# Find Values of $h$ so that the Given Vectors are Linearly Independent

## Problem 194

Find the value(s) of $h$ for which the following set of vectors

\[\left \{ \mathbf{v}_1=\begin{bmatrix}

1 \\

0 \\

0

\end{bmatrix}, \mathbf{v}_2=\begin{bmatrix}

h \\

1 \\

-h

\end{bmatrix}, \mathbf{v}_3=\begin{bmatrix}

1 \\

2h \\

3h+1

\end{bmatrix}\right\}\]
is linearly independent.

(*Boston College, Linear Algebra Midterm Exam Sample Problem*)

Add to solve later

We give two solutions. The first one uses the homogeneous system and the second one uses a determinant.

## Solution 1.

Let us consider the linear combination

\[x_1\mathbf{v}_1+x_2\mathbf{v}_2+x_3\mathbf{v}_3=\mathbf{0}. \tag{*}\]
If this homogeneous system has only zero solution $x_1=x_2=x_3=0$, then the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent.

We reduce the augmented matrix for the system as follows.

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

1 & h & 1 & 0 \\

0 &1 & 2h & 0 \\

0 & -h & 3h+1 & 0

\end{array}\right]
\xrightarrow{R_3+hR_2}

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

1 & h & 1 & 0 \\

0 &1 & 2h & 0 \\

0 & 0 & 2h^2+3h+1 & 0

\end{array}\right].

\]
From this, we see that the homogeneous system (*) has only the zero solution if and only if

\[2h^2+3h+1 \neq 0.\]
Since we have

\[2h^2+3h+1=(2h+1)(h+1),\]
if $h \neq -\frac{1}{2}, -1$, then $2h^2+3h+1 \neq 0$.

In summary, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent for any $h$ except $-\frac{1}{2}, -1$.

## Solution 2.

Note that the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent if and only the matrix

\[A:=[\mathbf{v}_1 \mathbf{v}_2 \mathbf{v}_3]=\begin{bmatrix}

1 & h & 1 \\

0 &1 &2h \\

0 & -h & 3h+1

\end{bmatrix}\]
is nonsingular.

Also, the matrix $A$ is nonsingular if and only if the determinant $\det(A)$ is nonzero.

So we compute the determinant of the matrix $A$ by the first column cofactor expansion and obtain

\begin{align*}

\det(A)&=\begin{vmatrix}

1 & h & 1 \\

0 &1 &2h \\

0 & -h & 3h+1

\end{vmatrix}\\[6pt]
&=\begin{vmatrix}

1 & 2h\\

-h& 3h+1

\end{vmatrix}\\[6pt]
&=2h^2+3h+1\\

&=(2h+1)(h+1).

\end{align*}

Hence $\det(A)\neq 0$ if and only if $h\neq -\frac{1}{2}, -1$.

Therefore, the vectors $\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3$ are linearly independent for any values of $h$ except $-\frac{1}{2}, -1$.

Add to solve later