Question 1, Exercise 2.5

Solutions of Question 1 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.

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]$.

Solution.

\begin{align*} & \quad \left[\begin{array}{ccc}1 & 3 & 5 \\ -6 & 8 & 3 \\ -4 & 6 & 5\end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 26 & 33 \\ 0 & 18 & 25 \end{array}\right]\quad R_2 + 6R_1 \quad R_3 + 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 18 & 25 \end{array}\right] \quad \frac{1}{26}R_2 \\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & \frac{217}{26} \end{array}\right]\quad R_3 - 18R_2\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 5 \\ 0 & 1 & \frac{33}{26} \\ 0 & 0 & 1 \end{array}\right]\quad \frac{26}{217}R_3\end{align*} This is echelon form \begin{align*} \sim & \text{R} \left[\begin{array}{ccc} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\quad R_1 - 5R_3 \quad R_2 - \frac{33}{26}R_3\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\quad R_1 - 3R_2 \end{align*} This is reduce echelon form.

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ll}2 & 1 \\ 3 & 2 \\ 1 & 9\end{array}\right]$.

Solution. \begin{align*} & \quad \left[\begin{array}{cc} 2 & 1 \\ 3 & 2 \\ 1 & 9 \end{array}\right] \\ \sim &\text{R} \left[\begin{array}{cc} 2 & 1 \\ 0 & \frac{1}{2} \\ 1 & 9 \end{array}\right] \quad R_2 - \frac{3}{2}R_1 \\ \sim & \text{R} \left[\begin{array}{cc} 2 & 1 \\ 0 & \frac{1}{2} \\ 0 & \frac{17}{2} \end{array}\right] \quad R_3 - \frac{1}{2}R_1 \\ \sim & \text{R} \left[\begin{array}{cc} 2 & 1 \\ 0 & 1 \\ 0 & \frac{17}{2} \end{array}\right] \quad 2R_2 \\ \sim &\text{R} \left[\begin{array}{cc} 2 & 1 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \quad R_3 - \frac{17}{2}R_2 \\ \sim & \text{R}\left[\begin{array}{cc} 1 & \frac{1}{2} \\ 0 & 1 \\ 0 & 0 \end{array}\right] \quad \frac{1}{2}R_1 \end{align*} This is echelon form. \begin{align*} \sim &\text{R} \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \quad R_1 - \frac{1}{2}R_2 \end{align*} The matrix is now in reduced row-echelon form: \begin{align*} \left[\begin{array}{cc} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{array}\right] \end{align*}

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ccc}2 & -1 & 0 \\ 4 & 7 & 8 \\ -3 & 1 & 3\end{array}\right]$.

Solution. \begin{align*} & \quad \left[\begin{array}{ccc} 2 & -1 & 0 \\ 4 & 7 & 8 \\ -3 & 1 & 3 \end{array}\right] \\ \sim &\text{R} \left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & 9 & 8 \\ -3 & 1 & 3 \end{array}\right] \quad R_2 - 2R_1 \\ \sim & \text{R}\left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & 9 & 8 \\ 0 & \frac{1}{2} & 3 \end{array}\right] \quad R_3 + \frac{3}{2}R_1 \\ \sim & \text{R}\left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & 9 & 8 \\ 0 & 1 & 6 \end{array}\right] \quad 2R_3 \\ \sim & \text{R}\left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & 9 & 8 \\ 0 & 0 & \frac{46}{9} \end{array}\right] \quad R_3 - \frac{1}{9}R_2 \\ \sim &\text{R} \left[\begin{array}{ccc} 2 & -1 & 0 \\ 0 & 9 & 8 \\ 0 & 0 & 1 \end{array}\right] \quad \frac{9}{46}R_3 \\ \sim & \text{R}\left[\begin{array}{ccc} 1 & -\dfrac{1}{2} & 0 \\ 0 & 1 & \frac{8}{9} \\ 0 & 0 & 1 \end{array}\right]\quad \frac{1}{2}R_1 \quad \frac{1}{9}R_2 \end{align*} This is echelon form.

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{ccc}2 & -4 & 3 \\ 4 & 1 & 8 \\ 7 & 3 & 0\end{array}\right]$.

Solution. \begin{align*} &\quad\left[\begin{array}{ccc} 2 & -4 & 3 \\ 4 & 1 & 8 \\ 7 & 3 & 0 \end{array}\right]\\ \sim & \text{R}\left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 4 & 1 & 8 \\ 7 & 3 & 0 \end{array}\right]\quad \frac{1}{2}R_1\\ \sim & \text{R}\left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 9 & 2 \\ 7 & 3 & 0 \end{array}\right]\quad R_2 - 4R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 9 & 2 \\ 0 & 17 & -\dfrac{21}{2} \end{array}\right]\quad R_3 - 7R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 17 & -\dfrac{21}{2} \end{array}\right]\quad \frac{1}{9}R_2\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & -\dfrac{257}{18} \end{array}\right]\quad R_3 + 17R_3\\ \sim & \text{R}\left[\begin{array}{ccc} 1 & -2 & \dfrac{3}{2} \\ 0 & 1 & \frac{2}{9} \\ 0 & 0 & 1 \end{array}\right]\quad -\dfrac{18}{257}R_3 \end{align*}

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{lll}3 & 1 & 2 \\ 2 & 9 & 8\end{array}\right]$.

Solution. \begin{align*} &\quad\left[\begin{array}{ccc} 3 & 1 & 2 \\ 2 & 9 & 8 \end{array}\right]\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 2 & 9 & 8 \end{array}\right]\quad \frac{1}{3}R_1 \\ \sim & \text{R} \left[\begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 9 - 2 \cdot \frac{1}{3} & 8 - 2 \cdot \frac{2}{3} \end{array}\right]\quad R_2 - 2R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 0 & \frac{25}{3} & \frac{20}{3} \end{array}\right]\quad R_2 - 2R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{20}{25} \end{array}\right]\quad \frac{3}{25}R_2 \\ \sim & \text{R} \left[\begin{array}{ccc} 1 & \frac{1}{3} & \frac{2}{3} \\ 0 & 1 & \frac{4}{5} \end{array}\right]\quad \frac{3}{25}R_2 \end{align*}

First reduce each of the matrix into echelon form then into reduced echelon form $\left[\begin{array}{lll}0 & 2 & 4 \\ 0 & 3 & 6 \\ 0 & 1 & 2\end{array}\right]$

Solution. \begin{align*} &\quad\left[\begin{array}{ccc} 0 & 2 & 4 \\ 0 & 3 & 6 \\ 0 & 1 & 2 \end{array}\right]\\ \sim & \text{R}\left[\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 3 & 6 \\ 0 & 2 & 4 \end{array}\right]\quad R_1 \leftrightarrow R_3 \\ \sim & \text{R} \left[\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 2 & 4 \end{array}\right]\quad R_2 - 3R_1\\ \sim & \text{R} \left[\begin{array}{ccc} 0 & 1 & 2 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{array}\right]\quad R_3 - 2R_1\ \end{align*} The matrix in row echelon form.

Question 2 >