# Solve the system of equations # 2x + 3y = 5 # and # x + y = 1 # using matrices?

##### 1 Answer

#### Explanation:

We have:

# 2x + 3y = 5 #

# x + y = 1 #

Which we can write in vector matrix form:

# ( (2,3), (1,1) ) ( (x), (y) ) = ( (5), (1) ) #

Or:

# bb(A) bb(ul x) = bb(ul b) => bb(ul x) = bb(A)^(-1) bb(ul b) #

Where

# bb(A) = ( (2,3), (1,1) ) # ;# bb(ul x) = ( (x), (y) ) # ;# bb(ul b) ( (5), (1) ) #

We can find

**Method 2 - Matrix Inversion**

A matrix,

- Calculating the Matrix of Minors,
- Form the Matrix of Cofactors,
#cof(bb(A))# - Form the adjoint matrix,
#adj(bb(A))# - Multiply
#adj(bb(A))# by#1/abs(bb(A))# to form the inverse#bb(A)^-1#

At some point we need to calculate

# bb(A) = ( (2,3), (1,1)) #

If we expand about the first row;

# abs(bb(A)) = (2)(1) - (1)(3) #

# \ \ \ \ \ = 2-3 #

# \ \ \ \ \ = -1 #

As

#"minors"(bb(A)) = ( (1, 1), (3, 2 ))#

We now form the matrix of cofactors,

# ( (+, -), (-, +) )#

Where we change the sign of those elements with the minus sign to get;

# cof(bbA) = ( (1, -1), (-3, 2 )) #

Then we form the adjoint matrix by transposing the matrix of cofactors,

#adj(A) = cof(A)^T#

#\ \ \ \ \ \ \ \ \ \ \ = ( (1, -1), (-3, 2 ))^T #

#\ \ \ \ \ \ \ \ \ \ \ = ( (1, -3), (-1, 2 )) #

And then finally we multiply by the reciprocal of the determinant to get:

#bb(A)^-1 = 1/abs(bb A) adj(bb A)#

#\ \ \ \ \ \ \ = (1.(-1)) ( (1, -3), (-1, 2 )) #

#\ \ \ \ \ \ \ = ( (-1, 3), (1, -2 )) #

So then we get the solution the linear equations as:

# bb(ul x) = bb(A)^(-1) bb(ul b) # .....#[star]#

# :. ( (x), (y) ) = ( (-1, 3), (1, -2 )) ( (5), (1) ) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( ((-1)(5)+(3)(1) ), ((1)(5)+(-2)(1) ) ) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( (-5+3), (5-2) ) #

# \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = ( (-2), (3) ) #

Thus we have a unique solution:

# x=-2, y=3#