Use the matrix method to solve
5x+3y=35
3x-4y=-8
http://www.gregthatcher.com/Mathematics/GaussJordan.aspx
in the end
1 0 4
0 1 5
check
20 + 15 = 35
12 - 20 = -8 whew !
another matrix way with determinants
det of
5 3
3 -4 = -20 -9 = -29
for y put column 3 in for col 2
det of
5 35
3 - 8 = -40 - 105 = -145
-145 / -29 = 5 which is y
To solve the system of equations using the matrix method, we can represent the system of equations in matrix form.
Let's define the coefficient matrix A as:
A = [[5, 3], [3, -4]]
And the variable matrix X as:
X = [[x], [y]]
We can also define the constant matrix B as:
B = [[35], [-8]]
The matrix equation becomes AX = B.
To find the solution matrix X, we need to find the inverse of matrix A and then multiply both sides of the equation by the inverse of A. The solution matrix will be given by X = A^(-1) * B.
1. Finding the inverse of matrix A:
To find the inverse of A, we can use the formula:
A^(-1) = (1/det(A)) * Adj(A)
Where det(A) denotes the determinant of matrix A and Adj(A) denotes the adjugate of matrix A.
2. Determinant of matrix A:
det(A) = (5 * (-4)) - (3 * 3)
= (-20) - 9
= -29
3. Adjugate of matrix A:
The adjugate of a 2x2 matrix is obtained by swapping the main diagonal elements (i.e., the top-left and bottom-right elements) and changing the signs of the other diagonal elements.
Adj(A) = [[-4, -3], [-3, 5]]
4. Inverse of matrix A:
A^(-1) = (1/det(A)) * Adj(A)
= (1/-29) * [[-4, -3], [-3, 5]]
= [[4/29, 3/29], [3/29, -5/29]]
5. Multiplying A^(-1) with B:
X = A^(-1) * B
= [[4/29, 3/29], [3/29, -5/29]] * [[35], [-8]]
= [[4/29 * 35 + 3/29 * (-8)], [3/29 * 35 + (-5/29) * (-8)]]
= [[140/29 - 24/29], [105/29 + 40/29]]
= [[116/29], [145/29]]
The solution to the system of equations is x = 116/29 and y = 145/29.