 & 2 & 2 \\ 7 & 5 & 3 \end{vmatrix} = x^3 \cdot 1(15 - 14) = x^3 = \text{R.H.S.}\)

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

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

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

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} = (a-b)(b-c)(c-a)(bc+ca+ab) \]
Answer: L.H.S. = \( \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} \)
Applying \( R_1 \to aR_1, R_2 \to bR_2, R_3 \to cR_3 \), we get
\( = \frac{1}{abc} \begin{vmatrix} a^2 & a^3 & abc \\ b^2 & b^3 & abc \\ c^2 & c^3 & abc \end{vmatrix} \)
Taking \( abc \) common from \( C_3 \), we get
\( = \frac{abc}{abc} \begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix} = \begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get
\( = \begin{vmatrix} a^2-b^2 & a^3-b^3 & 0 \\ b^2-c^2 & b^3-c^3 & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
\( = \begin{vmatrix} (a-b)(a+b) & (a-b)(a^2+ab+b^2) & 0 \\ (b-c)(b+c) & (b-c)(b^2+bc+c^2) & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Taking \( (a-b) \) common from \( R_1 \) and \( (b-c) \) common from \( R_2 \), we get
\( = (a-b)(b-c) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ b+c & b^2+bc+c^2 & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1 \), we get
\( = (a-b)(b-c) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ c-a & (c-a)(b+c+a) & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Taking \( (c-a) \) common from \( R_2 \), we get
\( = (a-b)(b-c)(c-a) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ 1 & a+b+c & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Expanding along \( C_3 \), we get
\( = (a-b)(b-c)(c-a) [ 1 \cdot \{ (a+b)(a+b+c) - (a^2+ab+b^2) \} ] \)
\( = (a-b)(b-c)(c-a) [ a^2+ab+ac+ba+b^2+bc - a^2-ab-b^2 ] \)
\( = (a-b)(b-c)(c-a) (bc+ca+ab) = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc \]
Answer: L.H.S. = \( \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_2 - R_3 \), we get
\( \begin{vmatrix} 0 & -2c & -2b \\ b & c+a & b \\ c & c & a+b \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( = 0 - (-2c) \begin{vmatrix} b & b \\ c & a+b \end{vmatrix} + (-2b) \begin{vmatrix} b & c+a \\ c & c \end{vmatrix} \)
\( = 2c [ b(a+b) - bc ] - 2b [ bc - c(c+a) ] \)
\( = 2c [ ab+b^2-bc ] - 2b [ bc-c^2-ac ] \)
\( = 2abc + 2b^2c - 2bc^2 - 2bc^2 + 2bc^2 + 2abc \)
\( = 4abc = \text{R.H.S.} \)

Question. Show that \( \Delta = \Delta_1 \), where \[ \Delta = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix}, \Delta_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix} \]
Answer: We have \( \Delta_1 = \begin{vmatrix} A & B & C \\ x & y & z \\ zy & zx & xy \end{vmatrix} \)
Multiplying \( C_1, C_2 \) and \( C_3 \) by \( x, y \) and \( z \) respectively, we get
\( \Delta_1 = \frac{1}{xyz} \begin{vmatrix} Ax & By & Cz \\ x^2 & y^2 & z^2 \\ xyz & xyz & xyz \end{vmatrix} \)
Taking \( xyz \) common from \( R_3 \), we get
\( \Delta_1 = \frac{xyz}{xyz} \begin{vmatrix} Ax & By & Cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} = \begin{vmatrix} Ax & By & Cz \\ x^2 & y^2 & z^2 \\ 1 & 1 & 1 \end{vmatrix} \)
Interchanging corresponding rows and columns (taking transpose), we get
\( \Delta_1 = \begin{vmatrix} Ax & x^2 & 1 \\ By & y^2 & 1 \\ Cz & z^2 & 1 \end{vmatrix} = \Delta \)

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: L.H.S. = \( \begin{vmatrix} x & x+y & x+2y \\ x+2y & x & x+y \\ x+y & x+2y & x \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} 3x+3y & x+y & x+2y \\ 3x+3y & x & x+y \\ 3x+3y & x+2y & x \end{vmatrix} \)
Taking \( (3x+3y) \) common from \( C_1 \), we get
\( 3(x+y) \begin{vmatrix} 1 & x+y & x+2y \\ 1 & x & x+y \\ 1 & x+2y & x \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_2 \), we get
\( 3(x+y) \begin{vmatrix} 1 & x+y & x+2y \\ 0 & -y & -y \\ 0 & 2y & -y \end{vmatrix} \)
Taking \( y \) common from \( R_2 \) and \( R_3 \) both, we get
\( 3y^2(x+y) \begin{vmatrix} 1 & x+y & x+2y \\ 0 & -1 & -1 \\ 0 & 2 & -1 \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = 3y^2(x+y) [ 1(1 - (-2)) ] \)
\( = 3y^2(x+y)(3) = 9y^2(x+y) = \text{R.H.S.} \)

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{vmatrix} = 3(x+y+z)(xy+yz+zx) \]
Answer: L.H.S. = \( \begin{vmatrix} 3x & -x+y & -x+z \\ x-y & 3y & z-y \\ x-z & y-z & 3z \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 + C_3 \), we get
\( \begin{vmatrix} x+y+z & -x+y & -x+z \\ x+y+z & 3y & z-y \\ x+y+z & y-z & 3z \end{vmatrix} \)
Taking \( (x+y+z) \) common from \( C_1 \), we get
\( (x+y+z) \begin{vmatrix} 1 & -x+y & -x+z \\ 1 & 3y & z-y \\ 1 & y-z & 3z \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get
\( (x+y+z) \begin{vmatrix} 1 & -x+y & -x+z \\ 0 & 2y+x & x-y \\ 0 & x-z & 2z+x \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = (x+y+z) [ (2y+x)(2z+x) - (x-z)(x-y) ] \)
\( = (x+y+z) [ (4yz + 2xy + 2zx + x^2) - (x^2 - xy - zx + yz) ] \)
\( = (x+y+z) (3xy + 3yz + 3zx) \)
\( = 3(x+y+z)(xy+yz+zx) = \text{R.H.S.} \)

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

Question. Using properties of determinants, prove that \[ \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ bc & ca & ab \end{vmatrix} = (a-b)(b-c)(c-a)(ab+bc+ca) \]
Answer: Let \( \Delta = \begin{vmatrix} a & b & c \\ a^2 & b^2 & c^2 \\ bc & ca & ab \end{vmatrix} \)
Transposing the determinant (which does not change its value):
\( \Delta = \begin{vmatrix} a & a^2 & bc \\ b & b^2 & ca \\ c & c^2 & ab \end{vmatrix} \)
Applying \( R_1 \to aR_1, R_2 \to bR_2, R_3 \to cR_3 \), we get
\( = \frac{1}{abc} \begin{vmatrix} a^2 & a^3 & abc \\ b^2 & b^3 & abc \\ c^2 & c^3 & abc \end{vmatrix} \)
Taking \( abc \) common from \( C_3 \), we get
\( = \begin{vmatrix} a^2 & a^3 & 1 \\ b^2 & b^3 & 1 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Applying \( R_1 \to R_1 - R_2, R_2 \to R_2 - R_3 \), we get
\( = \begin{vmatrix} a^2-b^2 & a^3-b^3 & 0 \\ b^2-c^2 & b^3-c^3 & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Taking \( (a-b) \) common from \( R_1 \) and \( (b-c) \) common from \( R_2 \), we get
\( = (a-b)(b-c) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ b+c & b^2+bc+c^2 & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1 \), we get
\( = (a-b)(b-c) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ c-a & (c-a)(b+c+a) & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Taking \( (c-a) \) common from \( R_2 \), we get
\( = (a-b)(b-c)(c-a) \begin{vmatrix} a+b & a^2+ab+b^2 & 0 \\ 1 & a+b+c & 0 \\ c^2 & c^3 & 1 \end{vmatrix} \)
Expanding along \( C_3 \), we get
\( = (a-b)(b-c)(c-a) [ (a+b)(a+b+c) - (a^2+ab+b^2) ] \)
\( = (a-b)(b-c)(c-a) (bc+ca+ab) = \text{R.H.S.} \)

Question. Using properties of determinants, prove that \[ \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} = 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} \]
Answer: L.H.S. = \( \begin{vmatrix} b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \)
Applying \( R_1 \to R_1 + R_2 + R_3 \), we get
\( \begin{vmatrix} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \)
Taking \( 2 \) common from \( R_1 \), we get
\( 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get
\( 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ -b & -q & -y \\ -c & -r & -z \end{vmatrix} \)
Taking \( -1 \) common from both \( R_2 \) and \( R_3 \), we get
\( 2 \begin{vmatrix} a+b+c & p+q+r & x+y+z \\ b & q & y \\ c & r & z \end{vmatrix} \)
Applying \( R_1 \to R_1 - (R_2 + R_3) \), we get
\( 2 \begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = \text{R.H.S.} \)

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

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix} = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) \]
Answer: L.H.S. = \( \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix} \)
Applying \( R_3 \to R_3 + R_1 \), we get
\( \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma \end{vmatrix} \)
Taking \( (\alpha+\beta+\gamma) \) common from \( R_3 \), we get
\( (\alpha+\beta+\gamma) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix} \)
Applying \( C_1 \to C_1 - C_2, C_2 \to C_2 - C_3 \), we get
\( (\alpha+\beta+\gamma) \begin{vmatrix} \alpha-\beta & \beta-\gamma & \gamma \\ \alpha^2-\beta^2 & \beta^2-\gamma^2 & \gamma^2 \\ 0 & 0 & 1 \end{vmatrix} \)
Taking \( (\alpha-\beta) \) and \( (\beta-\gamma) \) common from \( C_1 \) and \( C_2 \) respectively, we get
\( (\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma) \begin{vmatrix} 1 & 1 & \gamma \\ \alpha+\beta & \beta+\gamma & \gamma^2 \\ 0 & 0 & 1 \end{vmatrix} \)
Expanding along \( R_3 \), we get
\( = (\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma) [ 1 \cdot \{ (\beta+\gamma) - (\alpha+\beta) \} ] \)
\( = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) = \text{R.H.S.} \)

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

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} = 3abc-a^3-b^3-c^3 \]
Answer: L.H.S. = \( \begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix} \)
Transposing the determinant, we get
\( = \begin{vmatrix} b+c & c+a & a+b \\ a-b & b-c & c-a \\ a & b & c \end{vmatrix} \)
Interchanging \( R_1 \leftrightarrow R_3 \), we get
\( = - \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} \)
From the previous result, we know that \( \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = a^3+b^3+c^3-3abc \).
Therefore, \( = -(a^3+b^3+c^3-3abc) = 3abc-a^3-b^3-c^3 = \text{R.H.S.} \)

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

Question. Using properties of determinants, prove the following : \[ \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} = (a+b+c)(a^2+b^2+c^2) \]
Answer: L.H.S. = \( \begin{vmatrix} a & b-c & c+b \\ a+c & b & c-a \\ a-b & b+a & c \end{vmatrix} \)
Applying \( C_1 \to aC_1 \), we get
\( \frac{1}{a} \begin{vmatrix} a^2 & b-c & c+b \\ a^2+ac & b & c-a \\ a^2-ab & b+a & c \end{vmatrix} \)
Applying \( C_1 \to C_1 + bC_2 + cC_3 \), we get
\( \frac{1}{a} \begin{vmatrix} a^2+b^2+c^2 & b-c & c+b \\ a^2+b^2+c^2 & b & c-a \\ a^2+b^2+c^2 & b+a & c \end{vmatrix} \)
Taking \( (a^2+b^2+c^2) \) common from \( C_1 \), we get
\( \frac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 1 & b & c-a \\ 1 & b+a & c \end{vmatrix} \)
Applying \( R_2 \to R_2 - R_1, R_3 \to R_3 - R_1 \), we get
\( \frac{a^2+b^2+c^2}{a} \begin{vmatrix} 1 & b-c & c+b \\ 0 & c & -a-b \\ 0 & a & a \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = \frac{a^2+b^2+c^2}{a} \cdot 1[ ac - a(-a-b) ] \)
\( = \frac{a^2+b^2+c^2}{a} [ ac + a^2 + ab ] \)
\( = (a^2+b^2+c^2)(a+b+c) = \text{R.H.S.} \)

Question. Using properties of determinants, prove that \[ \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} = 4a^2b^2c^2 \]
Answer: L.H.S. = \( \begin{vmatrix} -a^2 & ab & ac \\ ba & -b^2 & bc \\ ca & cb & -c^2 \end{vmatrix} \)
Taking \( a, b, c \) common from \( R_1, R_2 \) and \( R_3 \), respectively, we get
\( abc \begin{vmatrix} -a & b & c \\ a & -b & c \\ a & b & -c \end{vmatrix} \)
Taking \( a, b, c \) common from \( C_1, C_2 \) and \( C_3 \), respectively, we get
\( a^2b^2c^2 \begin{vmatrix} -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & -1 \end{vmatrix} \)
Applying \( C_1 \to C_1 + C_2 \), we get
\( a^2b^2c^2 \begin{vmatrix} 0 & 1 & 1 \\ 0 & -1 & 1 \\ 2 & 1 & -1 \end{vmatrix} \)
Expanding along \( C_1 \), we get
\( = a^2b^2c^2 \cdot 2(1 - (-1)) \)
\( = 4a^2b^2c^2 = \text{R.H.S.} \)

Question. Using properties of determinants, prove that \[ \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x) \]
Answer: L.H.S. = \( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} \)
Taking \( x, y, z \) common from \( C_1, C_2, C_3 \), respectively, we get
\( xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} \)
Applying \( C_2 \to C_2 - C_1, C_3 \to C_3 - C_1 \), we get
\( xyz \begin{vmatrix} 1 & 0 & 0 \\ x & y-x & z-x \\ x^2 & y^2-x^2 & z^2-x^2 \end{vmatrix} \)
Taking \( (y-x) \) and \( (z-x) \) common from \( C_2 \) and \( C_3 \), respectively, we get
\( xyz(y-x)(z-x) \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ x^2 & y+x & z+x \end{vmatrix} \)
Expanding along \( R_1 \), we get
\( = xyz(y-x)(z-x) [ 1(z+x - (y+x)) ] \)
\( = xyz(y-x)(z-x)(z-y) \)
\( = xyz(x-y)(y-z)(z-x) = \text{R.H.S.} \)