17 The Determinant
We have seen that a $2\times 2$ matrix $\begin{pmatrix} a&b\\ c&d \end{pmatrix}$ has an inverse if and only if $ad-bc\ne 0$. The quantity $ad-bc$ is the "determinant" of the matrix. In fact, a determinant is defined for any $n\times n$ matrix. However, for larger matrices, it is not given by such a simple formula. The determinant of a square matrix $A$ can be written as $det(A)$ or as $|A|$. When the entries of the matrix are written in full, the determinant is indicated by using vertical bars instead of parentheses: $\begin{vmatrix} a&b \\ c&d \end{vmatrix}=ad-bc$. (It's best not to use the latter notation for the determinant of a $1\times 1$ matrix $(a)$ since the determinant is $det\big((a)\big) = a$, not the absolute value of $a$.)
The determinant of an $n\times n$ matrix can be thought of as a function of the rows of $A$. So if the rows of the matrix $A$ are $\vec r_1,\vec r_2,\dots,\vec r_n$, we might write $det(A) = det(\vec r_1,\vec r_2,...,\vec r_n)$. As a function of $n$ row vectors, the determinant has certain properties. In particular, it is multilinear. This means that it is linear in each input. So, if $\vec u,\vec v\in \R^n$ and $\alpha\in\R$, we have $$\begin{align*} det(\vec v + \vec u, \vec r_2, \dots, \vec r_n) &= det(\vec v, \vec r_2, \dots, \vec r_n) + det(\vec u, \vec r_2, \dots, \vec r_n)\\ det(\alpha\cdot \vec r_1, \vec r_2, \dots, \vec r_n) &= \alpha\cdot det(\vec r_1, \vec r_2, \dots, \vec r_n)\\[6pt] det(\vec r_1, \vec v + \vec u, \dots, \vec r_n) &= det(\vec r_1, \vec v, \dots, \vec r_n) + det(\vec r_1, \vec u,, \dots, \vec r_n)\\ det(\vec r_1, \alpha\cdot \vec r_2, \dots, \vec r_n) &= \alpha\cdot det(\vec r_1, \vec r_2, \dots, \vec r_n)\\[6pt] &\vdots\\[6pt] det(\vec r_1, \vec r_2, \dots, \vec v + \vec u) &= det(\vec r_1, \vec r_2, \dots, \vec v) + det(\vec r_1, \vec r_2, \dots, \vec u)\\ det(\vec r_1, \vec r_2, \dots, \alpha\cdot \vec r_n) &= \alpha\cdot det(\vec r_1, \vec r_2, \dots, \vec r_n) \end{align*}$$ Another important property of the determinant function is that it is antisymmetric. This means that when two rows are swapped, the sign of the determinant is changed: $$\begin{align*} det(\vec r_1,\dots,\vec r_{i-1},\vec u,\vec r_{i+1}&\dots,\vec r_{j-1},\vec v,\vec r_{j+1},\dots\vec r_n)\\ &= -det(\vec r_1,\dots,\vec r_{i-1},\vec v,\vec r_{i+1},\dots,\vec r_{j-1},\vec u,\vec r_{j+1},\dots\vec r_n) \end{align*}$$ Finally, the determinant of the identity matrix $I_n$ is 1. In fact, it can be shown that there is one and only one function of $n\times n$ matrices that satisfies all of these properties: It is multilinear in the $n$ row vectors, antisymmetric in the $n$ row vectors, and maps the identity matrix to $1.$ However, we will not do that proof.
Several other important properties follow from the basic defining properties of the determinant. In particular, the effect of a row operation on a matrix is easy to prove.
Theorem: Let $A$ be an $n\times n$ matrix. Then
- If $A$ has a zero row, then $det(A)=0$.
- If $A$ has two rows that are identical, then $det(A)=0$.
- If $A$ has a row that is a scalar multiple of another row, then $det(A)=0$.
- If the row operation $\rho_i\leftrightarrow \rho_j$ is applied to $A$, then sign of the determinant is changed.
- If the row operation $k\rho_i$ is applied to $A$, then the determinant is multiplied by $\frac1k$.
- If the row operation $k\rho_i + \rho_j$ is applied to $A$, then the determinant is unchanged.
Proof:
- This follows easily from multilinearity: $$\begin{align*}det(\vec r_1,\dots,\vec r_{i-1},\vec 0,\vec r_{i+1},\dots,\vec r_n) &= det(\vec r_1,\dots,\vec r_{i-1},0\cdot \vec 0,\vec r_{i+1},\dots,\vec r_n)\\ &=0\cdot det(\vec r_1,\dots,\vec r_{i-1},\vec 0,\vec r_{i+1},\dots,\vec r_n)\\ &=0\end{align*}$$
- This follows from antisymmetry. If any two rows are swapped, the sign of the determinant must change. However, if the two rows of $A$ that are swapped are identical, then resulting matrix is identical to $A$. So, we must have $det(A) = -det(A)$, which implies that $det(A)=0$.
- This follows from part (2) of this theorem and the multilinearity property of the determinant: $$\begin{align*} det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1}&,\vec k\cdot r_i,\vec r_{j+1},\dots,\vec r_n)\\ &= k\cdot det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1},\vec r_i,\vec r_{j+1},\dots,\vec r_n)\\ &= k\cdot 0\\ &= 0 \end{align*}$$
- This simply restates the antisymmetry property.
- This follows from the part of the mulitilinearity property that deals with scalar multiples. (Suppose that $B$ is obtained from $A$ by applying the row operation $k\rho_i$. Then $det(B) = k \cdot det(A)$ by multilinearity. But note that this also means that $det(A) = \frac 1k \cdot det(B).)$
- This follows from multilinearity and part (2) of this theorem: $$\begin{align*} det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1}&,\vec k\cdot r_i + \vec r_j,\vec r_{j+1},\dots,\vec r_n)\\ &= k\cdot det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1},\vec r_i,\vec r_{j+1},\dots,\vec r_n)\\ &\hskip 1 in + det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1},\vec r_j,\vec r_{j+1},\dots,\vec r_n)\\ &= k\cdot 0 + det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1},\vec r_j,\vec r_{j+1},\dots,\vec r_n)\\ &= det(\vec r_1,\dots,\vec r_i,\dots,\vec r_{j-1},\vec r_j,\vec r_{j+1},\dots,\vec r_n)\\ &= det(A) \end{align*}$$
Knowing the effect of row operations on the determinant of a matrix allows us to easily calculate the determinant of an $n\times n$ matrix that is in echelon form.
Theorem: Let $A$ be an $n\times n$ matrix that is in echelon form. Then the determinant of $A$ is the product of the entries on the diagonal of $A$.
Proof: If $A$ has less than $n$ leading variables, then the last row of $A$ is a row of zeros. By part (1) of the previous theorem, $det(A)=0$. Since the $n^{\rm th}$ diagonal entry is zero, the product of the diagonal entries will also be 0, so the determinant is equal to the product of the diagonal entries in this case.
Now, suppose that all of the diagonal entries are 1, so that the product of the diagonal entries is 1. We must show that $det(A)=1$. Then $A$ can be row-reduced to the identity matrix by applying row operations of the form $k\rho_i+\rho_j$. By part (6) of the previous theorem, applying those row operations does not change the determinant. By a basic property of determinants, the determinant of the resulting identity matrix is 1. So the determinant of the original matrix, $det(A)$, is also 1.
If $A$ has $n$ leading variables, then all of the diagonal entries are leading entries and hence are non-zero. Say the diagonal entries are $d_1,d_2,\dots,d_n$. We can apply the scalar multiple part of multilinearity to write $det(A) = d_1d_2\cdots d_n\cdot det(B)$ where $B$ is a matrix in which all of the diagonal entries are 1. Since $det(B) = 1$, we see that $det(A) = d_1d_2\cdots d_n$.
Using these results, we can calculate the determinant of a matrix by applying row operations to put the matrix into echelon form, as long as we keep track of the effect of the row operations on the determinant. For example, $$\begin{align*} \begin{vmatrix} 0 & 1 & 2 & 3 \\ -2 & 4 & 0 & 2 \\ 2 & -2 & 3 & -3 \\ 0 & 3 & -1 & 0 \end{vmatrix} &\rowop{\rho_1\leftrightarrow\rho_2} \begin{vmatrix} -2 & 4 & 0 & 2 \\ 0 & 1 & 2 & 3 \\ 2 & -2 & 3 & -3 \\ 0 & 3 & -1 & 0 \end{vmatrix}\\[8pt] &\rowop{\frac12\rho_1} 2\cdot\begin{vmatrix} -1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 2 & -2 & 3 & -3 \\ 0 & 3 & -1 & 0 \end{vmatrix}\\[8pt] &\rowop{2\rho_1+\rho_3} -2\cdot\begin{vmatrix} -1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 2 & 3 & -1 \\ 0 & 3 & -1 & 0 \end{vmatrix}\\[8pt] &\rowop{-2\rho_2+\rho_3\\ -3\rho_2+\rho_4} -2\cdot\begin{vmatrix} -1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & -1 & -7 \\ 0 & 0 & -7 & -9 \end{vmatrix}\\[8pt] &\rowop{-7\rho_3+\rho_4} -2\cdot\begin{vmatrix} -1 & 2 & 0 & 1 \\ 0 & 1 & 2 & 3 \\ 0 & 0 & -1 & -7 \\ 0 & 0 & 0 & 40 \end{vmatrix} \end{align*}$$ The final determinant can then be computed as $-2(-1)(1)(-1)(40)= -80$.
An $n\times n$ matrix has an inverse if and only if its determinant is non-zero. This is now easy to see: An $n\times n$ matrix $A$ is invertible if and only if it has no free variables, that is, if and only if when put into echelon form, all of the diagonal entries are non-zero. But that will be true if and only if the determinant is non-zero, since the determinant of echelon form matrix is the product of the diagonal entries and the determinant of $A$ will be that value times a non-zero constant that comes from any row operations of the form $\rho_i\leftrightarrow\rho_j$ and $k\rho_i$ that are used in putting the matrix into echelon form.
The determinant also has important properties related to matrix multiplication: $det(AB) = det(A)det(B)$ and for an invertible matrix, $det(A^{-1})= \frac{1}{det(A)}$.