Matrix help
find
1) lA^Tl
2) lA^2l
3) lAA^Tl
4) l2Al
5) lA^(-1)l
for:
-4 10
5 6

Question

Matrix help
find 
1) lA^Tl
2) lA^2l
3) lAA^Tl
4) l2Al
5) lA^(-1)l
for:
-4   10
 5      6

anonymous · Answer

Determinant of a 2x2 matrix $\begin{bmatrix}a&b\c&d\end{bmatrix}$ is $ad-bc$. $A^T$ of that same matrix is $\begin{bmatrix}a&c\b&d\end{bmatrix}$ . That should be enough info to solve the first few, assuming you know how to multiply matrices. It'd be useful if you could mention specifically where you're getting stuck.

anonymous · Answer

Here I'm assuming that by using that notation you mean the determinants of each of those matrices. If not, then please clarify.

maheshmeghwal9 · Answer

For a inverse of 2 x 2 matrix if $$[A]=\left[\begin{matrix}a & b \ c & d\end{matrix}\right]$$Then $$[A^{-1}]=\frac{1}{|A|}\left[\begin{matrix}d & -b \ -c & a \end{matrix}\right]$$

maheshmeghwal9 · Answer

Where |A| = Determinant of [A]

anonymous · Answer

could you work them out one at the time

maheshmeghwal9 · Answer

Matrix is a vector
If 'k' is a scalar multiplied to matrix then;
$$[A]=\left[\begin{matrix}a & b \ c & d\end{matrix}\right]; \space \space k[A]=\left[\begin{matrix}ka & kb \ kc & kd\end{matrix}\right]$$

anonymous · Answer

Which part are you having trouble with? You just plug the numbers into the formula.

maheshmeghwal9 · Answer

I think all the theories for ur question is almost completed
but where do u have ur acual problem now?
I think u must do as @nbouscal says:)

anonymous · Answer

Here's an example. For your matrix A=$\begin{bmatrix}-4&10\5&6\end{bmatrix}$, |A| is $-4\times6-10\times5=-24-50=-74$

maheshmeghwal9 · Answer

Here a =-4
b= 10
c= 5
d=6.

maheshmeghwal9 · Answer

now just put the value for the remaining:)

maheshmeghwal9 · Answer

in the formulae; we have given:)

anonymous · Answer

First calculate each of the matrices, you can check with us that you calculate them right, then after that, calculate their determinants. :)

zarkon · Answer

you really only need to compute one determinant.  The others can be computed using properties of the determinant

anonymous · Answer

@Zarkon While that is true, I think the exercise was designed to simultaneously give the student practice computing 2x2 determinants and provide a concrete reference point for how the properties of the determinant work.