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#