To solve the system using matrices, we can rewrite the system as follows:
| -3 2 | | x | = | 10 |
| -4 3 | | y | | 2 |
Let's represent the coefficient matrix as A, the variable matrix as X, and the constant matrix as B. So the system can be represented as AX = B.
A = | -3 2 |
| -4 3 |
X = | x |
| y |
B = | 10 |
| 2 |
We can solve for X by multiplying each side of the equation by the inverse of A:
A^(-1) * A * X = A^(-1) * B
Since A^(-1) * A is the identity matrix (I), we have:
I * X = A^(-1) * B
Simplifying gives us:
X = A^(-1) * B
Now let's find the inverse of matrix A:
A = | -3 2 |
| -4 3 |
To find A^(-1), we need to calculate the determinant of A first:
det(A) = (-3 * 3) - (2 * -4) = -9 + 8 = -1
If the determinant is not equal to zero, the inverse exists.
Now let's find the inverse of A using the following formula:
A^(-1) = (1/det(A)) * adj(A)
Where adj(A) is the adjugate matrix of A, obtained by transposing the cofactor matrix of A.
The cofactor matrix of A is obtained by taking the determinants of the submatrices of A with alternating signs (+ and -).
The adjugate matrix of A is obtained by interchanging rows and columns of the cofactor matrix.
First, let's find the cofactor matrix of A:
C = | 3 -2 |
| 4 -3 |
Since A is a 2x2 matrix, we only have one submatrix.
The adjugate matrix of A is obtained by transposing the cofactor matrix:
adj(A) = | 3 4 |
| -2 -3 |
Now, let's calculate A^(-1):
A^(-1) = (1/det(A)) * adj(A) = (1/(-1)) * | 3 4 |
| -2 -3 |
A^(-1) = | -3 -4 |
| 2 3 |
Now, let's calculate X:
X = A^(-1) * B = | -3 -4 | * | 10 |
| 2 3 | | 2 |
X = | (-3 * 10) + (-4 * 2) |
| (2 * 10) + (3 * 2) |
X = | -30 - 8 |
| 20 + 6 |
X = | -38 |
| 26 |
Therefore, the solution to the system of equations is x = -38 and y = 26.
What is the solution of the system? Solve using matrices.
left double bracket negative 3x plus 2y equals 10 negative 4x plus 3y equals 2
1 answer