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|

a₂₂ = -7
Alternative Method:
Given \(A =\left| \begin{matrix} 1 & 2 & 3 \\ 2 & 0 & 1 \\ 5 & 8 & 8 \end{matrix} \right|\)
a₂₂ = \(\left| \begin{matrix} 1 & 3 \\ 5 & 8 \end{matrix} \right|\)
⇒ a₂₂ = 8-15
⇒ a₂₂ = -7

|A| = ±15
Alternative Method:
|adjA| = |A|^n-1, where n is the order of the matrix.
|A|² = 15¹²
⇒ ±15
⇒ |A'| = ±15

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

\(\left| \begin{matrix} 3x & 7 \\ -2 & 4 \end{matrix} \right| =\left| \begin{matrix} 8 & 7 \\ 6 & 4 \end{matrix} \right| \)
\(\Rightarrow x=-2 \)
Alternative Method:
\(\Rightarrow 12x+14=32-42 \)
\(\Rightarrow 12x=-10-14 \)
\(\Rightarrow x=-\frac { 24 }{ 12 } \)
\(\Rightarrow x=-2\)

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 \(\frac { \pi }{ 2 }\)

10 - 12 = 5x - 6x ⇒ x = 2

|2A|=2³ |A| = 8 x 4 = 32

We have
\(\left.\Delta=\left|\begin{array}{lll} 4 & a & a+b+c \\ 4 & b & b+c+a \\ 4 & c & c+a+b \end{array}\right| \text { [by performing } C_{3} \rightarrow C_{3}+C_{2}\right] \)
\(=4(a+b+c)\left|\begin{array}{lll} 1 & a & 1 \\ 1 & b & 1 \\ 1 & c & 1 \end{array}\right|\)
[by taking 4 and 0 + b + c common from C₁ and C₃]
= 4(0 + b + c) x 0 = 0 [C₁ and C₃ are identical]

\(\Delta =\left| \begin{matrix} x+2 & x+6 & x-1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6 \end{matrix} \right| \)
R₁-->R₁+R₂+R₃
\(\Delta =\left| \begin{matrix} 3x+7 & 3x+7 & 3x+7 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6 \end{matrix} \right| \)
\(\Delta =3x+7\left| \begin{matrix} 1 & 1 & 1 \\ x+6 & x-1 & x+2 \\ x-1 & x+2 & x+6 \end{matrix} \right| \)
C₁-->C₁-C₂  C₂-->C₂-C₃
\(\Delta =(3x+7)\left| \begin{matrix} 0 & 0 & 1 \\ 7 & -3 & x+2 \\ -3 & -4 & x+6 \end{matrix} \right| \)
expanding by R1
\(\Delta \)= (3x+7)[-28-9]
\(\Delta \)= 0
=> 3x + 7 = 0
x = -7/3

LHS = \(\frac { 1 }{ abc } \left| \begin{matrix} { (a+b) }^{ 2 } & c & c \\ a & { (b+c) }^{ 2 } & a \\ b & b & { (c+a) }^{ 2 } \end{matrix} \right| \)
\({ C }_{ 1 }\rightarrow { C }_{ 1 }-{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 }\)
\(=\frac { 1 }{ abc } \left| \begin{matrix} (a+b+c)(a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c+a)(b+c-a) & { a }^{ 2 } \\ (b+c+a)(b-c-a) & (b+c+a)(b-c-a) & { (c+a) }^{ 2 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (a+b-c) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 }-{ R }_{ 2 }\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abc } \left| \begin{matrix} (ac+bc-{ c }^{ 2 }) & 0 & { c }^{ 2 } \\ 0 & (b+c-a) & { a }^{ 2 } \\ (-2a) & (-2c) & { 2ca } \end{matrix} \right| \)
\(({ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }+{ C }_{ 3 })\)
\(=\frac { { (a+b+c) }^{ 2 } }{ abcca } \left| \begin{matrix} (ac+bc) & { c }^{ 2 } & { c }^{ 2 } \\ { a }^{ 2 } & (ba+ca) & { a }^{ 2 } \\ 0 & 0 & { 2ca } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2{ c }^{ 2 }{ a }^{ 2 } }{ abcca } \left| \begin{matrix} (a+b) & c & c \\ a & (b+c) & a \\ 0 & 0 & { 1 } \end{matrix} \right| \)
\(=\frac { { (a+b+c) }^{ 2 }2 }{ b } (ab+ac+{ b }^{ 2 }+bc-ac)\)
\(=2{ (a+b+c) }^{ 3 }\)

Given \(A=\left\lceil \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right\rceil \) and \(B=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] \) 
then \(AB=\left[ \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right] \left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] =\left[ \begin{matrix} -1 & 5 \\ 5 & -14 \end{matrix} \right] \)
Taking L.H.S = \((AB)^{ -1 }=\frac { adj(AB) }{ \left| AB \right| } \)
Here adj(AB) \(=\left[ \begin{matrix} -14 & -5 \\ -5 & -1 \end{matrix} \right] \) and |AB| = 14 - 25 = 11
\(\left| AB \right| =14-25=-11\)
\((AB)^{ -1 }=-\frac { 1 }{ 11 } \left[ \begin{matrix} -14 & -5 \\ -5 & -1 \end{matrix} \right] =\frac { 1 }{ 11 } \left[ \begin{matrix} 14 & 5 \\ 5 & 1 \end{matrix} \right] \)
\(adjB=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \)
\(|B|=3-2=1\)
\(adjA=\left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(\left| A \right| =-8-3=-11\)
\(B^{ -1 }=\frac { 1 }{ \left| B \right| } (adjB)=\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \)
\(A^{ -1 }=\frac { 1 }{ 11 } \left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
Taking R.H.S = B^-1A^-1
\(=\left[ \begin{matrix} 1 & -2 \\ -1 & 3 \end{matrix} \right] ^{ -1 }\left[ \begin{matrix} 2 & 3 \\ 1 & -4 \end{matrix} \right] ^{ -1 }\)
\(=1\left[ \begin{matrix} 3 & 2 \\ 1 & 1 \end{matrix} \right] \times \frac { 1 }{ 11 } \left[ \begin{matrix} -4 & -3 \\ -1 & 2 \end{matrix} \right] \)
\(=\frac { 1 }{ 11 } \left[ \begin{matrix} 14 & 5 \\ 5 & 1 \end{matrix} \right] \)
\(L.H.S=R.H.S\)

Expanding along first row, we get
\(\Delta\)= 3 (6 – 6) – 2 (6 – 9) + 3 (4 – 6)
= 0 – 2 (–3) + 3 (–2) = 6 – 6 = 0

x=500, y=2000, z=3500

\(\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab &1-a^2+b^2 & 2a \\ 2b & -2a &1-a^2-b^2 \end{vmatrix}\)
= \(\begin{vmatrix} 1+a^2-b^2 & 0 & -b(1+a^2+b^2) \\ 2ab &1-a^2+b^2 & 2a \\ 2b & -2a &1-a^2-b^2 \end{vmatrix}\)
= (1+a²+b²)
\(=\left|\begin{matrix}1&0&-b\\2ab&1-a^2+b^2&2a\\2b&-2a&1-a^2-b^2\end{matrix}\right|\)
= (1+a²+b²)
\(=\left|\begin{matrix}1&0&-b\\0&1&a\\2b&-2a&1-a^2-b^2\end{matrix}\right|\)
= (1+a²+b²)²
\(\left[ (1)\left| \begin{matrix} 1 & a \\ -2a & 1-{ a }^{ 2 }-{ b }^{ 2 } \end{matrix} \right| -b\left| \begin{matrix} 0 & 1 \\ 2b & -2a \end{matrix} \right| \right] \)
= \((1+a^2+b^2)^2[(1-a^2-b^2+2ab)-b(0-2b)]\)
= \((1+a^2+b^2)^2(1+a^2+b^2)=(1+a^2+b^2)^3\)
which is true.

The given equations are:
x+3y=5
2x+6y=8
Here \(A=\begin{bmatrix}1&3\\2&6 \end{bmatrix}and\ B=\begin{bmatrix}5\\8 \end{bmatrix}\)
Now |A|=\(\begin{vmatrix}1&3\\2&6 \end{vmatrix}=6-6=0\)
Now  \(adj\ A=\begin{bmatrix} 6&-2\\-3&1\end{bmatrix}'=\begin{bmatrix}6&-3\\-2&1 \end{bmatrix}\) and \(\begin{bmatrix} 5\\8\end{bmatrix}\)
\(\therefore(adj\ A)B=​​​​\begin{bmatrix} 6&-3\\-2&1\end{bmatrix}\begin{bmatrix}5\\8 \end{bmatrix}\)
\(=\begin{bmatrix}30-24\\-10+8 \end{bmatrix}=\begin{bmatrix}6\\-2 \end{bmatrix}\neq0\)
Hence the given system of equation is inconsistent

Let A be invertible matrix of order n and I be the identity matrix of order n.
Then, there exists a square matrix B of order n such that AB = BA = I
Now AB = I. So AB = I or A B = 1 (since I = 1, AB = A B)
This gives A ≠ 0. Hence A is nonsingular.
Conversely, let A be nonsingular. Then A ≠ 0
Now A (adj A) = (adj A) A = A I
or \( \mathrm{A}\left(\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\right)=\left(\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\right) \mathrm{A}=\mathrm{I}\)
or \(\mathrm{AB}=\mathrm{BA}=\mathrm{I} \text {, where } \mathrm{B}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A} \)
Thus A is invertible and A–1 \(=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}\)

Area of \(\Delta ABC={1\over 2}\left|\begin{matrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{matrix}\right|\)
\(={1\over2}\begin{vmatrix} a&a+b&1\\b&c+a&1\\c&a+b&1 \end{vmatrix}\)
\(={1\over2}\begin{vmatrix}a+b+C&b+c&1\\a+b+c&c+a&1\\a+b+c&a+b&1 \end{vmatrix}\)
\({1\over2}(a+b+c)\begin{vmatrix}1&b+c&1\\1&c+a&1\\1&a+b&1 \end{vmatrix}\)
\(={1\over2}(a+b+c)(0)=0\)

= \(\left| \begin{matrix} 1 & x & x^{ 2 } \\ x^{ 2 } & 1 & x \\ x & x^{ 2 } & 1 \end{matrix} \right| \)
= \(\left| \begin{matrix} 1+x+x^2 & x & x^{ 2 } \\1+x+ x^{ 2 } & 1 & x \\ 1+x+x^2 & x^{ 2 } & 1 \end{matrix} \right| \)
\(=(1+x+x^2)\left| \begin{matrix} 1 & x & x^{ 2 } \\1 & 1 & x \\ 1 & x^{ 2 } & 1 \end{matrix} \right| \)
\(=(1+x+x^2)\left| \begin{matrix} 1 & x & x^{ 2 } \\0& 1-x & x-x^2 \\ 0 & x^{ 2 }-1 & 1-x \end{matrix} \right| \)
\(=(1+x+x^2)\left|\begin{matrix}1-x&x-x^2\\x^2-1&1-x\end{matrix}\right|\)
\(=(1+x+x^2)(1-x)^2\left|\begin{matrix}1&x\\-1-x&1\end{matrix}\right|\)
= (1-x)²(1+x+x²)(1x+x²)
= (1-x)²(1+x+x²)²
= (1-x³)²

\(\Delta=\left| \begin{matrix} 0 & a & -b \\ -a & 0 & -c \\ b & c & 0 \end{matrix} \right| \)
=\(\left| \begin{matrix} 0 & -a & b \\a & 0 &c \\- b & -c & 0 \end{matrix} \right| \)
=\((-1)^3=\left| \begin{matrix} 0 & a &- b \\-a & 0 &-c \\ b & c & 0 \end{matrix} \right| \)
\(=-\left|\begin{matrix}0&a&-b\\ -a&0&-c\\b&c&0\end{matrix}\right|\)
\(\Rightarrow2\Delta=0\ \Rightarrow \Delta=0\)

\(A=\begin{vmatrix} 1 & 2 \\ 4 & 2 \end{vmatrix}\)
\(\therefore 2 A=2\left[\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right]=\left[\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right] \)
\(\therefore \text { L.H.S. }=|2 A|=\left|\begin{array}{ll} 2 & 4 \\ 8 & 4 \end{array}\right|=2 \times 4-4 \times 8=8-32=-24 \)
\(\text { Now, }|A|=\left|\begin{array}{ll} 1 & 2 \\ 4 & 2 \end{array}\right|=1 \times 2-2 \times 4=2-8=-6 \)
\(\therefore \text { R.H.S. }=4|A|=4 \times(-6)=-24 \)
\(\therefore \text { L.H.S. }=\text { R.H.S. } \)

By definition of minors and cofactors, we have
\(\text {Minor of } a_{11}=\mathrm{M}_{11}=\left|\begin{array}{ll} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array}\right|=a_{22} a_{33}-a_{23} a_{32}\)
\(\text {Cofactor of } a_{11}=\mathrm{A}_{11}=(-1)^{1+1} \quad \mathrm{M}_{11}=a_{22} a_{33}-a_{23} a_{32}\)
\(\text {Minor of } a_{21}=\mathrm{M}_{21}=\left|\begin{array}{ll} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array}\right|=a_{12} a_{33}-a_{13} a_{32}\)
\(\text {Cofactor of } a_{21}=\mathrm{A}_{21}=(-1)^{2+1} \mathrm{M}_{21}=(-1)\left(a_{12} a_{33}-a_{13} a_{32}\right)=-a_{12} a_{33}+a_{13} a_{32}\)

Note that in the third column, two entries are zero. So expanding along third column (C₃), we get
\(\Delta =\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| -0\left| \begin{matrix} 1 & 2 \\ 4 & 1 \end{matrix} \right| +0\left| \begin{matrix} 1 & 2 \\ -1 & 3 \end{matrix} \right| \)
= 4 (–1 – 12) – 0 + 0 = – 52