Solve the following system of equations algebraically. Verify your solution by using matrices.
3x-y=0
5x+2y=22
-y=-3x
y=3x
5x+2(3x)=22
5x+6x=22
11x=22
x=2
y=3(2)
y=6
(3 -1)(2)
(5 2)(6)
=
(0)
(22)
thank you steve!
To solve the system of equations algebraically, we can use either the substitution method or the elimination method. Let's solve it using the elimination method.
Step 1: Multiply the first equation by 2 and the second equation by 3 to make the coefficients of y terms the same but with opposite signs:
2*(3x - y) = 2*0
3*(5x + 2y) = 3*22
Simplifying, we get:
6x - 2y = 0
15x + 6y = 66
Step 2: Now, we can eliminate the y variable by adding the two equations together:
(6x - 2y) + (15x + 6y) = 0 + 66
Simplifying, we get:
21x + 4y = 66
Step 3: Solve the newly formed equation for one variable (x or y). Let's solve for x:
21x = 66 - 4y
x = (66 - 4y) / 21
Step 4: Substitute the value of x back into one of the original equations to solve for y. Let's use the first equation:
3x - y = 0
Substituting, we get:
3((66 - 4y) / 21) - y = 0
Simplifying, we get:
(198 - 12y) / 21 - y = 0
Step 5: Solve for y:
(198 - 12y) - 21y = 0
198 - 12y - 21y = 0
198 - 33y = 0
33y = 198
y = 198 / 33
y = 6
Step 6: Substitute the value of y into the equation we found in Step 3 to solve for x:
x = (66 - 4 * 6) / 21
x = 42 / 21
x = 2
So the solution to the system of equations is x = 2 and y = 6.
Now, let's verify the solution by using matrices:
We can write the system of equations in matrix form as:
| 3 -1 | | x | | 0 |
| 5 2 | * | y | = | 22 |
Let's create the coefficient matrix (A) and the right-hand side matrix (B):
A = | 3 -1 |
| 5 2 |
B = | 0 |
| 22 |
Now, let's find the inverse of matrix A:
A^(-1) = (1 / det(A)) * adj(A)
To find the determinant of A:
det(A) = 3 * 2 - (-1) * 5
= 6 + 5
= 11
To find the adjoint of A:
adj(A) = | 2 1 |
| -5 3 |
Therefore, A^(-1) = (1 / 11) * | 2 1 |
| -5 3 |
Multiplying the inverse of A with matrix B:
A^(-1) * B = (1 / 11) * | 2 1 | * | 0 |
| -5 3 | | 22 |
= (1 / 11) * | 2 * 0 + 1 * 22 |
| -5 * 0 + 3 * 22 |
= (1 / 11) * | 0 + 22 |
| 0 + 66 |
= (1 / 11) * | 22 |
| 66 |
= | 2 |
| 6 |
The result of A^(-1) * B is equal to the solution we have obtained algebraically, confirming that our solution is correct.
Therefore, the solution to the system of equations is x = 2 and y = 6.