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.

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 $3$.

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 $2$.

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 $3$.

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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