Question

Use Gaussian Elimination Method. 2X + Y + 1 = 4 0. IX -0. 1Y+0. 1Z = 0. 4 3x + 2Y + 1 = 2 X-Y+Z = 4 -2X + 2Y - 22 = - 8 + = 2. ) Find the values of X, Y, and Z. (3+i)X - 3Y+(2+i)Z = 3+4i 2X + Y - Z = 2 +į 3X + (1+i)Y -4Z = 5 + 21 = + =

Answer 1

To solve the given system of equations using Gaussian elimination, let's rewrite the equations in matrix form:

```

[ 2 1 1 ] [ X ] [ 4 ]

[ 0 1 -0.1] * [ Y ] = [ 0.4]

[ 3 2 1 ] [ Z ] [ 2 ]

```

Performing Gaussian elimination:

1. Row 2 = Row 2 - 0.1 * Row 1

```

[ 2 1 1 ] [ X ] [ 4 ]

[ 0 0 0 ] * [ Y ] = [ 0 ]

[ 3 2 1 ] [ Z ] [ 2 ]

```

2. Row 3 = Row 3 - (3/2) * Row 1

```

[ 2 1 1 ] [ X ] [ 4 ]

[ 0 0 0 ] * [ Y ] = [ 0 ]

[ 0 1/2 -1/2] [ Z ] [ -2 ]

```

3. Row 3 = 2 * Row 3

```

[ 2 1 1 ] [ X ] [ 4 ]

[ 0 0 0 ] * [ Y ] = [ 0 ]

[ 0 1 -1 ] [ Z ] [ -4 ]

```

Now, we have reached an upper triangular form. Let's solve the system of equations:

From the third row, we have Z = -4.

Substituting Z = -4 into the second row, we have 0 * Y = 0, which implies that Y can take any value.

Finally, substituting Z = -4 and Y = k (where k is any arbitrary constant) into the first row, we can solve for X:

2X + 1k + 1 = 4

2X = 3 - k

X = (3 - k) / 2

Therefore, the solution to the system of equations is:

X = (3 - k) / 2

Y = k

Z = -4

Note: The given system of equations in the second part of your question is not clear due to missing operators and formatting issues. Please provide the equations in a clear and properly formatted manner if you need assistance with solving that system.