We have:
\(A=\left| \begin{matrix} 1 & -2 & 3 \\ 0 & -1 & 4 \\ -2 & 2 & 1 \end{matrix} \right| \)
\(\therefore\ B=A'=\begin{bmatrix} 1&0&-2\\-2&-1&2\\3&4&1\end{bmatrix}\)
Now \(|B|=\begin{bmatrix}1&0&-2\\-2&-1&2\\3&4&1 \end{bmatrix}\)
= (1)(-1-8)-0-2(-8+3)
= -9 + 10 = 1 ≠ 0
= B^-1 Exsits
Now \(B_{11}=\begin{vmatrix}-1&2\\4&1 \end{vmatrix}=-1-8=-9;\)
\(B_{12}=-\begin{vmatrix}-2&2\\3&1 \end{vmatrix}=-(-2-6)=8;\)
\(B_{13}=\begin{vmatrix}-2&-1\\3&4 \end{vmatrix}=-8+3=-5\)
\(B_{21}=-\begin{vmatrix} 0&-2\\4&1\end{vmatrix}=-(0+8)=-8;\)
\(B_{22}=\begin{vmatrix} 1&-2\\3&1\end{vmatrix}=1+6=7;\)
\(B_{23}=-\begin{vmatrix} 1&0\\3&4\end{vmatrix}=-(4-0)=-4\)
\(B_{31}=\begin{vmatrix}0&-2\\-1&2 \end{vmatrix}=0-2=-2;\)
\(B_{32}=-\begin{vmatrix} 1&-2\\-2&2\end{vmatrix}=-(2-4)=2;\)
\(B_{33}=\begin{vmatrix} 1&0\\-2&-1\end{vmatrix}=-1+0=-1\)
\(\therefore\ adj.B=\begin{bmatrix}B_{11 }&B_{12}&B_{13}\\B_{21}&B_{22}&B_{23}\\B_{31}&B_{32}&B_{33}\end{bmatrix}=\begin{bmatrix} -9&8&-5\\-8&7&-4\\-2&2/&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(\therefore\ B^{-1}={adj.B\over |B|}={1\over1}\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
\(=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)
Hence, \((A')^{-1}=B^{-1}=\begin{bmatrix} -9&-8&-2\\8&7&2\\-5&-4&-1\end{bmatrix}\)

Area of the triangle = \(\frac { 1 }{ 2 } \left| \begin{matrix} { at }_{ 1 }^{ 2 } & { 2at }_{ 1 } & 1 \\ { at }_{ 2 }^{ 2 } & { 2at }_{ 2 } & 1 \\ { at }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(\frac { 1 }{ 2 } (a)(2a)\left| \begin{matrix} { t }_{ 1 }^{ 2 } & { t }_{ 1 } & 1 \\ { t }_{ 2 }^{ 2 } & { t }_{ 2 } & 1 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \({ a }^{ 2 }\left| \begin{matrix} { t }_{ 1 }^{ 2 }-{ t }_{ 2 }^{ 2 } & { t }_{ 1 }-{ t }_{ 2 } & 0 \\ { t }_{ 2 }^{ 2 }-{ t }_{ 3 }^{ 2 } & { t }_{ 2 }-{ t }_{ 3 } & 0 \\ { t }_{ 3 }^{ 2 } & { t }_{ 3 } & 0 \end{matrix} \right| \)
= \(a^2(t_2-t_2(t_2-t_3)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 & 0 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 & 0 \\ t_{ 3 }^{ 2 } & { t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(1)\left| \begin{matrix} { t }_{ 1 }+{ t }_{ 2 } & 1 \\ { t }_{ 2 }+{ t }_{ 3 } & 1 \end{matrix} \right| \)
= \(a^2(t_1-t_2)(t_2-t_3)(t_1+t_2-t_2-t_3)\)
= \(a^2(t_1-t_2)(t_2-t_3)(t_3-t_1)\)

By the equation, \(\frac { 1 }{ 2 } \left| \begin{matrix} 2 & -6 & 1 \\ 5 & 4 & 1 \\ k & 4 & 1 \end{matrix} \right| =\pm 35\)
\({1\over2}[2(4-4)+6(5-k)+(1)920-4k)]=\le35\)
\({1\over2}[0+30-6k+20-4k]=\le35\)
\(50-10k=\le70\)
\(10k=50\mp 70\)
10k = -20, 120.
Hence k = -2, 12

LHS:\(\begin{vmatrix} 2y&y-z-x&2y\\2z&2z&z-x-y\\x-y-z&2x&2x\end{vmatrix}\)
= \(\begin{vmatrix}x+y+z&x+y+z&x+y+z\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix}1&1&1\\2z&2z&z-x-y\\x-y-z&2x&2x \end{vmatrix}\)
= \((x+y+z)\begin{vmatrix} 1&0&0\\2z&0&-(x+y+z)\\x-y-z&x+y+z&0\end{vmatrix}\)
= \((x+y+z)(1)\begin{vmatrix}0&-(x+y+z)\\x+y+z&0 \end{vmatrix}\)
= \((x+y+z)[0+(x+y+z)^2]\)
= \((x+y+z)^3\)= RHS

\(\begin{vmatrix}a&b&c\\a^2&b^2&c^2\\b+c&c+a&a+b \end{vmatrix}\)
= \(\begin{vmatrix}a-b&b-c&c\\(a-b)(a+b)&(b-c)(b+c)&c^2\\b-a&c-b&a+b \end{vmatrix}\)
= \((a-b)(b-c)\begin{vmatrix} 1&1&c\\a+b&b+c&c^2\\-1&-1&a+b\end{vmatrix}\)
= \((a-b)(b-c)\begin{vmatrix} 0&1&c\\a-c&b+c&c^2\\0&-1&a+b\end{vmatrix}\)
= \((a-b)(b-c)(c-a)\begin{vmatrix} 1&c\\-1&a+b\end{vmatrix}\)
= \((a-b)(b-c)(c-a)[a+b+c]\)
= \((a+b+c)(a-b)(b-c)(c-a)\) 

LHS: \(\begin{vmatrix}a+b+2c&a&b\\c&b+c+2a&b\\c&c&c+a+2b \end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1&0\\0&1&-1\\c&a&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix}1&0&0\\0&1&-1\\ c&a+c&c+a+2b\end{vmatrix}\)
= \((a+b+c)^2\begin{vmatrix} 1&-1\\a+c&c+a+2b\end{vmatrix}\)
= (a+b+c)²[(c+a+2b)+(a+c)]
= (a+b+c)²(2a+2b+2c)
= 2(a+b+c)²(a+b+c)
= 2(a+b+c)³ = RHS

We have:\(f(x)=\begin{vmatrix} a&-1&0\\ax&a&-1\\ax^2&ax&a\end{vmatrix}\)
\(=a\begin{vmatrix}1&-1&0\\x&a&-1\\x^2&ax&a\end{vmatrix}\)
\(a\begin{vmatrix} 1&0&0\\x&a+x&-1\\x^2&ax+x^2&a\end{vmatrix}\)
\(a[a+x)a+(ax+x^2)]\)
\(a[a^2+ax+ax+x^2]=a(x^2+2ax+a^2)\)
\(=a(x+a)^2\)
\(f(2x)=a(2x+a)^2\)
\(f(2x)-f(x)=a[(2x+a)^2-(x+a)^2]\)
\(=a[(4x^2+4ax+a^2)-(x^2+2ax +a^2)]\)
\(=a(3x^2+2ax)=ax(3x+2a)\)

\(\begin{vmatrix}1!&2!&3!\\2!&3!&4!\\3!&4!&5! \end{vmatrix}=1!\begin{vmatrix}3!&41\\41&5! \end{vmatrix}-2!\begin{vmatrix} 2!&4!\\3!&5!\end{vmatrix}+31\begin{vmatrix} 2!&3!&\\3!&4!\end{vmatrix}\)
= (1)[3!5!-(4!)²]-(2)(240-144)+6(48-36)
= 142(96) + 6(12) = 144 - 192 + 72
= 216 - 192 = 24 = 4! which is true