====== Question 2, Exercise 2.5 ======
Solutions of Question 2 of Exercise 2.5 of Unit 02: Matrices and Determinants. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 

=====Question 2(i)=====
Find the rank of each of the matrix $\left[\begin{array}{ccc}5 & 9 & 3 \\ 3 & -5 & 6 \\ 2 & 10 & 6\end{array}\right]$

** Solution. **

\begin{align*}&\quad\left[ \begin{array}{ccc}
5 & 9 & 3 \\ 
3 & -5 & 6 \\ 
2 & 10 & 6 
\end{array} \right]\\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
3 & -5 & 6 \\ 
2 & 10 & 6 
\end{array} \right]\quad \frac{1}{5} R1\\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & -\frac{52}{5} & \frac{15}{5} \\ 
2 & 10 & 6 
\end{array} \right]\quad R2 - 3 \cdot R1\\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & -\frac{52}{5} & \frac{15}{5} \\ 
0 & \frac{32}{5} & \frac{24}{5} 
\end{array} \right]\quad R3 - 2 \cdot R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & 1 & -\frac{15}{52} \\ 
0 & \frac{32}{5} & \frac{24}{5} 
\end{array} \right]\quad -\frac{5}{52} R2\\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & 1 & -\frac{15}{52} \\ 
0 & 0 & \frac{648}{52} 
\end{array} \right]\quad R3 - \frac{32}{5} R2 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & 1 & -\frac{15}{52} \\ 
0 & 0 & 1 
\end{array} \right]\quad \frac{52}{648} R3\\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{9}{5} & \frac{3}{5} \\ 
0 & 1 & -\frac{15}{52} \\ 
0 & 0 & 1 
\end{array} \right]
\end{align*}
There are $3$ non-zero rows.\\
The rank of the matrix is $3$.



=====Question 2(ii)=====
Find the rank of each of the matrix $\left[\begin{array}{ccc}-1 & -2 & 3 \\ -1 & 2 & -1 \\ -5 & 2 & 3\end{array}\right]$

** Solution. **
\begin{align*}
&\quad \left[ \begin{array}{ccc}
-1 & -2 & 3 \\ 
-1 & 2 & -1 \\ 
-5 & 2 & 3 
\end{array} \right] \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & 2 & -3 \\ 
-1 & 2 & -1 \\ 
-5 & 2 & 3 
\end{array} \right] \quad (-1)R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & 2 & -3 \\ 
0 & 4 & -4 \\ 
-5 & 2 & 3 
\end{array} \right] \quad R2 + R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & 2 & -3 \\ 
0 & 4 & -4 \\ 
0 & 12 & -12 
\end{array} \right] \quad R3 + 5R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & 2 & -3 \\ 
0 & 1 & -1 \\ 
0 & 12 & -12 
\end{array} \right] \quad \frac{1}{4}R2 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & 2 & -3 \\ 
0 & 1 & -1 \\ 
0 & 0 & 0 
\end{array} \right] \quad R3 - 12R2 \\
\end{align*}
There are $2$ non-zero rows.\\
The rank of the matrix is $2$.

=====Question 2(iii)=====
Find the rank of each of the matrix $\left[\begin{array}{ccc}3 & 2 & 4 \\ 2 & 1 & 6 \\ 4 & -1 & 0\end{array}\right]$

** Solution. **
\begin{align*}
&\quad \left[ \begin{array}{ccc}
3 & 2 & 4 \\ 
2 & 1 & 6 \\ 
4 & -1 & 0 
\end{array} \right] \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
2 & 1 & 6 \\ 
4 & -1 & 0 
\end{array} \right] \quad \frac{1}{3}R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
0 & \frac{1}{3} & \frac{14}{3} \\ 
4 & -1 & 0 
\end{array} \right] \quad R2 - 2R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
0 & \frac{1}{3} & \frac{14}{3} \\ 
0 & -\frac{10}{3} & -\frac{16}{3} 
\end{array} \right] \quad R3 - 4R1 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
0 & 1 & 14 \\ 
0 & -\frac{10}{3} & -\frac{16}{3} 
\end{array} \right] \quad 3R2 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
0 & 1 & 14 \\ 
0 & 0 & \frac{124}{3} 
\end{array} \right] \quad R3 + \frac{10}{3}R2 \\
\sim & \text{R}\left[ \begin{array}{ccc}
1 & \frac{2}{3} & \frac{4}{3} \\ 
0 & 1 & 14 \\ 
0 & 0 & 1 
\end{array} \right] \quad \frac{3}{124}R3 
\end{align*}
There are $3$ non-zero rows.\\
The rank of the matrix is $3$.

=====Question 2(iv)=====
Find the rank of each of the matrix $\left[\begin{array}{ll}1 & 3 \\ 2 & 9 \\ 1 & 6\end{array}\right]$.

** Solution. **
\begin{align*}
&\quad \left[ \begin{array}{cc}
1 & 3 \\ 
2 & 9 \\ 
1 & 6 
\end{array} \right] \\
\sim & \text{R}\left[ \begin{array}{cc}
1 & 3 \\ 
0 & 3 \\ 
1 & 6 
\end{array} \right] \quad R2 - 2R1 \\
\sim & \text{R}\left[ \begin{array}{cc}
1 & 3 \\ 
0 & 3 \\ 
0 & 3 
\end{array} \right] \quad R3 - R1 \\
\sim & \text{R}\left[ \begin{array}{cc}
1 & 3 \\ 
0 & 1 \\ 
0 & 3 
\end{array} \right] \quad \frac{1}{3}R2 \\
\sim & \text{R}\left[ \begin{array}{cc}
1 & 0 \\ 
0 & 1 \\ 
0 & 0 
\end{array} \right] \quad R1 - 3R2, \, R3 - 3R2
\end{align*}
There are $2$ non-zero rows.\\
The rank of the matrix is $2$








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