# Find the value(s) of h for which the vectors are linearly dependent. Justify your answer.

The main objective of this question is to determine which of the following vectors are linearly dependent.

This question uses the concept of linearly dependent. If the non-trivial linear combination of vectors is equal to zero, then that set of vectors is said to be linearly dependent while the vectors are said to be linearly independent if there is no such linear combination.

Given that:

$\begin{bmatrix} 1 \\ 5 \\ -3 \end{bmatrix} \space , \space \begin{bmatrix} -2 \\ -9 \\ -6 \end{bmatrix} \space , \space \begin{bmatrix} 3 \\ h \\ -9 \end{bmatrix}$

We have to show that the given vectors are linearly dependent.

We know that:

$Ax \space = \space 0$

$A \space = \space \begin{bmatrix} 1 & -2 & 3 \\ 5 & -9 & h \\ -3 & h & -9\end{bmatrix}$

$x \space = \space \begin{bmatrix} x_1 \\ x_2 \\ -x_3 \end{bmatrix}$

$R_2 \space \rightarrow \space R_2 \space – \space 5R_1$

$R_3 \space \rightarrow \space R_1 \space + \space 2R_2$

$\begin{bmatrix} 1 & -2 & 3 & | 0 \\ 5 & -9 & h & | 0 \\ -3 & h & -9 & | 0\end{bmatrix} \space = \space \begin{bmatrix} 1 & -2 & 3 & | 0 \\ 0 & 1 & h – 15 & | 0 \\ 0 & 0 & 0 & | 0\end{bmatrix}$

$R_1 \space \rightarrow \space R_1 \space + \space 2R_2$

$\begin{bmatrix} 1 & 0 & -27 + 2h & | 0 \\ 0 & 1 & h – 15 & | 0 \\ 0 & 0 & 0 & | 0\end{bmatrix}$

$\begin{bmatrix} x_1 \\ x_2 \\ -x_3 \end{bmatrix} \space = \space \begin{bmatrix} (27 – 2h)x_3 \\ (15-h)x_3 \\ x_3 \end{bmatrix} \space = \space x_3 \space \begin{bmatrix} 27 – 2h \\ 15-h \\ 1\end{bmatrix}$

The given vectors are linearly independent for all the values of $h$ as the last coordinate does not depend on $h$.

## 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 or linearly dependent.

First, we have to transform the given matrix in reduced echelon as:

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

$R_2\to R_2-2R_1$

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

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

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

$R_1\to R_1-3R_2$

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

$R_3\to R_3-3R_2$

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

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

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

$R_1\to R_1-7R_3$

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

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

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

This is an identity matrix and hence, it is proved that the given vectors are linearly dependent.