\(\text { LHS }=\left|\begin{array}{ccc} a^2 & b c & a c+c^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right|\)
Take a, b, c from C₁ C₂, C₃ respectively
\(=a b c\left|\begin{array}{ccc} a & c & a+c \\ a+b & b & a \\ b & b+c & c \end{array}\right|\)
\(\text { Applying } C_1 \rightarrow C_1+C_2-C_3\)
\(=a b c\left|\begin{array}{ccc} 0 & c & a+c \\ 2 b & b & a \\ 2 b & b+c & c \end{array}\right|\)
\(\text { Applying } R_2 \rightarrow R_2-R_3\)
\(=a b c\left|\begin{array}{ccc} 0 & c & a+c \\ 0 & -c & a-c \\ 2 b & b+c & c \end{array}\right|\)
\(=a b c[2 b[c(a-c)+c(a+c)]]\)
\(=2 a b^2 c\left[a c-c^2+a c+c^2\right]\)
\(=2 a b^2 c(2 a c)=4 a^2 b^2 c^2=\text { RHS }\)
Hence Proved.

Since all the entries below the principal diagonal are zero, the value of the determinant is (x - 1) (x - 2) (x - 3) = 0 which gives x = 1, 2, 3.

|A| = [3(−1)(6) + 4(2)(5) + 1(0)(−2)] −[5(−1)(1) + (−2)(2)3 + 6(0)(4)]
= [−18 + 40 + 0]−[−5 −12 + 0] = 22 + 17 = 39.

The orders of the two matrices are same. Thus by comparing the corresponding elements, we get
\(4 x=x+6\)
\(4 x-x=6\)
\(3 x=6\)
\(x=2\)
\(2 x+y=7\)
Substituting  x = 2
\(2(2)+y=7\)
\(4+y=7\)
\(y=7-4\)
\(y=3\)
\(\therefore x+y=2+3\)
\(x+y=5\)

A² = AA =\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)\(\begin{bmatrix} 0 &c &b \\ c & 0 &a \\ b & a & 0 \end{bmatrix}\)=\(\begin{bmatrix} C_{11} &C_{12} &C_{13} \\ C_{21} & C_{22} &C_{23} \\ C_{31} & C_{32} & C_{33} \end{bmatrix}\)

Let \(\triangle =\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 2 & 3+2p & 4+4p+2q \\ 3 & 6+3p & 10+6p+3q \end{matrix} \right| \)
Applying C₂ ⟶ C₂ - pC₁ and C₃ ⟶ C₃ - qC₁, we get
\(\triangle =\left| \begin{matrix} 1 & 1 & 1+p \\ 2 & 3 & 4+3p \\ 3 & 6 & 10+6p \end{matrix} \right| \)
Applying C₃ ⟶ C₃ - pC₂, we get
= \(\left| \begin{matrix} 1 & 1 & 1 \\ 2 & 3 & 4 \\ 3 & 6 & 10 \end{matrix} \right| \)
Applying C₂ ⟶ C₂ - Cland C₃ ⟶ C₃ - C, we get,
\(\triangle =\left| \begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 3 & 3 & 7 \end{matrix} \right| \)
Expanding along |Rl we get,
\(\triangle =1\left| \begin{matrix} 1 & 2 \\ 3 & 7 \end{matrix} \right| \)+ 0 + = 1(7-6) = 1
\(\therefore \triangle =1\)

The matrix A is singular if |A| = 0
\(\left| \begin{matrix} 1 & -2 & 3 \\ 1 & 2 & 1 \\ x & 2 & -3 \end{matrix} \right| =0\)
\(1\left| \begin{matrix} 2 & 1 \\ 2 & -3 \end{matrix} \right| +2\left| \begin{matrix} 1 & 1 \\ x & -3 \end{matrix} \right| +3\left| \begin{matrix} 1 & 2 \\ x & 2 \end{matrix} \right| =0\)
\(\Rightarrow\)(-6 - 2) + 2(-3 - x) + 3(2 - 2x) = 0
\(\Rightarrow\) 
\(\Rightarrow\) -8x = 8 => x = -1

Let A be a symmetric matrix and \(n\in N\)
Then, An A.A.A ..........A upto n-times
\(\Rightarrow\) (Aⁿ)^T [A.A........... A upto n-times]T
\(\Rightarrow\)(Aⁿ)^T A^T.A^T.A^T A^Tupto n-times
\(\Rightarrow\) (Aⁿ)^T = (A^T)ⁿ = Aⁿ [\(\therefore\)  A^T = A]
\(\Rightarrow\) (Aⁿ)^T Aⁿ
Hence Aⁿ is also a symmetric matrix.

Consider [1 \(\times\) 1]\(\left[ \begin{matrix} 1 & 3 & 2 \\ 2 & 5 & 1 \\ 15 & 3 & 2 \end{matrix} \right] \left[ \begin{matrix} 1 \\ 2 \\ x \end{matrix} \right] =0\)
\(\Rightarrow \left[ 1\quad x\quad 1 \right] \left[ \begin{matrix} 7+ & 2x \\ 12 & +x \\ 21 & +2x \end{matrix} \right] =0\)
[7 + 2x + 12x + x² + 21 + 2x] = 0
\(\Rightarrow \) x² + 16 x +28 = 0
\(\Rightarrow \) (x + 14)(x + 2) = 0
\(\Rightarrow \) x = -2 or x = -14

To prove the given points are collinear, it suffices to prove |A| = \(\begin{vmatrix} a&b+c &1 \\ b &c+a &1 \\ c & a+b &1 \end{vmatrix}=0\)
Applying C₁\(\rightarrow\) C₁ + C₂, we deduce that 
|A| = \(\begin{vmatrix} a+b+c&b+c &1 \\ a+b+c &c+a &1 \\ a+b+c & a+b &1 \end{vmatrix}=(a+b+c)\begin{vmatrix} 1&b+c &1 \\ 1 &c+a &1 \\ 1 & a+b &1 \end{vmatrix}=(a+b+c)\times0=0\)
which shows that the given points are collinear.

Area of the triangle = absolute value of \(\frac { 1 }{ 2 } \begin{vmatrix} x_1 &y_1 &1 \\ x_2 &y_2 &1 \\ x_3 &y_3 & 1 \end{vmatrix} \)
\(9=\left|\frac{1}{2}\right| \begin{array}{ccc} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{array}|=| \frac{1}{2}(-k)(-3-3) \mid\)
\(\Rightarrow 9=3|k| \text { and hence, } k=\pm 3 \text {. }\)

\(\begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix}\) = (2 \(\times\) 2) - (- 1 \(\times\) 4) = 4 + 4 = 8.

Given \(a_{ij}={(i-2j)^2\over 2}with \ m=2,n=3\)
We need to construct a 2 \(\times\) 3 matrix.
\(A=\left[\begin{array}{lll} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \end{array}\right]\)
a₁₁ = \(\frac { { (1-2(1)) }^{ 2 } }{ 2 } =\frac { { (-1) }^{ 2 } }{ 2 } =\frac { 1 }{ 2 } \)
a₁₂ = \(\frac { { (1-2(2)) }^{ 2 } }{ 2 } =\frac { { (-3) }^{ 2 } }{ 2 } =\frac { 9 }{ 2 } \)
a₁₃ = \(\frac { { (1-2(3)) }^{ 2 } }{ 2 } =\frac { { (-5) }^{ 2 } }{ 2 } =\frac { 25 }{ 2 } \)
a₂₁ = \(\frac { { (2-2(1)) }^{ 2 } }{ 2 } =\frac { 0 }{ 2 } =0\)
a₂₂ = \(\frac { { (2-2(2)) }^{ 2 } }{ 2 } =\frac { { (-2) }^{ 2 } }{ 2 } =\frac { 4 }{ 2 } \)
a₂₃ = \(\frac { { (2-2(3)) }^{ 2 } }{ 2 } =\frac { { (-4) }^{ 2 } }{ 2 } =\frac { 16 }{ 2 } \)
\(\therefore\) A = \(\left( \begin{matrix} { a }_{ 11 } & a_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \end{matrix} \right) =\left( \begin{matrix} 1/2 & 9/2 & 25/2 \\ 0 & 4/2 & 16/2 \end{matrix} \right) =\frac { 1 }{ 2 } \left( \begin{matrix} 1 & 9 & 25 \\ 0 & 4 & 16 \end{matrix} \right) \)

\(\mathrm{LHS}=\left|\begin{array}{ccc} x+2 a & y+2 b & z+2 c \\ x & y & z \\ a & b & c \end{array}\right|\)
\(\text {Applying } R_1 \rightarrow R_1-R_2\)
\(=\left|\begin{array}{ccc} 2 a & 2 b & 2 c \\ x & y & z \\ a & b & c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ x & y & z \\ a & b & c \end{array}\right|=2(0)=0\) [Since R & R, are Proportional]
= RHS. Hence Proved.