Let's solve the system of linear equations using the method of matrix inversion. The system of linear equations is given by:
5p + 2q = 10
4p + q = 4
We'll represent the system in matrix form, where the coefficient matrix (A) multiplies the variable matrix (X) to yield the results matrix (B):
| 5 2 | | p | | 10 |
| 4 1 | | q | = | 4 |
Let's call the matrix of coefficients A, the variable matrix X, and the result matrix B, such that:
A = | 5 2 |
| 4 1 |
X = | p |
| q |
B = | 10 |
| 4 |
To solve for X, you can multiply both sides of the equation by the inverse of matrix A (denoted as A^-1), given that A^-1 exists. This can be calculated using the formula for the inverse of a 2x2 matrix:
If A = | a b |
| c d |
Then, A^-1 = (1/|A|) * | d -b |
| -c a |
where |A| is the determinant of A given by |A| = ad - bc.
Now, let's find the determinant of matrix A.
|A| = 5 * 1 - 4 * 2 = 5 - 8 = -3
Since the determinant is non-zero, the inverse of A exists. Now we can compute the inverse of A using the formula:
A^-1 = (1/|A|) * | 1 -2 |
| -4 5 |
A^-1 = (-1/3) * | 1 -2 |
| -4 5 |
A^-1 = | -1/3 2/3 |
| 4/3 -5/3 |
Now we can find X by multiplying A^-1 with B.
X = A^-1 * B
X = | -1/3 2/3 | * | 10 |
| 4/3 -5/3 | | 4 |
X = | (-1/3)*10 + (2/3)*4 |
| (4/3)*10 - (5/3)*4 |
X = | -10/3 + 8/3 |
| 40/3 - 20/3 |
X = | -2/3 |
| 20/3 |
Therefore, the values of p and q are:
p = -2/3
q = 20/3
So the solution to the system of equations is p = -2/3 and q = 20/3.