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)

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 $-\frac{1}{2}, -1$.

Let $A$ be a $3 \times 3$ matrix. Let $\mathbf{x}, \mathbf{y}, \mathbf{z}$ are linearly independent $3$-dimensional vectors. Suppose that we...