Historically, determinants preceded matrices--a curious fact in light of the way linear algebra is taught today, with matrices before determinants. Nevertheless, determinants arose independently of matrices in the solution of many practical problems, and the theory of determinants was well developed almost two centuries before matrices were deemed worthy of study in and of themselves.
Recall that the determinant of a 2 \times 2 matrices A = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] is det A = a_{11} a_{22} - a_{12} a_{21}.
We first encountered this expression when we determined ways to compute the inverse of a matrix. In particular we found that
\left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]^{-1} = \frac{1}{a_{11}a_{22}-a{12}a_{21}} \left[ \begin{array}{cc} a_{22} & -a_{12} \\ -a_{21} & a_{11} \end{array} \right].
The determinant of a matrix A is sometimes also denoted by |A|, so for the 2 \times 2 matrix A = \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] we may also write
|A| = \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| = a_{11}a_{22}-a_{12}a_{21}.
Warning: The notation for the determinant is reminiscent of absolute value nontation. It is easy to mistake \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right|, the notation for the determinant, for \left[ \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right] the notation for the matrix itself. Do no confuse these. Fortunately, it will usually be clear from the context which is intended.
We define the determinant of a 1 \times 1 matrix A = [a] to be det A = |a| = a. (Note that here |a| does not denote the absolute value of a.)
How should we define the determinant of a 3 \times 3 matrix? If we compute the inverse of
A = \left[ \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}\right] we get
A^{-1} = \frac{1}{\Delta} \left[ \begin{array}{ccc} ei-fh & ch-bi & bf-ce \\ fg-di & ai-cg & cd-af \\ dh-eg & bg-ah & ae-bd \end{array} \right]
where \Delta = aei-afh-bdi+bfg+cdh-ceg.
Observe that \Delta = aei-afh-bdi+bfg+cdh-ceg = a(ei-fh)-b(di-fg)+c(dh-eg) = a \left| \begin{array}{cc} e & f \\ h & i \end{array} \right| - b \left| \begin{array}{cc} d & f \\ g & i \end{array} \right| + c \left| \begin{array}{cc} d & e \\ g& h \end{array} \right|.
and that each of the entries in the matrix portion of A^{-1} appears to be the determinant of a 2 \times 2 submatrix of A. In fact, this is true, and it is the basis of the definition of the determinant of a 3 \times 3matrix. The definition is recursive in the sense that the determinant of a 3\times 3 matrix is defined in terms of determinants of 2\times2 matrices.
Let A = \left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right]. Then the determinant of A is the scalar
det A = |A| = a_{11} \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| - a_{12} \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| + a_{12} \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right|.
Notice that each of the 2 \times 2 determinants is obtained by deleting the row and column of A containing the entry the determinant is being multiplied by. If we denote by A_{ij} the submatrix of A obtained by deleting row i and column j then we may write
det A = a_{11}det A_{11}-a_{12}detA_{12}+a_{13}detA_{13} = \sum_{j=1}^3(-1)^{1+j}a_{1j}det A_{1j}.
For any square matrix A, det A_{ij} is called the (i,j)-minor of A.
Example: Compute the determinant of
A = \left[ \begin{array}{ccc} 1 & 2 & 3 \\ 4 & 5 & 0 \\ 1 & 0 & 1 \end{array} \right].
The definition of the determinant of a 3 \times 3 matrix extends naturally to arbitrary square matrices.
Let A = [a_{ij}] be an n \times n matrix, where n \geq 2. Then the determinant of A is the scalar
det A = |A| = a_{11} det A_{11} - a_{12} det A_{12} + \cdots +(-1)^{1+n}a_{1n} det A_{1n} = \sum_{j =1}^{n}(-1)^{1+j}a_{1j} det A_{1j}.
It is convenient to combine a minor with its plus or minus sign. To this end, we define the (i,j)-cofactor of A to be
C_{ij} = (-1)^{i+j}det A_{ij}
With this notation,
det A = \sum_{j = 1}^{n} a_{1j}C_{ij}
This definition is often referred to as the cofactor expansion along the first row. In fact, we get exactly the same result bky expanding along any row (or even any column)! We summarize this fact as the following theorem:
The determinant of an n \times n matrix A = [a_{ij}], where n \geq 2, can be computed as
det A = a_{i1}C_{i1}+a_{i2}C_{i2}+\cdots+a_{in}C_{in} = \sum_{j = 1}^n a_{ij}C_{ij}
(which is the cofactor expansion along the ith row) and also as
det A = a_{1j}C_{1j}+a_{2j}C_{2j}+\cdots+a_{nj}C_{nj} = \sum_{i = 1}^n a_{ij}C_{ij}
(the cofactor expansion along the jth column).
Example: Compute the determinant of the following matrix
A = \left[ \begin{array}{rrr} 5 & -3 & 2 \\ 1& 0 & 2 \\ 2 & -1 & 3 \end{array} \right]
by (a) cofactor expansion along the third row and (b) cofactor expansion along the second column.
For (a) we compute
det A = a_{31}C_{31}+a_{32}C_{32}+a_{33}C_{33}
= 2\left| \begin{array}{rr} -3 & 2 \\ 0 & 2 \end{array} \right| +(-1)(-1)\left|\begin{array}{rr} 5 & 2 \\ 1 & 2 \end{array} \right| + 3 \left| \begin{array}{rr} 5 & -3 \\ 1 & 0 \end{array} \right|
= 2(-6)+8+3(3)
=5
Exercise: Compute (b), the cofactor expansion along the second column.
Notice that in part (b) we do fewer computations than in (a) because we expanded along a column which contained a zero entry, namely a_{22}; so we didn't need to compute C_{22}. It follows that the Laplace Expansion Theorem is most useful when the matrix contains a row or column with lots of zero, since, by choosing to expand along that row or column, we minimize the number of cofactors we need to compute.
Exercise: Compute the determinant of
A = \left[ \begin{array}{rrrr} 2 & -3 & 0 & 1 \\ 5 & 4 & 2 & 0 \\ 1 & -1 & 0 & 3 \\ -2 & 1 & 0 & 0 \end{array} \right]
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The Laplace Expansion is particularly useful when the matrix is upper or lower triangular.
Example: Compute the determinant of
A = \left[ \begin{array}{rrrrr} 2 & -3 & 1 & 0 & 4 \\ 0 & 3 & 2 & 5 & 7 \\ 0 & 0 & 1 & 6 & 0 \\ 0 & 0 & 0 & 5 & 2 \\ 0 & 0 & 0 & 0 & -1 \end{array} \right]
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The determinant of a triangular matrix is the product of the entries on its main diagonal. Specifically, if A = [a_{ij}] is an n \times n triangular matrix, then
det A = a_{11}a_{22}\cdots a_{nn}
The most efficient way to compute determinants is to use row reduction. However, not every elementary row operation leaves the determinant of a matrix unchanged. The next theorem summarizes the main properties you need to understand in order to use row reduction effectively.
Let A = [a_{ij}] be a square matrix.
a. If A has a zero row (or column), then det A = 0.
b. If B is obtained by interchanging to rows (or columns) of A, then det B = - det A.
c. If A has two identical rows (or columns), then det A = 0.
d. If B is obtained by multiplying a row (or column) of A by k, then det B = k det A.
e. If A,B, and C are identical except that the ith row (or column) of C is the sum of the ith rows (or columns) of A and B, then det C = det A + det B.
f. If B is obtained by adding a multiple of one row (or column) of A to an other row (or column), then det B = det A.
Example: Compute the determinants of the following matrices:
(a) A = \left[ \begin{array}{rrr} 2 & 3 & -1 \\ 1 & 5 & 0 \\ -2 & -3 & 1 \end{array} \right]
(b) B = \left[ \begin{array}{rrrr} 0 & 1 & 1 & 3 \\ 4 & 7 & 2 & 1 \\ 5 & 3 & 4 & 0 \\ 1 & 3 & -2 & 4 \end{array} \right]
(a) Using property (f) and then (a), we have
det A = \left| \begin{array}{rrr} 2 & 3 & -1 \\ 1 & 5 & 0 \\ -2 & -3 & 1 \end{array} \right| =^{R_3+R_1} \left| \begin{array}{rrr} 2 & 3 & -1 \\ 1 & 5 & 0 \\ 0 & 0 & 0 \end{array} \right| = 0
(b) We reduce B to echelon form as follows (there are other possible ways to do this):
det B = \left| \begin{array}{rrrr} 0 & 1 & 1 & 3 \\ 4 & 7 & 2 & 1 \\ 5 & 3 & 4 & 0 \\ 1 & 3 & -2 & 4 \end{array} \right| = ^{R_1 \leftrightarrow R_4} - \left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 4 & 7 & 2 & 3 \\ 5 & 3 & 4 & 0 \\ 0 & 1 & 1 & 3 \end{array}\right| = ^{R_2-4R_1} -\left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & -5 & 10 & -13 \\ 5 & 3 & 4 & 0 \\ 0 & 1 & 1 & 3 \end{array}\right|
=^{R_3-5R_1}-\left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & -5 & 10 & -13 \\ 0 & -12 & 14 & -20 \\ 0 & 1 & 1 & 3 \end{array}\right|=^{R_2 \leftrightarrow R_4} \left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & -12 & 14 & -20 \\ 0 & -5 & 10 & -13 \end{array}\right| =^{R_3+12R_2} \left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 26 & 16 \\ 0 & -5 & 10 & -13 \end{array}\right|
=^{R_4+5R_2}\left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 26 & 16 \\ 0 & 0 & 15 & 2 \end{array}\right| =^{R_4-15/26R_3} \left|\begin{array}{rrrr} 1 & 3 & -2 & 4 \\ 0 & 1 & 1 & 3 \\ 0 & 0 & 26 & 16 \\ 0 & 0 & 0 & -\frac{94}{13} \end{array}\right|
Now that B has been row reduced to an upper triangular matrix, we see that the determinant is 26(-94/13) = -188.
Remark: By the previous theorem, we can also use column operations in the process of finding the determinant. Further, we can use both column and row operations in our computations (i.e., you don't have to pick either or). Keep this in mind when working determinants by hand.
Let E be an n \times n matrix.
a. If E results from interchanging two rows of I_n, then det E = -1.
b. If E results from multiplying one row of I_n by k, then det E = k.
c. If E results from adding a multiple of one row of I_n to another row, then det E = 1.
Let B be an n \times n matrix and E be an n \times n elementary matrix. Then
det(EB) = det E det B.
A square matrix A is invertible if and only if det A \neq 0.
If A is an n \times n matrix and k is a scalar,
det (kA) = k^n det A.
So we have a nice formula for scalar multiplication. Unfortunately there is no nice formula for matrix addition; in general det(A+B) \neq det A + det B. However, matrix multiplication behaves nicely:
If A and B are n \times n matrices, then
det(AB) = det A \cdot det B,
i.e., this works for all matrices, not only when one is an elementary matrix.
Matrix inverses also work nicely,
If A is invertible, then
det A^{-1} = \frac{1}{det A}.
as do transposes.
For any square matrix A,
det A = det A^{T}.
In this section, we derive two useful formulas relating determinants to the solution of linear systems and the inverse of a matrix. The first of these, Cramer's Rule, gives a formula for describing the solution of certain systems of n linear equations in n variables entirely in terms of determinants.
We need some new notation. For an n \times n matrix A and a vector \vec{b} in \mathbb{R}^n, let A_i(\vec{b}) denote the matrix obtained by replacing the ith column of A by \vec{b}. That is,
A_i(\vec{b}) = [a_1, \cdots a_{i-1}, \vec{b}, a_{i+1} \cdots a_n].
Let A be an invertible n \times n matrix and let \vec{b} be a vector in \mathbb{R}^n. Then the unique solution \vec{x} of the system A\vec{x} = \vec{b} is given by
x_i = \frac{det(A_i(\vec{b}))}{det A} for i = 1,...,n.
Let A = [a_{ij}] be an invertible n \times n matrix. Let C_{ij} be the (i,j)th-cofactor.
Then the adjoint (or adjugate) matrix of A is adj A = [C_{ji}] = [C_{ij}]^{T}.
Let A be an invertible n \times n matrix. Then
A^{-1} = \frac{1}{det A}adj A.
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