When Triangularization method is is applied to x + y + z = 1 , 4x + 3y - z = 6; 3x + 5y + 3z = 4 Then the value L (lower triangular matrix is)
Answers
Step-by-step explanation:
A matrix can serve as a device for representing and solving a system of equations. To express a system in matrix form, we extract the coefficients of the variables and the constants, and these become the entries of the matrix. We use a vertical line to separate the coefficient entries from the constants, essentially replacing the equal signs. When a system is written in this form, we call it an augmented matrix.
For example, consider the following 2 × 2system of equations.
3x + 4y = 7
4x − 2y = 5
We can write this system as an augmented matrix (please note that a more common formatting for augmented matrices has a solid vertical line running through the matrix, rather than a line on each row):
[34|74−2|5][34|74−2|5]
We can also write a matrix containing just the coefficients. This is called the
coefficient matrix.
[344−2][344−2]
A three-by-three system of equations such as
3x − y − z = 0
x + y = 5
2x − 3z = 2
has a coefficient matrix
⎡⎢⎣3−1−111020−3⎤⎥⎦[3−1−111020−3]
and is represented by the augmented matrix
⎡⎢⎣3−1−1∣0110∣5
How To
Given a system of equations, write an augmented matrix.
Write the coefficients of the x-terms as the numbers down the first column.
Write the coefficients of the y-terms as the numbers down the second column.
If there are z-terms, write the coefficients as the numbers down the third column.
Draw a vertical line and write the constants to the right of the line.
Answer:
Solve equation [1] for the variable z
[1] z = -x - y + 1
// Plug this in for variable z in equation [2]
[2] 4x + 3y - (-x -y +1) = 6
[2] 5x + 4y = 7
// Plug this in for variable z in equation [3]
[3] 3x + 5y + 3•(-x -y +1) = 4
[3] 2y = 1
// Solve equation [3] for the variable y
[3] 2y = 1
[3] y = 1/2
// Plug this in for variable y in equation [2]
[2] 5x + 4•(1/2) = 7
[2] 5x = 5
// Solve equation [2] for the variable x
[2] 5x = 5
[2] x = 1
// By now we know this much :
x = 1
y = 1/2
z = -x-y+1
// Use the x and y values to solve for z
z = -(1)-(1/2)+1 = -1/2
{x,y,z} = {1,1/2,-1/2}