Blogs, Linear Algebra

Finding a Determinant (Part 2)

In Finding a Determinant (Part 1) we were able to find the determinant by working with progressively smaller matrices. In this post, we will try to find the determinant by working with row operations.

Find the determinant

As we did last time, we will again find the determinant of
\[A=\begin{bmatrix}
1 & 2 & 1 & -1 \\
0 & 1 & 0 & 2 \\
0 & 3 & -1 & 1\\
1 & 2 & 5 & 0
\end{bmatrix}.\]

In this case, however, we need to know a little bit more before calculating this. In particular, we want to know how row operation affect the determinant of a matrix. We know that

  • If you obtain \(A’\) from \(A\) by performing the row operation \(cR_{i}+R_{j} \to R_{j}\), then \(|A|=|A’|\).
  • If you obtain \(A’\) from \(A\) by performing the row operation of \(R_{i} \leftrightarrow R_{j}\), then \(|A|=-|A’|\).
  • If you obtain \(A’\) from \(A\) by performing the row operation of \(cR_{i} \to R_{i}\), then \(|A|=\frac{1}{c}|A’|\), that is \(c|A|=|A’|\).

We are now ready to find the determinant of the above matrix.

Performing operations

We begin by letting
\[A=\begin{bmatrix}
1 & 2 & 1 & -1 \\
0 & 1 & 0 & 2 \\
0 & 3 & -1 & 1\\
1 & 2 & 5 & 0
\end{bmatrix},\] then work on putting this in row reduced echelon form. To do this, we begin by letting \(A_{1}\) be given by starting with \(A\) and performing the operation \(-R_{1}+R_{4} \to R_{4}\). Note that this will leave \(|A_{1}|=|A|\). We now have
\[A_{1}=\begin{bmatrix}
1 & 2 & 1 & -1 \\
0 & 1 & 0 & 2 \\
0 & 3 & -1 & 1\\
0 & 0 & 4 & 1
\end{bmatrix}.\]

We next want the the first non-zero term in the second row to be 1. Since it already is, we do not have to do anything for this. Therefore, we will work on turning the other terms in the second column into a 0. In order to do this, we will perform the \(-2R_{2}+R_{1} \to R_{1}\) and \(-3R_{2}+R_{3} \to R_{3}\). Note that neither of these will change the determinant. Therefore, if our resulting matrix is denoted \(A_{2}\), we have that \(|A_{2}|=|A|\). We then get
\[A_{2}=\begin{bmatrix}
1 & 0 & 1 & -5 \\
0 & 1 & 0 & 2 \\
0 & 0 & -1 & -5\\
0 & 0 & 4 & 1
\end{bmatrix}.\]

We now want a \(1\) to be the leading non-zero term in the third row. We, therefore, perform the row operation \(-R_{3} \to R_{3}\). Note that this will change the determinant so that, if \(A_{3}\) is the new matrix, we have that \(|A|=-|A_{3}|\). This give us
\[A_{3}=\begin{bmatrix}
1 & 0 & 1 & -5 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 5\\
0 & 0 & 4 & 1
\end{bmatrix}.\]

Next, we want to turn the other entries in the third column into \(0\)s. We, therefore, perform the row operations \(-R_{3}+R_{1} \to R_{1}\) and \(-4R_{3}+R_{4} \to R_{4}\). Note that this will not change the determinant. Therefore, if we let \(A_{4}\) be the resulting matrix, we get that \(|A_{4}|=|A_{3}|\). Furthermore, since \(|A|=-|A_{3}|\), we get \(|A|=-|A_{4}|\). We now have
\[A_{4}=\begin{bmatrix}
1 & 0 & 0 & -10 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 5\\
0 & 0 & 0 & -19
\end{bmatrix}.\]

Now, in order to turn the leading non-zero number in the fourth row into a \(1\), we must perform the row operation \(\frac{-1}{19}R_{4} \to R_{4}\). With the resulting matrix denoted \(A_{5}\), we find the \(-19|A_{5}|=|A_{4}|\). Therefore, \(|A|=-|A_{4}|=19|A_{5}|\). The matrix, \(A_{5}\), is now
\[A_{5}=\begin{bmatrix}
1 & 0 & 0 & -10 \\
0 & 1 & 0 & 2 \\
0 & 0 & 1 & 5\\
0 & 0 & 0 & 1
\end{bmatrix}.\]

We can find the determinant of this matrix if we would like. However, we will continue to row reduce the matrix to make the calculations a little easier. If we then perform the row operation \(10R_{4}+R_{1} \to R_{1}\), \(-2R_{4}+R_{2} \to R_{2}\) and \(-5R_{4}+R_{3}\), we won’t change the determinant at all. Letting the resulting matrix be denoted \(A_{6}\), we find the \(|A_{6}|=|A_{5}|\), so \(|A|=19|A_{6}|\). We now have
\[A_{6}=\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1
\end{bmatrix}.\]

Since \(A_{6}\) is just the identity matrix, we can note that \(|A_{6}|=1\). Hence, \(|A|=19*1=19\). At this point, it is worth looking back at the work in Finding the Determinant (Part 1) for two reasons. First, we should check to make sure that we got the same answer using both techniques. If we didn’t, this means that one of our techniques or calculations with them was incorrect. We should also determine which of the two techniques was easier to work with. Your answer to the which one was easier will likely determine which technique you will use in the future.

Conclusion

As always, I hope that this helped you learn and that you enjoyed the process. If you did, make sure to share the post on Social Media with other people that may need some help with linear algebra. Also, be sure to subscribe to our YouTube channel for updates on the videos we are making.

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