Use an inverse matrix to solve each system of linear equations.
A) x+2y=-1 B)x+2y=10 C)x+2y=-3
x-2y=3 x-2y=-6 x-2y=0
To solve each system of linear equations using an inverse matrix, we can follow the following steps:
Step 1: Write the system of equations in matrix form. Let's call the coefficient matrix A and the variable matrix X.
The system of equations can be written as:
A) | 1 2 | | x | = | -1 |
| 1 -2 | | y | | 3 |
B) | 1 2 | | x | = | 10 |
| 1 -2 | | y | | -6 |
C) | 1 2 | | x | = | -3 |
| 1 -2 | | y | | 0 |
Step 2: Calculate the inverse of matrix A, denoted as A^(-1).
Step 3: Multiply both sides of the equation by the inverse of matrix A.
A^(-1) * A * X = A^(-1) * B
Since A^(-1) * A is the identity matrix, the equation simplifies to:
X = A^(-1) * B
Step 4: Compute the product A^(-1) * B to find the values of the variables x and y.
Now, let's calculate the inverse matrix for matrix A using the following formula:
A^(-1) = (1 / (ad - bc)) * | d -b |
| -c a |
where a, b, c, and d are the elements of matrix A.
For matrix A:
a = 1, b = 2, c = 1, and d = -2.
We can now calculate the inverse matrix A^(-1) using the formula:
A^(-1) = (1 / ((1)(-2) - (2)(1))) * | -2 -2 |
| -1 1 |
Simplifying further:
A^(-1) = (1 / (-4 - 2)) * | -2 -2 |
| -1 1 |
A^(-1) = (1 / (-6)) * | -2 -2 |
| -1 1 |
A^(-1) = | 2/6 2/6 |
| 1/6 -1/6 |
Simplifying the fractions:
A^(-1) = | 1/3 1/3 |
| 1/6 -1/6 |
Now, let's calculate X by multiplying A^(-1) with B for each system of equations.
A) X = A^(-1) * B
| x | | 1/3 1/3 | | -1 |
| y | = | 1/6 -1/6 | * | 3 |
Multiplying matrices:
| x | | (1/3)(-1) + (1/3)(3) |
| y | = | (1/6)(-1) + (-1/6)(3) |
Simplifying further:
| x | | 0 |
| y | = | 0 |
Therefore, the solution to the system of equations in A) is x = 0 and y = 0.
Similarly, you can now calculate X for systems B) and C) using the inverse matrix A^(-1) and matrix B for each respective system.