Solve Using an Inverse Matrix 26a+8b=28 , 8a+3b=9

Problem

{[26*a+8*b=28],[8*a+3*b=9])

Solution

1. Write the system in matrix form A*X=B where A is the coefficient matrix, X is the variable matrix, and B is the constant matrix.

A=[[26,8],[8,3]]

X=[[a],[b]]

B=[[28],[9]]

2. Calculate the determinant of matrix A denoted as det(A) or |A| using the formula a*d−b*c

det(A)=(26)*(3)−(8)*(8)

det(A)=78−64

det(A)=14

3. Find the inverse matrix A(−1) using the formula A(−1)=1/det(A)*[[d,−b],[−c,a]]

A(−1)=1/14*[[3,−8],[−8,26]]

4. Solve for X by multiplying the inverse matrix A(−1) by the constant matrix B such that X=A(−1)*B

X=1/14*[[3,−8],[−8,26]]*[[28],[9]]

5. Perform matrix multiplication to find the values of a and b

X=1/14*[[(3)*(28)+(−8)*(9)],[(−8)*(28)+(26)*(9)]]

X=1/14*[[84−72],[−224+234]]

X=1/14*[[12],[10]]

6. Simplify the fractions to obtain the final values for the variables.

a=12/14=6/7

b=10/14=5/7

Final Answer

[[a],[b]]=[[6/7],[5/7]]

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