Given vertices are (0, 0), (1, 2) and (4, 3)
Area of the triangle \(=\left|\frac{1}{2}\right| \begin{array}{lll} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \mid\)
\(=\left|\frac{1}{2}\right| \begin{array}{lll} 0 & 0 & 1 \\ 1 & 2 & 1 \\ 4 & 3 & 1 \end{array}|=| \frac{1}{2}[1(3-8)] \mid\)
\(=\left|\frac{1}{2}(-5)\right|=\frac{5}{2}=2.5\)
Area of the triangle = 2.5 sq units

Given |A| = 2
\(|A|=\left|A^T\right| \text { (by property } 1 \text { ) }\)
\(\therefore\left|A^T\right|=2 .\)
|AA^T= |A|A^T| = |A|.|A|  [\(\therefore\) |A|^T= |A|]
\(=2 \times 2=4 .\)

Since all the entries below the principal diagonal are zero, the value of the determinant is (x - 1) (x - 2) (x - 3) = 0 which gives x = 1, 2, 3.

\(\mathrm{LHS}=\left|\begin{array}{ccc} x+2 a & y+2 b & z+2 c \\ x & y & z \\ a & b & c \end{array}\right|\)
\(\text {Applying } R_1 \rightarrow R_1-R_2\)
\(=\left|\begin{array}{ccc} 2 a & 2 b & 2 c \\ x & y & z \\ a & b & c \end{array}\right|=2\left|\begin{array}{lll} a & b & c \\ x & y & z \\ a & b & c \end{array}\right|=2(0)=0\) [Since R & R, are Proportional]
= RHS. Hence Proved.

LHS = \(\begin{bmatrix} sec^2 \theta & tan ^2 \theta & 1 \\ tan^2 \theta & sec^2 \theta & -1 \\ 38 & 36 & 2 \end{bmatrix}\)
Applying C₂ ⟶ C₂ + C₃ we get,
= \(\left| \begin{matrix} { sec }^{ 2 }\theta & 1+{ tan }^{ 2 }\theta & 1 \\ { tan }^{ 2 }\theta & { sec }^{ 2 }\theta & -1 \\ 38 & 36 & 2 \end{matrix} \right| \)
Applying C₂ ⟶ C₂ + C₃ we get,
= \(\left| \begin{matrix} { sec }^{ 2 }\theta & 1+{ tan }^{ 2 }\theta & 1 \\ { tan }^{ 2 }\theta & { -1+sec }^{ 2 }\theta & -1 \\ 38 & 38 & 2 \end{matrix} \right| =\left| \begin{matrix} { sec }^{ 2 }\theta & { sec }^{ 2 }\theta & 1 \\ { tan }^{ 2 }\theta & { tan }^{ 2 }\theta & -1 \\ 38 & 38 & 2 \end{matrix} \right| \)
[\(\therefore\) 1 + tan²\(\theta\) = sec²\(\theta\) and sec²\(\theta\) - 1 = tan²\(\theta\)]
= 0 [\(\therefore\) C₁ ≡ C₂] = RHS
Hence proved.

Given B is singular
\(\therefore|B|=0\)
\(\left|\begin{array}{ccc} b-1 & 2 & 3 \\ 3 & 1 & 2 \\ 1 & -2 & 4 \end{array}\right|=0\)
\((b-1)(4+4)-2(12-2)+3(-6-1)=0\)
\((b-1)(8)-2(10)+3(-7)=0\)
\(8 b-8-20-21=0\)
\(8 b-49=0\)
\(8 b=49\)
\(b=\frac{49}{8}\)

Given A is singular
\(\therefore|A|=0\)
\(\left|\begin{array}{cc} 7 & 3 \\ -2 & a \end{array}\right|=0\)
7a + 6 = 0
7a = -6
\(a=\frac{-6}{7} .\)

 \(\left|\begin{array}{ccc} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 6 x & 9 x & 12 x \end{array}\right|\)
Take 3x from R₃. We get
\(=3 x\left|\begin{array}{lll} 2 & 3 & 4 \\ 5 & 6 & 8 \\ 2 & 3 & 4 \end{array}\right|\)
\(=3 x(0)=0\)
Since R₁ and R₃, are proportional.

\(|A|=\left|\begin{array}{ccc} 0 & a-b & k \\ b-a & 0 & 5 \\ -k & -5 & 0 \end{array}\right|\)
\(=0-(a-b)[0+5 k]+k(-5(b-a)-0)\)
\(=(-a+b)(5 k)+k(-5 b+5 a)\)
\(=-5 k a+5 k b-5 k b+5 a k\)
= 0
\(|A|=0\)
\(\therefore\) A is singular

\(|A|=\left|\begin{array}{ccc} 2 & -3 & 5 \\ 6 & 0 & 4 \\ 1 & 5 & -7 \end{array}\right|\)
\(=2(0-20)+3(-42-4)+5(30-0)\)
\(=2(-20)+3(-46)+5(30)\)
\(=-40-138+150\)
\(=-28 \neq 0\)
\(|A| \neq 0\)
\(\therefore\) A is non singular