Question

[1 2 0 1] 10. Let A 2 3 1 1 13 5 1 2 (a). Find the reduced row echelon form of A. (b). Using the answer for (a), find rank(A), and find a basis for Col(A).

Answer 1

Answer 10:

Data given:

1 2 20 Let A= 2 3 1 3 5 1 2 1 1

(a). Now, we can find the reduced row-echelon form of matrix A by performing some elementary row operations as -

$A = \begin{bmatrix} 1&2 &0 & 1 \\ 2&3 &1&1 \\ 3& 5& 1& 2 \end{bmatrix}$

On performing $R_{2}\rightarrow R_{2}-2R_{1}$ , we get -

$A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & -1 &1 & -1 \\ 3 & 5 & 1 & 2 \end{bmatrix}$

On performing $R_{3}\rightarrow R_{3}-3R_{1}$ , we get -

$A = \begin{bmatrix} 1 & 2 & 0 & 1 \\ 0 & -1 &1 & -1 \\ 0& -1 & 1 & -1 \end{bmatrix}$

On performing $R_{1}\rightarrow R_{1}+2R_{2}$ , we get -

$A = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & -1 &1 & -1 \\ 0& -1 & 1 & -1 \end{bmatrix}$

On performing $R_{3}\rightarrow R_{3}-R_{2}$ ​​​​​​​ , we get -

$A = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & -1 &1 & -1 \\ 0& 0 & 0 & 0 \end{bmatrix}$

On performing $R_{2}\rightarrow (-1)R_{2}$ ​​​​​​​ , we get -

$A = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 &-1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}$ (in the reduced row-echelon form)

$\therefore \ \ \ rref(A) = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 &-1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}$

(b).

$rref(A) = \begin{bmatrix} 1 & 0 & 2 & -1 \\ 0 & 1 &-1 & 1 \\ 0& 0 & 0 & 0 \end{bmatrix}$

(i). From the reduced row-echelon form of matrix A obtained in part(a), we can see that the matrix has two non-zero rows. Therefore, the rank of the matrix is 2.

$\therefore \ rank(A) = 2$

(ii). Since the reduced row-echelon form of the matrix A has leading 1's in the first and second columns, therefore we can conclude that the first and second vectors of A forms the basis of the column space of A.

Thus, $\begin{Bmatrix} \begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix} , \begin{bmatrix} 2\\ 3\\ 5 \end{bmatrix} \end{Bmatrix}$ ​​​​​​​ is a basis of the column space of A.