# Determine if the columns of the matrix form a linearly independent set. Justify each answer.

$$\begin{bmatrix}1&4&-3&0\\-2&-7&4&1\\-4&-5&7&5\end{bmatrix}$$

The main objective of this question is to determine whether the columns of the given matrix form a linearly independent or dependent set.

If the non-trivial linear combination of vectors equals zero, then the set of vectors is said to be linearly dependent. The vectors are said to be linearly independent if there is no such linear combination.

Mathematically, assume that $B=\{v_1,v_2,v_3,\cdots\}$ is the set of vectors. Then $B$ will be linearly independent if the vector equation $y_1v_1+y_2v_2+\cdots+y_kv_k=0$ possesses the trivial solution such that $y_1=y_2=\cdots=y_k=0$.

Let $A$ be a matrix, then columns of $A$ will be linearly independent if the equation $Ax=0$ possesses the trivial solution. In other words, The row space of matrix $A$ is the span of its rows. The column space denoted by $C(A)$ is the span of $A$’s columns. The dimension of the row and column spaces is always the same, which is known as the rank of $A$. Suppose that $r=$ rank$(A)$, then $r$ represents the maximum number of linearly independent row vectors and column vectors. As a result, if $r <n$, the columns are linearly dependent; if $r< m$, the rows are linearly dependent, where $n$ and $m$ denote the number of columns and rows, respectively.

The columns of the given matrix will form a linearly independent set if the equation $Ax=0$ has the trivial solution.

For this purpose, transform the matrix in reduced echelon form using elementary row operations as:

$\begin{bmatrix}1&4&-3&0\\-2&-7&5&1\\-4&-5&7&5\end{bmatrix}$

$R_2\to R_2+2R_1$

$\begin{bmatrix}1 & 4 & -3 & 0 \\0 & 1 & -1 & 1\\-4 & -5 & 7 & 5 \end{bmatrix}$

$R_3\to R_3+4R_1$

$\begin{bmatrix}1 & 4 & -3 & 0 \\0 & 1 & -1 & 1\\0 & 11 & -5 & 5 \end{bmatrix}$

$R_1\to R_1-4R_2$

$\begin{bmatrix}1 & 0 & 1 & -4 \\0 & 1 & -1 & 1\\0 & 11 & -5 & 5 \end{bmatrix}$

$R_3\to R_3-11R_2$

$\begin{bmatrix}1 & 0 & 1 & -4 \\0 & 1 & -1 & 1\\0 & 0 & 6 & -6 \end{bmatrix}$

$R_3\to\dfrac{1}{6}R_3$

$\begin{bmatrix}1 & 0 & 1 & -4 \\0 & 1 & -1 & 1\\0 & 0 & 1 & -1 \end{bmatrix}$

$R_1\to R_1-R_3$

$\begin{bmatrix}1 & 0 & 0 & -3 \\0 & 1 & -1 & 1\\0 & 0 & 1 & -1 \end{bmatrix}$

$R_2\to R_2+R_3$

$\begin{bmatrix}1 & 0 & 0 & -3 \\0 & 1 & 0 & 0\\0 & 0 & 1 & -1 \end{bmatrix}$

Since the given matrix does not have a trivial solution, the columns of the given matrix form a linearly dependent set.

## Example

Let $A=\begin{bmatrix}1 & 3 & 9 \\2 & -6 & 10\\0 & 3 & 9 \end{bmatrix}$. Determine whether the vectors in $A$ are linearly independent.

### Solution

First, transform the matrix in reduced echelon form using elementary row operations as:

$\begin{bmatrix}1 & 3 & 9 \\2 & -6 & 10\\0 & 3 & 9 \end{bmatrix}$

$R_2\to R_2-2R_1$

$\begin{bmatrix}1 & 3 & 9 \\0 & -12 & -8\\0 & 3 & 9 \end{bmatrix}$

$R_2\to -\dfrac{1}{12}R_2$

$\begin{bmatrix}1 & 3 & 9 \\0 & 1 & \dfrac{2}{3}\\0 & 3 & 9 \end{bmatrix}$

$R_1\to R_1-3R_2$

$\begin{bmatrix}1 & 0 & 7 \\0 & 1 & \dfrac{2}{3}\\0 & 3 & 9 \end{bmatrix}$

$R_3\to R_3-3R_2$

$\begin{bmatrix}1 & 0 & 7 \\0 & 1 & \dfrac{2}{3}\\0 & 0 & 7 \end{bmatrix}$

$R_3\to \dfrac{1}{7}R_3$

$\begin{bmatrix}1 & 0 & 7 \\0 & 1 & \dfrac{2}{3}\\0 & 0 & 1 \end{bmatrix}$

$R_1\to R_1-7R_3$

$\begin{bmatrix}1 & 0 & 0 \\0 & 1 & \dfrac{2}{3}\\0 & 0 & 1 \end{bmatrix}$

$R_2\to R_2-\dfrac{2}{3}R_3$

$\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1 \end{bmatrix}$

Which is an identity matrix and hence shows that the vectors in $A$ are linearly independent.