Question

What is the method of solving Cramer's rule and matrix inversion ?​

Answer 1

Step-by-step explanation:

Cramer’s rule

Given a system of n equations with n variables we may try to solve it with Cramer’s rule. Let A be the matrix of this system without the column of free coefficients. Let A_{i} be the matrix A, in which instead of i-th column we put the column of free coefficients. Then:

if \det A\neq 0, the system has exactly one solution. The solution is given by the following formula: x_{i}=\frac{\det A_i}{\det A},

if \det A=0, and at least one of \det A_{i} is not equal to 0, the system has no solutions,

if \det A=0 and for every i, \det A_{i}=0, there can be zero or infinitely many solutions — Cramer’s method does not give any precise answer.

Answer 2

Step-by-step explanation:

Consider a system of two equations in two variables x1 and x2:

ax1 + bx2  = u

cx1 + dx2  = v,

where a, b, c, d, u, and v are numbers with a, b, c, and d nonzero.

One straightforward way to solve for x1 and x2 is to isolate one of the variables in one of the equations and substitute the result into the other equation. For example, from the second equation we have

x2 = (v − cx1)/d.

Substituting this expression for x2 into the first equation yields

ax1 + b(v − cx1)/d = u,

which we can write as

(a − bc/d)x1 + bv/d = u.

First suppose that a ≠ bc/d. Then

x1 =  

u − bv/d

a − bc/d

or

x1 =  

ud − bv

ad − bc

.

To find x2 we use the fact that x2 = (v − cx1)/d, to get

x2 =  

va − cu

ad − bc

.

Now suppose that a = bc/d. Then the equations are

ax1 + bx2  = u

(ad/b)x1 + dx2  = v

or, multiplying the first equation by d and the second by b,

adx1 + bdx2  = ud

adx1 + bdx2  = vb.

Thus in this case if ud ≠ vb then the equations have no solution and if ud = vb (in which case also va = uc) then the set of solutions is the set of pairs (x1, x2) satisfying ax1 + bx2 = u (that is, the set of pairs (x1, (u − ax1)/b) for any number x1).

In summary, the solutions of the system of equations have three possible forms.

If ad ≠ bc then the equations have a unique solution,

(x1, x2)  =  left parenthesis  

ud − bv

ad − bc

,  

va − cu

ad − bc

left parenthesis .

If ad = bc and ud = vb then the set of solutions of the equations is the set of pairs

left parenthesis x1,  

u − ax1

b

left parenthesis

for any number x1.

If ad = bc and ud ≠ vb then the equations have no solution.

This method of isolating a variable in one equation and substituting it into another equation is cumbersome when the system consists of more than two equations and two variables. I now describe a more elegant method.

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