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.

24.

$\mathbf{24}\mathbf{·}\left[\begin{array}{ccccc}4& 8& 1& 1& 6\\ 3& 6& 1& 2& 5\\ 2& 4& 1& 9& 10\\ 1& 2& 3& 2& 0\end{array}\right]$

Short answer

The reduced row-echelon form of the given matrix A is $\left[\begin{array}{ccccc}1& 2& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$.

The basis of the image of A =.$\left\{\left[\begin{array}{c}4\\ 3\\ 2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 9\\ 2\end{array}\right],\left[\begin{array}{c}6\\ 5\\ 10\\ 0\end{array}\right]\right\}$

The basis of the kernel of A = localid="1659421616740" $\left\{\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right]\right\}$.

Step-by-step solution

Finding the reduced row-echelon form of the given matrix

We have$A=\left[\begin{array}{ccccc}4& 8& 1& 1& 6\\ 3& 6& 1& 2& 5\\ 2& 4& 1& 9& 10\\ 1& 2& 3& 2& 0\end{array}\right]$

$$\begin{gathered} \Rightarrow A=\left[\begin{array}{ccccc}4& 8& 1& 1& 6\\ 0& 0& 1& 5& 2\\ 0& 0& 1& 17& 14\\ 0& 0& 11& 7& -6\end{array}\right]\left(\begin{array}{ccc}{R}_2\to & 4R_2-& 3R_1\\ R_3\to & 2R_3& R_1\\ R_4\to & 4R_4& R_1\end{array}\right) \\ \Rightarrow A=\left[\begin{array}{ccccc}4& 8& 1& 1& 6\\ 0& 0& 1& 5& 2\\ 0& 0& 0& 12& 12\\ 0& 0& 0& -48& -28\end{array}\right]\left(\begin{array}{ccc}\begin{array}{c}{R}_3\to \\ R_4\to \end{array}& \begin{array}{c}{R}_3-\\ R_4-\end{array}& \begin{array}{c}{R}_2\\ 11R_2\end{array}\end{array}\right) \\ \Rightarrow A=\left[\begin{array}{ccccc}4& 8& 1& 0& 0\\ 0& 0& 1& 4& -4\\ 0& 0& 0& 12& 12\\ 0& 0& 0& -48& -28\end{array}\right]\left(\begin{array}{cc}{C}_4\to & C_4-C_3\\ C_4\to & C_5-C_3\end{array}\right) \\ \Rightarrow A=\left[\begin{array}{ccccc}4& 8& 0& 0& 0\\ 0& 0& 4& 4& -4\\ 0& 0& 0& 24& 0\\ 0& 0& 0& 0& 20\end{array}\right]\left(\begin{array}{cc}{C}_3\to & 4C_3-C_1\\ R_3\to & R_3+3R_2\\ R_4\to & R_4+4R_3\end{array}\right) \end{gathered}$$

$$\begin{gathered} A\Rightarrow \left[\begin{array}{ccccc}1& 2& 0& 0& 0\\ 0& 0& 4& 0& 0\\ 0& 0& 0& 24& 0\\ 0& 0& 0& 0& 20\end{array}\right]\left(\begin{array}{ccc}{R}_1& \frac{1}{4}{R}_1& \\ C_4\to & C_4& C_3\\ C_5\to & C_5& C_3\end{array}\right) \\ A\Rightarrow \left[\begin{array}{ccccc}1& 2& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]\left(\begin{array}{cc}{C}_3\to & \frac{1}{4}{C}_3\\ C_4\to & \frac{1}{24}{C}_4\\ C_5\to & \frac{1}{20}{C}_5\end{array}\right) \end{gathered}$$

Thus the reduced row-echelon form of A is localid="1659421713608" $\left[\begin{array}{ccccc}1& 2& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$.

  Finding the basis of the image of A

Let $\overrightarrow{v_1}=\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right],\overrightarrow{v_2}=\left[\begin{array}{c}2\\ 0\\ 0\\ 0\end{array}\right],\overrightarrow{v_3}=\left[\begin{array}{c}0\\ 1\\ 0\\ 0\end{array}\right],\overrightarrow{v_4}=\left[\begin{array}{c}0\\ 0\\ 1\\ 0\end{array}\right],\overrightarrow{v_5}=\left[\begin{array}{c}0\\ 0\\ 0\\ 1\end{array}\right]$

Here we have $v{⃗}_2=2v{⃗}_1$

$\Rightarrow \overrightarrow{v_2}$is a redundant vector $v{⃗}_1,v{⃗}_3,v{⃗}_{4}andv{⃗}_5$are non-redundant vectors.

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

Since column vectors $v{⃗}_1,v{⃗}_3,v{⃗}_{4}andv{⃗}_5$are non-redundant vectors of matrix A, thus the basis of the image A =.$\left\{\left[\begin{array}{c}4\\ 3\\ 2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 9\\ 2\end{array}\right],\left[\begin{array}{c}6\\ 5\\ 10\\ 0\end{array}\right]\right\}$

  Finding the basis of the kernel of A

Since the vector$v{⃗}_2$ is redundant vector such that

$$\begin{gathered} v{⃗}_2=2v{⃗}_1 \\ \Rightarrow 2v{⃗}_1-v{⃗}_2=0 \end{gathered}$$

Thus the vector in the kernel of A is$\overrightarrow{w}=\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right]$

Final Answer

The reduced row-echelon form of the given matrix A is $\left[\begin{array}{ccccc}1& 2& 0& 0& 0\\ 0& 0& 1& 0& 0\\ 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 1\end{array}\right]$.

The basis of the image of A =.$\left\{\left[\begin{array}{c}4\\ 3\\ 2\\ 1\end{array}\right],\left[\begin{array}{c}1\\ 1\\ 1\\ 3\end{array}\right],\left[\begin{array}{c}1\\ 2\\ 9\\ 2\end{array}\right],\left[\begin{array}{c}6\\ 5\\ 10\\ 0\end{array}\right]\right\}$

The basis of the kernel of A =.$\left[\begin{array}{c}2\\ -1\\ 0\\ 0\\ 0\end{array}\right]$