Section 3.5 The determinant
The determinant is a map that assigns to a square matrix \(A\) a scalar \(\det A\in\R \text{.}\) The definition given below of the determinant is far from intuitive, and we will do little to motivate it up front. Instead, we allow its various properties to speak for themselves by way of retroactive motivation. In particular, we will see that
making the determinant an important tool for investigating invertibility.
Subsection 3.5.1 Definition of the determinant
Our definition of the determinant is a recursive one; given an \(n\times n\) matrix \(A\) its determinant is defined in terms of the determinant of certain submatrices of dimension \((n-1)\times (n-1)\text{.}\) This necessitates some notation to help our discussion along.
Definition 3.5.1. Submatrix notation.
Let \(A\) be an \(n\times n\) matrix with \(n\geq 2\text{.}\) Given \(1\leq i, j\leq n\text{,}\) the submatrix of \(A\) obtained by removing the \(i\)-th row and \(j\)-th column of \(A\) is denoted \(A_{i k}\text{.}\)
Warning 3.5.2.
Do not conflate the submatrix notation \(A_{ij}\) with matrix entry notation \([A]_{ij}\text{:}\) the former returns the submatrix of \(A\) obtained by deleting the \(i\)-th row and \(j\)-th column; the latter returns the \(ij\)-th entry of \(A\text{.}\)
Definition 3.5.3. The determinant.
Let \(A=[a_{ij}]_{n\times n}\text{.}\) The determinant is defined as follows:
- Base case: \(n=1\)
When \(n=1\) we have \(A=[a_{11}]\) and we define \(\det A=a_{11}\text{.}\)
- Recursive case: \(n\geq 2 \)
-
When \(n\geq 2\) we define
\begin{equation*} \det A=\sum_{j=1}^n(-1)^{1+j}a_{1j}\det A_{1 j}=a_{11}\det A_{11}-a_{12} A_{12}+\cdots +(-1)^{1+n}a_{1n}\det A_{1 n}\text{.} \end{equation*}
Remark 3.5.4. Small \(n\) cases.
Let's look at determinant formulas for the \(n=2,3\) cases. You may remember the formula for \(2\times 2\) matrices from Theorem 3.3.5; we will make the connection more explicit in Theorem 3.5.16.
Given \(A=\abcdmatrix{a}{b}{c}{d}\text{,}\) we have
The formula for the \(n=2\) case is simple enough to serve as a “second base case”, allowing us to end the recursive process of computing a general \(n\times n\) matrix once we get to expressions involving \(2\times 2\) matrices.
Given
we have
The recursive nature of the determinant definition makes induction arguments particularly useful when proving properties of the determinant, as illustrated by the next theorem.
Theorem 3.5.5. Determinant of triangular matrices.
Let \(A=[a_{ij}]_{n\times n}\) be triangular (upper, lower, or diagonal). Then
In other words, the determinant of a triangular matrix is the product of its diagonal entries.
Proof.
We only give the proof for lower triangular matrices; the proof in the upper triangular case is nearly identical.
For any \(n\geq 1\) let \(P(n)\) denote the proposition: “The determinant of any \(n\times n\) lower triangular matrix is the product of its diagonal entries”. We prove by induction that \(P(n)\) is true for all \(n\geq 1\text{.}\)
Base step: show \(P(1)\) is true.
In this case \(A=[a_{11}]\text{,}\) and \(\det A=a_{11} \) is indeed the product of the diagonal entries of \(A\text{.}\)
Induction step: show \(P(n)\implies P(n+1)\) for all \(n\geq 1\).
Let \(A=[a_{ij}]_{(n+1)\times (n+1)}\) be a lower triangular matrix. Then \(a_{1j}=0\) for all \(j\geq 2\text{,}\) and hence the determinant of \(A\) is given by
Claim: \(A_{i k}\) is lower triangular. Indeed, first observe that we have
for all \(1\leq i,j\leq n\text{;}\) by deleting the first row and first column we effectively bump each index up by one. Since \(A\) is lower triangular we have \(a_{ij}=0\) for all \(1\leq i \lt; j\leq n+1\text{,}\) and hence also
for all \(1\leq i\lt; j\leq n \text{,}\) proving the claim.
Lastly, assuming \(P(n)\) is true (the induction hypothesis) we have
as desired.
Corollary 3.5.6. Determinant of identity matrices.
Let \(I\) be the \(n\times n\) identity matrix. Then \(\det I=1\text{.}\)
Proof.
This follows directly from Theorem 3.5.5 since the diagonal entries of \(I\) are all ones.
Subsection 3.5.2 Expansion along rows and columns
Morally speaking, we should give some examples of higher-dimensional determinants, but we first introduce some theory that affords us more leeway in our computations.
Definition 3.5.7. Minors and expansions along rows/columns.
Given an \(n\times n\) matrix \(A\text{,}\) for any pair \(1\leq i,j\leq n\) the \(ij\)-th minor of \(A\) is defined as
For any \(1\leq i\leq n\) the expression
is called the expansion along the \(i\)-th row of \(A\).
For any \(1\leq j\leq n\text{,}\) the expression
is called the expansion along the \(j\)-th column of \(A\).
Theorem 3.5.8. Expansion along rows.
Let \(A=[a_{ij}]_{n\times n}\text{.}\) For any \(1\leq i\leq n\) we have
In other words, we can compute \(\det A\) by expanding along any row of \(A\text{.}\)
Proof.
The proof is by induction on the size \(n\) of the matrix.
Base step: \(n=1, 2\).
For \(n=1\) there is nothing to prove. Given
expanding along either row yields \(ad-bc\text{,}\) as one easily verifies.
Induction step.
Assume the claim is true of any \((n-1)\times (n-1)\) matrix. Given \(A=[a_{ij}]_{n\times n}\) we have
Expanding along the \(i\)-th row of \(A\) for any \(2\leq i\leq n\text{,}\) on the other hand, we get
To show these two expressions are equal we use the induction hypothesis to compute each \(\det A_{1 j}\) by expanding along its \((i-1)\)-th row:
The matrix \((A_{1j})_{(i-1) k}\) is the result of first deleting row 1 and column \(j\) from \(A\text{,}\) and then deleting row \((i-1)\) and column \(k\) of the resulting matrix. To deal with such iterated submatrices, we make some simple observations relating the rows and columns of \(A_{1 j}\) and \(A_{i k}\) with those of \(A\text{.}\)
The \((i-1)\)-th row of \(A_{1 j}\) corresponds to the \(i\)-th row of \(A\text{,}\) and the first row of \(A_{i k}\) corresponds to the first row of \(A\text{.}\)
If \(k\lt j\text{,}\) then the \(k\)-th column of \(A_{1 j}\) corresponds to the \(k\)-th column of \(A\text{;}\) if \(k\geq j\text{,}\) then the \(k\)-th column of \(A_{1 j}\) corresponds to the \((k+1)\)-th column of \(A\text{.}\)
If \(j\lt k\text{,}\) then the \(j\)-th column of \(A_{ik}\) corresponds to the \(j\)-th column of \(A\text{;}\) if \(j\geq k\text{,}\) then the \(j\)-th column of \(A_{ik}\) corresponds to the \((j+1)\)-th column of \(A\text{.}\)
From these observations we derive the following table of formulas:
We now begin to unpack (3.5.1):
This completes the induction step, and thus the proof is finished.
Suprisingly, it turns out that we can compute the determinant of a matrix by expanding along any column (Corollary 3.5.10). This is a consequence of the following theorem, which is useful in its own right. The proof uses induction and a wonderful trick starting from the observation that \(c=\frac{1}{n}\sum_{i=1}^n c\) for any \(c\in\R\text{.}\)
Theorem 3.5.9. Determinant and transposition.
Let \(A\) be an \(n\times n\) matrix. Then
Proof.
The proof is by induction on \(n\text{.}\) The base case (\(n=1\)) is trivial since \([a]^T=[a]\) for any \(1\times 1\) matrix \([a]\text{.}\)
For induction we assume that for all \(n\geq 2\) we have \(\det B^T=\det B\) for any \((n-1)\times (n-1)\) matrix. Suppose \(A\) is an \(n\times n\) matrix. We have
This completes the proof by induction. (Note how in the second equality in the chain above we compute \(\det A^T\) in the \(i\)-th term of \(\sum_{i=1}^n\det A^T\) by expanding along the \(i\)-th row of \(\det A^T\text{.}\) A similar observation applies to the penultimate equality.)
Corollary 3.5.10. Expansion along columns.
Let \(A=[a_{ij}]_{n\times n}\text{.}\) For any \(1\leq j\leq n\) we have
In other words, we can compute \(\det A\) by expanding along any row of \(A\text{.}\)
Proof.
For any \(1\leq j\leq n\text{,}\) we have
Example 3.5.11.
Compute \(\det A\) for
First we compute \(A\) by expanding along the second row. The only nonzero term of this expansion is the last one, yielding
We have
To compute its determinant we expand along its third column:
We conclude that
Remark 3.5.12. Matrix of signs.
When expanding along a row or column, it is easy to get tripped up by the sign \((-1)^{i+j}\) in front of the \(ij\)-th coefficient. A “matrix of signs” is a sort of mnemonic device to help you in this regard. It is easily generated by observing that the sign in front of the \((1,1)\)-th entry is always a \(+\) (since \(1=(-1)^{1+1}\)), and that any horizontal or vertical step within the matrix is accompanied by a change of sign. As an example, for \(n=4\) we have the following matrix of signs:
The freedom to compute the determinant by expanding along any row or column gives rise to the following intuitive property.
Theorem 3.5.13. Zero rows/columns, swapping rows/columns, identical rows/columns.
Let \(A\) be an \(n\times n\) matrix.
If \(A\) has a zero row or zero column, then \(\det A=0\)
-
Assume \(n\geq 2\text{.}\) Let \(A'\) be the matrix obtained by swapping two rows (or two columns) of \(A\text{.}\) Then
\begin{equation*} \det A'=-\det A\text{.} \end{equation*} Assume \(n\geq 2\text{.}\) If \(A\) has two identical rows or two identical columns, then \(\det A=0\text{.}\)
Proof.
The first statement is obvious since according to Theorem 3.5.8 and Corollary 3.5.10 we may compute the determinant by expanding along the zero row or zero column in question.
The third statement follows from the second. Indeed, if \(A\) has two identical rows or columns, then the matrix \(A'\) obtained from \(A\) by swapping the rows (or columns) in question is \(A\) itself. Thus \(\det A=-\det A\) by the second statement, and we conclude that \(\det A=0\text{.}\)
It remains only to show the second statement. We prove only the statement regarding swapping rows; the corresponding statement about columns follows from Theorem 3.5.9. The proof is by induction.
Base step: \(n=2\).
Let \(A=\begin{bmatrix}a\amp b\\c\amp d \end{bmatrix}\text{.}\) Then \(A'=\abcdmatrix{c}{d}A{b}\text{,}\) and \(\det(A')=cb-ad=-(ad-bc)=-\det A\text{.}\)
Induction step.
We assume by induction that the result holds for any \(n\times n\) matrices, \(n\geq 2\text{,}\) and show the same is true for any \((n+1)\times (n+1)\) matrix.
Let \(A\) be an \((n+1)\times (n+1)\) matrix, and suppose \(A'\) is the result of swapping the \(i\)-th and \(k\)-th rows of \(A\text{.}\) We compute the determinants of \(A\) and \(A'\) by expanding along the \(\ell\)-th row, where \(\ell\ne i\) and \(\ell\ne k\text{.}\) This is possible since \((n+1)\geq 3\text{.}\)
Moving along the \(\ell\)-th row, notice that each submatrix \({A'}_{\ell j}\) is the result of swapping the two rows of \(A_{\ell j}\) that originally corresponded to the \(i\)-th and \(k\)-th rows of \(A\text{.}\) Since these submatrices are of dimension \(n\times n\text{,}\) we have \(\det {A'}_{\ell j}=-\det A_{\ell j}\) by induction. Lastly, since the \(\ell\)-th rows of \(A\) and \(A'\) are the same we have
As a further consequence of Theorem 3.5.8 and Corollary 3.5.10, we can derive the adjoint formula.
Definition 3.5.14. Adjoint matrix.
Let \(A\) be \(n\times n\text{.}\) The adjoint of \(A\text{,}\) denoted \(\adj A\text{,}\) is the \(n\times n\) matrix whose \(ij\)-th entry is defined as follows:
Remark 3.5.15.
Be careful of the order reversal in this definition. The \(ij\)-th entry of \(\adj A\) is equal to plus or minus the \(ji\)-th minor of \(A\text{.}\) Let's see this in action for some small matrices.
For
we have
For
we have
Theorem 3.5.16. Adjoint formula.
Let \(A\) be an \(n\times n\) matrix. Then
As a consequence, if \(\det A\ne 0\text{,}\) then \(A\) is invertible and
Proof.
First observe that the second statement regarding invertibility follows directly from (3.5.7), since in this case setting \(B=\frac{1}{\det A}\, \adj A\) we have
Thus it suffices to prove (3.5.7). To do so, we must show that
Case: \(i=j\).
In this case we have
A similar argument shows that \((\adj A\, A)_{ii}=\det A\text{,}\) though in this case we use expansion along a column.
Case: \(i\ne j\).
When \(i\ne j\) we have
where \(C\) is the matrix obtained by replacing the \(j\)-th row of \(A\) with a copy of its \(i\)-th row. Since \(C\) has two identical rows Theorem 3.5.13 implies
as desired. Once again, a similar argument using expansion along a column shows that \([\adj A\, A]_{ij}=0\text{.}\)
Example 3.5.17.
Use the adjoint formula to compute \(A^{-1}\text{,}\) where
First compute \(\det A=\) by expanding along the third row:
Next, compute
Then we have
Remark 3.5.18.
Before you get too excited about the adjoint formula, you should know that as \(n\) grows, this procedure becomes much more costly in terms of number of arithmetic operations involved than our inverse algorithm based on Gauss-Jordan elimination. You get a sense of this already from the previous \(3\times 3\) example. In general, the Gauss-Jordan inverse algorithm is the way to go.
Subsection 3.5.3 Row operations and determinant
Suppose the square matrix \(A\) can be row reduced to \(B\) via sequence of row operations. In general we do not have \(\det A=\det B\text{,}\) but we can compute \(\det A\) from \(\det B\) by keeping track of which operations are used.
Theorem 3.5.19. Row operations and determinant.
Let \(A\) be an \(n\times n\) matrix. Using the notation from Definition 3.4.1 we have:
\(\displaystyle \det(\underset{c\cdot r_i}{E}\, A)=c\det(A)\)
\(\displaystyle \det(\underset{r_i\leftrightarrow r_j}{E}\, A)=-\det(A)\)
\(\det(\underset{r_i+cr_j}{E}\, A)=\det(A)\text{.}\)
In particular, taking \(A=I\text{,}\) we have
Proof.
The first statement follows easily by computing \(\det (\underset{cr_i}{E}\, A)\) by expanding along the \(i\)-th row. The second statement is in fact a rephrasing of the second statement of Theorem 3.5.13. It remains to prove the third statement.
Let \(A=[a_{ij}]_{n\times n}\text{,}\) and set \(B=\underset{r_i+cr_j}{E}\cdot A\text{.}\) Then \(B\) is identical to \(A\) with the exception of the \(i\)-th row, whose \(k\)-th entry is
It follows that
where \(C\) is the matrix obtained by replacing the \(i\)-th row of \(A\) with a row identical with its \(j\)-th row. By Theorem 3.5.13 we conclude \(\det C=0\text{,}\) and thus
as desired.
Remark 3.5.20.
In the language of row operations, Theorem 3.5.19 translates as follows:
Scaling a row of a matrix by \(c\) has the effect of scaling the determinant by \(c\text{.}\)
Swapping two rows of a matrix changes the sign of the determinant.
Performing a row addition operation on a matrix has no effect on the determinant.
Remark 3.5.21. Column operations and the determinant.
As shown in Exercise 3.5.4.22 the determinant behaves in a similar manner with respect to elementary column operations: i.e., scaling a column by a nonzero constant scales the determinant by \(c\text{,}\) swapping columns multiplies the determinant by \(-1\text{,}\) adding a multiple of one column to another leaves the determinant unchanged.
Corollary 3.5.22. Determinant and products of elementary matrices.
Let \(A\) be an \(n\times n\) matrix, and suppose we have
for some collection of elementary matrices \(E_i\text{.}\) Then
Proof.
This is an easy proof by induction on the number \(r\) of elementary matrices involved, the base case (\(r=1\)) of which is covered by Theorem 3.5.19.
Corollary 3.5.22 has both computational and theoretical applications.
On the computational side, it suggests an alternative method of computing \(\det A\text{:}\) first row reduce \(A\) to a simpler matrix \(B\text{,}\) making sure to keep track of the operations you use; set up an equation as in (3.5.9) representing the row reduction; then solve the corresponding equation (3.5.10) for \(\det A\) in terms of \(\det B\) and the \(\det E_i\text{.}\)
Example 3.5.23. Determinant via row reduction.
Suppose the matrix \(A\) can be row reduced to
by perfomring the following sequence of row operations:
First swap the second and third rows.
Then scale the first row by \(3\)
Then replace the second row with the second row plus the first row.
Compute \(\det A\text{.}\)
In terms of elementary matrices we have
and hence
We conclude that
On the theoretical side, Corollary 3.5.22 implies both Theorem 3.5.24 and Theorem 3.5.25.
Theorem 3.5.24. Determinant and invertibility.
Let \(A\) be an \(n\times n\) matrix. Then \(A\) is invertible if and only if \(\det A\ne 0\text{.}\)
Proof.
The implication \(\det A\ne 0\implies A \text{ invertible}\) was proved in Theorem 3.5.16.
For the other direction, assume \(A\) is invertible. Then Theorem 3.4.5 implies \(A\) is a product of elementary matrices:
Then Corollary 3.5.22 implies
Since \(\det E_i\ne 0\) for all \(i\) (Theorem 3.5.19), we conclude \(\det A\ne 0\text{.}\)
Theorem 3.5.25. Determinant is multiplicative.
Let \(A\) and \(B\) be \(n\times n\) matrices. Then
Proof.
We consider two cases based on the invertibility of \(A\) and/or \(B\text{.}\)
\(A\) or \(B\) is not invertible.
In this case \(AB\) is not invertible (Corollary 3.4.12), and hence \(\det AB=0\) by Theorem 3.5.24. By the same reasoning we must have \(\det A=0\) or \(\det B=0\text{.}\) It follows that
in this case.
\(A\) and \(B\) both invertible.
In this case we can write
for some elementary matrices \(E_i\) and \(E'_j\) (Theorem 3.4.5). Then
We end this section (and chapter) by adding the results of Theorem 3.5.24 and one of our homework exercises to our invertibility theorem.
Theorem 3.5.26. Invertibility theorem (extended cut).
Let \(A=\) be an \(n\times n\) matrix. The following statements are equivalent.
\(A\) is invertible.
-
The matrix equation
\begin{equation*} A\underset{n\times 1}{\boldx}=\underset{n\times 1}{\boldb} \end{equation*}has a unique solution for any column vector \(\boldb\text{.}\)
-
The matrix equation
\begin{equation*} A\underset{n\times 1}{\boldx}=\underset{n\times 1}{\boldb} \end{equation*}has a solution for any column vector \(\boldb\text{.}\)
-
The matrix equation
\begin{equation*} A\underset{n\times 1}{\boldx}=\underset{n\times 1}{\boldzero} \end{equation*}has a unique solution: namely, \(\boldx=\boldzero_{n\times 1}\text{.}\)
\(A\) is row equivalent to \(I_n\text{,}\) the \(n\times n\) identity matrix.
\(A\) is a product of elementary matrices.
\(\det A\ne 0\text{.}\)
Exercises 3.5.4 Exercises
1.
Let
Compute \(\det(A)\) by expanding along the second row.
Compute \(\det(A)\) by expanding along the third column.
Exercise Group.
Compute the determinant of the given matrix. Indicate which row or column you expand along.
2.
3.
4.
Exercise Group.
For each matrix, find all values of \(c\) (if any) making the matrix invertible. Use the determinant.
5.
6.
7.
8.
Exercise Group.
Use the adjoint formula to compute the inverse of each matrix.
9.
10.
11.
Let
Show \(\det A=0\) without computing the determinant directly. In other words, use a row reduction technique or Theorem 3.5.26.
12.
Let \(A\) be an \(n\times n\) matrix, and let \(c\) be a scalar. State and prove a formula relating \(\det(cA)\) with \(\det(A)\text{.}\)
Warning: in general \(\det(cA)\ne c\det(A)\text{!}\) We claim in fact that if \(A\) is \(n\times n\text{,}\) then \(\det(cA)=c^n\det(A)\text{.}\) We give two proofs: one uses only induction and the definition of the determinant; the other uses the multiplicative property of the determinant.
Proof by induction.
We show that for all \(n\geq 1\text{,}\) if \(A\) is an \(n\times n\) matrix, then \(\det(cA)=c^n\det A\text{.}\)
Base step: \(n=1\).
Here we have \(A=[a]\text{,}\) \(cA=[ca]\text{,}\) and thus \(\det(cA)=ca=c\det A\text{.}\)
Induction step.
We assume that \(\det(cB)=c^n\det B\) for any \(n\times n\) matrix \(B\text{,}\) and prove \(\det(cA)=c^{n+1}\det A\) for any \((n+1)\times (n+1)\) matrix \(A\text{.}\) To this end, let \(A=[a_{ij}]_{(n+1)\times (n+1)}\text{.}\) Then
Proof using determinant properties.
We have
Then
13.
Assume \(A\) is a \(5\times 5\) matrix satisfying \(\det A=2\text{.}\) Compute the determinant of the given matrix.
\(\displaystyle \det(-A)\)
\(\displaystyle \det(A^{-1})\)
\(\displaystyle \det(2A^T)\)
\(\displaystyle \det(A^3)\)
14.
Let \(A\) and \(B\) be \(n\times n\) matrices, and suppose \(B\) is invertible. Prove the following:
\(\ds \det(B^{-1})=\frac{1}{\det(B)}\text{.}\)
\(\ds \det(B^{-1}AB)=\det(A)\text{.}\)
We prove (a).
Furthermore, since \(B\) is invertible, we know \(\det B\ne 0\) (Theorem 3.5.26), allowing us to solve the last equation for \(\det B^{-1}\text{:}\)
15.
Assume the square matrix \(A\) satisfies \(A^TA=I\text{.}\) Show that \(\det A=\pm 1\text{.}\)
16.
Prove that a square matrix \(A\) is invertible if and only if \(A^TA\) is invertible.
Exercise Group.
The following exercises explore the relationship between a square matrix \(A\) and its adjoint \(\adj A\text{.}\)
17.
Give an explicit example of a square matrix \(A\) satisfying \(A\ne \boldzero_{n\times n}\) and \(\adj A=\boldzero_{n\times n}\text{.}\)
In other words, show that it is possible for a nonzero matrix to have a zero adjoint matrix.
18.
Let \(A\) be an \(n\times n\) matrix. Prove:
19.
Prove: \(A\) is invertible if and only if \(\adj A\) is invertible.
20.
Assume \(A\) is invertible. Prove:
21.
In our proof of statement (2) of Theorem 3.5.13 we only showed that if \(A\) is a square matrix with two identical rows, then \(\det A=0\text{.}\) Assuming this, show that the same is true if \(A\) has two identical columns.
22.
State and prove an analogue to Theorem 3.5.19 that describes how the corresponding column operations (i.e., scale a column by \(c\text{,}\) swap two columns, column addition) affect the determinant of a matrix. (See Remark 3.5.21).
Express each of these types of column operations as multiplication on the right by an elementary matrix.
23.
Let \(A\) be the \(n\times n\) matrix with \(a's\) along the diagonal and \(b\)'s elsewhere: i.e.,
State and prove a formula for \(\det A\text{.}\) (Your formula will involve \(a\text{,}\) \(b\text{,}\) and \(n\text{.}\))
Look at the \(n=2\) and \(n=3\) cases first. To prove the formula in the general case you may want to make use of row reduction and Theorem 3.5.19.
24.
Given scalars \(r_1, r_2, \dots, r_n\in \R\) the Vandermonde matrix is defined as
In other words, we have
Prove: