Questions: For what values of h are the vectors [-1, 4, 6], [5, 2, -3], [4, -4, -3], and [-16, 16, h] linearly dependent? A. The vectors are linearly dependent for h=12. B. The vectors are linearly dependent for h ≠ 12. C. The vectors are linearly dependent for all h. D. The vectors are linearly independent for all h.

For what values of h are the vectors [-1, 4, 6], [5, 2, -3], [4, -4, -3], and [-16, 16, h] linearly dependent?
A. The vectors are linearly dependent for h=12.
B. The vectors are linearly dependent for h ≠ 12.
C. The vectors are linearly dependent for all h.
D. The vectors are linearly independent for all h.

Transcript text: For what values of $h$ are the vectors $\left[\begin{array}{r}-1 \\ 4 \\ 6\end{array}\right],\left[\begin{array}{r}5 \\ 2 \\ -3\end{array}\right],\left[\begin{array}{r}4 \\ -4 \\ -3\end{array}\right]$, and $\left[\begin{array}{r}-16 \\ 16 \\ h\end{array}\right]$ linearly dependent? A. The vectors are linearly dependent for $\mathrm{h}=12$. B. The vectors are linearly dependent for $\mathrm{h} \neq 12$. C. The vectors are linearly dependent for all h . D. The vectors are linearly independent for all $h$.

Solution

Solution Steps

To determine the values of $ h $ for which the vectors are linearly dependent, we need to set up a matrix with these vectors as columns and find the determinant. The vectors are linearly dependent if the determinant of this matrix is zero. We will compute the determinant of the 3x3 matrix formed by the first three vectors and the vector with $ h $ as the third component of the last vector. Then, solve for $ h $ when the determinant is zero.

Step 1: Understanding Linear Dependence

Vectors are linearly dependent if there exist scalars, not all zero, such that a linear combination of the vectors equals the zero vector. For the vectors given:

\[ \begin{bmatrix} -1 \\ 4 \\ 6 \end{bmatrix}, \begin{bmatrix} 5 \\ 2 \\ -3 \end{bmatrix}, \begin{bmatrix} 4 \\ -4 \\ -3 \end{bmatrix}, \begin{bmatrix} -16 \\ 16 \\ h \end{bmatrix} \]

we need to find if there exist scalars $a, b, c, d$, not all zero, such that:

\[ a \begin{bmatrix} -1 \\ 4 \\ 6 \end{bmatrix} + b \begin{bmatrix} 5 \\ 2 \\ -3 \end{bmatrix} + c \begin{bmatrix} 4 \\ -4 \\ -3 \end{bmatrix} + d \begin{bmatrix} -16 \\ 16 \\ h \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} \]

Step 2: Forming the Matrix

To determine linear dependence, we can form a matrix with these vectors as columns and find its determinant:

\[ \begin{bmatrix} -1 & 5 & 4 & -16 \\ 4 & 2 & -4 & 16 \\ 6 & -3 & -3 & h \end{bmatrix} \]

Step 3: Determinant Calculation

For the vectors to be linearly dependent, the determinant of the matrix must be zero. We will calculate the determinant using the first row:

\[ \text{det} = -1 \cdot \begin{vmatrix} 2 & -4 & 16 \\ -3 & -3 & h \end{vmatrix} - 5 \cdot \begin{vmatrix} 4 & -4 & 16 \\ 6 & -3 & h \end{vmatrix} + 4 \cdot \begin{vmatrix} 4 & 2 & 16 \\ 6 & -3 & h \end{vmatrix} - 16 \cdot \begin{vmatrix} 4 & 2 & -4 \\ 6 & -3 & -3 \end{vmatrix} \]

Calculate each of the 2x2 determinants:

$\begin{vmatrix} 2 & -4 \\ -3 & -3 \end{vmatrix} = (2)(-3) - (-4)(-3) = -6 - 12 = -18$
$\begin{vmatrix} 2 & 16 \\ -3 & h \end{vmatrix} = (2)(h) - (16)(-3) = 2h + 48$
$\begin{vmatrix} 4 & -4 \\ 6 & -3 \end{vmatrix} = (4)(-3) - (-4)(6) = -12 + 24 = 12$
$\begin{vmatrix} 4 & 16 \\ 6 & h \end{vmatrix} = (4)(h) - (16)(6) = 4h - 96$
$\begin{vmatrix} 4 & 2 \\ 6 & -3 \end{vmatrix} = (4)(-3) - (2)(6) = -12 - 12 = -24$
$\begin{vmatrix} 4 & -4 \\ 6 & -3 \end{vmatrix} = 12$ (already calculated)
$\begin{vmatrix} 4 & 2 \\ 6 & -3 \end{vmatrix} = -24$ (already calculated)
$\begin{vmatrix} 4 & 2 \\ -4 & -3 \end{vmatrix} = (4)(-3) - (2)(-4) = -12 + 8 = -4$
$\begin{vmatrix} 6 & -3 \\ -4 & -3 \end{vmatrix} = (6)(-3) - (-3)(-4) = -18 - 12 = -30$

Now substitute back into the determinant expression:

\[ \text{det} = -1(-18) - 5(12) + 4(-24) - 16(-4) \]

Simplify:

\[ = 18 - 60 - 96 + 64 \]

\[ = 18 - 60 - 96 + 64 = -74 \]

Step 4: Solving for $h$

The determinant simplifies to a constant value, indicating that the determinant is not zero for any $h$. Therefore, the vectors are linearly independent for all $h$.