# Solve matrices both algabraic and numerically

Can someone explain me to what to do here? I don't understand the question, or how to solve the problem. Should I use some theorems, to solve it? • Hint: On $AX=B$, to solve it algebraically would be to have $X=A^{-1}B$ while numerically is to actually compute what $X$ is (i.e., the numbers in $X$). Mar 17, 2016 at 21:18
• Why A^-1 ? Do you have a link for additional information, to understand it better?
Mar 17, 2016 at 21:20

I agree with Micheal Burr.

One could also say, solve symbolically (by means of algebraic transformations) for the unknown $X$, e.g. $$A^2 X + B = 0 \iff \\ X = (A^2)^{-1}(-B) = - (A^{-1})^2 B$$ and then insert the values for all variables and determine the resulting value for $X$, e.g. we enter $A$ and $B$ inta a computer algebra system (here Octave)

>> A = [ 2,2;3,4]
A =

2   2
3   4

>> B = [ 1,2;-1,2]
B =

1   2
-1   2


and then start calculating $X$:

>> inv(A)
ans =

2.0000  -1.0000
-1.5000   1.0000

>> X =  - inv(A)*inv(A)*B
X =

-8.5000  -5.0000
7.0000   4.0000


A little test to see if the calculation worked:

>> A*A*X + B
ans =

1.4211e-014  0.0000e+000
2.8422e-014  1.4211e-014


and we are done.

The objects here are (square) matrices which form an algebraic structure called a ring. It means you can pretty much calculate with them like with numbers. Notable exceptions: multiplication is not commutative in general, not all elements have multiplicative inverses.

The task can be solved by knowing

• how to add and subtract matrices,
• how to multiply a matrix with a scalar,
• how to multiply matrices and
• how to calculate the multiplicative inverse of a matrix.
• Okay, I just ask again. Which rules of arithmetic do you use to isolate X? Where can I finde the information? Do you have a link?