Determine whether {v 1 , v 2 , v 3 } is a basis for R 3
-$$\mathbf { v } _ { 1 } = \left[ \begin{array} { r }
- 2 \
4 \
- 8
\end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { l }
1 \
0 \
3
\end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r }
4 \
- 4 \
14
\end{array} \right]$$
A) Yes
B) No

Question

Determine whether {v 1 , v 2 , v 3 } is a basis for R 3 
-$$\mathbf { v } _ { 1 } = \left[ \begin{array} { r } 
- 2 \
4 \
- 8
\end{array} \right] , \mathbf { v } _ { 2 } = \left[ \begin{array} { l } 
1 \
0 \
3
\end{array} \right] , \mathbf { v } _ { 3 } = \left[ \begin{array} { r } 
4 \
- 4 \
14
\end{array} \right]$$
A) Yes
B) No

Quizplus · Accepted Answer

To determine if the set of vectors forms a basis for $\mathbb{R}^3$, they must be linearly independent and span $\mathbb{R}^3$. By setting up a matrix with these vectors as columns and attempting to reduce it to reduced row echelon form (RREF), we can check for linear independence. If the matrix reduces to the identity matrix, the vectors are linearly independent and span $\mathbb{R}^3$, forming a basis. However, if there is a row of zeros in the RREF, it indicates the vectors are linearly dependent, and thus, they do not form a basis for $\mathbb{R}^3$.Given vectors:$$\mathbf{v}_1 = \begin{bmatrix} -2 \ 4 \ -8 \end{bmatrix}, \quad\mathbf{v}_2 = \begin{bmatrix} 1 \ 0 \ 3 \end{bmatrix}, \quad\mathbf{v}_3 = \begin{bmatrix} 4 \ -4 \ 14 \end{bmatrix}$$Forming a matrix and attempting to reduce it:$$\begin{bmatrix}-2 & 1 & 4 \4 & 0 & -4 \-8 & 3 & 14\end{bmatrix}$$Without performing the full reduction, we can observe that the third vector $\mathbf{v}_3$ can be expressed as a linear combination of $\mathbf{v}_1$ and $\mathbf{v}_2$. Specifically, $\mathbf{v}_3 = 2\mathbf{v}_1 + \mathbf{v}_2$. This indicates that the vectors are linearly dependent, and therefore, they do not form a basis for $\mathbb{R}^3$.

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