Question

In Exercises 21 through 25, find the reduced row-echelon form of the given matrix A. Then find a basis of the image of A and a basis of the kernel of A.

21.$\left[\begin{array}{ccc}1& 3& 9\\ 4& 5& 8\\ 7& 6& 3\end{array}\right]$

Short answer

The reduced row-echelon form of the given matrix A is $\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\\ 0& 0& 0\end{array}\right]$.

The basis of the image of A = $\left\{\left[\begin{array}{c}1\\ 4\\ 7\end{array}\right],\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]\right\}$.

The basis of the kernel of A = $\left\{\left[\begin{array}{c}-3\\ 4\\ -1\end{array}\right]\right\}$.

Step-by-step solution

Finding the reduced row-echelon form of the given matrix

We have$A=\left[\begin{array}{ccc}1& 3& 9\\ 4& 5& 8\\ 7& 6& 3\end{array}\right]$

Thus the reduced row-echelon form of A is $A=\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\\ 0& 0& 0\end{array}\right]$.

  Finding the basis of the image of A

Let ${\overrightarrow{v}}_1=\left[\begin{array}{c}1\\ 0\\ 0\end{array}\right],{\overrightarrow{v}}_2=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],{\overrightarrow{v}}_3=\left[\begin{array}{c}-3\\ 4\\ 0\end{array}\right]$

Here we have role="math" localid="1659411936043" ${\overrightarrow{v}}_3=-3{\overrightarrow{v}}_1+4{\overrightarrow{v}}_2$.

$\Rightarrow {\overrightarrow{v}}_3$is redundant vector ${\overrightarrow{v}}_1$and role="math" localid="1659411791029" ${\overrightarrow{v}}_2$are non-redundant vectors.

The non-redundant column vectors of A form the basis of the image of A.

Since column vectors ${\overrightarrow{v}}_1\mathrm{and}{\overrightarrow{v}}_2$are non-redundant vectors of matrix A, thus the basis of the image A = role="math" localid="1659412047823" $\left\{\left[\begin{array}{c}1\\ 4\\ 7\end{array}\right],\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]\right\}$.

  Finding the basis of the kernel of A

Since the vector is redundant vector such that

$$\begin{gathered} {\overrightarrow{v}}_3=-3{\overrightarrow{v}}_1+4{\overrightarrow{v}}_2 \\ \Rightarrow -3{\overrightarrow{v}}_1+4{\overrightarrow{v}}_2-{\overrightarrow{v}}_3=0 \end{gathered}$$

Thus the vector in the kernel of A is $\overrightarrow{w}=\left[\begin{array}{c}-3\\ 4\\ -1\end{array}\right]$.

Final Answer

The reduced row-echelon form of the given matrix A is $\left[\begin{array}{ccc}1& 0& -3\\ 0& 1& 4\\ 0& 0& 0\end{array}\right]$.

The basis of the image of A = $\left\{\left[\begin{array}{c}1\\ 4\\ 7\end{array}\right],\left[\begin{array}{c}3\\ 5\\ 6\end{array}\right]\right\}$.

The basis of the kernel of A = $\left\{\left[\begin{array}{c}-3\\ 4\\ -1\end{array}\right]\right\}$.