\[\color {black} {1.}\ Solve\ the\ following\ equations\ using\ Cramers\ Rule\ \hspace{20cm}\]
\[x\ +\ y\ +\ z\ =\ 3,\ 2\ x\ -\ y\ +\ z\ =\ 2\ and\ 3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}\]
\[\color {black}{Solution:}\ \hspace{20cm}\]
\[x\ +\ y\ +\ z=3\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ +\ z\ =\ 2\ \hspace{15cm}\]
\[3\ x\ +\ 2\ y\ -\ 2\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 1 & 1 \\
2 & -1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
-1 & 1 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(2\ -\ 2)\ – 1 (-4\ -\ 3)\ +\ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta\ =\ 1(0)\ – 1 (-\ 7)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\Delta\ =\ 14\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
3 & 1 & 1 \\
2 & -1 & 1 \\
3 & 2 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =3\begin{vmatrix}
-1 & 1 \\
2 & -2 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 3(2\ -\ 2) – 1 (-\ 4\ -\ (3))\ +\ 1(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x\ =\ 3(0)\ – 1 (-\ 7)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta_x = 0\ +\ 7\ +\ 7\
\hspace{14cm}\]
\[\Delta_x\ =\ 14\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & 3 & 1 \\
2 & 2 & 1 \\
3 & 3 & -2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 1\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
2 & 1 \\
3 & -2 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & 2\\
3 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 1(-\ 4\ -\ 3)\ -\ 3 (-4\ -\ 3)\ +\ 1(6\ -\ 6)\
\hspace{9cm}\]
\[\Delta_y\ =\ 1(- \ 7)\ -\ 3 (-\ 7)\ +\ 1(0)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 7\ +\ 21\ +\ 0\
\hspace{14cm}\]
\[\Delta_y\ =\ 14\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 1 & 3 \\
2 & -1 & 2 \\
3 & 2 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 1\begin{vmatrix}
-1 & 2 \\
2 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & 2 \\
3 & 3 \\
\end{vmatrix}\ +\ 3\begin{vmatrix}
2 & – 1\\
3 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 1(-\ 3\ -\ 4)\ -\ 1 (6\ -\ 6)\ +\ 3(4\ +\ 3)\
\hspace{9cm}\]
\[\Delta_z\ =\ 1(-\ 7)\ -\ 1 (0)\ +\ 3(7)\
\hspace{13cm}\]
\[\Delta_z\ =\ -\ 7\ -\ 0\ +\ 21\
\hspace{14cm}\]
\[\Delta_z\ =\ 14\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{14}{14} =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ 2\ and\ z = 3\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 1 + 2 – 3\]\[ = 3 + 2 – 3 = 2\]\[ = RHS\]
\[2.\ Find\ the\ inverse\ of\ \begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\color {black}{Solution:}\ A\ =\begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Let\ A=\begin{bmatrix}
2 & 3 & 4 \\
4 & 3 & 1 \\
1 & 2 & 4 \\
\end{bmatrix}\ \hspace{15cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =2\begin{vmatrix}
3 & 1 \\
2 & 4 \\
\end{vmatrix}\ -\ 3\begin{vmatrix}
4 & 1 \\
1 & 4 \\
\end{vmatrix}\ +\ 4\begin{vmatrix}
4 & 3\\
1 & 2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[ =2(12\ -\ 2)\ – 3 (16\ -\ 1) + 4(8\ -\ 3)\
\hspace{9cm}\]
\[ =2(10)\ – 3 (15) + 4(5)\
\hspace{13cm}\]
\[ =20\ -45 + 20\
\hspace{14cm}\]
\[\begin{vmatrix}
A \\
\end{vmatrix}\ =-5\ \neq\ 0\
\hspace{17cm}\]
\[\therefore\ Inverse\ of\ A\ exist\ \hspace{10cm}\]
\[\color {black}{Cofactors\ of\ Matrix\ A:}\ \hspace{14cm}\]
\[cofactor\ of\ 2 = (-1)^{1\ +\ 1}\ \begin{vmatrix}
3 & 1 \\
2 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^2 (12 – 2)\ \hspace{15cm}\]
\[= (1) (10)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 10\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{1\ +\ 2}\ \begin{vmatrix}
4 & 1 \\
1 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (16 – 1)\ \hspace{15cm}\]
\[= (-1) (15)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = -15\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{1\ +\ 3}\ \begin{vmatrix}
4 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (8- 3)\ \hspace{15cm}\]
\[= (1) (5)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = 5\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{2\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
2 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^3 (12- 8)\ \hspace{15cm}\]
\[= (-1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = -4\ \hspace{15cm}\]
\[cofactor\ of\ 3 = (-1)^{2\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
1 & 4 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (8- 4)\ \hspace{15cm}\]
\[= (1) (4)\ \hspace{15cm}\]
\[cofactor\ of\ 3 = 4\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{2\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
1 & 2 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (4- 3)\ \hspace{15cm}\]
\[= (-1) (1)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -1\ \hspace{15cm}\]
\[cofactor\ of\ 1 = (-1)^{3\ +\ 1}\ \begin{vmatrix}
3 & 4 \\
3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^4 (3- 12)\ \hspace{15cm}\]
\[= (1) (-9)\ \hspace{15cm}\]
\[cofactor\ of\ 1 = -9\ \hspace{15cm}\]
\[cofactor\ of\ 2 = (-1)^{3\ +\ 2}\ \begin{vmatrix}
2 & 4 \\
4 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^5 (2- 16)\ \hspace{15cm}\]
\[= (-1) (-14)\ \hspace{15cm}\]
\[cofactor\ of\ 2 = 14\ \hspace{15cm}\]
\[cofactor\ of\ 4 = (-1)^{3\ +\ 3}\ \begin{vmatrix}
2 & 3 \\
4 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[= (-1)^6 (6- 12)\ \hspace{15cm}\]
\[= (1) (-6)\ \hspace{15cm}\]
\[cofactor\ of\ 4 = -6\ \hspace{15cm}\]
\[Cofactor\ matrix=\begin{bmatrix}
10 & -15 & 5 \\
-4 & 4 & -1 \\
-9 & 14 & -6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[Adj.\ A=\begin{bmatrix}
10 & -4 & -9 \\
-15 & 4 & 14 \\
5 & -1 & -6 \\
\end{bmatrix}\ \hspace{15cm}\]
\[A^{-1} = \frac{1}{\begin{vmatrix} A \\ \end{vmatrix}}\ adj.\ A\ \hspace{5cm}\]
\[A^{-1} = \frac{1}{-5}\ \begin{bmatrix}
10 & -4 & -9 \\
-15 & 4 & 14 \\
5 & -1 & -6 \\
\end{bmatrix}\ \hspace{2cm}\]
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