I have a few questions I need help with!
1. Solve the system by triangularizing the augmented matrix and using back substitution. If the system is linearly dependent, give the solution in terms of a parameter.
-3x+y-z=-4
6x-2y+2z=8
-12x+4y-4z=-16
2. Solve the system by triangularizing the augmented matrix and using back substitution.
6x=y-4
3y=-13-7x
3. What row operation would produce zero beneath the first entry in the diagonal?
1 2 | -1
8 -2 | -2
Sure, I'd be happy to help you with your questions! Let's take them one by one:
1. To solve the system by triangularizing the augmented matrix and using back substitution, we can follow these steps:
Step 1: Write down the augmented matrix for the system of equations. The augmented matrix is formed by the coefficients of the variables on the left-hand side and the constants on the right-hand side. For this system, the augmented matrix would be:
|-3 1 -1 -4|
| 6 -2 2 8|
|-12 4 -4 -16|
Step 2: Perform row operations to transform the matrix into triangular form. There are different ways to do this, but one common approach is Gaussian elimination. Subtract twice the first row from the second row, and add four times the first row to the third row. This way, the element in the second row, first column becomes zero.
|-3 1 -1 -4|
| 0 -4 0 16|
| 0 8 -8 0|
Step 3: Now, we have a triangular matrix. We can start with the bottom row and solve for the variables using back substitution. In this case, we get:
8z = 0, which gives z = 0.
-4y + 0z = 16, which gives y = -4.
-3x + y - z = -4, plugging in the values of y and z, we get -3x - 4 - 0 = -4, which gives x = 0.
Therefore, the solution to the system is x = 0, y = -4, and z = 0.
2. Let's follow similar steps to solve the second system:
Step 1: Write down the augmented matrix for the system of equations:
| 6 -1 0 -4|
|-7 3 1 -13|
Step 2: Perform row operations to transform the matrix into triangular form. To eliminate the x-term in the second equation, multiply the first equation by 7 and add it to the second equation:
| 6 -1 0 -4|
| 0 20 1 15|
Step 3: Solve the system using back substitution. Starting from the bottom row, we get:
20y + z = 15, which gives z = -15 + 20y.
6x - y = -4, plugging in the value of y, we get 6x - (-15 + 20y) = -4, which simplifies to 6x + 15 - 20y = -4. Rearranging this equation gives us 20y - 6x = 19.
Therefore, the system has infinitely many solutions, as it can be expressed in terms of the parameter y: x = (19 - 20y)/6 and z = -15 + 20y.
3. To produce a zero beneath the first entry in the diagonal, we can use the row operation of multiplying a row by a constant and adding it to another row. In this case, we can multiply the first row by -8 and add it to the second row:
|-8 -16 | 8|
| 8 -2 | -2|
This operation will result in a zero beneath the first entry in the diagonal.
I hope that helps! Let me know if you have any more questions.