Math Snap

STEP 1

Assumptions1. The vectors are $\mathbf{v}_{1}=\left[\begin{array}{r} 0 \\ 0 \\ -3\end{array}\right], \mathbf{v}_{}=\left[\begin{array}{r}
0 \\
-5 \\
6\end{array}\right]$and$\mathbf{v}_{3}=\left[\begin{array}{r}
5 \\
-3 \\
9\end{array}\right]$
. We need to determine if these vectors span $\mathbb{R}^{3}$
3. Spanning $\mathbb{R}^{3}$ means that any vector in $\mathbb{R}^{3}$ can be written as a linear combination of these vectors.

STEP 2

We will first write these vectors as the columns of a matrix $A$.
$$A = \begin{bmatrix} 0 &0 &5 \\ 0 & -5 & - \\ - &6 &9\end{bmatrix}$$

STEP 3

To determine if the vectors span $\mathbb{R}^{3}$, we need to find the rank of the matrix $A$. The rank of a matrix is the maximum number of linearly independent rows or columns. If the rank of the matrix is3, then the vectors span $\mathbb{R}^{3}$.

STEP 4

We will perform Gaussian elimination to find the row echelon form of the matrix. We start by swapping the first and third rows to bring a non-zero element to the top.
$$A = \begin{bmatrix} -3 &6 &9 \\ 0 & - & -3 \\ 0 &0 &\end{bmatrix}$$

STEP 5

Next, we multiply the first row by $-\frac{1}{3}$ to make the leading element of the first row1.
$$A = \begin{bmatrix} 1 & -2 & -3 \\ 0 & -5 & -3 \\ 0 &0 &5\end{bmatrix}$$

STEP 6

Then, we multiply the second row by $-\frac{1}{5}$ to make the leading element of the second row1.
$$A = \begin{bmatrix} 1 & -2 & -3 \\ 0 &1 & \frac{3}{5} \\ 0 &0 &5\end{bmatrix}$$

STEP 7

Finally, we multiply the third row by $\frac{1}{5}$ to make the leading element of the third row1.
$$A = \begin{bmatrix} 1 & -2 & -3 \\ 0 &1 & \frac{3}{5} \\ 0 &0 &1\end{bmatrix}$$

Now that we have the row echelon form of the matrix, we can see that the rank of the matrix is3, since there are3 non-zero rows. This means that the vectors $\mathbf{v}_{1}, \mathbf{v}_{2}, \mathbf{v}_{3}$ span $\mathbb{R}^{3}$.
So, the correct answer is D. Yes. When the given vectors are written as the columns of a matrix $A, A$ has a pivot position in every row.