using gauss jordan method find the jnverse of the matrix {(3,-3,4),(2,-3,4),(0,-1,1)}

To find the inverse of a matrix using the Gauss-Jordan method, we need to follow these steps:

1. Write the matrix in an augmented form, appending the identity matrix to it. For your matrix:
   
   \[
   \left( \begin{array}{ccc|ccc}
   3 & -3 & 4 & 1 & 0 & 0 \\
   2 & -3 & 4 & 0 & 1 & 0 \\
   0 & -1 & 1 & 0 & 0 & 1 \\
   \end{array} \right)
   \]

2. Perform row operations to transform the left part (original matrix) into the identity matrix. The right part will become the inverse matrix.

Let's proceed with the row operations step by step.

### Step 1: Make the first element of the first row a 1.
We divide the entire first row by 3 to get a leading 1.

\[
\left( \begin{array}{ccc|ccc}
1 & -1 & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\
2 & -3 & 4 & 0 & 1 & 0 \\
0 & -1 & 1 & 0 & 0 & 1 \\
\end{array} \right)
\]

### Step 2: Make the first element of the second row 0.
We subtract 2 times the first row from the second row.

\[
\left( \begin{array}{ccc|ccc}
1 & -1 & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\
0 & -1 & \frac{4}{3} & -\frac{2}{3} & 1 & 0 \\
0 & -1 & 1 & 0 & 0 & 1 \\
\end{array} \right)
\]

### Step 3: Make the first element of the third row 0.
We subtract 0 times the first row from the third row (no change needed in this case).

\[
\left( \begin{array}{ccc|ccc}
1 & -1 & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\
0 & -1 & \frac{4}{3} & -\frac{2}{3} & 1 & 0 \\
0 & -1 & 1 & 0 & 0 & 1 \\
\end{array} \right)
\]

### Step 4: Make the second element of the second row a 1.
We multiply the second row by -1 to simplify.

\[
\left( \begin{array}{ccc|ccc}
1 & -1 & \frac{4}{3} & \frac{1}{3} & 0 & 0 \\
0 & 1 & -\frac{4}{3} & \frac{2}{3} & -1 & 0 \\
0 & -1 & 1 & 0 & 0 & 1 \\
\end{array} \right)
\]

### Step 5: Make the second element of the first row 0.
We add the second row to the first row to eliminate the second element of the first row.

\[
\left( \begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 0 \\
0 & 1 & -\frac{4}{3} & \frac{2}{3} & -1 & 0 \\
0 & -1 & 1 & 0 & 0 & 1 \\
\end{array} \right)
\]

### Step 6: Make the second element of the third row 0.
We add the second row to the third row.

\[
\left( \begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 0 \\
0 & 1 & -\frac{4}{3} & \frac{2}{3} & -1 & 0 \\
0 & 0 & \frac{-1}{3} & \frac{2}{3} & -1 & 1 \\
\end{array} \right)
\]

### Step 7: Make the third element of the third row a 1.
We divide the third row by \(-\frac{1}{3}\).

\[
\left( \begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 0 \\
0 & 1 & -\frac{4}{3} & \frac{2}{3} & -1 & 0 \\
0 & 0 & 1 & -2 & 3 & -3 \\
\end{array} \right)
\]

### Step 8: Make the third element of the second row 0.
We add \(\frac{4}{3}\) times the third row to the second row.

\[
\left( \begin{array}{ccc|ccc}
1 & 0 & 0 & 1 & -1 & 0 \\
0 & 1 & 0 & \frac{-2}{3} & 3 & -4 \\
0 & 0 & 1 & -2 & 3 & -3 \\
\end{array} \right)
\]

### Final Step: The inverse matrix is now complete. The inverse of the given matrix is:

\[
\left( \begin{array}{ccc}
1 & -1 & 0 \\
-\frac{2}{3} & 3 & -4 \\
-2 & 3 & -3 \\
\end{array} \right)
\]

Do you want further details or have any questions?

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### Related Questions:
1. What is the Gauss-Jordan method, and how is it different from Gaussian elimination?
2. How does row reduction work in finding matrix inverses?
3. Can every matrix be inverted using the Gauss-Jordan method?
4. What conditions must a matrix satisfy to have an inverse?
5. What are the practical applications of finding a matrix inverse?

### Tip:
Always check the determinant of a matrix before attempting to find its inverse. If the determinant is zero, the matrix is singular and does not have an inverse.

using gauss jordan method find the inverse of the matrix {(3,-3,4),(2,-3,4),(0,-1,1)}

Discover how to find the inverse of the matrix {(3,-3,4),(2,-3,4),(0,-1,1)} using the Gauss-Jordan elimination method. Follow a detailed step-by-step solution that covers key concepts in linear algebra, including matrix inversion and row operations. Ideal for students in Grades 10-12.

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