====== 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. =====Question 1(i)===== 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. =====Question 1(ii)===== 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*} =====Question 1(iii)===== 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. =====Question 1(iv)===== 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*} =====Question 1(v)===== 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*} =====Question 1(vi)===== 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. ====Go to ==== [[math-11-nbf:sol:unit02:ex2-5-p2|Question 2 >]]