Analyse and Model Engineering System

Analyse and Model Engineering
System

TASK 2.0
2.1 Determinant of matrix
1 2 3
0 -4 1
0 3 -1
Solution
Determinant of given matrix = 1*[(-4)*(-1) – (1)*(3)] – 2*[(0)*(-1) – (1)*(0)] + 3*[(0)*(3) – (-
4)*(0)]
Determinant = 1
2.21 Current equations obtained from closed loop
2x + 3y – 4z = 26
x – 5y -3z = -87
-7x +2y + 6z = 12
Solution
2 3 -4 | 26
1 -5 -3 | -87
-7 2 6 | 12
Using Gaussian elimination form:
R1 < --->R3
R2 ---> R2 + (1/7)*R1
R3 ---> R3 + (2/7)*R1
R3 ---> R3 + (25/33)*R2
-7 2 6 | 12
0 (-33/7) (-15/7) | -85.28
0 0 (-43/11) | -35.18
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Again converting this into equation form:
-7x +2y +6z = 12
(-33/7) y + (-15/7) z = -85.28
(-43/11) z = -35.18
By third equation we have z = 8.99
On putting this value in second equation we get y = 14
Similarly by substituting values of y and z, we have x = 10
x = 10
y = 14
z = 8.99
2.22 Inverse matrix method
The matrix form of equations is:
2 3 -4 x 26
1 -5 -3 y = -87
-7 2 6 z 12
Solution
The inverse of coefficient matrix A is determined as follows:
2 3 -4 | 1 0 0
1 -5 -3 | 0 1 0
-7 2 6 | 0 0 1
On reducing the matrix to echelon form we have the inverse matrix as:
(-8/43) (-26/129) (-29/129)
(5/43) (-16/129) (2/129)
(-11/43) (-25/129) (-13/129)
From the given equation we have
AA ̄ X = A ̄ B
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