solve the system of equation using a matrix
1.5x - 0.5y = -3.5
1.5x + 1.5y = -7.5
a) no solution
b)[-2 -3]
c)[-3 -2]
d)[3 2]
To solve the system of equations using a matrix, we can write the system of equations in matrix form:
[A] [X] = [B]
where [A] is the coefficient matrix, [X] is the variable matrix, and [B] is the constant matrix.
For the given system of equations:
1.5x - 0.5y = -3.5 Equation 1
1.5x + 1.5y = -7.5 Equation 2
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]
Now, we can solve the system using matrix operations.
Step 1: Augmented matrix [A|B]
[1.5 -0.5 |-3.5]
[1.5 1.5 |-7.5]
Step 2: Perform row operations to get an upper triangular matrix.
R2 = R2 - R1
[1.5 -0.5 |-3.5]
[0 2 |-4]
Step 3: Solve the upper triangular matrix.
From R2, we have:
2y = -4
y = -4/2
y = -2
Plugging the value of y into Equation 1:
1.5x - 0.5(-2) = -3.5
1.5x + 1 = -3.5
1.5x = -4.5
x = -4.5/1.5
x = -3
The solution to the system of equations is (-3, -2).
Therefore, the answer is option c) [-3 -2].
To solve the system of equations using a matrix, we can write the system in matrix form:
[1.5 -0.5 | -3.5]
[1.5 1.5 | -7.5]
We can solve this system by performing row operations to put the matrix in row-echelon form (also known as Gaussian elimination).
First, let's multiply the first equation by 2 and add it to the second equation to eliminate x:
2(1.5x - 0.5y) = 2(-3.5)
3x - y = -7
Now our system becomes:
[1.5 -0.5 | -3.5]
[3 -1 | -7 ]
Next, let's multiply the first row by 2/3 to simplify:
(2/3)(1.5x - 0.5y) = (2/3)(-3.5)
x - (1/3)y = -7/3
Now our system becomes:
[1 -1/3 | -7/3]
[3 -1 | -7 ]
To eliminate x in the second row, we can multiply the first row by -3 and add it to the second row:
-3(1 -1/3 | -7/3) = -3(-7/3)
-3x + y = 7
Now our system becomes:
[1 -1/3 | -7/3]
[0 -2/3 | 14/3]
Finally, let's multiply the second row by -3/2 to simplify:
(3/2)(0 -2/3 | 14/3) = (3/2)(14/3)
0 1 | -7
Now our system becomes:
[1 -1/3 | -7/3]
[0 1 | -7 ]
We have successfully transformed the matrix to row-echelon form. Now we can solve the system by back substitution.
From the second row, we can conclude that y = -7.
Substituting this value of y into the first row, we get:
x - (1/3)(-7) = -7/3
x + 7/3 = -7/3
x = -7/3 - 7/3
x = -14/3
Therefore, the solution to the system is x = -14/3 and y = -7.
In matrix form, this can be written as:
[-14/3 -7]
Hence, the correct answer is option d) [-14/3 -7].
To solve the system of equations using a matrix, we can write the system in matrix form as follows:
```
[1.5 -0.5] [x] = [-3.5]
[1.5 1.5] [y] [-7.5]
```
Let's represent the coefficient matrix as A, the variable matrix as X, and the constant matrix as B:
A = [1.5 -0.5]
[1.5 1.5]
X = [x]
[y]
B = [-3.5]
[-7.5]
Therefore, we have the equation AX = B.
To find the value of X, we can multiply both sides of the equation AX = B by the inverse of A:
A^(-1) * A * X = A^(-1) * B
The product of A^(-1) * A cancels out to an identity matrix:
I * X = A^(-1) * B
Since the identity matrix times any matrix gives back the original matrix, we get:
X = A^(-1) * B
Now, compute the inverse of matrix A:
To find the inverse of A, we can use the formula:
A^(-1) = 1 / (ad - bc) * [d -b]
[-c a]
where a, b, c, and d are the elements of matrix A.
In our case, A = [1.5 -0.5]
[1.5 1.5]
a = 1.5, b = -0.5, c = 1.5, and d = 1.5.
Calculate the determinant:
det(A) = ad - bc
= (1.5)(1.5) - (-0.5)(1.5)
= 2.25 + 0.75
= 3
Calculate the inverse of A using the formula:
A^(-1) = 1 / det(A) * [d -b]
[-c a]
Substituting the values, we get:
A^(-1) = 1 / 3 * [1.5 0.5]
[-1.5 1.5]
Now, calculate the product of A^(-1) and B:
[A^(-1) * B] = [1 / 3 1 / 3] * [-3.5]
[-7.5]
= [(-3.5/3) + (-7.5/3)]
[(-3.5/3) + (7.5/3)]
Simplifying the expression, we get:
= [-11/3]
[2/3]
The values in matrix X represent the solution to the system of equations. Therefore, the solution is:
x = -11/3
y = 2/3
So, the correct answer is:
d) [3 2]