Let A = \(\left[ \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right] \)
|A| = \(\left| \begin{matrix} 1 & 2 \\ 2 & 4 \end{matrix} \right| \)
= 4 – 4 = 0
\(\therefore\) A is a singular matrix

\(\left| \begin{matrix} x-1 & x & x-2 \\ 0 & x-2 & x-3 \\ 0 & 0 & x-3 \end{matrix} \right| =0\)
⇒(x – 1)(x – 2)(x – 3) = 0
x = 1, x = 2, x = 3

\(\left| \overset { x }{ 2x\underset { a }{ + } 2a } \quad \overset { y }{ 2y\underset { b }{ + } 2b } \quad \overset { z }{ 2z\underset { c }{ + } 2c } \right|\)
\(=\left| \begin{matrix} x & y & z \\ 2x & 2y & 2z \\ a & b & c \end{matrix} \right| +\left| \begin{matrix} x & y & z \\ 2a & 2b & 2c \\ a & b & c \end{matrix} \right| \)
= 0 + 0
= 0

\(\left| \begin{matrix} 1 & 2 & 4 \\ -1 & 3 & 0 \\ 4 & 1 & 0 \end{matrix} \right| \) = 1 (Minor of 1) –2 (Minor of 2) + 4 (Minor of 4)
\(=1\left| \begin{matrix} 3 & 0 \\ 1 & 0 \end{matrix} \right| -2\left| \begin{matrix} -1 & 0 \\ 4 & 0 \end{matrix} \right| +4\left| \begin{matrix} -1 & 3 \\ 4 & 1 \end{matrix} \right| \)
= 0 – 0 – 52 = –52.

\(\left| \begin{matrix} 2 & 4 \\ -1 & 4 \end{matrix} \right| \) = (2)(4) - (-1)(4)
= 8 + 4 = 12

Let A = \(\begin{vmatrix} x &a &x+a \\ y & b &y+b \\z & c & z+c \end{vmatrix}\)
Applying the elementary transformation, \(C_1\rightarrow C_1+C_2\) we get,
\(A=\begin{vmatrix} x+a&a&a+x\\y+b&b&y+b\\z+c&c&z+c\end{vmatrix}=0[C_1\equiv C_3]\)
\(\therefore\) |A| = 0.

Let |A| = \(\begin{vmatrix} 6 &5 &12 \\ 2 & 4 &4 \\2 & 1 & 4 \end{vmatrix}\)
Taking 2 common from C₁ and 4 common from C₃, we get,
\(|A|=2\times4\begin{vmatrix} 3 & 5&3 \\ 1 & 4 & 1\\1 &1 &1\end{vmatrix}=8\times 0\ [\because C_1\equiv C_3]=0\)

Given \(\begin{vmatrix}2 & 4 \\5 & 1 \end{vmatrix}=\begin{vmatrix} 2x & 4 \\ 6 & x \end{vmatrix}\)
\(\Rightarrow\)   2 - 20 = 2x² - 24
\(\Rightarrow\)  -18 = 2x² - 24
\(\Rightarrow\)  -18 + 24 = 2x²
\(\Rightarrow\)  6 = 2x²
\(\Rightarrow\) x² = 3 
\(\Rightarrow\) x = \(\pm\sqrt{3}\)

Let \(|A|=\begin{vmatrix} 2&-1&-2 \\ 0 &2&-1\\3&-5&0 \end{vmatrix}\)
Expanding along R₁ we get, 
\(|A|=2\begin{vmatrix} 2 & -1\\ -5 & 0\end{vmatrix}+1\begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix}-2\begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}\)
= 2 ( 0 - 5 ) + 1 ( 0 + 3 )  - 2 ( 0 - 6 )
= - 10 + 3 + 12 = 5
\(\therefore\) |A| = 5.

AB = \(\begin{bmatrix} 1 \\ -4 \\3 \end{bmatrix}\begin{bmatrix} -1 &2 & 1 \end{bmatrix}=\begin{bmatrix} -1 & 2 & 1 \\ 4 & -8 & -4 \\-3 & 6 & 3 \end{bmatrix}\)
\(\therefore\)  \({(AB)}^{T}=\begin{bmatrix} -1 &4&-3 \\ 2 & -8&6\\1&-4&3 \end{bmatrix}\)  ....(1)
\({B}^{T}=\begin{bmatrix} -1 & 2 & 1 \end{bmatrix}^{T}=\begin{bmatrix} -1\\2\\1\end{bmatrix}\)and \({A}^{T}={\begin{bmatrix} 1\\-4\\3\end{bmatrix}}^{T}=\begin{bmatrix} 1&-4&3 \end{bmatrix}\)
\(\therefore\)  \({B}^{T}{A}^{T}=\begin{bmatrix} -1\\2\\1 \end{bmatrix}\begin{bmatrix} 1&-4&3 \end{bmatrix}=\begin{bmatrix} -4 & 4&-3 \\ 2&-8 &6\\1&-4&3 \end{bmatrix}\)  ....(2)
From (1) and (2), (AB)^T = B^T . A^T

Let A = \(\begin{vmatrix}5&20\\ 0&-1 \end{vmatrix}\)
Minor of 5 = M₁₁ = -1
Minor of 20 = M₁₂ = 0
Minor of 0 = M₂₁ = 20
Minor of -1 = M₂₂ = 5
Co-factor of 5 = A₁₁ = -1
Co-factor of 20 = A₁₂ = 0
Co-factor of 0 = A₂₁ = -20
Co-factor of -1 = A₂₂ = 5

B = \(\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}\) 
I - B = \(\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}-\begin{bmatrix} 0.6 & 0.9 \\ 0.20 & 0.80 \end{bmatrix}=\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
|I - B| =\(\begin{bmatrix} 0.4 & -0.9 \\ -0.20 & 0.20 \end{bmatrix}\)
= 0.08 - 0.18 = - 0.1 < 0
Since |I - B| is negative, Hawkins - Simon conditions are not satisfied.
Therefore the given system is not viable.

\(|A|=\left| \begin{matrix} -2 & 6 \\ 3 & -9 \end{matrix} \right| =0\)
Since A is a singular matrix, A ^-1 does not exist.

\(AB=\begin{bmatrix} 3&-1\\2&1 \end{bmatrix}\begin{bmatrix} 3&0\\1&-2 \end{bmatrix}=\begin{bmatrix} 9-1&0+2\\6+1&0-2 \end{bmatrix}=\begin{bmatrix} 8&2\\7&-2 \end{bmatrix}\)
= -16 - 14 = -30
\(\therefore\) |AB| = -30

Expanding along R₁ we get,
\(\left|\begin{array}{lll} 2 & x & 3 \\ 4 & 1 & 6 \\ 1 & 2 & 7 \end{array}\right|=0\)
⇒ 2 (7 -12) -x (28- 6) + 3 (8 - 1) = 0
⇒ -10 - 22x + 21 = 0
⇒ 11 = 22x
⇒ \(x={11\over 22}={1\over 2}\)