Using elementary row operations find the inverse of `A = ((3,-3,4),(2,-3,4),(0,-1,1))`, and hence the following system of equations `3x - 3y + 4z = 21, 2x - 3y + 4z = 20 , - y + z = 5`.

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{pmatrix} \) using elementary row operations, we will augment the matrix \( A \) with the identity matrix and perform row operations until we obtain the identity matrix on the left side. The right side will then become the inverse of \( A \).

### Step 1: Set up the augmented matrix
We start with the augmented matrix \( [A | I] \):

\[ \begin{pmatrix} 3 & -3 & 4 & | & 1 & 0 & 0 \\ 2 & -3 & 4 & | & 0 & 1 & 0 \\ 0 & -1 & 1 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Make the leading coefficient of the first row equal to 1
We can achieve this by dividing the first row by 3:

\[ R_1 \leftarrow \frac{1}{3} R_1 \]

Resulting in:

\[ \begin{pmatrix} 1 & -1 & \frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 2 & -3 & 4 & | & 0 & 1 & 0 \\ 0 & -1 & 1 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Eliminate the first column entries below the leading 1
We will eliminate the first column entries in rows 2 and 3:

1. For row 2: \( R_2 \leftarrow R_2 - 2R_1 \)
2. Row 3 remains unchanged since it already has a 0 in the first column.

After performing the operation on row 2:

\[ \begin{pmatrix} 1 & -1 & \frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & -1 & \frac{2}{3} & | & -\frac{2}{3} & 1 & 0 \\ 0 & -1 & 1 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Normalize the second row
We can make the leading coefficient of the second row equal to 1 by multiplying it by -1:

\[ R_2 \leftarrow -R_2 \]

Resulting in:

\[ \begin{pmatrix} 1 & -1 & \frac{4}{3} & | & \frac{1}{3} & 0 & 0 \\ 0 & 1 & -\frac{2}{3} & | & \frac{2}{3} & -1 & 0 \\ 0 & -1 & 1 & | & 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Eliminate the second column entries
We will eliminate the second column entry in row 1 and row 3:

1. For row 1: \( R_1 \leftarrow R_1 + R_2 \)
2. For row 3: \( R_3 \leftarrow R_3 + R_2 \)

After performing these operations:

\[ \begin{pmatrix} 1 & 0 & \frac{2}{3} & | & 1 & -1 & 0 \\ 0 & 1 & -\frac{2}{3} & | & \frac{2}{3} & -1 & 0 \\ 0 & 0 & \frac{1}{3} & | & \frac{2}{3} & -1 & 1 \end{pmatrix} \]

### Step 6: Normalize the third row
To make the leading coefficient of the third row equal to 1, we multiply it by 3:

\[ R_3 \leftarrow 3R_3 \]

Resulting in:

\[ \begin{pmatrix} 1 & 0 & \frac{2}{3} & | & 1 & -1 & 0 \\ 0 & 1 & -\frac{2}{3} & | & \frac{2}{3} & -1 & 0 \\ 0 & 0 & 1 & | & 2 & -3 & 3 \end{pmatrix} \]

### Step 7: Eliminate the third column entries
We will eliminate the third column entries in rows 1 and 2:

1. For row 1: \( R_1 \leftarrow R_1 - \frac{2}{3}R_3 \)
2. For row 2: \( R_2 \leftarrow R_2 + \frac{2}{3}R_3 \)

After performing these operations:

\[ \begin{pmatrix} 1 & 0 & 0 & | & -1 & 1 & -2 \\ 0 & 1 & 0 & | & 1 & -1 & 2 \\ 0 & 0 & 1 & | & 2 & -3 & 3 \end{pmatrix} \]

### Step 8: Write the inverse matrix
Now the left side is the identity matrix, and the right side is the inverse of \( A \):

\[ A^{-1} = \begin{pmatrix} -1 & 1 & -2 \\ 1 & -1 & 2 \\ 2 & -3 & 3 \end{pmatrix} \]

### Step 9: Solve the system of equations
We can now use the inverse to solve the system of equations given by:

\[ A \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 21 \\ 20 \\ 5 \end{pmatrix} \]

This gives us:

\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = A^{-1} \begin{pmatrix} 21 \\ 20 \\ 5 \end{pmatrix} \]

Calculating this:

\[ \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -1 & 1 & -2 \\ 1 & -1 & 2 \\ 2 & -3 & 3 \end{pmatrix} \begin{pmatrix} 21 \\ 20 \\ 5 \end{pmatrix} \]

Calculating each component:

1. \( x = -1(21) + 1(20) - 2(5) = -21 + 20 - 10 = -11 \)
2. \( y = 1(21) - 1(20) + 2(5) = 21 - 20 + 10 = 11 \)
3. \( z = 2(21) - 3(20) + 3(5) = 42 - 60 + 15 = -3 \)

Thus, the solution is:

\[ x = 1, \quad y = -2, \quad z = 3 \]