# Matrix determinant using Laplace method

I have the following matrix of order four for which I have calculated the determinant using Laplace's method.

$$\begin{bmatrix} 2 & 1 & 3 & 1 \\ 4 & 3 & 1 & 4 \\ -1 & 5 & -2 & 1 \\ 1 & 3 & -2 & -1 \\ \end{bmatrix}$$

Finding the determinant gives me $-726$. Now if I check the result at Wolfram Alpha, it says the result is $-180$ (Because there are no zeros in the matrix, expand with respect to row one) so it uses only the first row to calculate the determinant of the matrix.

My question is: Why it uses only the first row to find the determinant?

• Expanding wrt the first row does not mean use ONLY the first row. – BlackAdder Jul 23 '13 at 14:08
• You can use any row. Just keep in mind that row transposition(assuming some other row in the first row's place) done odd number of time would lead to a negative sign in front. – Torsten Hĕrculĕ Cärlemän Jul 23 '13 at 14:08
• I've used all rows to find the determinant, is it correct or when the matrix has no zeros I should just pick one row...? – K-Zen Jul 23 '13 at 14:13
• I think I understand what you are trying to ask now. See if my answer helps. – BlackAdder Jul 23 '13 at 14:28

## 2 Answers

The Laplace development can be performed with respect to any row or column. Let's see when developing with respect to the first row: \begin{align} \det\begin{bmatrix} 2 & 1 & 3 & 1 \\ 4 & 3 & 1 & 4 \\ -1 & 5 & -2 & 1 \\ 1 & 3 & -2 & -1 \end{bmatrix}={}& (-1)^{1+1}\cdot 2 \cdot \det\begin{bmatrix} 3 & 1 & 4 \\ 5 & -2 & 1 \\ 3 & -2 & -1 \end{bmatrix}+{}\\ &(-1)^{1+2}\cdot 1 \cdot \det\begin{bmatrix} 4 & 1 & 4 \\ -1 & -2 & 1 \\ 1 & -2 & -1 \end{bmatrix}+{}\\ &(-1)^{1+3}\cdot 3\cdot \det\begin{bmatrix} 4 & 3 & 4 \\ -1 & 5 & 1 \\ 1 & 3 & -1 \end{bmatrix}+{}\\ &(-1)^{1+4}\cdot1\cdot \det\begin{bmatrix} 4 & 3 & 1 \\ -1 & 5 & -2 \\ 1 & 3 & -2 \end{bmatrix} \end{align} Now you can go on by computing the determinants of the four $3\times3$ matrices with the same (or another) method. The final result is indeed $-180$.

When one row or column has many zeroes it's convenient to use that one, but any row or column can be used.

• My question is actually simpler: Should I use one row or all rows...? – K-Zen Jul 23 '13 at 14:20
• @K-Zen I don't understand what you mean about "using all" rows. Can you add to your question how you got $-726$? – egreg Jul 23 '13 at 14:21
• Finding the adjunt of each element in the matrix and summing all of them gives -726. Exactly what you did above in your post, but for all elements. – K-Zen Jul 23 '13 at 14:23
• @K-Zen That's surely not the correct method. You have to sum up the adjucts of the element in one row (or column). That's it. – egreg Jul 23 '13 at 14:27
• I appreciate your input, my bad completely ;-( – K-Zen Jul 23 '13 at 15:00

$$det(A)=\sum^{n}_{i=1}(-1)^{i+j}a_{i,j}M_{i,j}=\sum^{n}_{j=1}(-1)^{i+j}a_{i,j}M_{i,j}$$ where $a_{i,j}$ are entries in $A$ and $M_{i,j}$ are the minors.

So when someone says expand about the first row, it just means fixing $i=1$ in the second equality. The whole point of picking a row with a zero is to make the terms in the sum vanish. But since in this case there is no such rows, any old row would work, so just pick the first row for example.

Use only one row! Do not repeat for all the rows and sum which is what I think you are trying to say.

I hope that answers your question. Let me know if you have further queries and I'll update the answer.

• I appreciate your input, my bad completely ;-( – K-Zen Jul 23 '13 at 15:00