We have, \(A=\left| \begin{matrix} 6 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4 \end{matrix} \right| \)
|A| = 6(4-0)-0(0-0)+1(0-0)
|2A| = \(\left| \begin{matrix} 12 & 0 & 2 \\ 0 & 2 & 4 \\ 0 & 0 & 8 \end{matrix} \right| \)
= 12(16-0)-0(0-0)+2(0-0)
= 192
|2A| = 8 x 24 = 8|A|

We have, \(A=\left| \begin{matrix} 2 & 3 & -1 \\ 4 & 1 & 0 \\ 3 & 3 & 2 \end{matrix} \right| \)
\(M_{ 12 }\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =8-0=8\)
\(M_{ 21 }\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =6+3=9\)
\(C_{ 21 }=-\left| \begin{matrix} 3 & -1 \\ 3 & 2 \end{matrix} \right| =-(6+3)=-9\)
\(C_{ 12 }=-\left| \begin{matrix} 4 & 0 \\ 3 & 2 \end{matrix} \right| =-(8-0)=-8\)
\( M_{ 12 }\times M_{ 21 }+C_{ 21 }\times C_{ 12 }\)
\( 8\times (9)+(-9)(-8)=72+72\)
\(=144\)

a₃₂ = -11
Alternative Method:
Given \(A =\left| \begin{matrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{matrix} \right|\)
⇒ a₃₂ \(=\left| \begin{matrix} 5 & 8 \\ 2 & 1 \end{matrix} \right|\)
⇒ a₃₂ = 5-16 = -11

k=2³=8

\(A^1=\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right|\)
Alternative Method:
\(A=\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right| \)
\(A^{ 1 }=\frac { 1 }{ \left| A \right| } (adjA) \)
\(\left| A \right| =\left| \begin{matrix} 3 & 10 \\ 2 & 7 \end{matrix} \right| =21-20=1 \)
\(adjA=\left| \begin{matrix} 7 & -10 \\ -2 & 3 \end{matrix} \right| \Rightarrow A^{ 1 }=\left| \begin{matrix} 7 & -10 \\ -2 & 3 \end{matrix} \right|\)

Let \(\Delta=\left|\begin{array}{ccc}2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7\end{array}\right|\)
Here, \(a_{32}=5\)
Given, \(A_{i j}\) is the cofactor of the element \(a_{i j}\) of \(\Delta\).
\(\therefore  A_{32}=(-1)^{3+2}\left|\begin{array}{ll} 2 & 5 \\ 6 & 4 \end{array}\right|=-1(8-30)=22\)
a ₃₂ A₃₂ = 110
Alternative Method:
\(a_{ 32 }A_{ 32 }=-5\left| \begin{matrix} 2 & 5 \\ 6 & 4 \end{matrix} \right|\)
= 5(8-30)
= -5(-22) = 110

x = 2
Alternative Method:
\(\left| \begin{matrix} x+1 & x-1 \\ x-3 & x+2 \end{matrix} \right| =\left| \begin{matrix} 4 & -1 \\ 1 & 3 \end{matrix} \right| \)
\(\Rightarrow (x+1)(x+2-(x-3)(x-1)=12+1 \)
\(\Rightarrow \ x^{ 2 }+3x+2-x^{ 2 }+4x-3=13 \)
\(\Rightarrow 7x-1=13 \)
\(\Rightarrow 7x=14 \)
\(\Rightarrow x=2\)

\(\left| \begin{matrix} 2x & x+3 \\ 2(x+1) & x+1 \end{matrix} \right| =\left| \begin{matrix} 1 & 5 \\ 3 & 3 \end{matrix} \right| \Rightarrow x=1\)
Alternative Method:
\(\left| \begin{matrix} 2x & x+3 \\ 2(x+1) & x+1 \end{matrix} \right| =\left| \begin{matrix} 1 & 5 \\ 3 & 3 \end{matrix} \right| \)
\(\Rightarrow 2x(x+1)-(x+3)\left[ 2(x+1) \right] =3-15 \)
\(\Rightarrow 2x^{ 2 }+2x-2x^{ 2 }-8x-6=-12 \)
\(\Rightarrow -6x-6=-12 \)
\(\Rightarrow 6x=-12+6 \)
\(\Rightarrow x=1\)

|2AA^-1|=|2|
=8

k = 7
Alternative Method:
Since, |kA| = kⁿ|A|, where n is the order of matrix.
|3A| = 3³(A)
= 27|A|
⇒ k = 27

\(\left| \begin{matrix} 2x & 5 \\ 8 & x \end{matrix} \right| =\left| \begin{matrix} 6 & -2 \\ 7 & 3 \end{matrix} \right| \)
\(\Rightarrow x=\pm 6 \)
Alternative Method:
\(\left| \begin{matrix} 2x & 5 \\ 8 & x \end{matrix} \right| =\left| \begin{matrix} 6 & -2 \\ 7 & 3 \end{matrix} \right| \)
2x²-40=18+14
\(\Rightarrow 2x^{ 2 }=32+40 \)
\(\Rightarrow 2x^{ 2 }=72 \)
\(\Rightarrow x^{ 2 }=36 \)
\(\Rightarrow x=\pm 6\)

Let \(A=\left[ \begin{matrix} 2sinx \\ 1 \end{matrix}\begin{matrix} 3 \\ 2sinx \end{matrix} \right]\)
For singular matrix
|A| = 0
4sin 2x - 3 = 0
\(\Rightarrow sinx=\pm \frac { \sqrt { 3 } }{ 2 } \)
\(\Rightarrow x=\frac { 2\pi }{ 3 }\)
as \(\frac { \pi }{ 2 }\)

|adjA|=|A|²=(−4)²=16

x = 2

\(\text { If } A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right], \text { then adj } A=\left[\begin{array}{rr} d & -b \\ -c & a \end{array}\right] \text { . }\)
So, Adj \(A=\begin{bmatrix} 3 & 1 \\ -4 & 2 \end{bmatrix}\)

Here \(\Delta\)is a skew-symmetric determinant of 3rd (odd) order
\(\Delta\)=0

\(\Delta=\left| \begin{matrix} 1 & 1 & 1 \\ 1+sinA & 1+sinB & 1+sinC \\ sinA+sin^{ 2 }A & sinB+sin^{ 2 }B & sinC+sin^{ 2 }C \end{matrix} \right| \)
\(=\begin{vmatrix}1&1&1\\1+sin\ A&1+sin\ B&1+sin\ C\\-cos^2A&-cos^2B &-cos^2C \end{vmatrix}\) [Operating R₃ \(\rightarrow\)R₃​​\(\rightarrow\)R₂]
\(\begin{vmatrix} 1&0&0\\1+sin\ A&sin\ B-sin\ A&sin\ C-sin\ B\\-cos^2&cos^2A-cos^2B&cos^2B-cos^2C\end{vmatrix}\)  [Operating R₃ \(\rightarrow\) R₃\(\rightarrow\)R₂]
\(=(1)(sinB-sinA)(cos^2B-cos^2C)-(cos^2A-cos^2B)(sinC-sinB)\)     [Expanding by R₁]
\(=(sinB-sinA)(1-sin^2B-1+sin^2C)-(1-sin^2A-1+sin^2B)(sinC-sinB)\)
\(=(sinB-sinA)(sin^2C-sin^2B)-(sinC-sinB)(sin^2B-sin^2A)\)
\(=(sinB-sinA)(sinC-sinB)(sinC+sinB-sinB-sinA)\)
\(=(sinB-sinA)(sinC-sinB)(sinC-sinA)\)
Now \(\Delta=0\Rightarrow sinB=0\ or\ sinC-sinA=0\)
\(\Rightarrow A=B\ or\ B=C\ or\ C=A\)
\(\Rightarrow\)Triangle ABC is isosceles.

\(\Delta_1=\begin{vmatrix} 1&1&1\\yz&zx&xy\\x&y&z\end{vmatrix}\)
\(=\begin{vmatrix}1&yz&x&\\1&zx&y\\1&xy&z \end{vmatrix}\) [Inter-changing rows and columns]
\(={1\over xyz}\begin{vmatrix} x&xyz&x^2\\y&xyz&y^2\\z&xyz&z^2\end{vmatrix}\)[Multiplying R₁ by x, R₂ by y and R₃ by z and taking \(1\over xyz\) outside]
\(={xyz\over xyz}\begin{vmatrix} x&1&x^2\\y&1&y^2\\z&1&z^2\end{vmatrix}\)[Taking xyz common from C₂]\

\(=(-1)\begin{vmatrix}1 &x&x^2\\1&y&y^2\\1&z&z^2 \end{vmatrix}\)[Operating C₁↔C₂]
\(=-\Delta\)
Hence, \(\Delta+\Delta_1\)

\(\Delta =\left| \begin{matrix} 1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 3 & 7+3p \end{matrix} \right| \)
\((1)\left|\begin{matrix}1&2+p\\2&7+3p\end{matrix}\right|\)
= (7+3p)-(6+3p) = 7-6 = 1

\(\left|\begin{matrix}a&a^2&\beta+\gamma\\ \beta&\beta^2&\gamma+a\\ \gamma&\gamma^2&a+\beta\end{matrix}\right|\)
\(\begin{vmatrix}\alpha&\alpha^2& \alpha+\beta+\gamma\\ \beta&\beta^2& \alpha+\beta+\gamma\\\gamma&\gamma^2& \alpha+\beta+\gamma\end{vmatrix} \)
\((\alpha+\beta+\gamma)\begin{vmatrix} \alpha&\alpha^2&1\\\beta&\beta^2&1\\\gamma&\gamma^2&1\end{vmatrix}\)
\(=(\alpha+\beta+\gamma)\begin{vmatrix}\alpha&\alpha^2&1\\\beta-\alpha&\beta^2-\alpha^2&0\\\gamma& \gamma^2-\beta^2&0\end{vmatrix}\)
\(=(\alpha+\beta+\gamma)\begin{vmatrix} \beta-\alpha&\beta^2-\alpha^2\\\gamma-\beta&\gamma^2-\beta^2\end{vmatrix}\)
\((\alpha+\beta+\gamma)(\beta-\alpha)(\gamma-\beta)\begin{vmatrix}1&\beta+\alpha\\1&\gamma+\beta \end{vmatrix}\)
\(=(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma+\beta-\beta-\alpha)\)
\((\beta-\gamma)(\gamma-\alpha)(a-\beta)(a+\beta+\gamma).\)

\(M_{11}=\begin{vmatrix} 0&4\\5&-7\end{vmatrix}=-0-20=-20\)
\(A_{11}=(-1)^{1+1}M_{11}=(-1)^2(-20)=-20\)
\(M_{12}=\begin{vmatrix}6&4\\1&-7 \end{vmatrix}=-42-4=-46\)
\(A_{12}=(-1)^{1+2}M_{12}=(-1)^3(-46)=(-1)(-46)=46\)
\(M_{13}=\begin{vmatrix}6&0\\1&5 \end{vmatrix}=30-0=30\)
\(A_{13}=(-1)^{1+3}M_{13}=(-1)^4(30)=30\)
\(M_{21}=\begin{vmatrix} -3&5\\5&-7\end{vmatrix}=21-25=-4\)
\(A_{21}=(-1)^{ 2+1}M_{21}=(-1)^3(-4=(-1 )(-4)=4)\)
\(M_{22}=\begin{vmatrix} 2&5\\1&-7\end{vmatrix}=-14-15=-19\)
\(A_{22}=(-1)^{2+2}M_{22}=(1)^4(-19)=-19\)
\(M_{23}=\begin{vmatrix} 2&-3\\1&5\end{vmatrix}=10+3=13\)
\(A_{23}=(-1)^{2+3}M_{23}=(-1)^513=-13\)
\(M_{31}=\begin{vmatrix}-3&5\\0&4 \end{vmatrix}=-12-0=-12\)
\(A_{31}=(-1)^{3+1}M_{31}=(-1)^4(-12)=-12\)
\(M_{32}=\begin{vmatrix} 2&5\\6&4\end{vmatrix}=8-30=-22\)
\(A_{32}=(-1)^{3+2}M_{32}=(-1)^5(-22)=(-1)(-22)=22\)
\(M_{33}=\begin{vmatrix} 2&-3\\6&0\end{vmatrix}=0+18=18\)
\(A_{33}=(-1)^{3+3}M_{33}=(-1)^6(18)=18\)
(ii)\(a_{11}A_{31}+a_{12}A_{32}+a_{13}A_{32}\)
\(=(2)(-12)+(-3)(22)+(5)(18)=-24-66+90=0\)

Expanding R₁, we get:
\(\Delta =0\left| \begin{matrix} 0 & sin\beta \\ -sin\beta & 0 \end{matrix} \right| -sin\alpha \begin{vmatrix} -sin\alpha & sin\beta \\ cos\alpha & 0 \end{vmatrix}-cos\alpha \begin{vmatrix} -sin\alpha & 0 \\ cos\alpha & -sin\beta \end{vmatrix}\)
= 0 - sin \(\alpha\) (-0-sin \(\beta\) cos \(\alpha\))-cos \(\alpha\) (sin \(\alpha\) sin \(\beta\) - 0)
= sin \(\alpha\) sin \(\beta\) cos \(\alpha\) - cos \(\alpha\) sin \(\alpha\) sin \(\beta\) = 0

Applying \({ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 1 }\quad and\quad { C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 1 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA\cos ^{ 2 }{ B } +cosB & cosC-cosA\cos ^{ 2 }{ C } -cosC \\ \cos ^{ 2 }{ A } +cosA & \cos ^{ 2 }{ A } -cosA & -\cos ^{ 2 }{ A } -cosA \end{matrix} \right| =0\)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & cosB-cosA & cosC-cosA \\ \cos ^{ 2 }{ A } -1 & \cos ^{ 2 }{ B } -\cos ^{ 2 }{ A } & \cos ^{ 2 }{ C } -cosA \end{matrix} \right| =0\)
\({ C }_{ 3 }\rightarrow { C }_{ 3 }-{ C }_{ 2 }\)
(cos B - cos A ) x ( cos C -cos B)
\(\left| \begin{matrix} 1 & 0 & 0 \\ 1+cosA & 1 & 0 \\ \cos ^{ 2 }{ A } -1 & cosB+cosA & cosC-cosA \end{matrix} \right| =0\)
\(\therefore \quad (cosB-cosA)\times (cosC-cosA)\times (cosC-cosB)\)
[1-0]=0  (Expanding along C₃)
\(\therefore \ -cosB=cosA\quad or\quad cosC=cosAorcosC=cosB\)
\(\Rightarrow cosB=cosAorcosC=cosAorcosC=cosB\)
\(\Rightarrow \angle B=\angle Aor\angle C=\angle Aor\angle C=\angle B\)
\(\Rightarrow \triangle ABC\)  is an isosceles triangle.

Putting x-a = A, x - b = B, x - c = C we get
\(\Delta=\begin{vmatrix} 1&1&1\\A^2&B^2&C^2\\BC&CA&AB\end{vmatrix}={1\over ABC}\begin{vmatrix}A&B&C\\A^3&B^3&C3\\ABC&ABC&ABC \end{vmatrix}\)
\(={1\over ABC}.(ABC)\begin{vmatrix}A&B&C\\A^3&B^3&C^3\\1& 1&1\end{vmatrix}\)
\(=\begin{vmatrix}A&B&C\\A^3&B^3&C^3\\1&1&1 \end{vmatrix}=-\begin{vmatrix} A&B&C\\1&1&1\\A^3&B^3&C^3\end{vmatrix}\)
= \(\begin{vmatrix} A&B&C\\A&B&C\\A^3&B^3&C^3\end{vmatrix}=(A-B)(B-C)(C-A)(A+B+C)\)
= (x-ax-+b) (x-b-x+c) (x-c-x+a) (a-x+x-b+x-c)
= -(b-c) (c-a) (a-b) (3x-a-b-c)
Since a, b, c are different then, \(x={1\over3}(a+b+c)\)