# Express this 2x2 matrix as a linear combo of thoes other 2x2 matricies

1. Jul 14, 2008

### mohdhm

http://xmlearning.maths.ed.ac.uk/lecture_notes/vector_spaces/linear_span_set/linear_span_set.php [Broken]

Problem 1.29 #3

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Hello guys, i've been given a problem like the one above in my assignments and i'm not really sure what the final answer is supposed to be like. So far here is what i've done: just multiplied each 2x2 matrix by a scalar and added them up... this question is worth 20 marks and i get a sense that it is incomplete, what am i missing here?

http://g.imageshack.us/g.php?h=507&i=linalgar1.png [Broken]

Does this look right? i've never come across anything like that in my books and it's worth 20 marks so i have to be careful. any input you give will be appretiated

Last edited by a moderator: Apr 23, 2017 at 2:14 PM
2. Jul 14, 2008

### CompuChip

I must be looking at the wrong problem because I don't recognize any of the matrices you used in the linear combination. But in principle, you could do it that way and you'd get a system of four equations in three unknowns that should have a unique solution (e.g. one of them is dependent on the others), of course, you want to give that solution as well.

For example, if the question was, express the vector [3, 4, 2] as a linear combination of the vectors {[1, 0, 0], [0, 1, 0], [1, 0, -1]} you could of course write down an arbitrary linear combination
a [1, 0, 0] + b [0, 1, 0] + c [1, 0, -1]
but that would not do, of course. You would have to work it out:
[a, 0, 0] + [0, b, 0] + [c, 0, -c] = [a + c, b, - c]
should be equal to [3, 4, 2] so you get
a + c = 3
b = 4
- c = 2
Solving the system gives
a = 5, b = 4, c = -2
so the final answer would be:
[3, 4, 2] = 5 [1, 0, 0] + 4 [0, 1, 0] - 2 [1, 0, -1].

Now try this with the matrices.

3. Jul 14, 2008

### HallsofIvy

Staff Emeritus
Your matrix equation is, as you give,
$$\left[\begin{array}{cc}h+k & 3j+ k \\ 3h+ k & j+ k\end{array}\right]= \left[\begin{array}{cc} 5 & -1 \\ 11 & 1\end{array}\right]$$
which, since two matrices are equal if and only if corresponding terms are equal, is the same as the four equations, h+ k= -5, 3j+ k= -1, 3h+ k= 11, and j+ k= 1.

Of course, you can't in general solve four equations for three unknowns- 3 matrices can't span this 4 dimensional space. But it is possible that the given matrix is in the subspace spanned by them.

Go ahead and use any three of the equations to solve for j, h, and k, then put the values into the fourth equation to see if there is a solution.

4. Jul 14, 2008

### mohdhm

ah thanks for your help guys.