Question

(4.1)Use the reduced row echelon form to determine

2 4 6

0 0 2

2 -1 5

(4.2)Use the reduced row echelon form to determine

1 2 -4

2 -3 1

0 0 2

Answer 1

4.1--

$\begin{pmatrix} 2 & 4 & 6 \\ 0 & 0 & 2 \\ 2 & -1 & 5 \end{pmatrix}$

(i) $R_1 \rightarrow \dfrac{R_1}{2}$

$\Rightarrow$ $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 2 \\ 2 & -1 & 5 \end{pmatrix}$

(ii) $R_3\rightarrow R_3-2R_1$

$\Rightarrow$ $\begin{pmatrix} 1 & 2 & 3 \\ 0 & 0 & 2 \\ 0 & -5 & -1 \end{pmatrix}$

(iii) Interchange 2 and 3 row

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & -5 & -1 \\ 0 & 0 & 2 \end{pmatrix}$

(iv) $R_2 \rightarrow \dfrac{R_2}{-5}$

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 0.2 \\ 0 & 0 & 2 \end{pmatrix}$

(v) $R_3 \rightarrow \dfrac{R_3}{2}$

$\begin{pmatrix} 1 & 2 & 3 \\ 0 & 1 & 0.2 \\ 0 & 0 & 1 \end{pmatrix}$

This is the Row Echelon Form of the matrix.

Since there are 3 non -zero rows hence the Rank is = 3

4.2--

$\begin{pmatrix} 1 & 2 & -4 \\ 2 & -3 & 1 \\ 0 & 0 & 2 \end{pmatrix}$

(i) $R_2\rightarrow R_2 -2R_1$

$\Rightarrow$ $\begin{pmatrix} 1 & 2 & -4 \\ 0 & -7 & 9 \\ 0 & 0 & 2 \end{pmatrix}$

(ii) $R_2 \rightarrow \dfrac{R_2}{-7}$ $\begin{pmatrix} 1 & 2 & -4 \\ 0 & -1 & -9/7 \\ 0 & 0 & 2 \end{pmatrix}$

$\Rightarrow$ $\begin{pmatrix} 1 & 2 & -4 \\ 0 & -1 & -9/7 \\ 0 & 0 & 2 \end{pmatrix}$

(iii) $R_3 \rightarrow \dfrac{R_3}{2}$

$\Rightarrow$ $\begin{pmatrix} 1 & 2 & -4 \\ 0 & -1 & -9/7 \\ 0 & 0 & 1 \end{pmatrix}$

This is the Row Echelon Form of the matrix.

There are 3 non -zero rows hence the rank of the matrix = 3