Determinant Calculation Exercise 2

Let's find the determinant of the following matrix:

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

We'll use Gaussian elimination to simplify the calculation.

The goal is to transform the matrix into an upper triangular form using row operations:

If the number of row swaps is odd, the determinant’s sign must be flipped.

Step 1

Swap the first and fourth rows:

$$ R1 \leftrightarrow R4 $$

The matrix is now:

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

Step 2

Subtract the third row from the fourth row:

$$ R4 = R4 - R3 $$

Updating the matrix:

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

Step 3

Swap the second and third rows:

$$ R2 \leftrightarrow R3 $$

Now the matrix looks like this:

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

At this point, the matrix is in upper triangular form, meaning the determinant is simply the product of the diagonal elements.

Since we performed an even number of row swaps, the determinant’s sign remains unchanged:

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

Thus, the determinant of the matrix is 0.

And that’s it!