## 2.4 Determinants

You can check directly that a 2✕2 matrix $$A=\begin{pmatrix} a&b\\c&d \end{pmatrix}$$ is invertible if and only if $$ad-bc\neq 0$$. The quantity $$ad-bc$$ is called the determinant of A, and we want to generalize it to larger matrices.

Definition 2.12 Let $$A=(a_{ij})$$ be n✕n. The determinant of A, written $$\det (A)$$, is the number

$\sum_{\sigma \in S_n} \sgn(\sigma) a_{1,\sigma(1)}a_{2,\sigma(2)}\cdots a_{n,\sigma(n)}.$

Example 2.6

• $$S_2=\{\id, (1,2)\}$$ so the determinant of $$A=(a_{ij})$$ is $a_{11}a_{22} - a_{12}a_{21}$ where the first term is the one arising from $$\sigma=\id$$ and the second is from $$\sigma=(1,2)$$. This is equivalent to the expression $$ad-bd$$ at the start of this subsection.
• $$S_3=\{\mathrm{id}, (1,2,3), (1,3,2), (1,2), (1,3), (2,3)\}$$ and the determinant of $$A=(a_{ij})$$ is $a_{11}a_{22}a_{33}+a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32} -a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31} - a_{11}a_{23}a_{32}.$

Our aims in this section are to show that for a square matrix A of any size, A is invertible if and only if $$\det A\neq 0$$, and to investigate other ways of computing $$\det A$$. Summing over all elements $$\sigma \in S_n$$ quickly becomes unmanageable as $$|S_n|=n!$$ grows very quickly with n.

Proposition 2.7 Let $$A=(a_{ij})$$ be a square matrix and $$\lambda$$ a number.

• If A has a row or a column of zeroes, then $$\det A=0$$.
• $$\det A = \det A^T$$.
• If A has two equal columns or two equal rows, then $$\det A=0$$.
• $$\det I_n = 1$$.

Proof.

• Suppose the ith row of A is zero, so $$a_{ij}=0$$ for all j. Every term in the sum defining $$\det A$$ involves some $$a_{ij}$$ as a factor, so is zero. Thus $$\det A=0$$. The argument for a column of zeroes is similar.
• The i, j entry of $$A^T$$ is $$a_{ji}$$, so $\det A^T = \sum_{\sigma \in S_n} \sgn(\sigma) a_{\sigma(1),1}\cdots a_{\sigma(n),n}.$ Notice that $a_{\sigma(1),1}\cdots a_{\sigma(n),n} = a_{1,\sigma^{-1}(1)}\cdots a_{n,\sigma^{-1}(n)}$ as the terms on each side are the same, just written in a different order. Therefore \begin{align*} \det A^T& = \sum_{\sigma\in S_n} \sgn(\sigma) a_{\sigma(1),1}\cdots a_{\sigma(n),n} \\ &= \sum_{\sigma \in S_n} \sgn(\sigma) a_{1,\sigma^{-1}(1)}\cdots a_{n,\sigma^{-1}(n)} & \text{as above} \\ &= \sum_{\sigma \in S_n} \sgn(\sigma^{-1})a_{1,\sigma^{-1}(1)}\cdots a_{n,\sigma^{-1}(n)} & \text{as } \sgn(\sigma)=\sgn(\sigma^{-1}) \\ &= \sum_{\sigma \in S_n} \sgn(\sigma)a_{1,\sigma(1)}\cdots a_{n,\sigma(n)} \\ &= \det A \end{align*} where the second-to-last equality is because the permutations $$\sigma^{-1}$$ for $$\sigma \in S_n$$ are just the permutations in $$S_n$$, written in a different order.
• Suppose $$p<q$$ and $$a_{pj}=a_{qj}$$ for all j. Let $$\sigma \in S_n$$ and consider the terms in the sum defining $$\det A$$ corresponding to $$\sigma$$ and $$\sigma (p,q)$$. Since $$a_{p,\sigma(p)}=a_{q, \sigma(p)}= a_{q, \sigma (p,q) (q)}$$ and $$a_{q,\sigma(q)}=a_{p,\sigma(p,q)(p)}$$ the product of entries of A is exactly the same for $$\sigma (p,q)$$ as it is for $$\sigma$$, but $$\sgn \sigma = - \sgn (\sigma (p,q))$$. It follows that the two terms add to zero. Pairing up all terms in the sum this way, we see that it is zero. If A has two equal columns, then $$\det A=\det A^T$$ by the last part and $$A^T$$ has two equal rows so has determinant zero by the argument above.
• Let $$I_n = (\delta_{ij})$$, so $$\delta_{ij}=1$$ if $$i=j$$ and 0 otherwise. For $\delta_{1,\sigma(1)}\cdots \delta_{n,\sigma(n)} \neq 0$ we must have $$\sigma(1)=1$$, $$\sigma(2)=2$$, and so on, so the only nonzero term in the sum defining $$\det I_n$$ is the $$\sigma=\id$$ term, which has the value $\sgn(\id)\delta_{11}\cdots \delta_{nn}=1.$

Proposition 2.8 Let $$A=(a_{ij})$$ be a square matrix and r be a row op. Then

• If r is $$r_i \mapsto \lambda r_i$$ then $$\det r(A) = \lambda \det A$$,
• if r is $$r_i \mapsto r_i + \lambda r_j$$ then $$\det r(A)=\det A$$, and
• if r swaps rows i and j then $$\det r(A)=-\det A$$.

The crucial point here is that doing row operations can change the determinant (in a predictable way), but it can never change the determinant from being zero to being nonzero, or from being nonzero to being zero.

Proof.

• Each term in the sum defining $$\det r(A)$$ is the same as the corresponding term in $$\det A$$, but with exactly one factor of $$\lambda$$. Taking these factors outside the sum we get $$\det r(A) = \lambda \det A$$.
• Let $$r(A)=(b_{ij})$$, so $$b_{pq}=a_{pq}$$ if $$p \neq j$$ and $$b_{jq}=a_{jq}+\lambda a_{iq}$$. The term $b_{1,\sigma(1)} \cdots b_{n,\sigma(n)}$ from the sum defining the determinant of $$r_{ij}(\lambda)(A)$$ then equals $\begin{multline*} a_{1,\sigma(1)}\cdots a_{j-1, \sigma(j-1)} (a_{j, \sigma(j)}+\lambda a_{i, \sigma(j)}) a_{j+1,\sigma(j+1)}\cdots a_{n,\sigma(n)} =\\ a_{1,\sigma(1)}\cdots a_{n,\sigma(n)} + \lambda a_{1,\sigma(1)}\cdots a_{j-1,\sigma(j-1)} a_{i,\sigma(j)} a_{j+1,\sigma(j+1)}\cdots a_{n,\sigma(n)}.\end{multline*}$ The first term on the right of this equation is the term in the sum defining $$\det A$$ coming from the permutation $$\sigma$$. The second term on the right of this equation is $$\lambda$$ times the term coming from the permutation $$\sigma$$ in the sum defining the determinant of a matrix which is the same as A, except with the ith row of A repeated in place of the jth row. Such a matrix has two repeated rows so its determinant is zero. Summing over all $$\sigma\in S_n$$ gives $\det r(A)= \det A + \det B$ where B is a matrix with two repeated rows, so $$\det B=0$$.
• $$r(A)$$ is the matrix whose p, q entry is $$a_{pq}$$ if $$p\neq i,j$$, $$a_{jq}$$ if $$p=i$$, and $$a_{iq}$$ if $$p=j$$. The term in the sum defining $$\det r(A)$$ corresponding to $$\sigma\in S_n$$ is $\sgn(\sigma) a_{1, \sigma(1)} \cdots a_{i-1, \sigma(i-1)} a_{j,\sigma(i)} a_{i+1,\sigma(i+1)}\cdots a_{j-1, \sigma(j-1)} a_{j, \sigma(i)} a_{j+1,\sigma(j+1)} \cdots a_{n,\sigma(n)}.$ The product of $$a_{rs}$$s appearing here is the same as the one in the term of the sum defining $$\det A$$ corresponding to $$\sigma (i,j)$$. But $$\sgn (\sigma) = - \sgn (\sigma (i,j))$$, so each term in the sum for $$\det r(A)$$ appears with the opposite sign in $$\det A$$.

Theorem 2.3 Let A be a square matrix. Then A is invertible if and only if $$\det A \neq 0$$.

Proof. Proposition 2.8 shows that doing a row operation to A doesn’t change whether or not its determinant is zero, so used repeatedly shows that $$\det A \neq 0$$ if and only if the determinant of the RRE form of A is not 0.

If A is invertible its RRE form is $$I_n$$ which has nonzero determinant, so A has nonzero determinant. If A is not invertible its RRE form has a row of zeroes, so its determinant is zero by proposition 2.7, so $$\det A=0$$.

The next lemma is a small step on the way to proving that $$\det(AB) = \det(A)\det(B)$$ for square matrices A and B.

Lemma 2.7 Let A be an n ✕ n matrix and r a row op. Then $$\det (r(I_n) A)=\det r(I_n) \det A$$.

Proof. The special case $$A=I_n$$ of Proposition 2.8 says

• $$\det r(I_n)=\lambda\det I_n = \lambda$$ if $$r=r_i(\lambda)$$
• $$\det r(I_n)=\det I_n=1$$ if $$r=r_{ij}(\lambda)$$, and
• $$\det r(I_n)=-\det I_n=-1$$ if $$r=r_{ij}$$.

This means we could reformulate proposition 2.8 as saying $$\det r(A) = \det(r(I_n))\det(A)$$.

Now proposition 2.3 says $$r(I_n)A = r(A)$$, so $$\det(r(I_n)A) = \det(r(A)) = \det(r(I_n))\det(A)$$.

Theorem 2.4 Let A and B be nn matrices. Then $$\det (AB)=\det A \det B$$.

Proof.

• Case 1: A and B are both invertible. As in the proof of Theorem 2.2, we can write A and B as products of elementary matrices: \begin{align*} A &= r_1(I)\cdots r_l(I) \\ B &= s_1(I)\cdots s_m(I). \end{align*} Now by Lemma 2.7, for any two row ops r and s we have $$\det (r(I)s(I))=\det r(I) \det s(I)$$. Using this repeatedly: \begin{align*} \det A &= \det r_1(I)\cdots \det r_l(I) \\ \det B &= \det s_1(I)\cdots \det s_m(I) \\ \det (AB)&= \det (r_1(I)\cdots r_l(I)s_1(I)\cdots s_m(I))\\ &= \det r_1(I) \cdots \det r_l(I) \det s_1(I) \cdots \det s_m(I) \\ &= \det A \det B. \end{align*}
• Case 2: B isn’t invertible. By Corollary 2.2, there is a nonzero vector $$\mathbf{v}$$ with $$B \mathbf{v}=\mathbf{0}$$. Then $$AB \mathbf{v}=A\mathbf{0}=\mathbf{0}$$, so $$AB$$ isn’t invertible. Thus $$\det AB=0$$ and $$\det B=0$$ so certainly $$\det AB=\det A \det B$$.
• Case 3: B is invertible but A isn’t. By Corollary 2.2 again, there is a nonzero vector $$\mathbf{v}$$ with $$A\mathbf{v}=\mathbf{0}$$. Then $$B^{-1}\mathbf{v}$$ is nonzero too, and $$AB (B^{-1}\mathbf{v})=\mathbf{0}$$ so $$AB$$ isn’t invertible and $$\det AB=\det A \det B$$ as in case 2.

Corollary 2.3

• If A is invertible then $$\det(A^{-1})=(\det A)^{-1}$$.
• If a square matrix A has a left or right inverse then it is invertible.

Proof.

• Apply the previous result to $$AA^{-1}=I_n$$.
• If $$AB=I_n$$ then $$\det(AB)=\det(A)\det(B)=1$$, so $$\det(A)\neq 0$$, so A is invertible. The case of a left-inverse is similar.

Finally we look at computing determinants by row and column expansions.

Definition 2.13 The i, j minor of a square matrix A, written $$A_{ij}$$, is the determinant of the matrix obtained from A by deleting row i and column j from A.

Definition 2.14 For $$1 \leq i \leq n$$ the ith row expansion of an n✕n matrix $$A= (a_{ij})$$ is $\begin{equation*} r_i(A) =\sum_{r=1}^n (-1)^{i+r}a_{ir}A_{ir}. \end{equation*}$

Definition 2.15 For $$1 \leq j \leq n$$ the jth column expansion of an n✕n matrix $$A= (a_{ij})$$ is $\begin{equation*} c_j(A)= \sum_{r=1}^n (-1)^{r+j} a_{rj} A_{rj}. \end{equation*}$

It turns out that each of the row and column expansions actually equal $$\det A$$.

Example 2.7 Let’s look at the 2✕2 case to try and convince ourselves that these definitions really do all produce the determinant. Let $A = \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}.$ The first row expansion is \begin{align*} \sum_{r=1}^2 (-1)^{1+r}a_{1r}A_{1r} &= (-1)^{1+1}a_{11} A_{11} + (-1)^{1+2}a_{12}A_{12}\\ &= a_{11}a_{22}-a_{12}a_{21} \end{align*} which agrees with our old definition of the determinant. Now the second column expansion: \begin{align*} \sum_{r=1}^2 (-1)^{r+2} a_{r2} A_{r2} &= (-1)^{1+2}a_{12} A_{12} + (-1)^{2+2}a_{22}A_{22} \\ &= -a_{12}a_{21} + a_{22}a_{11} \end{align*} which again agrees with our previous definition.

The general proof that row and column expansions really do equal the determinant is slightly messy. Here’s a special case:

Proposition 2.9 Let A be a square matrix. Then $$c_n(A)=\det A$$.
Proof. \begin{align*} c_n(A)&= \sum_{r=1}^n (-1)^{n+r} a_{rn} A_{rn} \\ &= \sum_{r=1}^n (-1)^{n+r}a_{rn}\sum_{\sigma\in S_{n-1}} \sgn(\sigma) a_{1,\sigma(1)} \cdots a_{r-1,\sigma(r-1)}a_{r+1,\sigma(r)}\cdots a_{n,\sigma(n-1)} \\ &= \sum_{r=1}^n \sum_{\sigma\in S_{n-1}} (-1)^{n+r}\sgn(\sigma)a_{1,\sigma(1)}\cdots a_{r-1,\sigma(r-1)}a_{r,n}a_{r+1,\sigma(r)}\cdots a_{n,\sigma(n-1)} \end{align*} The permutation $\begin{pmatrix} 1 & 2 & \cdots & r-1 & r & r+1 & \cdots & n \\ \sigma(1) & \sigma(2) & \cdots & \sigma(r-1) & n & \sigma(r) & \cdots & \sigma(n-1) \end{pmatrix}$ obtained by inserting an n into the bottom row of $$\sigma$$ at position r, equals $$\sigma (n,n-1,\ldots,r+1,r)$$. The cycle has length $$n-r+1$$ so the sign of $$\sigma (n,n-1,\ldots, r+1,r)$$ is $$\sgn(\sigma) (-1)^{n-r}= \sgn(\sigma)(-1)^{n+r}$$. Every permutation in $$S_n$$ can be obtained in exactly one way by inserting n into the bottom row of a permutation from $$S_{n-1}$$. So our sum is $\sum _{\tau \in S_n} \sgn(\tau) a_{1,\tau(1)}\cdots a_{n,\tau(n)}$ which is $$\det A$$.