can some once show me how to solve this chemical equation with a matrix?

Start by assigning an algebraic variable to each species, which represents the coefficient to be found for each species in the balanced equation:

a K3Fe(SCN)6 + b Na2Cr2O7 + c H2SO4 →

d Fe(NO3)3 + e Cr2(SO4)3 + f Na2SO4 + g KNO3 + h CO2 + i H2O

Taking one element at a time, write equations for the balance of each element:

K: 3a = g

Fe: a = d

S: 6a + c = 3e + f

C: 6a = h

N: 6a = 3d + g

Na: 2b = 2f

Cr: 2b = 2e

O: 7b + 4c = 9d + 12e + 4f + 3g + 2h + i

H: 2c = 2i

Rearrange the equations so all the variables are on the left:

3a - g = 0

a - d = 0

6a + c - 3e - f = 0

6a - h = 0

6a - 3d - g = 0

2b - 2f = 0

2b - 2e = 0

7b + 4c - 9d - 12e - 4f - 3g - 2h - i = 0

2c - 2i = 0

At this point it becomes a problem in algebra:

Convert the equations to a matrix suitable for solving simultaneous equations. It's extremely difficult to show matrices here in YA because YA rearranges the columns, but as an example the second equation becomes a row in the matrix like this:

1 0 0 -1 0 0 0 0 0 0

and the longest equation becomes:

0 7 4 -9 -12 -4 -3 -2 -1 0

Then use the usual matrix methods for solving simultaneous equations.

It turns out these particular equations are not entirely independent mathematically speaking, so you don't get just numbers for answers, but answers in terms of one of the variables. Using my methods I got:

a = b/16, b = b, c = 29b/8, d = b/16, e = b, f = b, g = 3b/16, h = 3b/8, i = 29b/8

In order to achieve the lowest possible coefficients, set b = 16, so the smallest variables (a and d in this case) are equal to 1:

a = 1, b = 16, c = 58, d = 1, e = 16, f = 16, g = 3, h = 6, i = 58

Substitute these values for the variables back into the original chemical equation, then it's balanced:

K3Fe(SCN)6 + 16 Na2Cr2O7 + 58 H2SO4 →

Fe(NO3)3 + 16 Cr2(SO4)3 + 16 Na2SO4 + 3 KNO3 + 6 CO2 + 58 H2O

what do you mean by "solve this with a matrix"?