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Letbegin{equation*}mathbf{v}_1 := begin{pmatrix}1 2 3end{pmatrix}, quad mathbf{v}_2 := begin{pmatrix}4 5 6end{pmatrix}, quad mathbf{v}_3 := begin{pmatrix}7 8 9end{pmatrix}, quad mathbf{v}_4 := begin{pmatrix}10 11 12end{pmatrix}.end{equation*}
1. Find the reduced row echelon form $R = [mathbf{v}_1',mathbf{v}_2',mathbf{v}_3',mathbf{v}_4']$ of $A = [mathbf{v}_1,mathbf{v}_2,mathbf{v}_3,mathbf{v}_4]$.
Let $A = [mathbf{v}_1,mathbf{v}_2,mathbf{v}_3,mathbf{v}_4] = begin{pmatrix}1 & 4 & 7 & 10 2 & 5 & 8 & 11 3 & 6 & 9 & 12end{pmatrix}$. So begin{equation*}R = [mathbf{v}_1',mathbf{v}_2',mathbf{v}_3',mathbf{v}_4'] = begin{pmatrix}1 & 0 & -1 & -2 0 & 1 & 2 & 3 0 & 0 & 0 & 0end{pmatrix}.end{equation*}
2. Write $mathbf{v}_3'$ and $mathbf{v}_4'$ as linear combinations of $mathbf{v}_1'$ and $mathbf{v}_2'$.
We want to find $alpha$ and $beta$ such thatbegin{equation*}alphabegin{pmatrix}1 0 0end{pmatrix}+betabegin{pmatrix}0 1 0end{pmatrix} = begin{pmatrix}-1 2 0end{pmatrix}.end{equation*}This is when $alpha = -1$ and $beta = 2$. We also want to find $alpha$ and $beta$ such thatbegin{equation*}alphabegin{pmatrix}1 0 0end{pmatrix}+betabegin{pmatrix}0 1 0end{pmatrix} = begin{pmatrix}-2 3 0end{pmatrix}.end{equation*}This is when $alpha = -2$ and $beta = 3$.
3. Write $mathbf{v}_3$ and $mathbf{v}_4$ as linear combinations of $mathbf{v}_1$ and $mathbf{v}_2$. We want to find $alpha$ and $beta$ such thatbegin{equation*}alphabegin{pmatrix}1 2 3end{pmatrix}+betabegin{pmatrix}4 5 6end{pmatrix} = begin{pmatrix}7 8 9end{pmatrix}.end{equation*}This is when $alpha = -1$ and $beta = 2$. We also want to find $alpha$ and $beta$ such thatbegin{equation*}alphabegin{pmatrix}1 2 3end{pmatrix}+betabegin{pmatrix}4 5 6end{pmatrix} = begin{pmatrix}10 11 12end{pmatrix}.end{equation*}This is when $alpha = -2$ and $beta = 3$. Note the similarity to part 2.
4. Find all possible ways $mathbf{v}_4$ can be written as linear combinations of $mathbf{v}_1$, $mathbf{v}_2$ and $mathbf{v}_3$.
Having trouble with this one. Any help would be great!
For 4, think about this way. Let $A'$ be the $3times 3$ matrix whose columns are $mathbf v_1,mathbf v_2, mathbf v_3$. You want to solve the system$$A'mathbf x = mathbf v_4.$$Can you see how to get the solutions from the reduced echelon form $R$ you already have?
Ted ShifrinTed ShifrinNot the answer you're looking for? Browse other questions tagged linear-algebravector-spaces or ask your own question.
I had a failure of the O-rings which joins the heater matrix pipes to the pipes coming from the engine compartment. This led to anti-freeze/coolant leaking inot the driver's footwell.
Ford have no altered the spec and there is no longer a joint invovled. However they can (or will) not supply new O-rings or tell me the exact sixe of the O-rings. Their fix is to replace the whole heater matrick and pipe system at a cost of £700. They grudgingly agreed to contribute £200 after a lot of messing about.
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However an independnet mechanic that I know helped me with a fix costing £40 in labour and some old heater pipes/jubilee clips.
I have a pic of the fix if anyone wants to see it.
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I believe this will be a common failure as both seals let go within a couple of months of each other. My mechanic friend said it was a common problem on older Focuses but at least in the past Ford woudl supply the O-rings.
