Science:Math Exam Resources/Courses/MATH221/April 2010/Question 01 (f)

MATH221 April 2010

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Question 01 (f)

Let $\displaystyle A={\begin{bmatrix}1&2&3&-2&4\\2&5&8&-1&6\\1&4&7&5&2\\1&3&5&1&2\\\end{bmatrix}}$ and $\displaystyle U={\begin{bmatrix}1&2&3&-2&4\\0&1&2&3&-2\\0&0&0&1&2\\0&0&0&0&0\\\end{bmatrix}}$ Then the matrix U is an echelon form for A (you may assume this, you don't have to do the row reduction again.) Let $\displaystyle a_{1},a_{2},a_{3},a_{4},a_{5}$ be the columns of A. Express $\displaystyle a_{5}$ as a linear combination of the columns of $\displaystyle a_{1},a_{2},a_{4}$

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Hint

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 1.) To express $a_{5}$ as a linear combination of $a_{1},a_{2}$ and $a_{4}$, we take n multiples of each vector that will add up to $a_{5}$. 2.) One way to do this is by putting the vectors in a system of equations and then putting this system of equations into an augmented matrix: $a_{1}$ + 2$a_{2}$ - 2$a_{4}$ = 4$a_{5}$ 0+ $a_{2}$ + 3$a_{4}$ = -2$a_{5}$ 0 + 0 + $a_{4}$ = 2$a_{5}$ ${\begin{bmatrix}1&2&-2&4\\0&1&3&-2\\0&0&1&2\end{bmatrix}}$ 3.) Reduce to row echelon form through these steps: 1.) $R_{2}$ = $R_{2}$ -3$R_{3}$ 2.) $R_{1}$ = $R_{1}$ + 2$R_{3}$ 3.) $R_{1}$ = $R_{1}$ - 2$R_{2}$ and you will get: ${\begin{bmatrix}1&0&0&24\\0&1&0&-8\\0&0&1&2\end{bmatrix}}$ 4.) This shows the linear combination: $a_{1}$ = 24 $a_{2}$ = -8 $a_{4}$ = 2 $a_{5}$ = 24$a_{1}$ - 8$a_{2}$ + 2$a_{4}$ PROOF: ${\begin{bmatrix}4\\-2\\2\\0\end{bmatrix}}$ = 24${\begin{bmatrix}1\\0\\0\\0\end{bmatrix}}$ - 8${\begin{bmatrix}2\\1\\0\\0\end{bmatrix}}$+ 2${\begin{bmatrix}-2\\3\\1\\0\end{bmatrix}}$ ${\begin{bmatrix}4\\-2\\2\\0\end{bmatrix}}$ = ${\begin{bmatrix}24\\0\\0\\0\end{bmatrix}}$ - ${\begin{bmatrix}16\\8\\0\\0\end{bmatrix}}$+ ${\begin{bmatrix}-4\\6\\2\\0\end{bmatrix}}$