Computer Oriented Numerical Methods (BCA) 3rd Sem Previous Year Solved Question Paper 2022

To solve the system of linear simultaneous equations using the Gaussian Elimination method, we'll first represent the equations in an augmented matrix form and then perform row operations to transform it into row-echelon form and ultimately into reduced row-echelon form. Here are the steps:

The original system of equations:

1. 2x+2y+z=123

2. x+2y+2z=85

3. x+10y−8z=10

Now, let's represent these equations in an augmented matrix [A|B]:

[221∣123122∣85110−8∣10]⎣⎡​211​2210​12−8​∣∣∣​1238510​⎦⎤​

Step 1: Perform row operations to create zeros below the first element in the first column (i.e., make the first column entries below A[1,1] equal to zero):

1. Multiply Row 1 by -0.5 and add it to Row 2:

[221∣123011.5∣29.5110−8∣10]⎣⎡​201​2110​11.5−8​∣∣∣​12329.510​⎦⎤​

2. Multiply Row 1 by -0.5 and add it to Row 3:

[221∣123011.5∣29.509−8.5∣−113.5]⎣⎡​200​219​11.5−8.5​∣∣∣​12329.5−113.5​⎦⎤​

Step 2: Continue with row operations to create zeros below the second element in the second column:

3. Multiply Row 2 by -9 and add it to Row 3:

[221∣123011.5∣29.500−22∣−366]⎣⎡​200​210​11.5−22​∣∣∣​12329.5−366​⎦⎤​

Step 3: Perform back-substitution to find the values of x, y, and z:

From the last row, we have: −22=−366  ⟹  =−366−22=33−22z=−366⟹z=−22−366​=33

Now, substitute =33z=33 into the second row: +1.5⋅33=29.5  ⟹  +49.5=29.5  ⟹  =−20y+1.5⋅33=29.5⟹y+49.5=29.5⟹y=−20

Finally, substitute the values of y and z into the first row: 2+2⋅(−20)+33=123  ⟹  2−40+33=123  ⟹  2−7=123  ⟹  2=130  ⟹  =652x+2⋅(−20)+33=123⟹2x−40+33=123⟹2x−7=123⟹2x=130⟹x=65

So, the solution to the system of equations is =65x=65, =−20y=−20, and =33z=33.