Consider the product \(\begin{vmatrix} a_1 &b_1 &c_1 \\ a_2 & b_2 &c_2 \\ a_3 & b_3 & c_3 \end{vmatrix}\)\(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\)
= \(\begin{vmatrix} a_1A_1+b_1B_1+c_1C_1 & a_1A_2+b_1B_2+c_1C_2& a_1A_3+b_1B_3+c_1C_3\\ a_2A_1+b_2B_1+c_2C_1 & a_2A_2+b_2B_2+c_2C_2 &a_2A_3+b_2B_3+c_2C_3 \\ a_3 A_1+b_3B_1+c_3C_1 & a_3A_2+b_3B_2+c_3C_2 &a_3A_3+b_3B_3+ c_3C_3 \end{vmatrix}\)
= \(\begin{vmatrix} |A| &0 &0 \\ 0 & |A| &0 \\ 0 & 0 & |A| \end{vmatrix}\) = |A|³.
That is, |A| \(\times\) \(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\) = |A|³
\(\Rightarrow\)\(\begin{vmatrix} A_1 &B_1 &C_1 \\ A_2 & B_2 &C_2 \\ A_3 & B_3 & C_3 \end{vmatrix}\) = |A|².

\(|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right|\)
Put x = Y
\(|A|=\left|\begin{array}{ccc} 1 & 1 & 1 \\ y & y & z \\ y^2 & y^2 & z^2 \end{array}\right|=0 \quad\left(C_1 \cong C_2\right)\)
\(\therefore(x-y) \text { is a factor of }|A|\)
The given determinant is in cyclic symmetric form in x, y and z.
\(\therefore\) (y - 2) and (z - x) are also factors.
The degree of the product of the factors (x - y)y - z)(z - x) is 3 and the degree of the product of the leading diagonal elements 1 - y , z² = 3
The other factor is k
\(\therefore\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{array}\right|=k(x-y)(y-z)(z-x)\)
Put x = 1, y = -1, z = 0
\(\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -1 & 0 \\ 1 & 1 & 0 \end{array}\right|=k(2)(-1)(-1)\)
1(1 + 1) = 2k
2 = 2k
k = 1
Put in (1)
\(|A|=(x-y)(y-z)(z-x)\)
Hence proved.

\(|A|=\left|\begin{array}{ccc} 4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x \end{array}\right|\)
Put x = 0
\(|A|=\left|\begin{array}{lll} 4 & 4 & 4 \\ 4 & 4 & 4 \\ 4 & 4 & 4 \end{array}\right|=0 \quad\left(C_1 \cong C_2 \cong C_3\right)\)
Since all the three rows are identical.
\(\therefore(x-0)^2=x^2 \text { is a factor of }|A|\)
\(\text { Applying } C_1 \rightarrow C_1+C_2+C_3\)
\(|A|=\left|\begin{array}{ccc} 12+x & 4+x & 4+x \\ 12+x & 4-x & 4+x \\ 12+x & 4+x & 4-x \end{array}\right|\)
Put x = -12
\(=\left|\begin{array}{lll} 0 & 4+x & 4+x \\ 0 & 4-x & 4+x \\ 0 & 4+x & 4-x \end{array}\right|=0\)
\(\therefore\) (x+12) is an other factor of lA|.
The number of roots must be 3
\(\therefore\) x= 0,0,-12

\(|A|=\left|\begin{array}{lll} b+c & a & a^2 \\ c+a & b & b^2 \\ a+b & c & c^2 \end{array}\right|\)
Put a = b
\(|A|=\left|\begin{array}{ccc} a+c & a & a^2 \\ c+a & a & a^2 \\ a+a & c & c^2 \end{array}\right|=0 \quad R_1 \cong R_2\)
\(\therefore\)(a - b) is a factor.
Since |A| is in cyclic.symmetric form in a, b, c.
\(\therefore\)(b-c) and (c - a) are also factors.
The degree of the product of the leading diagonal elements (b+c)¹.b¹.c² is 4.
m = 4 - 3 = 1
\(\therefore\)The other factor is k(a+b+c)
\(\therefore\)\(\left|\begin{array}{lll} b+c & a & a^2 \\ c+a & b & b^2 \\ a+b & c & c^2 \end{array}\right|=k(a+b+c)(a-b)(b-c)(c-a)\)
Put a = 0, b = 1, c = 2
\(\left|\begin{array}{lll} 3 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 2 & 4 \end{array}\right|=k(3)(-1)(-1)(2)\)
3(4 - 2) =  6k
6 = 6k
k = 1
Put in (1)
\(\therefore|A|=(a+b+c)(a-b)(b-c)(c-a)\)
Hence proved.

\(|A|=\left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|\)
Put x = 0
\(|A|=\left|\begin{array}{lll} a & b & c \\ a & b & c \\ a & b & c \end{array}\right|=0 \quad\left(C_1 \cong C_2 \cong C_3\right)\)
Since all the three rows are identical
\(\therefore\)(x-0)² = x² is a factor of |A|
Put x = a - b - c
\(|A|=\left|\begin{array}{ccc} -a-b-c+a & b & c \\ a & -a-b-c+b & c \\ a & b & -a-b-c+c \end{array}\right|\)
\(=\left|\begin{array}{ccc} -b-c & b & c \\ a & -a-c & c \\ a & b & -a-b \end{array}\right|\)
\(=\left|\begin{array}{ccc} 0 & b & c \\ 0 & -a-c & c \\ 0 & b & -a-b \end{array}\right|\)
\(\left[\because C_1 \rightarrow C_1+C_2+C_3\right]\)
\(\therefore[x+(a+b+c)] is \ a \ factor \ of |A| \\ The \ number\ of\ roots\ must\ be\ 3\)
\(\therefore\) x = 0, 0, (a + b + c)

\(|A|=\left|\begin{array}{lll} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c-a & a+b \end{array}\right|\)
Put a = 0
\(|A|=\left|\begin{array}{ccc} b+c & -c & -b \\ b-c & c & b \\ c-b & c & b \end{array}\right|=0 \quad\left(C_2 \cong C_3\right)\)
\(\therefore\) (a - 0) is a factor. (i.c) a is a factor.
Since |A| is in cyclic symmetric form in a, b, c and hence b, c also factors.
The degree of the product of the factor a, b, c is 3. The delerminant is a culbic polynomial.
The olher faclor must be a constant k.
\(\left|\begin{array}{ccc} b+c & a-c & a-b \\ b-c & c+a & b-a \\ c-b & c+a & a+b \end{array}\right|=k(a b c)\)
Put a = 1, b = 1, c = 1
\(\left|\begin{array}{lll} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{array}\right|=k(1)(1)(1)\)
8 = k
\((1) \Rightarrow \quad|A|=8 a b c\)
Hence proved.

By putting sin A = sin B, we get
\(\begin{vmatrix} 1& 1 &1 \\1+sin A &1+sin B &1+sin C \\ sinA(1+sin A) &sin B(1+sin B) &sin C(1+sin C) \end{vmatrix}=0\)
That is, by putting sin A = sin B we see that, the given equation is satisfied.
Similarly by putting sin B = sin C and sin C = sin A, the given equation is satisfied.
Thus, we have A = B or B = C or C = A.
In all cases atleast two angles are equal. Thus the triangle is isosceles.

Let |A| = \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) .
Putting x = y gives |A| = \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) = 0 (since R₁ \(\equiv\) R₂).
Therefore (x - y) is a factor.
The given determinant is in cyclic symmetric form in x, y and z. Therefore (y - z) and (z - x) are also factors.
The degree of the product of the factors (x − y)( y − z)(z − x) is 3 and the degree of the product of the leading diagonal elements 1× y² × z³ is 5.
Therefore the other factor is k(x² + y² + z² ) + l(xy + yz + zx) .
Thus \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\)=[k(x² + y² + z² ) + l(xy + yz + zx) ] × (x − y)( y − z)(z − x) .
Putting x = 0, y = 1 and z = 2, we get
\(\begin{vmatrix} 1 &0 &0 \\ 1 & 1 & 1 \\ 1 &4 &8 \end{vmatrix}\)= [k(0+1+4 ) +l(0+2+0) ](-1)( 1 - 2)(2 - 0)
\(\Rightarrow\) (8 - 4) = [(5k + 2l)](−1)(−1)(2)
4 = 10k + 4l \(\Rightarrow\) 5k + 2l = 2.
Putting x = 0, y = −1 and z =1, We get
\(\begin{vmatrix} 1 &0 &0 \\ 1 & 1 & -1 \\ 1 &4 &8 \end{vmatrix}\) = [k(2) + l(−1)](1)(−2)(1)
\(\Rightarrow\) [(2k − l)(−2)] = 2
2k − l = - 1.
Solving (1) and (2), we get k = 0, l =1.
Thus \(\begin{vmatrix} 1 &x^2 &x^3 \\ 1 & y^2 &y^3 \\1 &z^2 &z^3 \end{vmatrix}\) = (x - y)(y - z)(z - x)(xy + yz + zx).

\(|A|=\left|\begin{array}{lll} x & a & a \\ a & x & a \\ a & a & x \end{array}\right|\)
\(\text {Put } x=a,|A|=\left|\begin{array}{lll} a & a & a \\ a & a & a \\ a & a & a \end{array}\right|=0\)
Since all the three rows are identical
\(\therefore(x-a)^2 \text { is a factor of }|A|\)
Put x = -2a in |A|
\(|A|=\left|\begin{array}{ccc} -2 a & a & a \\ a & -2 a & a \\ a & a & -2 a \end{array}\right|\)
\(=\left|\begin{array}{ccc} 0 & a & a \\ 0 & -2 a & a \\ 0 & a & -2 a \end{array}\right| C_1 \rightarrow C_1+C_2+C_3\)
= 0
\(\therefore\) (x + 2a) is a factor of |4|.
The degree of the product of the factor (x - a)² (x + 2a) is 3.
The determinant is a cubic polynomial in x
\(\therefore\) The other factor must be a constant k.
\(\left|\begin{array}{lll} x & a & a \\ a & x & a \\ a & a & x \end{array}\right|=k(x-a)^2(x+2 a)\)
Equating x³ term on both sides, we get
1 = k
Put in (1)
\(|A|=(x-a)^2(x+2 a)\)
Hence proved.