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The problem involves two matrices, \( J \) and \( K \), and asks us to compute:

a) The product \( JK \)  
b) The inverse of matrix \( J \), denoted as \( J^{-1} \)

Given matrices:
\[
J = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & -2 \\ 0 & 0 & -1 \end{bmatrix}, \quad K = \begin{bmatrix} 3 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & -3 \end{bmatrix}
\]

Let's solve each part step-by-step.

### a) Calculating the Product \( JK \)

To find the product \( JK \), we multiply matrix \( J \) by matrix \( K \):

\[
JK = \begin{bmatrix} 1 & 2 & -1 \\ 0 & 3 & -2 \\ 0 & 0 & -1 \end{bmatrix} \times \begin{bmatrix} 3 & -2 & 1 \\ 0 & 1 & -2 \\ 0 & 0 & -3 \end{bmatrix}
\]

### b) Finding the Inverse \( J^{-1} \)

The inverse of matrix \( J \) can be computed by using the formula for the inverse of a 3x3 matrix:

\[
J^{-1} = \frac{1}{\det(J)} \cdot \text{adj}(J)
\]

where:
- \( \det(J) \) is the determinant of matrix \( J \).
- \( \text{adj}(J) \) is the adjugate of matrix \( J \), which is the transpose of the cofactor matrix.

Let's compute these now. I will perform these calculations step-by-step.### Solutions:

a) **The Product \( JK \):**
\[
JK = \begin{bmatrix} 3 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}
\]

b) **The Inverse of Matrix \( J \), \( J^{-1} \):**
\[
J^{-1} = \begin{bmatrix} 1 & -\frac{2}{3} & \frac{1}{3} \\ 0 & \frac{1}{3} & -\frac{2}{3} \\ 0 & 0 & -1 \end{bmatrix}
\]

This completes the computation for both parts of the problem.

### Do you have any questions or would you like more details on any of the steps?

Here are some related questions that could help deepen your understanding:
1. How is the inverse of a matrix computed using elementary row operations?
2. What properties do the matrices \( J \) and \( K \) have that result in the product \( JK \) being a scalar multiple of the identity matrix?
3. How can determinants be used to check if a matrix is invertible?
4. What is the significance of matrix multiplication in different fields of study?
5. How do different methods of finding a matrix inverse compare in terms of complexity and applications?

**Tip:** Understanding matrix multiplication and inversion can help solve systems of linear equations efficiently.

Given matrices J = [ [1, 2, -1], [0, 3, -2], [0, 0, -1] ] and K = [ [3, -2, 1], [0, 1, -2], [0, 0, -3] ], compute (a) the product JK and (b) the inverse of J.

Explore how to compute the product of matrices JK and the inverse of matrix J. The problem involves matrix multiplication and inversion, key concepts in linear algebra, suitable for undergraduate students.

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