Science:Math Exam Resources/Courses/MATH221/December 2009/Question 05 (a)

MATH221 December 2009

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Question 05 (a)
Let $\displaystyle W={\text{span}}\left\{{\begin{bmatrix}1\\0\\3\\0\end{bmatrix}},{\begin{bmatrix}0\\1\\4\\-1\end{bmatrix}},{\begin{bmatrix}2\\-3\\-6\\3\end{bmatrix}},{\begin{bmatrix}4\\-1\\8\\2\end{bmatrix}}\right\}$ Find a basis for $\displaystyle W$.

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Hint

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Solution

Whoops, no solution has been written yet. Be the first to submit a suggestion. We begin by forming $W=\left[{\begin{matrix}1&0&2&4\\0&1&-3&-1\\3&4&-6&8\\0&-1&3&2\end{matrix}}\right]$ First thing that we need to do to matrix $W$ is find the rref: The steps needed to take to obtain the rref are as follows: 1.) Add -3 times the 1st row to the 3rd row ($R_{3}=R_{3}-3R_{1}$) 2.) Add -4 times the 2nd row to the 3rd row ($R_{3}=R_{3}-4R_{2}$) 3.) Add the 2nd row to the 4th row ($R_{4}=R_{2}+R_{2}$) 4.) Swap row 3 and row 4 ($R_{3}\Leftrightarrow R_{4}$) 5.) Add 3rd row to 2nd row ($R_{2}=R_{2}+R_{3}$) 6.) And finally add -4 times the 3rd row to the 1st row ($R_{1}=R_{1}-4R_{3}$) We should obtain that: $rref(W)=\left[{\begin{matrix}1&0&2&0\\0&1&-3&0\\0&0&0&1\\0&0&0&0\end{matrix}}\right]$ Lastly to obtain the basis all we need to do is to interpret $rref(W)$. The columns that are linearly independent in $rref(W)$ show us that those columns in the original equation are the vectors that form the basis for the span of $W$. We can see that vectors: $v_{1},v_{2},v_{4}$ are the basis vectors. $\therefore B={\begin{Bmatrix}{\begin{bmatrix}1\\0\\3\\0\end{bmatrix}},&{\begin{bmatrix}0\\1\\4\\-1\end{bmatrix}},&{\begin{bmatrix}4\\-1\\8\\2\end{bmatrix}}\end{Bmatrix}}$ Where $B$ forms a basis for the span of $W$