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

10 - 12 = 5x - 6x ⇒ x = 2

x = ±2√2

\(2 x^{2}-15 =32-15 \Rightarrow x^{2}=16 \)
\(\Rightarrow x =\pm 4 \Rightarrow x=4( > 0)\)

(a + ib) (a - ib) + (c + id) (c - id) = a²+b²+c²+d²

Here, \(\left| A \right| =\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
= 1(0+0) + 1(9+2) + 2(0-0)
= 11
\(\Rightarrow \left| A \right| =\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ...(i)\)
\(adj\quad A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \quad \)
\(A(adj\quad A)=\left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| \)
\((adj\quad A)A=\left| \begin{matrix} 0 & 3 & 2 \\ -11 & 1 & 8 \\ 0 & 1 & 3 \end{matrix} \right| \left| \begin{matrix} 1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3 \end{matrix} \right| \)
\(=\left| \begin{matrix} 11 & 0 & 0 \\ 0 & 11 & 0 \\ 0 & 0 & 11 \end{matrix} \right| ....(i)\)
Thus, it is verified that A(adj A) = (adj A)A = |A|I

\(|A| =\left|\begin{array}{ccc} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right| \)
\(=2(0)+3(-2)+5(1)=-1 \neq 0 \)
\(\operatorname{Adj} A =\left[\begin{array}{rrr} 0 & 2 & 1 \\ -1 & -9 & -5 \\ 2 & 23 & 13 \end{array}\right]^{\top} \)
\(=\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(A^{-1} =\frac{1}{|A|} \operatorname{adj} A=-\frac{1}{1}\left[\begin{array}{rrr} 0 & -1 & 2 \\ 2 & -9 & 23 \\ 1 & -5 & 13 \end{array}\right] \)
\(=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & -13 \end{array}\right]\)
Consider equations,
\(2 x-3 y+5 z=16 \)
\(3 x+2 y-4 z=-4 \)
\(x+y-2 z=-3\)
Corresponding matrix equation is
\(\left[\begin{array}{rrr} 2 & -3 & 5 \\ 3 & 2 & -4 \\ 1 & 1 & -2 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 16 \\ -4 \\ -3 \end{array}\right]\)
i.e., AX = B, Its solution is X = A ^-1B
\(\text { [from {i} } X=\left[\begin{array}{rrr} 0 & 1 & -2 \\ -2 & 9 & -23 \\ -1 & 5 & 13 \end{array}\right]\left[\begin{array}{r} 16 \\ -4 \\ -3 \end{array}\right] \)
\(\text { or } \left.\quad \begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 0-4+6 \\ -32-36+69 \\ -16-20+39 \end{array}\right]=\left[\begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right] \)
\(\therefore \quad x=2, y=1, z=3\)

\(\begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab &1-a^2+b^2 & 2a \\ 2b & -2a &1-a^2-b^2 \end{vmatrix}\)
= \(\begin{vmatrix} 1+a^2-b^2 & 0 & -b(1+a^2+b^2) \\ 2ab &1-a^2+b^2 & 2a \\ 2b & -2a &1-a^2-b^2 \end{vmatrix}\)
= (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.

Since 6 lies in the second row and third column, its minor M₂₃ is given by
\(M_{23}=\left|\begin{array}{ll} 1 & 2 \\ 7 & 8 \end{array}\right|=8-14=-6\)(obtained by deleting R₂ and C₃ in Δ).

We have 
\(\begin{vmatrix} 2 &4 \\-1 & 2 \end{vmatrix}\) = 2(2) – 4(–1)
= 4 + 4
= 8.