Question 1, Exercise 2.3

Solutions of Question 1 of Exercise 2.3 of Unit 02: Matrices and Determinants. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.

Reduce the matrices to the echelon form: $\begin{bmatrix}1 & 3 & -1 \\2 & 1 & 4 \\3 & 4 & -5\end{bmatrix}$.

\begin{align}&\begin{bmatrix} 1 & 3 & -1 \\ 2 & 1 & 4 \\ 3 & 4 & -5 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-3R_1 \\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & -5 & -2 \end{bmatrix} \text{ by } R_3-R_2 \\ \underset{\sim}{R}&\begin{bmatrix} 1 & 3 & -1 \\ 0 & -5 & 6 \\ 0 & 0 & -8 \end{bmatrix}\end{align}

Reduce the matrices to the reduce echelon form: $\begin{align} \begin{bmatrix}2 & 3 & -1 & 9 \\1 & -1 & 2 & -3 \\3 & \quad 1 & 3 & \quad 2\end{bmatrix} \end{align}$.

\begin{align} &\begin{bmatrix} 2 & 3 & -1 & 9 \\ 1 & -1 & 2 & -3 \\ 3 & \quad & 3 & \quad 2 \end{bmatrix} \\ \underset{\sim}{R}& \begin{bmatrix} 1 & -1 & 2 & -3 \\ 2 & 3 & -1 & 9 \\ 3 & \quad 1 & 3 & \quad 2 \end{bmatrix}\text{ by } R_1\leftrightarrow R_2\\ \underset{\sim}{R}& \begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 5 & -5 & 15 \\ 0 & \,\,4 & -3 & 11 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_3-3R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1 & -1 & 2 & -3 \\ 0 & 1 & -1 & 3 \\ 0 & \,\,4 & -3 & 11 \end{bmatrix} \text{by} R_3-4R_2 \text{ and } R_1+R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & \quad 1 & 0 \\ 0 & 1 & -1 & 3 \\ 0 & \,\,0 & 1 & -1 \end{bmatrix} \text{by} R_2+R_3 \text{ and } R_1-R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 0 & 2\\ 0 & 0 & 1 & -1 \end{bmatrix}\end{align}

Reduce the matrices to the reduce echelon form: $\begin{bmatrix}2 & -3 & 1 \\1 & 1 & 2 \\4 & 1 & 7\end{bmatrix}$.

\begin{align}&\begin{bmatrix} 2 & -3 & 1 \\ 1 & 1 & 2 \\ 4 & 1 & 7 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 & 2 \\ 2 & -3 & 1 \\ 4 & 1 & 7 \end{bmatrix} \text{by}R_1\leftrightarrow R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 & 2 \\ 0 & -5 & -3 \\ 0 & -3 & -1 \end{bmatrix} \text{ by } R_2-2R_1 \text{ and } R_3-4R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 1 & 2 \\ 0 & -5 & -3 \\ 0 & 1 & -1 \end{bmatrix} \text{ by }-2R_3+R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 3 \\ 0 & -5 & -3 \\ 0 & 1 & -1 \end{bmatrix} \text{ by }R_1-R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -9 \\ 0 & 1 & -1 \end{bmatrix}\text{ by }R_2+6R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -9 \\ 0 & 0 & 8 \end{bmatrix} \text{ by }R_3-R_2\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 3 \\ 0 & 1 & -9 \\ 0 & 0 & 1 \end{bmatrix} \text{ by }\dfrac{1}{8}R_3\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \text{ by }R_2+9R_3 \text{ and } R_1-3R_3\end{align}

Reduce the matrices to the echelon form: $\begin{align} \begin{bmatrix}1 & 0 & -2 \\2 & 1 & 1 \\3 & 2 & 3\end{bmatrix}\end{align}$.

\begin{align} &\begin{bmatrix} 1 & 0 & -2 \\ 2 & 1 & 1 \\ 3 & 2 & 3 \end{bmatrix}\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 5 \\ 0 & 2 & 9 \end{bmatrix} \text{ by }R_2-2R_1 \text{ and } R_3-3R_1\\ \underset{\sim}{R}&\begin{bmatrix} 1 & 0 & -2 \\ 0 & 1 & 5 \\ 0 & 0 & -1 \end{bmatrix} \text{ by }R_3-2R_2\end{align}

Question 2 >