\(|A|=\left|\begin{array}{ccc} x+a & b & c \\ a & x+b & c \\ a & b & x+c \end{array}\right|\)
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)

LHS = \(\begin{vmatrix} 1& a & a^2-bc \\1 &b &b^2-ca \\ 1 & c & c^2-ab \end{vmatrix}\)\(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\left| \begin{matrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{matrix} \right| \) [By proverty 7]
Multiplying and dividing R₁, R₂ and R₃ of second determinant by a, b, c respectively.
LHS \(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\frac { 1 }{ abc } \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| \)
In II determinant, Take abe from C₃
\(=\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\frac { abc }{ abc } \left| \begin{matrix} a & { a }^{ 2 } & 1 \\ b & { b }^{ 2 } & 1 \\ c & { c }^{ 2 } & 1 \end{matrix} \right| \)
Applying C₁ ↔️ C₃ in the second determinant,
Applying C₃ ↔️ C₂ in the second determinant
= \(\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { { b }^{ 2 } } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| -\left| \begin{matrix} 1 & a & { a }^{ 2 } \\ 1 & b & { b }^{ 2 } \\ 1 & c & { c }^{ 2 } \end{matrix} \right| \) = 0 = RHS
Hence proved.

Let the points be A(a, b + c), B(b, c + a) and C(c, a + b)
Area of the \(\triangle\)ABC = absolute value of \(\frac { 1 }{ 2 } \left| \begin{matrix} { x }_{ 1 } & { y }_{ 1 } & 1 \\ { x }_{ 2 } & { y }_{ 2 } & 1 \\ { x }_{ 3 } & { y }_{ 3 } & 1 \end{matrix} \right| \)
= absolute value of \(\frac { 1 }{ 2 } \left| \begin{matrix} a & b+c & 1 \\ b & c+a & 1 \\ c & a+b & 1 \end{matrix} \right| \)
Applying C₁⟶C₁ + C₂ we get,
Area of \(\triangle\)ABC = \(\frac { 1 }{ 2 } \left| \begin{matrix} a+b+c & b+c & 1 \\ a+b+c & c+a & 1 \\ a+b+c & a+b & 1 \end{matrix} \right| \)
= \(\frac { 1 }{ 2 } (a+b+c)\left| \begin{matrix} 1 & b+c & 1 \\ 1 & c+a & 1 \\ 1 & a+b & 1 \end{matrix} \right| \) [Taking out(a + b + c) common from C₁]
= \(\frac{1}{2}\)(a + b + c)(0) = 0
Since area of \(\triangle\)ABC = 0, the given points are collinear.

Let \(\triangle\) = \(\left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{matrix} \right| \)
Putting a = -b in (1) we get,
\(\triangle =\left| \begin{matrix} 2b & 0 & -b+c \\ 0 & -2b & b+c \\ c-b & c+b & -2c \end{matrix} \right| \)
Expanding along R₁ we get,
\(\triangle\) = 2b (4bc - (b + cp (- b + c) (2b(c - b))
= 2b (4bc - b2 - c2 - 2bc) + (c - b) (2bc - 2b2)
= 2b (2bc - b2 - c2) + (c - b) (2bc - 2b2)
=
\(\therefore\) (a + b) is a factor of A.
Similarly (b + c) and (c + a) are factors of \(\triangle\).
Since the leading diagonal is of degree 3, their will be a constant k and 3 factors.
\(\therefore\) \(\triangle\) = k (a + b)(b + c)(c + a)
\(\triangle =\left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+ & c+b & -2c \end{matrix} \right| \)=k (a + b)(b + c)(c + a)
Put a = 0, b = 1 and c = 2 we get,
\(\left| \begin{matrix} 0 & 1 & 2 \\ 1 & -2 & 3 \\ 2 & 3 & -4 \end{matrix} \right| \)=k(1)(3)(2)
\(\Rightarrow\) - 1(-4 - 6) +2(3 + 4) = 6 k [Expanded along R₁]
\(\Rightarrow\) -1(-10) + 14 = 6k
\(\Rightarrow\) 24 = 6k
\(\Rightarrow\) k = 4
\(\therefore \left| \begin{matrix} -2a & a+b & a+c \\ b+a & -2b & b+c \\ c+a & c+b & -2c \end{matrix} \right| \) = 4(a + b)(b + c)(c + a)

Given A² - B² = (A - B)(A + B)
\(\Rightarrow\) A² - B² =(A - B)A + (A - B)B
[Distributive property of matrix multiplication over addition]
\(\Rightarrow\) A² - B² = A² - BA + AB - B^{2
\(\Rightarrow\)
\(\Rightarrow\)} BA - AB = 0
^{\(\Rightarrow\)} AB = BA
Thus, the given matrix is true if the matrices A and B are commutative with each other.

Given A² = B
\(\Rightarrow \left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \end{matrix} \right] \left[ \begin{matrix} \alpha & 0 \\ 1 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { \alpha }^{ 2 }+0 & 0+0 \\ \alpha +1 & 0+1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { \alpha }^{ 2 } & 0 \\ \alpha +1 & 1 \end{matrix} \right] =\left[ \begin{matrix} 1 & 0 \\ 5 & 1 \end{matrix} \right] \)
\(\Rightarrow\) \({ \alpha }^{ 2 }\) = 1 or \(\alpha +1\) = 5
\(\Rightarrow\) = \(\pm \) 1 or \(\alpha\) = 4 which is not possible.
Hence, there is no value of  for which A² = B is true.

Given \(x\left[ \begin{matrix} 2x & 2 \\ 3 & x \end{matrix} \right] +2\left[ \begin{matrix} 8 & 5x \\ 4 & 4x \end{matrix} \right] =\left[ \begin{matrix} { x }^{ 2 }+8 & 24 \\ 10 & 6x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} { 2x }^{ 2 } & 2x \\ 3x & { x }^{ 2 } \end{matrix} \right] +\left[ \begin{matrix} 16 & 10x \\ 8 & 8x \end{matrix} \right] =\left[ \begin{matrix} { 2x }^{ 2 }+16 & 48 \\ 20 & 6x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 2{ { x }^{ 2 }+16 } & 2x+10x \\ 3x+8 & { x }^{ 2 }+8x \end{matrix} \right] =\left[ \begin{matrix} 2{ x }^{ 2 }+16 & 48 \\ 20 & 12x \end{matrix} \right] \)
\(\Rightarrow \left[ \begin{matrix} 2{ { x }^{ 2 }+16 } & 12x \\ 3x+8 & { x }^{ 2 }+8x \end{matrix} \right] =\left[ \begin{matrix} 2{ x }^{ 2 }+16 & 48 \\ 20 & 12x \end{matrix} \right] \)
Equating the corresponding entries on both sides, we get
12x = 48 \(\Rightarrow\) x = 4
and x² + 8x = 12x \(\Rightarrow\) x² - 4x = 0 \(\Rightarrow\) x(x - 4) = 0
\(\Rightarrow\) x = 0,4
Since x = 0 is not possible \(\Rightarrow\) x = 4.

Given a, b, c are p^th, q^th, r^th terms of an A.P.
\(p^{\text {th }} \text { term } \Rightarrow A+(p-1) R=a \Rightarrow A+p R-R=a\)
\(q^{\text {th }} \text { term } \Rightarrow A+(q-1) R=b \Rightarrow A+q R-R=b\)
\(r^{\text {th }} \text { term } \Rightarrow A+(r-1) R=c \Rightarrow A+r R-R=c\)
Here A → first term, R → Common difference.
\(\mathrm{LHS}=\left|\begin{array}{ccc} a & b & c \\ p & q & r \\ 1 & 1 & 1 \end{array}\right|\)
\(=\left|\begin{array}{ccc} A+p R-R & A+q R-R & A+r R-R \\ p & q & r \\ 1 & 1 & 1 \end{array}\right|\)
Multiply R₂ & R₃ by R respectively.
\(=\frac{1}{R^2}\left|\begin{array}{ccc} A+p R-R & A+q R-R & A+r R-R \\ p R & q R & r R \\ R & R & R \end{array}\right|\)
\(\text { Applying } R_1 \rightarrow R_1-R_2+R_3\)
\(=\frac{1}{R^2}\left|\begin{array}{ccc} A & A & A \\ p R & q R & r R \\ R & R & R \end{array}\right|\)
\(=\frac{1}{R^2}(0)=0\) [Since R₁ and R₂ are proportional.]

\(|A|=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right|\)
\(=2(0-20)+3(-42-4)+5(30-0)\)
\(=2(-20)+3(-46)+5(30)\)
\(=-40-138+150\)
\(=-28 \neq 0\)
\(|A| \neq 0\)
\(\therefore\) A is non singular

Let A =\(\begin{bmatrix} 1&2 &3 \\ 4 & 5 &6 \\ 7 & 8 & 9 \end{bmatrix}\)
= 1(45 - 48) -2(36 - 42) + 3(32 - 35)
= I(-3) - 2(-6) + 3(-3)
= -3 + 12 - 9
\(|A|=0\)
\(\therefore\) A is singular