\(\left|\begin{array}{ccc} a^2+x^2 & a b & a c \\ a b & b^2+x^2 & b c \\ a c & b c & c^2+x^2 \end{array}\right|\)
\(=\left(a^2+x^2\right)\left[b^2 c^2+b^2 x^2+c^2 x^2+x^4-b^2 c^2\right]\) \(-a b\left(a b c^2+a b x^2-a b c^2\right)+a c\left(a b^2 c-a c b^2-a c x^2\right)\)
\(=\left(a^2+x^2\right)\left(b^2 x^2+c^2 x^2+x^2\right)-a b\left(a b x^2\right)+a c\left(-a x^2 c\right)\)
\(=a^2 b^2 x^2+a^2 c^2 x^2+a^2 x^4+b^2 x^4+x^4 c^2+x^n\) \(-a^2 b^2 x^2-a^2 c^2 x^2\)
\(=x^4\left(a^2+b^2+c^2+x^2\right)\)
Which is divisible by x⁴
Hencè proved.

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.

Given \(\begin{vmatrix} a & b &a\alpha +b \\ b & c & b\alpha+c\\ a\alpha+b & b \alpha+c &0 \end{vmatrix}=0.\)
Expanding along R₃ we get,
\(a\alpha +b)\left| \begin{matrix} b & a\alpha +b \\ c & b\alpha +c \end{matrix} \right| +(b\alpha +c)\left| \begin{matrix} a & a\alpha +b \\ b & b\alpha +c \end{matrix} \right| +0=0\)
\(\Rightarrow\) - (a\(\alpha\) + b) (b²\(\alpha\) + bc - ac\(\alpha\) - bc) + (b\(\alpha\) + c)(ab\(\alpha\) + ac - ab\(\alpha\) - b²) = 0
\(\Rightarrow\) - (a\(\alpha\) + b)(b²\(\alpha\) - ac\(\alpha\)) + (b\(\alpha\) + c)(ac - b²) = 0
\(\Rightarrow\) \(\alpha\)(a\(\alpha\) + b)(ac - b²) + (b\(\alpha\) + c)(ac - b²) = 0
\(\Rightarrow\) (ac - b²)(a\({ \alpha }^{ 2 }\) + b + b + c) = 0
\(\Rightarrow\) (ac - b²)(a\({ \alpha }^{ 2 }\) + 2b + c) = 0
\(\Rightarrow\) ac - b² = 0 or a + 2b\(\alpha\) + c = 0
\(\Rightarrow\) ac = b² or a\({ \alpha }^{ 2 }\) + 2b\(\alpha\) + c = 0
\(\Rightarrow\) a, b, c are in G.P.(or)  is a root of ax² + 2bx + c = 0
Hence proved.

We have |B| = \(\begin{vmatrix} 2(a+b+c) & 2(a+b+c) &2(a+b+c) \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ c+a & a+b &b+c \\ a+b & b+c & c+a \end{vmatrix}\)
= 2\(\begin{vmatrix} a+b+c & a+b+c & a+b+c \\ -b & -c &-a \\ -c & -a & -b \end{vmatrix}\)\((R_2 \rightarrow R_2-R_1andR_3\rightarrow R_3-R_1)\)

= 2\(\begin{vmatrix}a &b &c \\ -b & -c & -a \\ -c & -a & -b \end{vmatrix}\)\((R_1 \rightarrow R_1+R_2+R_3)\)
= 2(-1)² \(\begin{vmatrix}a &b &c \\b &c & a \\ c & a & b \end{vmatrix}\)
= 2| A |.

Minors : M₁₁ = \(\begin{vmatrix} -5 & 6 \\ 5 & 2 \end{vmatrix}\) = -10 - 30 = -40
M₁₂ = \(\begin{vmatrix}4 & 6 \\ -3 & 2 \end{vmatrix}\) = 8 + 18 = 26
M₁₃ = \(\begin{vmatrix}4 & -5 \\ -3 & 5 \end{vmatrix}\) = 20 - 15 = 5
M₂₁ = \(\begin{vmatrix}3 & -2 \\ 5 & 2 \end{vmatrix}\) = 6 + 10 = 16
M₂₂ = \(\begin{vmatrix}1 & -2 \\ -3 & 2 \end{vmatrix}\) = 2 - 6 = -4
M₂₃ = \(\begin{vmatrix}1 & 3 \\ -3 & 5 \end{vmatrix}\) = 5 + 9 = 14
M₃₁ = \(\begin{vmatrix}3 & -2 \\ -5 & 6 \end{vmatrix}\) = 18 - 10 = 8
M₃₂ = \(\begin{vmatrix}1 & -2 \\ 4 & 6 \end{vmatrix}\) = 6 + 8 = 14
M₃₃ = \(\begin{vmatrix}1 & 3 \\ 4 & -5 \end{vmatrix}\) = -5 - 12 = -17
Cofactors:
A₁₁ = (−1)¹⁺¹(−40) = −40
A₁₂ = (−1)¹⁺² (+26) = −26
A₁₃ = (−1)¹⁺³ (5) = 5
A₂₁ = (−1)²⁺¹(16) = −16
A₂₂ = (−1)²⁺² (−4) = −4
A₂₃ = (−1)²⁺³ (14) = −14
A₃₁ = (−1)³⁺¹(8) = 8
A₃₂ = (−1)³⁺² (14) = −14
A₃₃ = (−1)³⁺³ (−17) = −17
Expanding along R₁ yields
|A| = a₁₁ A₁₁ + a₁₂ A₁₂ + a₁₃ A₁₃ .
|A| = 1(−40) + (3)(−26) + (−2)(5) = −128 ....(1)
Expanding along C₁ yields
|A| = a₁₁ A₁₁ + a₂₁ A₂₁ + a₃₁ A₃₁
= 1(−40) + 4(−16) + −3(8) = −128....(2)
From (1) and (2), we have
|A| obtained by expanding along R₁ is equal to expanding along C₁.

Given f(x) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)
f(x) \(\times\) f(y) = \(\begin{bmatrix} cos \ x & -sin \ x & 0 \\ sin x & cos x & 0 \\ 0 & 0 & 1 \end{bmatrix}\)\(\left[ \begin{matrix} cosy & -siny & 0 \\ siny & cosy & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cosxcosy-sinxcosy & -cosxsiny-sinxcosy & 0 \\ sin \ x cos \ y+cos \ xsin \ y & -sin \ x \ sin \ y+cos \ x \ cos \ y & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
= \(\left[ \begin{matrix} cos(x+y) & -sin(x+y) & 0 \\ sin(x+y) & cos(x+y) & 0 \\ 0 & 0 & 1 \end{matrix} \right] \)
[\(\therefore\)cos(x + y) = cos x cos y - sin x sin y]
sin(x + y) = sin x cos y + cos x sin y]
= f(x + y)

\(A=\left[\begin{array}{ccc} 1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5 \end{array}\right] \Rightarrow A^T=\left[\begin{array}{ccc} 1 & -6 & -4 \\ 3 & 8 & 6 \\ 5 & 3 & 5 \end{array}\right]\)
\( Let \ P=\frac{1}{2}\left(A+A^T\right)=\frac{1}{2}\left[\begin{array}{ccc}2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10\end{array}\right] \)
\(Now \ P^T=\frac{1}{2}\left[\begin{array}{ccc}2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10\end{array}\right]=P\)
\(\text { Thus, } P=\frac{1}{2}\left(A+A^T\right)\) is a symmetric matrix.
\(\text { Let } Q=\frac{1}{2}\left(A-A^T\right)\)
\(=\frac{1}{2}\left[\begin{array}{ccc} 0 & 9 & 9 \\ -9 & 0 & -3 \\ -9 & 3 & 0 \end{array}\right]\)
\(\text { Then } Q^T=\frac{1}{2}\left[\begin{array}{ccc} 0 & -9 & -9 \\ 9 & 0 & 3 \\ 9 & -3 & 0 \end{array}\right]=-Q\)
\(\text { Thus } Q=\frac{1}{2}\left(A-A^T\right)\) is a skew-symmetric matrix.
\(A=P+Q=\frac{1}{2}\left[\begin{array}{ccc} 2 & -3 & 1 \\ -3 & 16 & 9 \\ 1 & 9 & 10 \end{array}\right]+\frac{1}{2}\left[\begin{array}{ccc} 0 & 9 & 9 \\ -9 & 0 & -3 \\ -9 & 3 & 0 \end{array}\right]\)
Thus A is expressed as the sum of symmetric and skew-symmetric matrices.

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.]

\(\text { LHS }=\left|\begin{array}{ccc} a^2 & b c & a c+c^2 \\ a^2+a b & b^2 & a c \\ a b & b^2+b c & c^2 \end{array}\right|\)
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.

\(Given A=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3\end{array}\right]\)
\(A^2=A \times A=\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\)
\(A^2=\left[\begin{array}{lll} 1+0+4 & 0+0+0 & 2+0+6 \\ 0+0+2 & 0+4+0 & 0+2+3 \\ 2+0+6 & 0+0+0 & 4+0+9 \end{array}\right]\)
\(=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\)
\(A^3=A^2 \times A=\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]\)
\(=\left[\begin{array}{lll} 5+0+16 & 0+0+0 & 10+0+24 \\ 2+0+10 & 0+8+0 & 4+4+15 \\ 8+0+26 & 0+0+0 & 16+0+39 \end{array}\right]\)
\(=\left[\begin{array}{lll} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]\)
\(A^3-6 A^2+7 A+k I=0\)
\(\left[\begin{array}{ccc} 21 & 0 & 34 \\ 12 & 8 & 23 \\ 34 & 0 & 55 \end{array}\right]-6\left[\begin{array}{ccc} 5 & 0 & 8 \\ 2 & 4 & 5 \\ 8 & 0 & 13 \end{array}\right]+7\left[\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right]+k\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]=0\)
\(\left[\begin{array}{ccc} 21-30+7+k & 0+0+0+0 & 34-48+14+0 \\ 12-12+0+0 & 8-24+14+k & 23-30+7+0 \\ 34-48+14+0 & 0+0+0+0 & 55-78+21+k \end{array}\right]=0\)
\(\left[\begin{array}{ccc} -2+k & 0 & 0 \\ 0 & -2+k & 0 \\ 0 & 0 & -2+k \end{array}\right]=0\)
\(-2+k=0\)
\(\therefore k=2\)
Alternative Method:
W.K.T, Product of roots \(=|A|\) ...........(1)
\(\text { Given } A^3-6 A^2+7 A+k I=0\)
\(\text {Product }=\frac{-c}{a}=\frac{-k}{1}\)
\((1) \Rightarrow \frac{-k}{1}=\left|\begin{array}{lll} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{array}\right|\)
\(-k=1(6-0)-0+2(0-4)\)
\(-k=6-8\)
-k = -2
\(\therefore\) k = 2