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The image asks to find the inverse of matrix \( A = \begin{bmatrix} 1 & -2 & -3 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{bmatrix} \) using the Gauss-Jordan method.

Let's proceed to compute the inverse of the matrix \( A \) step by step using the Gauss-Jordan elimination method.

We start with the augmented matrix \( [A | I] \), where \( I \) is the identity matrix of the same dimension as \( A \):

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

Now, we'll apply row operations to transform the left side into the identity matrix, while performing the same operations on the right side to obtain the inverse.

### Step 1: Normalize Row 1
Row 1 is already in the form where the leading entry is 1, so no changes are needed.

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

### Step 2: Normalize Row 2
We divide Row 2 by 2 to make the leading entry 1:

\[
R_2 \rightarrow \frac{R_2}{2}
\]

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

### Step 3: Normalize Row 3
We divide Row 3 by 3 to make the leading entry 1:

\[
R_3 \rightarrow \frac{R_3}{3}
\]

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

### Step 4: Eliminate Off-Diagonal Entries
Now, we perform row operations to eliminate the non-zero entries above and below the pivot elements.

- To eliminate the -2 in Row 1, we add 2 times Row 2 to Row 1:

\[
R_1 \rightarrow R_1 + 2R_2
\]

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

- To eliminate the -3 in Row 1, we add 3 times Row 3 to Row 1:

\[
R_1 \rightarrow R_1 + 3R_3
\]

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

Now the left side is the identity matrix, and the right side is the inverse of \( A \).

### Final Inverse Matrix:
\[
A^{-1} = \begin{bmatrix}
1 & 1 & 1 \\
0 & \frac{1}{2} & 0 \\
0 & 0 & \frac{1}{3}
\end{bmatrix}
\]

Would you like any further clarification or details?

Here are five related questions to extend your understanding:

1. What are the key differences between the Gauss and Gauss-Jordan methods for matrix inversion?
2. How can we verify that the computed inverse is correct?
3. What types of matrices do not have an inverse, and why?
4. How can you compute the inverse of a matrix using determinants and cofactors instead?
5. How does the computational complexity of inverting a matrix change as the size of the matrix increases?

**Tip:** For symmetric matrices, special properties can often simplify the inversion process.

Find the inverse of the matrix A = [[1, -2, -3], [0, 2, 0], [0, 0, 3]] using the Gauss-Jordan method.

Learn how to find the inverse of matrix A = [[1, -2, -3], [0, 2, 0], [0, 0, 3]] using the Gauss-Jordan elimination method. This step-by-step solution explains the key concepts of matrix algebra and inversion.

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