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Introduction to Determinants

For any square matrix of order 2, we have found a necessary and sufficient condition for invertibility. Indeed, consider the matrix

$\begin{displaymath}A = \left(\begin{array}{cc} a&b\\ c&d\\ \end{array}\right).\end{displaymath}$

The matrix A is invertible if and only if $ad - bc \neq 0$ . We called this number the determinant of A. It is clear from this, that we would like to have a similar result for bigger matrices (meaning higher orders). So is there a similar notion of determinant for any square matrix, which determines whether a square matrix is invertible or not?

In order to generalize such notion to higher orders, we will need to study the determinant and see what kind of properties it satisfies. First let us use the following notation for the determinant

$\begin{displaymath}\mbox{determinant of $\left(\begin{array}{cc} a&b\\ c&d\\ \... ...begin{array}{cc} a&b\\ c&d\\ \end{array}\right\vert = ad -bc.\end{displaymath}$

Properties of the Determinant

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

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

This is interesting since it implies that whenever we use rows, a similar behavior will result if we use columns. In particular we will see how row elementary operations are helpful in finding the determinant. Therefore, we have similar conclusions for elementary column operations.
2.
The determinant of a triangular matrix is the product of the entries on the diagonal, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a&b\\ 0&d\\ \end{array}\right\v... ...rt\begin{array}{cc} a&0\\ b&d\\ \end{array}\right\vert = ad .\end{displaymath}$

3.
If we interchange two rows, the determinant of the new matrix is the opposite of the old one, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a&b\\ c&d\\ \end{array}\right\v... ...left\vert\begin{array}{cc} c&d\\ a&b\\ \end{array}\right\vert\end{displaymath}$

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, that is

$\begin{displaymath}\left\vert\begin{array}{cc} \lambda a&\lambda b\\ c&d\\ \en... ...rray}{cc} a&b\\ \lambda c&\lambda d\\ \end{array}\right\vert.\end{displaymath}$

In particular, if all the entries in one row are zero, then the determinant is zero.
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, that is

$\begin{displaymath}\left\vert\begin{array}{cc} a + \lambda c&b + \lambda d\\ c&... ...} a&b\\ c + \lambda a&d + \lambda b\\ \end{array}\right\vert.\end{displaymath}$

Note that whenever you want to replace a row by something (through elementary operations), do not multiply the row itself by a constant. Otherwise, you will easily make errors (due to Property 4).
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

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

If A and B are similar, then $\det(A) = \det(B)$ .

Let us look at an example, to see how these properties work.

Example. Evaluate

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

Let us transform this matrix into a triangular one through elementary operations. We will keep the first row and add to the second one the first multiplied by $\displaystyle \frac{1}{2}$ . We get

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ -1&3\\ \end{array}\right\... ...} 2&1\\ 0&\displaystyle \frac{7}{2}\\ \end{array}\right\vert.\end{displaymath}$

Using the Property 2, we get

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ 0&\displaystyle \frac{7}{2}\\ \end{array}\right\vert = 2 \cdot \frac{7}{2} = 7.\end{displaymath}$

Therefore, we have

$\begin{displaymath}\left\vert\begin{array}{cc} 2&1\\ -1&3\\ \end{array}\right\vert= 7\end{displaymath}$

which one may check easily.

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 = (a_ij), where a_ij 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 A_ij (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
. Determinant and Inverse of Matrices

Finding the inverse of a matrix is very important in many areas of science. For example, decrypting a coded message uses the inverse of a matrix. Determinant may be used to answer this problem. Indeed, let A be a square matrix. We know that A is invertible if and only if $\det(A) \neq 0$ . Also if A has order n, then the cofactor A_i,j is defined as 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. Recall

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

for any fixed i, and

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

for any fixed j. Define the adjoint of A, denoted adj(A), to be the transpose of the matrix whose ij^th entry is A_ij.

Example. Let

$\begin{displaymath}A = \left(\begin{array}{rrr} 1&3&2\\ -1&0&2\\ 3&1&-1\\ \end{array}\right).\end{displaymath}$

We have

$\begin{displaymath}adj(A) = \left(\begin{array}{rrr} -2&5&-1\\ 5&-7&8\\ 6&-4&3... ...{array}{rrr} -2&5&6\\ 5&-7&-4\\ -1&8&3\\ \end{array}\right).\end{displaymath}$

Let us evaluate $A \cdot adj(A)$ . We have

$\begin{displaymath}A \cdot adj(A) = \left(\begin{array}{rrr} 1&3&2\\ -1&0&2\\ ... ...n{array}{rrr} 11&0&0\\ 0&11&0\\ 0&0&11\\ \end{array}\right).\end{displaymath}$

Note that $\det(A) = 11$ . Therefore, we have

$\begin{displaymath}A \cdot adj(A) = \det(A) I_3.\end{displaymath}$

Is this formula only true for this matrix, or does a similar formula exist for any square matrix? In fact, we do have a similar formula.

Theorem. For any square matrix A of order n, we have

$\begin{displaymath}A \cdot adj(A) = \det(A) I_n.\end{displaymath}$

In particular, if $\det(A) \neq 0$ , then

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

For a square matrix of order 2, we have

$\begin{displaymath}adj \left(\begin{array}{rr} a&b\\ c&d\\ \end{array}\right) ... ...^T = \left(\begin{array}{rr} d&-b\\ -c&a\\ \end{array}\right)\end{displaymath}$

which gives

$\begin{displaymath}\left(\begin{array}{rr} a&b\\ c&d\\ \end{array}\right)^{-1}... ...bc} \left(\begin{array}{rr} d&-b\\ -c&a\\ \end{array}\right).\end{displaymath}$

Application of Determinant to Systems: Cramer's Rule

We have seen that determinant may be useful in finding the inverse of a nonsingular matrix. We can use these findings in solving linear systems for which the matrix coefficient is nonsingular (or invertible).
Consider the linear system (in matrix form)

A X = B

where A is the matrix coefficient, B the nonhomogeneous term, and X the unknown column-matrix. We have:

Theorem. The linear system AX = B has a unique solution if and only if A is invertible. In this case, the solution is given by the so-called Cramer's formulas:

$\begin{displaymath}x_i = \frac{\det(A_i)}{detA}\;,\;\; \mbox{for $i=1,\cdots,n$}\end{displaymath}$

where x_i are the unknowns of the system or the entries of X, and the matrix A_i is obtained from A by replacing the i^th column by the column B. In other words, we have

$\begin{displaymath}x_i = \frac{b_1 A_{1i} + b_2 A_{2i} + \cdots + b_n A_{ni}}{\det(A)}\end{displaymath}$

where the b_i are the entries of B.

In particular, if the linear system AX = B is homogeneous, meaning $B = {\cal O}$ , then if A is invertible, the only solution is the trivial one, that is $X = {\cal O}$ . So if we are looking for a nonzero solution to the system, the matrix coefficient A must be singular or noninvertible. We also know that this will happen if and only if $\det (A) = 0$ . This is an important result.
Example. Solve the linear system

$\begin{displaymath}\left(\begin{array}{rrr} 1&2&0\\ -1&1&1\\ 1&2&3\\ \end{arr... ...ht) = \left(\begin{array}{r} 0\\ 1\\ 0\\ \end{array}\right).\end{displaymath}$

Answer. First note that

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

which implies that the matrix coefficient is invertible. So we may use the Cramer's formulas. We have

$\begin{displaymath}x = \frac{1}{9} \left\vert\begin{array}{rrr} 0&2&0\\ 1&1&1\\... ...array}{rrr} 1&2&0\\ -1&1&1\\ 1&2&0\\ \end{array}\right\vert.\end{displaymath}$

We leave the details to the reader to find

$\begin{displaymath}x = \frac{-4}{9},\; y = \frac{3}{9} = \frac{1}{3},\;\mbox{and}\; z = 0.\end{displaymath}$

Note that it is easy to see that z=0. Indeed, the determinant which gives z has two identical rows (the first and the last). We do encourage you to check that the values found for x, y, and z are indeed the solution to the given system.

Remark. Remember that Cramer's formulas are only valid for linear systems with an invertible matrix coefficient

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