class 9th math 1.6 solution

Question#1: Use matrices, if possible to solve the following systems of linear equations.
(i) Inversion Method (ii) Cramer’s Rule

\(i\)

\[\begin{align}
 & 2x-2y=4 \\
 & 3x+2y=6 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & -2  \\
   3 & 2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   6  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & -2  \\
   3 & 2  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 2 \right)-\left( -2 \right)\left( 3 \right) \\
 & =4+6 \\
 & =10 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   2 & 2  \\
   -3 & 2  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{10}\left[ \begin{matrix}
   2 & 2  \\
   -3 & 2  \\
\end{matrix} \right]\left[ \begin{matrix}
   4  \\
   6  \\
\end{matrix} \right] \\
 & =\frac{1}{10}\left[ \begin{matrix}
   8+12  \\
   -12+12  \\
\end{matrix} \right] \\
 & =\frac{1}{10}\left[ \begin{matrix}
   20  \\
   0  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{20}{10}  \\
   \frac{0}{20}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   2  \\
   0  \\
\end{matrix} \right] \\
  x&=2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( 2,0 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & -2  \\
   3 & 2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   6  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & -2  \\
   3 & 2  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 2 \right)-\left( -2 \right)\left( 3 \right) \\
 & =4+6 \\
 & =10 \\
\left| A_x \right|&=\left| \begin{matrix}
   4 & -2  \\
   6 & 2  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( 2 \right)-\left( -2 \right)\left( 6 \right) \\
 & =8+12 \\
 & =20 \\
\left| A_y \right|&=\left| \begin{matrix}
   2 & 4  \\
   3 & 6  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 6 \right)-\left( 4 \right)\left( 3 \right) \\
 & =12-12 \\
 & =0 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{20}{10} \\
 & =2 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{0}{10} \\
 & =0 \\
x&=2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( 2,0 \right) \right\}
\end{align}\]

\(ii\)

\[\begin{align}
 & 2x+y=3 \\
 & 6x+5y=1 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & 1  \\
   6 & 5  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   3  \\
   1  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & 1  \\
   6 & 5  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 5 \right)-\left( 1 \right)\left( 6 \right) \\
 & =10-6 \\
 & =4 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   5 & -1  \\
   -6 & 2  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{4}\left[ \begin{matrix}
   5 & -1  \\
   -6 & 2  \\
\end{matrix} \right]\left[ \begin{matrix}
   3  \\
   1  \\
\end{matrix} \right] \\
 & =\frac{1}{4}\left[ \begin{matrix}
   15-1  \\
   -18+2  \\
\end{matrix} \right] \\
 & =\frac{1}{4}\left[ \begin{matrix}
   14  \\
   -16 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{14}{4}  \\
   -\frac{16}{4}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{7}{2}  \\
   -4  \\
\end{matrix} \right] \\
  x&=\frac{7}{2},\quad y=-4 \\
\text{Solution Set }&=\left\{ \left( \frac{7}{2},-4 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & 1  \\
   6 & 5  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   3  \\
   1  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & 1  \\
   6 & 5  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 5 \right)-\left( 1 \right)\left( 6 \right) \\
 & =10-6 \\
 & =4 \\
\left| A_x \right|&=\left| \begin{matrix}
   3 & 1  \\
   1 & 5  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 5 \right)-\left( 1 \right)\left( 1 \right) \\
 & =15-1 \\
 & =14 \\
\left| A_y \right|&=\left| \begin{matrix}
   2 & 3  \\
   6 & 1  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( 1 \right)-\left( 3 \right)\left( 6 \right) \\
 & =2-18 \\
 & =-16 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{14}{4} \\
 & =\frac{7}{2} \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{-16}{4} \\
 & =-4 \\
x&=\frac{7}{2},\quad y=-4 \\
\text{Solution Set }&=\left\{ \left( \frac{7}{2},-4 \right) \right\}
\end{align}\]

\(iii\)

\[\begin{align}
 & 4x+2y=8 \\
 & 3x-y=-1 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   4 & 2  \\
   3 & -1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   8  \\
   -1  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   4 & 2  \\
   3 & -1  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -1 \right)-\left( 2 \right)\left( 3 \right) \\
 & =-4-6 \\
 & =-10 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   -1 & -2  \\
   -3 & 4  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{-10}\left[ \begin{matrix}
   -1 & -2  \\
   -3 & 4  \\
\end{matrix} \right]\left[ \begin{matrix}
   8  \\
   -1  \\
\end{matrix} \right] \\
 & =\frac{1}{-10}\left[ \begin{matrix}
   -8+2  \\
   -24-4  \\
\end{matrix} \right] \\
 & =\frac{1}{-10}\left[ \begin{matrix}
   -6  \\
   -28 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{-6}{-10}  \\
   \frac{-28}{-10}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{3}{5}  \\
   \frac{14}{5}   \\
\end{matrix} \right] \\
  x&=\frac{3}{5},\quad y=\frac{14}{5} \\
\text{Solution Set }&=\left\{ \left( \frac{3}{5},\frac{14}{5} \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   4 & 2  \\
   3 & -1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   8  \\
   -1  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   4 & 2  \\
   3 & -1  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -1 \right)-\left( 2 \right)\left( 3 \right) \\
 & =-4-6 \\
 & =-10 \\
\left| A_x \right|&=\left| \begin{matrix}
   8 & 2  \\
   -1 & -1  \\
\end{matrix} \right| \\
 & =\left( 8 \right)\left( -1 \right)-\left( 2 \right)\left( -1 \right) \\
 & =-8+2 \\
 & =-6 \\
\left| A_y \right|&=\left| \begin{matrix}
   4 & 8  \\
   3 & -1  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -1 \right)-\left( 8 \right)\left( 3 \right) \\
 & =-4-24 \\
 & =-28 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-6}{-10} \\
 & =\frac{3}{5} \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{-28}{-10} \\
 & =\frac{14}{5} \\
x&=\frac{3}{5},\quad y=\frac{14}{5} \\
\text{Solution Set }&=\left\{ \left( \frac{3}{5},\frac{14}{5} \right) \right\}
\end{align}\]

\(iv\)

\[\begin{align}
 & 3x-2y=-6 \\
 & 5x-2y=-10 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   3 & -2  \\
   5 & -2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   -6  \\
   -10  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   3 & -2  \\
   5 & -2  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( -2 \right)-\left( -2 \right)\left( 5 \right) \\
 & =-6+10 \\
 & =4 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   -2 & 2  \\
   -5 & 3  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{4}\left[ \begin{matrix}
   -2 & 2  \\
   -5 & 3  \\
\end{matrix} \right]\left[ \begin{matrix}
   -6  \\
   -10  \\
\end{matrix} \right] \\
 & =\frac{1}{4}\left[ \begin{matrix}
   12-20  \\
   30-30  \\
\end{matrix} \right] \\
 & =\frac{1}{4}\left[ \begin{matrix}
   -8  \\
   0 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{-8}{4}  \\
   \frac{0}{4}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   -2  \\
   0   \\
\end{matrix} \right] \\
  x&=-2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( -2,0 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   3 & -2  \\
   5 & -2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   -6  \\
   -10  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   3 & -2  \\
   5 & 2  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 2 \right)-\left(-2 \right)\left( 5 \right) \\
 & =-6+10 \\
 & =4 \\
\left| A_x \right|&=\left| \begin{matrix}
   -6 & -2  \\
   -10 & -2  \\
\end{matrix} \right| \\
 & =\left( -6 \right)\left( -2 \right)-\left( -2 \right)\left( -10 \right) \\
 & =12-20 \\
 & =-8 \\
\left| A_y \right|&=\left| \begin{matrix}
   3 & -6  \\
   5 & -10  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( -10 \right)-\left( -6 \right)\left( 5 \right) \\
 & =-30+30 \\
 & =0 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-8}{4} \\
 & =-2 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{0}{4} \\
 & =0 \\
x&=-2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( -2,0 \right) \right\}
\end{align}\]

\(v\)

\[\begin{align}
 & 3x-2y=4 \\
 & -6x+4y=7 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   3 & -2  \\
   -6 & 4  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   7  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   3 & -2  \\
   -6 & 4  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 4 \right)-\left( -2 \right)\left( -6 \right) \\
 & =12-12 \\
 & =0\\
\text{Since }\left| A \right|&=0\text{, So, Solution is not possible }
\end{align}\]

\(vi\)

\[\begin{align}
 & 4x+y=9 \\
 & -3x-y=-5 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   4 & 1  \\
   -3 & -1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   9  \\
   -5  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   4 & 1  \\
   -3 & -1  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -1 \right)-\left( 1 \right)\left( -3 \right) \\
 & =-4+3 \\
 & =-1 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   -1 & -1  \\
   3 & 4  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{-1}\left[ \begin{matrix}
   -1 & -1  \\
   3 & 4  \\
\end{matrix} \right]\left[ \begin{matrix}
   9  \\
   -5  \\
\end{matrix} \right] \\
 & =\frac{1}{-1}\left[ \begin{matrix}
   -9+5  \\
   27-20  \\
\end{matrix} \right] \\
 & =\frac{1}{-1}\left[ \begin{matrix}
   -4  \\
   7 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{-4}{-1}  \\
   \frac{7}{-1}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   4  \\
   -7   \\
\end{matrix} \right] \\
  x&=4,\quad y=-7 \\
\text{Solution Set }&=\left\{ \left( 4,-7 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   4 & 1  \\
   -3 & -1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   9  \\
   -5  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   4 & 1  \\
   -3 & -1  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -1 \right)-\left(1 \right)\left( -3 \right) \\
 & =-4+3 \\
 & =-1 \\
\left| A_x \right|&=\left| \begin{matrix}
   9 & 1  \\
   -5 & -1  \\
\end{matrix} \right| \\
 & =\left( 9 \right)\left( -1 \right)-\left( 1 \right)\left( -5 \right) \\
 & =-9+5 \\
 & =-4 \\
\left| A_y \right|&=\left| \begin{matrix}
   4 & 9  \\
   -3 & -5  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -5 \right)-\left( 9 \right)\left( -3 \right) \\
 & =-20+27 \\
 & =7 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-4}{-1} \\
 & =4 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{7}{-1} \\
 & =-7 \\
x&=4,\quad y=-7 \\
\text{Solution Set }&=\left\{ \left( 4,-7 \right) \right\}
\end{align}\]

\(vii\)

\[\begin{align}
 & 2x-2y=4 \\
 & -5x-2y=-10 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & -2  \\
   -5 & -2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   -10  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & -2  \\
   -5 & -2  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( -2 \right)-\left( -2 \right)\left( -5 \right) \\
 & =-4-10 \\
 & =-14 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   -2 & 2  \\
   5 & 2  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{-14}\left[ \begin{matrix}
   -2 & 2  \\
   5 & 2  \\
\end{matrix} \right]\left[ \begin{matrix}
   4  \\
   -10  \\
\end{matrix} \right] \\
 & =\frac{1}{-14}\left[ \begin{matrix}
   -8-20  \\
   20-20  \\
\end{matrix} \right] \\
 & =\frac{1}{-14}\left[ \begin{matrix}
   -28  \\
   0 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{-28}{-14}  \\
   \frac{0}{-14}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   2  \\
   0   \\
\end{matrix} \right] \\
  x&=2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( 2,0 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   2 & -2  \\
   -5 & -2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   -10  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   2 & -2  \\
   -5 & -2  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( -2 \right)-\left( -2 \right)\left( -5 \right) \\
 & =-4-10 \\
 & =-14 \\
\left| A_x \right|&=\left| \begin{matrix}
   4 & -2  \\
   -10 & -2  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( -2 \right)-\left( -2 \right)\left( -10 \right) \\
 & =-8-20 \\
 & =-28 \\
\left| A_y \right|&=\left| \begin{matrix}
   2 & 4  \\
   -5 & -10  \\
\end{matrix} \right| \\
 & =\left( 2 \right)\left( -10 \right)-\left( 4 \right)\left( -5 \right) \\
 & =-20+20 \\
 & =0 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-28}{-14} \\
 & =2 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{0}{-14} \\
 & =0 \\
x&=2,\quad y=0 \\
\text{Solution Set }&=\left\{ \left( 2,0 \right) \right\}
\end{align}\]

\(viii\)

\[\begin{align}
 & 3x-4y=4 \\
 & x+2y=8 \\
\end{align}\]

Solution:
Inversion Method:

\[\begin{align}
   A&=\left[ \begin{matrix}
   3 & -4  \\
   1 & 2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   8  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   3 & -4  \\
   1 & 2  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 2 \right)-\left( -4 \right)\left( 1 \right) \\
 & =6+4 \\
 & =10 \\
  Adj\left( A \right) &=\left[ \begin{matrix}
   2 & 4  \\
   -1 & 3  \\
\end{matrix} \right] \\
  X&=\frac{1}{\left| A \right|}Adj\left( A \right)\times b \\
 & =\frac{1}{10}\left[ \begin{matrix}
   2 & 4  \\
   -1 & 3  \\
\end{matrix} \right]\left[ \begin{matrix}
   4  \\
   8  \\
\end{matrix} \right] \\
 & =\frac{1}{10}\left[ \begin{matrix}
   8+32  \\
   -4+24  \\
\end{matrix} \right] \\
 & =\frac{1}{10}\left[ \begin{matrix}
   40  \\
   20 \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   \frac{40}{10}  \\
   \frac{20}{10}  \\
\end{matrix} \right] \\
 & =\left[ \begin{matrix}
   4  \\
   2   \\
\end{matrix} \right] \\
  x&=4,\quad y=2 \\
\text{Solution Set }&=\left\{ \left( 4,2 \right) \right\}
\end{align}\]

Cramer’s Rule:

\[\begin{align}
   A&=\left[ \begin{matrix}
   3 & -4  \\
   1 & 2  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   4  \\
   8  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   3 & -4  \\
   1 & 2  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 2 \right)-\left( -4 \right)\left( 1 \right) \\
 & =6+4 \\
 & =10 \\
\left| A_x \right|&=\left| \begin{matrix}
   4 & -4  \\
   8 & 2  \\
\end{matrix} \right| \\
 & =\left( 4 \right)\left( 2 \right)-\left( -4 \right)\left( 8 \right) \\
 & =8+32 \\
 & =40 \\
\left| A_y \right|&=\left| \begin{matrix}
   3 & 4  \\
   1 & 8  \\
\end{matrix} \right| \\
 & =\left( 3 \right)\left( 8 \right)-\left( 4 \right)\left( 1 \right) \\
 & =24-4 \\
 & =20 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{40}{10} \\
 & =4 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{20}{10} \\
 & =2 \\
x&=4,\quad y=2 \\
\text{Solution Set }&=\left\{ \left( 4,2 \right) \right\}
\end{align}\]

Question#2: The length of a rectangle is 4 times its width. The perimeter of the rectangle is 150cm. Find the dimensions of the rectangle.

Solution:

\[\begin{align}
    &\text{Let } \\
  &\text{Width of rectangle}=x \\
  &\text{Length of rectangle}=y \\
  &\text{As per conditions of Question}&& \\
  &y=4x\qquad&&|\qquad2\left( x+y \right)=150 \\
  &-4x+y=0…\left( i \right) \qquad &&|\qquad x+y=\frac{150}{2} \\
 &\qquad &&| \qquad x+y=75…\left( ii \right) \\
\end{align}\]

\(\qquad \text{Changing Equations into matrix form}\)

\[\begin{align}
   A&=\left[ \begin{matrix}
   -4 & 1  \\
   1 & 1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   0  \\
   75  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   -4 & 1  \\
   1 & 1  \\
\end{matrix} \right| \\
 & =\left( -4 \right)\left( 1 \right)-\left( 1 \right)\left( 1 \right) \\
 & =-4-1 \\
 & =-5 \\
\left| A_x \right|&=\left| \begin{matrix}
   0 & 1  \\
   75 & 2  \\
\end{matrix} \right| \\
 & =\left( 0 \right)\left( 2 \right)-\left( 1 \right)\left( 75 \right) \\
 & =0-75 \\
 & =-75 \\
\left| A_y \right|&=\left| \begin{matrix}
   -4 & 0  \\
   1 & 75  \\
\end{matrix} \right| \\
 & =\left( -4 \right)\left( 75 \right)-\left( 0 \right)\left( 1 \right) \\
 & =-300-0 \\
 & =-300 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-75}{-5} \\
 & =15 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{-300}{-5} \\
 & =60 \\
x&=15,\quad y=60 \\
\text{Width of rectangle}&=15 \\
 \text{Length of rectangle}&=60 \\
\end{align}\]

Question#3: Two sides of a rectangle differ by 3.5cm. Find the dimension of the rectangle if its perimeter is 67cm.

Solution:

\[\begin{align}
    &\text{Let } \\
  &\text{Width of rectangle}=x \\
  &\text{Length of rectangle}=y \\
  &\text{As per conditions of Question}&& \\
  &y-x=3.5\qquad&&|\qquad2\left( x+y \right)=67 \\
  &-x+y=3.5…\left( i \right) \qquad &&|\qquad x+y=\frac{67}{2} \\
 &\qquad &&| \qquad x+y=33.5…\left( ii \right) \\
\end{align}\]

\(\qquad \text{Changing Equations into matrix form}\)

\[\begin{align}
   A&=\left[ \begin{matrix}
   -1 & 1  \\
   1 & 1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   3.5  \\
   33.5  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   -1 & 1  \\
   1 & 1  \\
\end{matrix} \right| \\
 & =\left( -1 \right)\left( 1 \right)-\left( 1 \right)\left( 1 \right) \\
 & =-1-1 \\
 & =-2 \\
\left| A_x \right|&=\left| \begin{matrix}
   3.5 & 1  \\
   33.5 & 1  \\
\end{matrix} \right| \\
 & =\left( 3.5 \right)\left( 1 \right)-\left( 1 \right)\left( 33.5 \right) \\
 & =3.5-33.5 \\
 & =-30 \\
\left| A_y \right|&=\left| \begin{matrix}
   -1 & 3.5  \\
   1 & 33.5  \\
\end{matrix} \right| \\
 & =\left( -1 \right)\left( 33.5 \right)-\left( 3.5 \right)\left( 1 \right) \\
 & =-33.5-3.5 \\
 & =-37 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-30}{-2} \\
 & =15 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{-37}{-2} \\
 & =18.5 \\
x&=15,\quad y=18.5 \\
\text{Width of rectangle}&=15 \\
 \text{Length of rectangle}&=18.5 \\
\end{align}\]

Question#4: The third angle of an isosceles triangle is 16 less than the sum of two equal angles. Find three angles of the triangle.

Solution:

\[\begin{align}
    &\text{Let } \\
  &\text{Each Equal angle}=x \\
  &\text{Third angle}=y \\
  &\text{As per conditions of Question}&& \\
  &y=x+x-16\qquad&&|\qquad x+x+y=180 \\
  &y=2x-16 \qquad &&|\qquad 2x+y=180…\left( ii \right) \\
&-2x+y=-16…\left( i \right) \qquad &&| \\
\end{align}\]

\(\qquad \text{Changing Equations into matrix form}\)

\[\begin{align}
   A&=\left[ \begin{matrix}
   -2 & 1  \\
   2 & 1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   -16  \\
   180  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   -2 & 1  \\
   2 & 1  \\
\end{matrix} \right| \\
 & =\left( -2 \right)\left( 1 \right)-\left( 1 \right)\left( 2 \right) \\
 & =-2-2 \\
 & =-4 \\
\left| A_x \right|&=\left| \begin{matrix}
   -16 & 1  \\
   180 & 1  \\
\end{matrix} \right| \\
 & =\left( -16 \right)\left( 1 \right)-\left( 1 \right)\left( 180 \right) \\
 & =-16-180 \\
 & =-196 \\
\left| A_y \right|&=\left| \begin{matrix}
   -2 & -16  \\
   2 & 180  \\
\end{matrix} \right| \\
 & =\left( -2 \right)\left( 180 \right)-\left( -16 \right)\left( 2 \right) \\
 & =-360+32 \\
 & =-328 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{-196}{-4} \\
 & =49 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{-328}{-4} \\
 & =82 \\
x&=49,\quad y=82 \\
\text{Length of each equal angle}&=49 \\
 \text{Length of third angle}&=82 \\
\end{align}\]

Question#5: One acute angle of a right triangle is 12o more than twice the other acute angle. Find the acute angles of the right triangle.

Solution:

\[\begin{align}
    &\text{Let } \\
  &\text{On acute angle}=x \\
  &\text{Second acute angle}=y \\
  &\text{As per conditions of Question}&& \\
  &x=2y+12\qquad&&|\qquad x+y+90=180 \\
  &x-2y=12 \qquad &&|\qquad x+y=180-90 \\
&x-2y=12…\left( i \right) \qquad &&| \qquad x+y=90…(ii) \\
\end{align}\]

\(\qquad \text{Changing Equations into matrix form}\)

\[\begin{align}
   A&=\left[ \begin{matrix}
   1 & -2  \\
   1 & 1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   12  \\
   90  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   1 & -2  \\
   1 & 1  \\
\end{matrix} \right| \\
 & =\left(1 \right)\left( 1 \right)-\left( -2 \right)\left( 1 \right) \\
 & =1+2 \\
 & =3 \\
\left| A_x \right|&=\left| \begin{matrix}
   12 & -2  \\
   90 & 1  \\
\end{matrix} \right| \\
 & =\left(12 \right)\left( 1 \right)-\left( -2 \right)\left( 90 \right) \\
 & =12+180 \\
 & =192 \\
\left| A_y \right|&=\left| \begin{matrix}
   1 & 12  \\
   1 & 90  \\
\end{matrix} \right| \\
 & =\left(1 \right)\left( 90 \right)-\left( 12 \right)\left( 1 \right) \\
 & =90-12 \\
 & =78 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{192}{3} \\
 & =64 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{78}{3} \\
 & =26 \\
x&=64,\quad y=26 \\
\text{Length of first acute angle}&=64 \\
 \text{Length of second acute angle}&=26 \\
\end{align}\]

Question#6: Two cars that are 600 km apart are moving towards each other. Their speed differs by 6 Km per hour and the cars are 123 km apart after \(4\frac{1}{2}\) hours. Find the speed of each car.

Solution:

\[\begin{align}
    &\text{Let } \\
  &\text{Speed of first car}=x \\
  &\text{Speed of second car}=y \\
  &\text{As per conditions of Question}&& \\
  &x-y=6…\left( i \right)\qquad&&|\qquad distance=600-123=477km \\
  & \qquad &&|\qquad \text{Speed of both cars}=\frac{\text{distance}}{\text{time}} \\
& \qquad &&| \qquad x+y=\frac{477}{\frac{9}{2}} \qquad \because 4\frac{1}{2}=\frac{9}{2}\\
& \qquad &&| \qquad x+y=\cancelto{53}{477}\times \frac{2}{\cancel{9}} \\
& \qquad &&| \qquad x+y=106…(ii) \\
\end{align}\]

\(\qquad \text{Changing Equations into matrix form}\)

\[\begin{align}
   A&=\left[ \begin{matrix}
   1 & -1  \\
   1 & 1  \\
\end{matrix} \right],\;b=\left[ \begin{matrix}
   6  \\
   106  \\
\end{matrix} \right],\;X=\left[ \begin{matrix}
   x  \\
   y  \\
\end{matrix} \right] \\
  \left| A \right|&=\left| \begin{matrix}
   1 & -1  \\
   1 & 1  \\
\end{matrix} \right| \\
 & =\left(1 \right)\left( 1 \right)-\left( -1 \right)\left( 1 \right) \\
 & =1+1 \\
 & =2 \\
\left| A_x \right|&=\left| \begin{matrix}
   6 & -1  \\
   106 & 1  \\
\end{matrix} \right| \\
 & =\left(6 \right)\left( 1 \right)-\left( -1 \right)\left( 106 \right) \\
 & =6+106 \\
 & =112 \\
\left| A_y \right|&=\left| \begin{matrix}
   1 & 6  \\
   1 & 106  \\
\end{matrix} \right| \\
 & =\left(1 \right)\left( 106 \right)-\left( 6 \right)\left( 1 \right) \\
 & =106-6 \\
 & =100 \\
   x&=\frac{\left| {{A}_{x}} \right|}{\left| A \right|} \\
& =\frac{112}{2} \\
 & =56 \\
 y&=\frac{\left| {{A}_{y}} \right|}{\left| A \right|} \\
 & =\frac{100}{2} \\
 & =50 \\
x&=56,\quad y=50 \\
\text{Speed of first car}&=56km/hr \\
 \text{Speed of second car}&=50km/hr \\
\end{align}\]

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