Solving Matrices with multiple variables

I am really struggling to work out how to solve this, can anyone help with any suggestions of how to solve this for a beginner?

I have tried to search how to solve but all the videos and tutorials I can find are solving simulatneous equations with only variables in one matrix.

I think I mostly need to know how to input variables within the matrices themselves, this should help me with solving the rest.

Answer 1

[Walter Roberson]

Example:

M = randn(6,6)

M = 6×6

0.0373 -1.0836 -1.1469 0.1980 0.1078 0.8221 0.5644 -0.8136 0.2201 0.2133 -0.6369 0.2241 0.9454 0.8694 1.8952 0.8920 -0.4400 0.4640 1.3960 -0.6729 -2.1750 -0.4647 0.5789 -0.9932 1.2544 -0.7392 -0.1625 0.1822 0.1910 -1.0507 -0.6796 -1.0151 -0.3467 -0.1365 -1.3496 0.4000

guesses = randn(1,6);
f = @(x) [x(1),0,1,2,x(2),x(3)].' - M*[0;x(4);x(5);x(6);0;0]

f = function_handle with value:
    @(x)[x(1),0,1,2,x(2),x(3)].'-M*[0;x(4);x(5);x(6);0;0]

fsolve(f, guesses)

Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.

ans = 1×6

2.8099 0.7598 -0.6862 0.7500 -2.2618 5.1956

[Emma Haines]

Sorry to be a pain, can you explain the steps? I am very new to using MATLAB and I am still getting up to speed with all the coding phrases etc.

[Walter Roberson]

The general idea is that when you have sets of equations, you can often rewrite them as sets of zero finding.

 L1(x) = R1(x)
 L2(x) = R2(x)

can be rewritten as

 L1(x) - R1(x) = 0
 L2(x) - R2(x) = 0

then you drop the = 0 part and build a vector out of the rest,

 [L1(x) - R1(x);
  L2(x) - R2(x)]

and wrap it in an anonymous function

 F = @(x) [L1(x) - R1(x);
  L2(x) - R2(x)]

and now you have a function that you can pass to a numeric root-finder such as fsolve .

fsolve() has several algorithms. Estimates of the jacobian might be used to figure out which direction to go in.