Determinant Calculation Exercise 1

Let's compute the determinant of the following matrix:

$$ A = \begin{pmatrix} 5 & 2 & 3 & 4 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 2 & 1 \\ 0 & 1 & 2 & 2 \end{pmatrix} $$

We'll use Gaussian elimination to simplify the computation.

The goal is to convert the matrix into an upper triangular form by applying Gaussian row operations:

• Adding a multiple of one row to another
• Multiplying a row by a scalar
• Swapping two rows

Note that swapping two rows flips the sign of the determinant.

Step 1

Swap the second and third rows:

$$ R2 \leftrightarrow R3 $$

The matrix is now:

$$ A = \begin{pmatrix} 5 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 1 & 2 & 2 \end{pmatrix} $$

Step 2

Subtract the second row from the fourth row:

$$ R4 = R4 - R2 $$

Updating the matrix:

$$ A = \begin{pmatrix} 5 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0-0 & 1-1 & 2-2 & 2-1 \end{pmatrix} $$

$$ A = \begin{pmatrix} 5 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} $$

The matrix is now in upper triangular form.

At this point, the determinant is simply the product of the diagonal elements.

Since we performed an odd number of row swaps, we need to flip the sign of the determinant:

$$ - \det(A) = - \begin{pmatrix} 5 & 2 & 3 & 4 \\ 0 & 1 & 2 & 1 \\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix} = - (5 \cdot 1 \cdot 1 \cdot 1 ) = - 5 $$

Thus, the determinant of the matrix is -5.

And so on.