\[\color {purple}{Example\ 1\ :}\ \color{red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[x + 2y – z=-1,\ 3x + 8y + 2z = 28\ and\ 4x + 9y – z = 14\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[x + 2y – z = – 1\ ————— (1) \hspace{10cm}\]
\[3x + 8y + 2z = 28\ \hspace{15cm}\]
\[4x + 9y – z = 14\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 2 & -1 \\
3 & 8 & 2 \\
4 & 9 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
8 & 2 \\
9 & – 1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
3 & 2 \\
4 & -1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
3 & 8\\
4 & 9 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(-8\ -\ 18)\ – 2 (-3\ -\ 8) – 1(27\ -\ 32)\
\hspace{9cm}\]
\[\Delta =1(-26)\ – 2 (-11) – 1(-5)\
\hspace{13cm}\]
\[\Delta =-26\ +22 + 5\
\hspace{14cm}\]
\[\boxed{\Delta =1}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
-1 & 2 & -1 \\
28 & 8 & 2 \\
14 & 9 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =-1\begin{vmatrix}
8 & 2 \\
9 & -1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
28 & 2 \\
14 & -1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
28 & 8\\
14 & 9 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =-1(-8\ -\ 18)\ – 2 (-28\ -\ 28) – 1(252\ -\ 112)\
\hspace{9cm}\]
\[\Delta_x =-1(-26)\ – 2 (-56) -1 (140)\
\hspace{13cm}\]
\[\Delta_x =26\ + 112 – 140\
\hspace{14cm}\]
\[\boxed{\Delta_x = -2}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & -1 & -1 \\
3 & 28 & 2 \\
4 & 14 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix}
28 & 2 \\
14 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
3 & 2 \\
4 & -1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
3 & 28\\
4 & 14 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(-28\ -\ 28)\ + 1 (-3\ -\ 8) – 1(42\ -\ 112)\
\hspace{9cm}\]
\[\Delta_y =1(-56)\ + 1 (-11) – 1(-70)\
\hspace{13cm}\]
\[\Delta_y = -56\ -11\ + 70\
\hspace{14cm}\]
\[\boxed{\Delta_y =3}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 2 & – 1 \\
3 & 8 & 28 \\
4 & 9 & 14 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix}
8 & 28 \\
9 & 14 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
3 & 28 \\
4 & 14 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
3 & 8\\
4 & 9 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(112\ -\ 252)\ – 2 (42\ -\ 112) – 1(27\ -\ 32)\
\hspace{9cm}\]
\[\Delta_z =1(-140)\ – 2 (-70) – 1(-5)\
\hspace{13cm}\]
\[\Delta_z =-140\ + 140 + 5\
\hspace{14cm}\]
\[\boxed{\Delta_z =5}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{-2}{1} =\ -2\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{3}{1} =\ 3\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{5}{1} =\ 5\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =-2\ y = 3\ z = 5\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = -2 + 2(3) – 5\]\[ = -2 + 6 – 5 = -1\]\[ = RHS\]
\[\color {violet}{Example\ 2\ :}\ \color {red} {Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6,\ 2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ and\ x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[4\ x\ +\ y\ +\ z\ =\ 6\ ——————-(1)\ \hspace{6cm}\]
\[2\ x\ -\ y\ -\ 2\ z\ =\ -\ 6\ \hspace{15cm}\]
\[x\ +\ y\ +\ z\ =\ 3\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
4 & 1 & 1 \\
2 & -1 & -2\\
1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =4\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =4(-1\ +\ 2)\ – 1 (2\ +\ 2)\ +\ 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta\ =\ 4(1)\ – 1 (4)\ +\ 1(7)\
\hspace{13cm}\]
\[\Delta =4\ -\ 4\ +\ 3\
\hspace{14cm}\]
\[\boxed{\Delta\ =\ 3}\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
6 & 1 & 1 \\
-6 & -1 & -2 \\
3 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =6\begin{vmatrix}
-1 & -2 \\
1 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
-6 & -2 \\
3 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
-6 & -1\\
3 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x\ =\ 6(-1\ +\ 2) – 1 (\ -6\ +\ 6)\ +\ 1(-6\ +\ 3)\
\hspace{9cm}\]
\[\Delta_x\ =\ 6(1)\ – 1 (0)\ +\ 1(-3)\
\hspace{13cm}\]
\[\Delta_x = 6\ +\ 0\ -3\
\hspace{14cm}\]
\[\boxed{\Delta_x\ =\ 3}\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
4 & 6 & 1 \\
2 & -6 & -2 \\
1 & 3 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y\ =\ 4\begin{vmatrix}
-6 & -2 \\
3 & 1\\
\end{vmatrix}\ -\ 6\begin{vmatrix}
2 & -2 \\
1 & 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -6\\
1 & 3 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y\ =\ 4(-\ 6\ +\ 6)\ -\ 6 (2\ +\ 2)\ +\ 1(6\ +\ 6)\
\hspace{9cm}\]
\[\Delta_y\ =\ 4(0)\ -\ 6 (4)\ +\ 1(12)\
\hspace{13cm}\]
\[\Delta_y\ =\ -\ 0\ -\ 24\ +\ 12\
\hspace{14cm}\]
\[\boxed{\Delta_y\ =\ -12}\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
4 & 1 & 6 \\
2 & -1 & -6 \\
1 & 1 & 3 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 4\begin{vmatrix}
-1 & -6 \\
1 & 3 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -6 \\
1 & 3 \\
\end{vmatrix}\ +\ 6\begin{vmatrix}
2 & – 1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 4(-\ 3\ +\ 6)\ -\ 1 (6\ +\ 6)\ +\ 6(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_z\ =\ 4(3)\ -\ 1 (12)\ +\ 6(3)\
\hspace{13cm}\]
\[\Delta_z\ =\ 12\ -\ 12\ +\ 18\
\hspace{14cm}\]
\[\boxed{\Delta_z\ =\ 18}\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-12}{3} =\ -4\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{18}{3} =\ 6\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x\ =\ 1,\ y\ =\ -4\ and\ z = 6\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 4(1) – 4 + 6\]\[ = 4 – 4 + 6 = 6\]\[ = RHS\]
\[\color {purple}{Example\ 3\ :}\ \color{red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[x\ +\ y\ +\ z\ =\ 2,\ 2x\ -\ y\ -\ 2z\ =\ -1\ and\ x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[x\ +\ y\ +\ z\ =\ 2\ ————— (1) \hspace{10cm}\]
\[2x\ -\ y\ -\ 2z\ =\ -1\ \hspace{15cm}\]
\[x\ -\ 2y\ -\ z\ =\ 1\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 1 & 1 \\
2 & -1 & -2 \\
1 & -2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
-1 & -2 \\
-2 & -1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -2 \\
1 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(1\ -\ 4)\ – 1 (-2\ +\ 2) + 1(-4\ +\ 1)\
\hspace{9cm}\]
\[\Delta =1(-3)\ – 1 (0) + 1(-3)\
\hspace{13cm}\]
\[\Delta =-3\ – 0 – 3\
\hspace{14cm}\]
\[\Delta =-\ 6\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
2 & 1 & 1 \\
-1 & -1 & -2 \\
1 & -2 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =\ 2\begin{vmatrix}
-1 & -2 \\
-2 & -1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
-1 & -2 \\
1 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
-1 & -1\\
1 & -2 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =\ 2(1\ -\ 4)\ – 1 (1\ +\ 2) + 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_x =\ 2(-3)\ – 1 (3) + 1(3)\
\hspace{13cm}\]
\[\Delta_x =-6\ -3 + 3\
\hspace{14cm}\]
\[\Delta_x =\ -6\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & 2 & 1 \\
2 & -1 & -2 \\
1 & 1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix}
-1 & -2 \\
1 & -1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
2 & -2 \\
1 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & -1\\
1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(1\ +\ 2)\ – 2 (-2\ +\ 2) + 1(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_y =1(3)\ – 2 (0) + 1(3)\
\hspace{13cm}\]
\[\Delta_y = 3\ -\ 0\ +\ 3\
\hspace{14cm}\]
\[\Delta_y =\ 6\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 1 & 2 \\
2 & -1 & -1 \\
1 & -2 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix}
-1 & -1 \\
-2 & 1 \\
\end{vmatrix}\ -\ 1\begin{vmatrix}
2 & -1 \\
1 & 1 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
2 & -1\\
1 & -2 \\ \end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(-1\ -\ 2)\ – 1 (2\ +\ 1) + 2(-4\ +\ 1)\
\hspace{9cm}\]
\[\Delta_z =1(-3)\ – 1 (3) + 2(-3)\
\hspace{13cm}\]
\[\Delta_z =-3\ -3 – 6\
\hspace{14cm}\]
\[\Delta_z =\ -12\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{-6}{-6} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{6}{-6} =\ -1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{-12}{-6} =\ 2\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =1\ y = -1\ z = 2\ in\ equation (1)\ \hspace{18cm}\]
\[LHS\ =\ 1 – 1 + 2 = 2\]\[ = RHS\]
\[\color {purple}{Example\ 4\ :}\ \color{red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[x + 2y + 5z=4,\ 3x + y + 4z = 6\ and\ -x + y +z = -1\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{22cm}\]
\[x + 2y + 5z=4\ ————— (1) \hspace{10cm}\]
\[3x + y + 4z = 6\ \hspace{15cm}\]
\[-x + y +z = -1\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
1 & 2 & 5 \\
3 & 1 & 4 \\
-1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =1\begin{vmatrix}
1 & 4 \\
1 & 1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
3 & 4 \\
-1 & 1 \\
\end{vmatrix}\ +\ 5\begin{vmatrix}
3 & 1\\
-1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =1(1\ -\ 4)\ – 2 (3\ -\ (-4)) + 5(3\ -\ (-1))\
\hspace{9cm}\]
\[\Delta =1(-3)\ – 2 (7) + 5(4)\
\hspace{13cm}\]
\[\Delta =-3\ -14 + 20\
\hspace{14cm}\]
\[\Delta =3\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
4 & 2 & 5 \\
6 & 1 & 4 \\
-1 & 1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =4\begin{vmatrix}
1 & 4 \\
1 & 1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
6 & 4 \\
-1 & 1 \\
\end{vmatrix}\ +\ 5\begin{vmatrix}
6 & 1\\
-1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =4(1\ -\ 4)\ – 2 (6\ -\ (-4)) + 5(6\ -\ (-1))\
\hspace{9cm}\]
\[\Delta_x =4(-3)\ – 2 (10) + 5(7)\
\hspace{13cm}\]
\[\Delta_x =-12\ -20 + 35\
\hspace{14cm}\]
\[\Delta_x =3\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
1 & 4 & 5 \\
3 & 6 & 4 \\
-1 & -1 & 1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =1\begin{vmatrix}
6 & 4 \\
-1 & 1 \\
\end{vmatrix}\ -\ 4\begin{vmatrix}
3 & 4 \\
-1 & 1 \\
\end{vmatrix}\ +\ 5\begin{vmatrix}
3 & 6\\
-1 & -1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =1(6\ -\ (-4))\ – 4 (3\ -\ (-4)) + 5(-3\ -\ (-6))\
\hspace{9cm}\]
\[\Delta_y =1(10)\ – 4 (7) + 5(3)\
\hspace{13cm}\]
\[\Delta_y = 10\ -28\ + 15\
\hspace{14cm}\]
\[\Delta_y =-3\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
1 & 2 & 4 \\
3 & 1 & 6 \\
-1 & 1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z =1\begin{vmatrix}
1 & 6 \\
1 & -1 \\
\end{vmatrix}\ -\ 2\begin{vmatrix}
3 & 6 \\
-1 & -1 \\
\end{vmatrix}\ +\ 4\begin{vmatrix}
3 & 1\\
-1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z =1(-1\ -\ 6)\ – 2 (-3\ -\ (-6)) + 4(3\ -\ (-1))\
\hspace{9cm}\]
\[\Delta_z =1(-7)\ – 2 (3) + 4(4)\
\hspace{13cm}\]
\[\Delta_z =-7\ -6 + 16\
\hspace{14cm}\]
\[\Delta_z =3\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{-3}{3} =\ -1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta} =\ \frac{3}{3} =\ 1\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =1\ y = -1\ z = 1\ in\ equation (1)\ \hspace{18cm}\]
\[LHS = 1 + 2(-1) + 5(1)\]\[ = 1 – 2 + 5 = 4\]\[ = RHS\]
\[\color {purple}{Example\ 5\ :}\ \color{red}{Solve\ the\ following\ equations\ using\ Cramers\ Rule}\ \hspace{20cm}\]
\[3x\ -\ y\ +\ 2z\ =\ 8,\ x\ +\ y\ +\ z\ =\ 2\ and\ 2x\ +\ y\ -\ z\ =\ -\ 1\ \hspace{15cm}\]
\[\color {blue}{Solution:}\ \hspace{20cm}\]
\[3x\ -\ y\ +\ 2z\ =\ 8\ ————— (1) \hspace{10cm}\]
\[x\ +\ y\ +\ z\ =\ 2\ \hspace{15cm}\]
\[2x\ +\ y\ -\ z\ =\ -\ 1\ \hspace{15cm}\]
\[\Delta = \begin{vmatrix}
3 & -1 & 2 \\
1 & 1 & 1 \\
2 & 1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta =\ 3\begin{vmatrix}
1 & 1 \\
1 & – 1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
1 & 1 \\
2 & -1 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
1 & 1\\
2 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta =\ 3(-\ 1\ -\ 1)\ +\ 1 (-1\ -\ 2)\ +\ 2(1\ -\ 2)\
\hspace{9cm}\]
\[\Delta =\ 3(-2)\ +\ 1 (-3)\ +\ 2(-1)\
\hspace{13cm}\]
\[\Delta =\ -6\ -\ 3\ -\ 2\
\hspace{14cm}\]
\[\Delta =\ -\ 11\
\hspace{17cm}\]
\[\Delta_x = \begin{vmatrix}
8 & -1 & 2 \\
2 & 1 & 1 \\
-1 & 1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_x =\ 8\begin{vmatrix}
1 & 1 \\
1 & -1 \\
\end{vmatrix}\ +\ 1\begin{vmatrix}
2 & 1 \\
-1 & -1 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
2 & 1\\
-1 & 1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_x =\ 8(-1\ -\ 1)\ +\ 1 (-2\ +\ 1) + 2(2\ +\ 1)\
\hspace{9cm}\]
\[\Delta_x =\ 8(-2)\ + 1 (-1)\ +\ 2 (3)\
\hspace{13cm}\]
\[\Delta_x =\ -\ 16\ -\ 1\ +\ 6\
\hspace{14cm}\]
\[\Delta_x = -\ 11\
\hspace{17cm}\]
\[\Delta_y = \begin{vmatrix}
3 & 8 & 2 \\
1 & 2 & 1 \\
2 & -1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_y =\ 3\begin{vmatrix}
2 & 1 \\
-1 & -1 \\
\end{vmatrix}\ -\ 8\begin{vmatrix}
1 & 1 \\
2 & -1 \\
\end{vmatrix}\ +\ 2\begin{vmatrix}
1 & 2\\
2 & -1 \\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_y =\ 3(-2\ +\ 1)\ – 8 (-1\ -\ 2) +\ 2(-1\ -\ 4)\
\hspace{9cm}\]
\[\Delta_y =\ 3(-1)\ -\ 8(-3)\ + 2(-5)\
\hspace{13cm}\]
\[\Delta_y =\ -3\ +\ 24\ -\ 10\
\hspace{14cm}\]
\[\Delta_y =\ 11\
\hspace{17cm}\]
\[\Delta_z = \begin{vmatrix}
3 & -1 & 8 \\
1 & 1 & 2 \\
2 & 1 & -1 \\
\end{vmatrix}\ \hspace{15cm}\]
\[\Delta_z\ =\ 3\begin{vmatrix}
1 & 2 \\
1 & -1\\
\end{vmatrix}\ +\ 1\begin{vmatrix}
1 & 2 \\
2 & -1 \\
\end{vmatrix}\ +\ 8\begin{vmatrix}
1 & 1\\
2 & 1\\
\end{vmatrix}\ \hspace{10cm}\]
\[\Delta_z\ =\ 3(-\ 1\ -\ 2)\ +\ 1 (-1\ -\ 4)\ +\ 8(1\ -\ 2)\
\hspace{9cm}\]
\[\Delta_z =\ 3(-3)\ +\ 1 (-5)\ +\ 8(-1)\
\hspace{13cm}\]
\[\Delta_z =\ -\ 9\ -\ 5\ -\ 8\
\hspace{14cm}\]
\[\Delta_z =\ -\ 22\
\hspace{17cm}\]
\[The\ Solution\ is\ \hspace{20cm}\]
\[x=\ \frac{\Delta_x}{\Delta}\ =\ \frac{-11}{-11} =\ 1\ \hspace{20cm}\]
\[y=\ \frac{\Delta_y}{\Delta} =\ \frac{11}{-11} =\ -\ 1\ \hspace{20cm}\]
\[z=\ \frac{\Delta_z}{\Delta}\ =\ \frac{-22}{-11} =\ 2\ \hspace{20cm}\]
\[For\ cross\ verification\ \hspace{20cm}\]
\[Put\ x =\ 1\ y\ =\ -\ 1\ z\ =\ 2\ in\ equation (1)\ \hspace{18cm}\]
\[LHS =\ 3(1)\ -\ (-1)\ +\ 2(2)\]\[ =\ 3\ +\ 1\ +\ 4\ =\ 8\]\[ = RHS\]
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