Determinants of Matrices of Higher Order

As we said before, the idea is to assume that previous properties satisfied by the determinant of matrices of order 2, are still valid in general. In other words, we assume:

1.
Any matrix A and its transpose have the same determinant, meaning

\begin{displaymath}\det A = \det A^T.\end{displaymath}

2.
The determinant of a triangular matrix is the product of the entries on the diagonal.
3.
If we interchange two rows, the determinant of the new matrix is the opposite of the old one.
4.
If we multiply one row with a constant, the determinant of the new matrix is the determinant of the old one multiplied by the constant.
5.
If we add one row to another one multiplied by a constant, the determinant of the new matrix is the same as the old one.
6.
We have

\begin{displaymath}\det(AB) = \det(A) \det(B).\end{displaymath}

In particular, if A is invertible (which happens if and only if $\det(A) \neq 0$), then

\begin{displaymath}\det(A^{-1}) = \frac{1}{\det(A)}.\end{displaymath}

So let us see how this works in case of a matrix of order 4.

Example. Evaluate

\begin{displaymath}\left\vert\begin{array}{cccc}
1&2&3&4\\
5&6&7&8\\
2&6&4&8\\
3&1&1&2\\
\end{array}\right\vert.\end{displaymath}

We have

\begin{displaymath}\left\vert\begin{array}{cccc}
1&2&3&4\\
5&6&7&8\\
2&6&4&8\\...
...3&4\\
5&6&7&8\\
1&3&2&4\\
3&1&1&2\\
\end{array}\right\vert.\end{displaymath}

If we subtract every row multiplied by the appropriate number from the first row, we get

\begin{displaymath}\left\vert\begin{array}{cccc}
1&2&3&4\\
5&6&7&8\\
1&3&2&4\\...
...-4&-8&-12\\
0&1&-1&0\\
0&-5&-8&-10\\
\end{array}\right\vert.\end{displaymath}

We do not touch the first row and work with the other rows. We interchange the second with the third to get

\begin{displaymath}\left\vert\begin{array}{rrrr}
1&2&3&4\\
0&-4&-8&-12\\
0&1&-...
...1&-1&0\\
0&-4&-8&-12\\
0&-5&-8&-10\\
\end{array}\right\vert.\end{displaymath}

If we subtract every row multiplied by the appropriate number from the second row, we get

\begin{displaymath}\left\vert\begin{array}{rrrr}
1&2&3&4\\
0&1&-1&0\\
0&-4&-8&...
...1&-1&0\\
0&0&-12&-12\\
0&0&-13&-10\\
\end{array}\right\vert.\end{displaymath}

Using previous properties, we have

\begin{displaymath}\left\vert\begin{array}{rrrr}
1&2&3&4\\
0&1&-1&0\\
0&0&-12&...
...
0&1&-1&0\\
0&0&1&1\\
0&0&-13&-10\\
\end{array}\right\vert.\end{displaymath}

If we multiply the third row by 13 and add it to the fourth, we get

\begin{displaymath}\left\vert\begin{array}{rrrr}
1&2&3&4\\
0&1&-1&0\\
0&0&1&1\...
...3&4\\
0&1&-1&0\\
0&0&1&1\\
0&0&0&3\\
\end{array}\right\vert\end{displaymath}

which is equal to 3. Putting all the numbers together, we get

\begin{displaymath}\left\vert\begin{array}{cccc}
1&2&3&4\\
5&6&7&8\\
2&6&4&8\\...
...\end{array}\right\vert = 2 \cdot (-1) \cdot (-12) \cdot 3 = 72.\end{displaymath}

These calculations seem to be rather lengthy. We will see later on that a general formula for the determinant does exist.

Example. Evaluate

\begin{displaymath}\left\vert\begin{array}{rrr}
1&2&0\\
-1&1&1\\
1&2&3\\
\end{array}\right\vert.\end{displaymath}

In this example, we will not give the details of the elementary operations. We have

\begin{displaymath}\left\vert\begin{array}{rrr}
1&2&0\\
-1&1&1\\
1&2&3\\
\end...
...ay}{rrr}
1&2&0\\
0&3&1\\
0&0&3\\
\end{array}\right\vert = 9.\end{displaymath}

Example. Evaluate

\begin{displaymath}\left\vert\begin{array}{rrr}
1&1&2\\
0&1&0\\
2&1&-1\\
\end{array}\right\vert.\end{displaymath}

We have

\begin{displaymath}\left\vert\begin{array}{rrr}
1&1&2\\
0&1&0\\
2&1&-1\\
\end...
...}{rrr}
1&1&2\\
0&1&0\\
0&0&-5\\
\end{array}\right\vert = -5.\end{displaymath}




General Formula for the Determinant Let A be a square matrix of order n. Write A = (aij), where aij is the entry on the row number i and the column number j, for $i=1,\cdots,n$ and $j=1,\cdots,n$. For any i and j, set Aij (called the cofactors) to be the determinant of the square matrix of order (n-1) obtained from A by removing the row number i and the column number j multiplied by (-1)i+j. We have

\begin{displaymath}\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}\end{displaymath}

for any fixed i, and

\begin{displaymath}\det(A) = \sum_{i=1}^{i=n} a_{ij} A_{ij}\end{displaymath}

for any fixed j. In other words, we have two type of formulas: along a row (number i) or along a column (number j). Any row or any column will do. The trick is to use a row or a column which has a lot of zeros.
In particular, we have along the rows

\begin{displaymath}\left\vert\begin{array}{rrr}
a&b&c\\
d&e&f\\
g&h&k\\
\end{...
...eft\vert\begin{array}{rrr}
d&e\\
g&h\\
\end{array}\right\vert\end{displaymath}

or

\begin{displaymath}\left\vert\begin{array}{rrr}
a&b&c\\
d&e&f\\
g&h&k\\
\end{...
...eft\vert\begin{array}{rrr}
a&b\\
g&h\\
\end{array}\right\vert\end{displaymath}

or

\begin{displaymath}\left\vert\begin{array}{rrr}
a&b&c\\
d&e&f\\
g&h&k\\
\end{...
...ft\vert\begin{array}{rrr}
a&b\\
d&e\\
\end{array}\right\vert.\end{displaymath}

As an exercise write the formulas along the columns.

Example. Evaluate

\begin{displaymath}\left\vert\begin{array}{rrr}
3&2&1\\
2&1&-3\\
4&0&1\\
\end{array}\right\vert.\end{displaymath}

We will use the general formula along the third row. We have

\begin{displaymath}\left\vert\begin{array}{rrr}
3&2&1\\
2&1&-3\\
4&0&1\\
\end...
...&2\\
2&1\\
\end{array}\right\vert = 4 (-6-1) + 1 (3-4) = -29.\end{displaymath}

Which technique to evaluate a determinant is easier ? The answer depends on the person who is evaluating the determinant. Some like the elementary row operations and some like the general formula. All that matters is to get the correct answer.

Note that all of the above properties are still valid in the general case. Also you should remember that the concept of a determinant only exists for square matrices.

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Author: M.A. Khamsi

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