# Describe the Range of the Matrix Using the Definition of the Range

## Problem 703

Using the definition of the range of a matrix, describe the range of the matrix

\[A=\begin{bmatrix}

2 & 4 & 1 & -5 \\

1 &2 & 1 & -2 \\

1 & 2 & 0 & -3

\end{bmatrix}.\]

## Solution.

By definition, the range $\calR(A)$ of the matrix $A$ is given by

\[\calR(A)=\left \{ \mathbf{b} \in \R^3 \quad \middle | \quad A\mathbf{x}=\mathbf{b} \text{ for some } \mathbf{x} \in \R^4 \right \}.\]

Thus, a vector $\mathbf{b}=\begin{bmatrix}

b_1 \\

b_2 \\

b_3

\end{bmatrix}$ in $\R^3$ is in the range $\calR(A)$ if and only if the system $A\mathbf{x}=\mathbf{b}$ is consistent.

So, let us find the conditions on $\mathbf{b}$ so that the system is consistent.

To do this, we consider the augmented matrix of the system and reduce it as follows.

\begin{align*}

\left[\begin{array}{rrrr|r}

2 & 4 & 1 & -5 & b_1\\

1 &2 & 1 & -2 & b_2 \\

1 & 2 & 0 & -3 &b_3

\end{array}\right]
\xrightarrow{R_1 \leftrightarrow R_2}

\left[\begin{array}{rrrr|r}

1 &2 & 1 & -2 & b_2 \\

2 & 4 & 1 & -5 & b_1\\

1 & 2 & 0 & -3 &b_3

\end{array}\right]
\xrightarrow[R_3-R_1]{R_2-2R_1}\\[6pt]
\left[\begin{array}{rrrr|r}

1 &2 & 1 & -2 & b_2 \\

0 & 0 & -1 & -1 & b_1-2b_2\\

0 & 0 & 0 & -1 & b_3-b_2

\end{array}\right]
\xrightarrow{-R_2}

\left[\begin{array}{rrrr|r}

1 &2 & 1 & -2 & b_2 \\

0 & 0 & 1 & 1 & -b_1+2b_2\\

0 & 0 & -1 & -1 & b_3-b_2

\end{array}\right]\\[6pt]
\xrightarrow[R_3+R_2]{R_1-R_2}

\left[\begin{array}{rrrr|r}

1 &2 & 0 & -3 & b_1-b_2 \\

0 & 0 & 1 & 1 & -b_1+2b_2\\

0 & 0 & 0 & 0 & -b_1 +b_2 +b_3

\end{array}\right].

\end{align*}

Note that if the $(3, 5)$-entry $-b_1+b_2+b_3$ is not zero, then the system $A\mathbf{x}=\mathbf{0}$ is inconsistent because this implies $0=1$.

On the other hand, if $-b_1+b_2+b_3=0$, then we see that the system is consistent.

Hence, the vector $\mathbf{b}$ is in the range $\calR(A)$ if and only if $-b_1+b_2+b_3=0$.

In summary, we have

\[\calR(A)=\left\{ \begin{bmatrix}

b_1 \\

b_2 \\

b_3

\end{bmatrix}\in \R^3 \quad \middle | \quad -b_1+b_2+b_3=0 \right \}.\]

### Spanning set for the range

With a little bit additional computation, we can find the spanning set for the range as follows.

Thus, $\mathbf{b} \in \calR(A)$ if and only if

\begin{align*}

\mathbf{b}=\begin{bmatrix}

b_1 \\

b_2 \\

b_3

\end{bmatrix}=\begin{bmatrix}

b_2+b_3 \\

b_2 \\

b_3

\end{bmatrix}=b_2\begin{bmatrix}

1 \\

1 \\

0

\end{bmatrix}+b_3\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix}.

\end{align*}

In summary, we have

\begin{align*}

\calR(A)&=\left\{ \mathbf{b} \in \R^3 \quad \middle | \quad \mathbf{b}=b_2\begin{bmatrix}

1 \\

1 \\

0

\end{bmatrix}+b_3\begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix} \right \}\\[6pt]
&=\Span\left\{ \begin{bmatrix}

1 \\

1 \\

0

\end{bmatrix}, \begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix} \right \}.

\end{align*}

Hence, the spanning set is

\[ \left\{ \begin{bmatrix}

1 \\

1 \\

0

\end{bmatrix}, \begin{bmatrix}

1 \\

0 \\

1

\end{bmatrix} \right \}.\]

This spanning set is linearly independent, hence it’s a basis for the range.

Thus, the dimension of the range is $2$.

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