To solve the system of equations using a matrix, we can rewrite 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.
The coefficient matrix [A] is:
1.5 -0.5
1.5 1.5
The variable matrix [X] is:
x
y
The constant matrix [B] is:
-3.5
-7.5
So, the system becomes:
1.5x - 0.5y = -3.5
1.5x + 1.5y = -7.5
Now, we can solve this system by finding the inverse of [A] and multiplying it with [B]:
[X] = [A]^-1 * [B]
Let's calculate the inverse of [A]:
1.5 -0.5 1 -1
1.5 1.5 = -1 1
Now, let's multiply the inverse of [A] with [B]:
[X] = [-1 -1] * [-3.5]
[1 1] [-7.5]
[X] = [(-1 * -3.5) + (-1 * -7.5)]
[(1 * -3.5) + (1 * -7.5)]
[X] = [3.5 + 7.5]
[-3.5 - 7.5]
[X] = [11]
[-11]
So, the solution to the system of equations is: x = 11, y = -11.
Solve the system of equations using a matrix%0D%0A%0D%0A1.5x−0.5y=−3.5%0D%0A1.5%0D%0A%0D%0A−%0D%0A0.5%0D%0A%0D%0A=%0D%0A−%0D%0A3.5%0D%0A %0D%0A1.5x+1.5y=−7.5%0D%0A1.5%0D%0A%0D%0A+%0D%0A1.5%0D%0A%0D%0A=%0D%0A−%0D%0A7.5%0D%0A (1 point)%0D%0AResponses%0D%0A%0D%0A[32]%0D%0A[%0D%0A3%0D%0A2%0D%0A]%0D%0Athe 2 by 1 column matrix 3 2%0D%0A%0D%0A[−2−3]%0D%0A[%0D%0A−%0D%0A2%0D%0A−%0D%0A3%0D%0A]%0D%0Athe 2 by 1 column matrix row 1 negative 2 row 2 negative 3%0D%0A%0D%0A[−3−2]%0D%0A[%0D%0A−%0D%0A3%0D%0A−%0D%0A2%0D%0A]%0D%0Athe 2 by 1 column matrix row 1 negative 3 row 2 negative 2%0D%0A%0D%0Ano solution
1 answer