# Thread: how to get 5x5 matrix determinat

1. ## how to get 5x5 matrix determinat

hi, im an engineering student and I am looking for an easier way to find the determinant of a 5x5 matrix, its actually an admittance matrix, so i cant use calculator plus it needs to be done in manual. thanks in advance

2. ## Re: how to get 5x5 matrix determinat

It depends on the nature of the matrix. Sparse matrices (with lots of zeros) can be dealt with by the cofactors. Other methods have their own domains of excellence. Row reduction is a reasonably good method in general, but it's always worth keeping an eye out for short-cuts by mixing methods.

3. ## Re: how to get 5x5 matrix determinat

Originally Posted by GrayHat
hi, im an engineering student and I am looking for an easier way to find the determinant of a 5x5 matrix, its actually an admittance matrix, so i cant use calculator plus it needs to be done in manual. thanks in advance
The real question here is “Why are you not allowed to use calculators and/or webbased mathematics?
In the past twenty-five mathematics education has undergone a complete revolution. As usual schools of engineering are some fifty years behind the curve.
Look at this example.
Why not use Wolframalpha for all engineering courses? I think that the answer is found in the fact that faculty members are aging and hence have not kept up with mathematics education. I think that in the next twenty years that no calculus textbook will have chapters on techniques of integrations.
Look at this. Can anyone argue that knowing how to get that result is more important than understanding what it means?
So @GrayHat;932729, ask you Dean(Head of school) about these issues. Why should you not be able able to use tools that make you study more understandable?

4. ## Re: how to get 5x5 matrix determinat

One of the thing you can do is use "row reduction" to reduce the matrix to an upper triangular matrix.

The "row operations" are:
1) Swap two rows. This multiplies the determinant by -1.
2) Multiply every term in a row by a number. This multiplies the determinant by that number.
3) Add a multiple of one row to another. This doesn't change the determinant.

As an easy example consider $\displaystyle \begin{bmatrix}1 & 0 & 3 \\ 0 & 2 & 1 \\ 2 & 4 & 3\end{bmatrix}$
Start by subtracting twice the first row from the third.
That leaves $\displaystyle \begin{bmatrix}1 & 0 & 3 \\ 0 & 2 & 1 \\ 0 & 4 & -3\end{bmatrix}$
and does not change the determinant.

Multiply each term in the second row by 1/2.
That leaves $\displaystyle \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 4 & -3\end{bmatrix}$
and multiplies the determinant by [math]\frac{1}{2}

Add -4 times the second row to the third row.
That leaves $\displaystyle \begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & -5\end{bmatrix}$
and does not change the determinant.

That is now an "upper triangular" matrix so its determinant is the product of the numbers on the main diagonal, -5.

Since the only row operation that changed the determinant multiplied it by 1/2, go back to the original determinant by multiplying by 2:
the determinant of the original matrix is -10.

5. ## Re: how to get 5x5 matrix determinat

I have changed the Latex on my previous response so it is now readable:

One of the thing you can do is use "row reduction" to reduce the matrix to an upper triangular matrix.

The "row operations" are:
1) Swap two rows. This multiplies the determinant by -1.
2) Multiply every term in a row by a number. This multiplies the determinant by that number.
3) Add a multiple of one row to another. This doesn't change the determinant.

As an easy example consider $\begin{bmatrix}1 & 0 & 3 \\ 0 & 2 & 1 \\ 2 & 4 & 3\end{bmatrix}$
Start by subtracting twice the first row from the third.
That leaves $\begin{bmatrix}1 & 0 & 3 \\ 0 & 2 & 1 \\ 0 & 4 & -3\end{bmatrix}$
and does not change the determinant.

Multiply each term in the second row by 1/2.
That leaves $\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 4 & -3\end{bmatrix}$
and multiplies the determinant by $\frac{1}{2}$.

Add -4 times the second row to the third row.
That leaves $\begin{bmatrix}1 & 0 & 3 \\ 0 & 1 & \frac{1}{2} \\ 0 & 0 & -5\end{bmatrix}$
and does not change the determinant.

That is now an "upper triangular" matrix so its determinant is the product of the numbers on the main diagonal, -5.

Since the only row operation that changed the determinant multiplied it by 1/2, go back to the original determinant by multiplying by 2:
the determinant of the original matrix is -10.