To determine the value of [tex]\( h \)[/tex] for which the set of vectors
[tex]\[
\begin{array}{l}
\left[
\begin{array}{r}
1 \\
-1 \\
4
\end{array}
\right],
\left[
\begin{array}{r}
3 \\
-5 \\
7
\end{array}
\right],
\left[
\begin{array}{r}
-1 \\
5 \\
h
\end{array}
\right]
\end{array}
\][/tex]
are linearly dependent, we need to calculate the determinant of the matrix formed by using these vectors as columns. If the determinant is zero, the vectors are linearly dependent.
The matrix formed is:
[tex]\[
\mathbf{A} = \begin{bmatrix}
1 & 3 & -1 \\
-1 & -5 & 5 \\
4 & 7 & h
\end{bmatrix}
\][/tex]
We need to find the determinant of this matrix and set it to zero:
[tex]\[
\det(\mathbf{A}) = \begin{vmatrix}
1 & 3 & -1 \\
-1 & -5 & 5 \\
4 & 7 & h
\end{vmatrix}
\][/tex]
The determinant of a [tex]\( 3 \times 3 \)[/tex] matrix [tex]\(\begin{bmatrix}
a & b & c \\
d & e & f \\
g & h & i
\end{bmatrix}\)[/tex] is given by:
[tex]\[
aei + bfg + cdh - ceg - bdi - afh
\][/tex]
For our matrix [tex]\( \mathbf{A} \)[/tex], we have:
[tex]\[
a = 1, \ b = 3, \ c = -1, \ d = -1, \ e = -5, \ f = 5, \ g = 4, \ h = 7, \ i = h
\][/tex]
Using the formula for the determinant:
[tex]\[
\det(\mathbf{A}) = 1 \cdot (-5) \cdot h + 3 \cdot 5 \cdot 4 + (-1) \cdot (-1) \cdot 7 - (-1) \cdot (-5) \cdot 4 - 3 \cdot (-1) \cdot h - 1 \cdot 5 \cdot 7
\][/tex]
Simplifying term by term:
[tex]\[
\det(\mathbf{A}) = (-5h) + (3 \cdot 5 \cdot 4) + (1 \cdot 7) - (1 \cdot 5 \cdot 4) - (3 \cdot -1 \cdot h) - (5 \cdot 7)
\][/tex]
[tex]\[
= (-5h) + 60 + 7 - 20 + 3h - 35
\][/tex]
[tex]\[
= (-5h + 3h) + (60 + 7 - 20 - 35)
\][/tex]
[tex]\[
= -2h + 12
\][/tex]
For the vectors to be linearly dependent, the determinant must equal zero:
[tex]\[
-2h + 12 = 0
\][/tex]
Solving for [tex]\( h \)[/tex]:
[tex]\[
-2h = -12
\][/tex]
[tex]\[
h = 6
\][/tex]
Thus, the value of [tex]\( h \)[/tex] that makes the vectors linearly dependent is:
[tex]\[
h = 6
\][/tex]
