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

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

The original system of equations:

1. 2x+y+6z=9

2. 8x+3y+2z=13

3. x+5y+z=7

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

[216∣9832∣13151∣7]⎣⎡​281​135​621​∣∣∣​9137​⎦⎤​

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. Divide Row 1 by 2 to make the leading coefficient 1:

[10.53∣4.5832∣13151∣7]⎣⎡​181​0.535​321​∣∣∣​4.5137​⎦⎤​

2. Subtract Row 1 from Row 2 and Row 3 to create zeros below A[1,1]:

[10.53∣4.50−1−22∣−3104.5−2∣2.5]⎣⎡​100​0.5−14.5​3−22−2​∣∣∣​4.5−312.5​⎦⎤​

Step 2: Perform row operations to create a 1 in the second row and second column (i.e., make A[2,2] equal to 1):

3. Multiply Row 2 by -1 to make A[2,2] equal to 1:

[10.53∣4.50122∣3104.5−2∣2.5]⎣⎡​100​0.514.5​322−2​∣∣∣​4.5312.5​⎦⎤​

4. Subtract 4.5 times Row 2 from Row 3 to create a zero below A[2,2]:

[10.53∣4.50122∣3100−101∣−140.5]⎣⎡​100​0.510​322−101​∣∣∣​4.531−140.5​⎦⎤​

Step 3: Perform row operations to create zeros above and below A[3,3] (i.e., make A[3,3] equal to 1):

5. Multiply Row 3 by -1/101 to make A[3,3] equal to 1:

[10.53∣4.50122∣31001∣1.39]⎣⎡​100​0.510​3221​∣∣∣​4.5311.39​⎦⎤​

6. Subtract 3 times Row 3 from Row 1 and 22 times Row 3 from Row 2 to create zeros above and below A[3,3]:

[10.50∣0.78010∣29.22001∣1.39]⎣⎡​100​0.510​001​∣∣∣​0.7829.221.39​⎦⎤​

Step 4: Perform row operations to create zeros above and below A[1,2]:

7. Subtract 0.5 times Row 2 from Row 1 to create a zero above A[1,2]:

[100∣0.78010∣29.22001∣1.39]⎣⎡​100​010​001​∣∣∣​0.7829.221.39​⎦⎤​

Step 5: Perform row operations to create zeros above and below A[2,1]:

8. Subtract 0.5 times Row 1 from Row 2 to create a zero above A[2,1]:

[100∣0.78010∣29.22001∣1.39]⎣⎡​100​010​001​∣∣∣​0.7829.221.39​⎦⎤​

Now, the augmented matrix is in the reduced row-echelon form. We can read the solutions directly:

From the last row, we have =1.39z=1.39.

From the second row, we have =29.22y=29.22.

From the first row, we have =0.78x=0.78.

So, the solution to the system of equations is =0.78x=0.78, =29.22y=29.22, and =1.39z=1.39.