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Linear algebra: the theory of vector spaces and linear transformations

Aaron Greicius

Section 2.5 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
\begin{equation*} A \text{ is invertible } \iff \det A \ne 0\text{,} \end{equation*}
making the determinant an important tool for investigating invertibility.

Subsection 2.5.1 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 2.5.1. 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 \(A_{i j}\text{.}\)

Warning 2.5.2.

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 2.5.3. 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 2.5.4. 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 2.3.6; we will make the connection more explicit in Theorem 2.5.17.
Given \(A=\abcdmatrix{a}{b}{c}{d}\text{,}\) we have
\begin{align*} \det A\amp = (-1)^{1+1}a\det A_{11}+(-1)^{1+2}b\det A_{12}\amp \text{(recursive case)}\\ \amp =a\det [d]-b\det [c] \amp \text{(base case)}\\ \amp =ad-bc \text{.} \end{align*}
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
\begin{equation*} A=\begin{amatrix}[ccc]a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{amatrix}\text{,} \end{equation*}
we have
\begin{align*} \det A \amp =a\det A_{11}-b\det A _{12}+c\det A_{13} \amp \text{(recursive case)}\\ \amp =a\det \abcdmatrix{e}{f}{h}{i}-b\det\abcdmatrix{d}{f}{g}{i}+c\det\abcdmatrix{d}{e}{g}{h}\\ \amp = a(ei-fh)-b(di-fg)+c(dh-eg) \amp \text{(\(n=2\) case)}\text{.} \end{align*}
The recursive nature of the determinant definition makes induction arguments particularly useful when proving properties of the determinant, as illustrated by the next theorem.
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
\begin{equation*} \det A=a_{11}\det A_{11}\text{.} \end{equation*}
Claim: \(A_{i k}\) is lower triangular. Indeed, first observe that we have
\begin{equation*} ( A_{11})_{ij}=a_{i+1,j+1} \end{equation*}
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
\begin{equation*} ( A_{11})_{ij}=a_{i+1,j+1}=0 \end{equation*}
for all \(1\leq i < j\leq n \text{,}\) proving the claim.
Lastly, assuming \(P(n)\) is true (the induction hypothesis) we have
\begin{align*} \det A \amp=a_{11}\det A_{11} \\ \amp =a_{11}(a_{22}a_{33}\cdots a_{n+1,n+1})\amp \text{(induction)} \\ \amp =a_{11}a_{22}\cdots a_{n+1,n+1} \text{,} \end{align*}
as desired.
This follows directly from Theorem 2.5.5 since the diagonal entries of \(I\) are all ones.

Subsection 2.5.2 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 2.5.7. 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
\begin{equation*} M_{ij}=\det A_{i k}\text{.} \end{equation*}
For any \(1\leq i\leq n\) the expression
\begin{equation*} \sum_{k=1}^n(-1)^{i+k}a_{ik}M_{ik} \end{equation*}
is called the expansion along the \(i\)-th row of \(A\).
For any \(1\leq j\leq n\text{,}\) the expression
\begin{equation*} \sum_{k=1}^n(-1)^{k+j}a_{kj}M_{kj} \end{equation*}
is called the expansion along the \(j\)-th column of \(A\).
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
\begin{equation*} A=\begin{bmatrix}a\amp b\\ c\amp d\end{bmatrix} \end{equation*}
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
\begin{equation} \det A=\sum_{j=1}^n (-1)^{1+j}a_{1 j}\det A_{1j}\text{.}\tag{2.5.1} \end{equation}
Expanding along the \(i\)-th row of \(A\) for any \(2\leq i\leq n\text{,}\) on the other hand, we get
\begin{equation} \sum_{k=1}^n (-1)^{i+k}a_{ik}\det A_{i k}\text{.}\tag{2.5.2} \end{equation}
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:
\begin{equation} \det A_{1j}=\sum_{k=1}^{n-1}(-1)^{i-1+k}[ A_{1j}]_{(i-1) k}\det {( A_{1j})}_{(i-1) k}\text{.}\tag{2.5.3} \end{equation}
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:
\begin{align} [ A_{1j}]_{(i-1) k} \amp =\begin{cases} a_{i k} \amp \text{if } k\lt j\\ a_{i (k+1)} \amp \text{if } k\geq j \end{cases} \tag{2.5.4}\\ [ A_{i k}]_{1 j} \amp =\begin{cases} a_{1 j} \amp \text{if } j\lt k\\ a_{1 (j+1)} \amp \text{if } k\leq j \end{cases} \tag{2.5.5}\\ {( A_{1j})}_{i-1 k}\amp=\begin{cases} {( A_{i k})}_{1 (j-1)} \amp\text{if } k\lt j \\ {( A_{i (k+1)})}_{1 j} \amp \text{if } k\geq j \end{cases}\text{.}\tag{2.5.6} \end{align}
We now begin to unpack (2.5.1):
\begin{align*} \det A \amp= \sum_{j=1}^n (-1)^{1+j}a_{1 j}\det A_{1j}\\ \amp = \sum_{j=1}^n (-1)^{1+j}a_{1 j}\sum_{k=1}^{n-1}(-1)^{i-1+k}[ A_{1j}]_{(i-1)k}\det {( A_{1j})}_{(i-1)k} \amp \knowl{./knowl/eq_det_submatrix.html}{\text{(2.5.3)}}\\ \amp = \underset{k\lt j \text{ terms}}{\underbrace{\sum_{j=2}^n \sum_{k=1}^{j-1}(-1)^{i+j+k}a_{1 j}a_{i k}\det (A_{i k})_{1 (j-1)}}}+\underset{k\geq j \text{ terms}}{\underbrace{\sum_{j=1}^{n-1} \sum_{k=j}^{n-1}(-1)^{i+j+k}a_{1 j}a_{i (k+1)}\det (A_{i (k+1)})_{1 j}}} \amp \knowl{./knowl/eq_entry_formula_1.html}{\text{(2.5.4)}}, \knowl{./knowl/eq_iterated_submatrix.html}{\text{(2.5.6)}}\\ \amp = \underset{j\geq k \text{ terms}}{\underbrace{\sum_{j=1}^{n-1} \sum_{k=1}^{j}(-1)^{i+j+k+1}a_{i k}[A_{ik}]_{1j}\det (A_{i k})_{1 j}}}+\underset{j\lt k \text{ terms}}{\underbrace{\sum_{j=1}^{n-1} \sum_{k=j+1}^{n}(-1)^{i+j+k-1}a_{ik}[A_{ik}]_{1 j}\det (A_{ik})_{1 j}}} \amp (\text{re-labeled } j=j-1, k=k+1)\\ \amp = \underset{j\geq k \text{ terms}}{\underbrace{\sum_{j=1}^{n-1} \sum_{k=1}^{j}(-1)^{i+j+k+1}a_{i k}[A_{ik}]_{1j}\det (A_{i k})_{1 j}}}+\underset{j\lt k \text{ terms}}{\underbrace{\sum_{j=1}^{n-1} \sum_{k=j+1}^{n}(-1)^{i+j+k+1}a_{ik}[A_{ik}]_{1 j}\det (A_{ik})_{1 j}}} \amp ((-1)^{c+2}=(-1)^c)\\ \amp =\sum_{k=1}^{n}(-1)^{i+k}a_{ik}\sum_{j=1}^{n-1}(-1)^{j+1}[A_{ik}]_{1j}\det (A_{ik})_{1j} \amp (\text{re-order terms})\\ \amp =\sum_{k=1}^{n}(-1)^{i+k}a_{ik}\det A_{ik} \text{.} \end{align*}
This completes the induction step, and thus the proof is finished.
Surprisingly, it turns out that we can compute the determinant of a matrix by expanding along any column (Corollary 2.5.10). This is a consequence of the following theorem, which is useful in its own right. The proof below is taken from Robert Beezer’s A First Course in Linear Algebra 1 . (See Theorem DT 2 .) It uses induction and a wonderful trick starting from the observation that \(c=\frac{1}{n}\sum_{i=1}^n c\) for any \(c\in\R\text{.}\)
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
\begin{align*} \det A^T \amp = \frac{1}{n}\sum_{i=1}^{n}\det A^T \\ \amp =\frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n(-1)^{i+j}[A^T]_{ij}\det \left((A^T)_{ij}\right) \amp (\knowl{./knowl/th_expansion_row.html}{\text{Theorem 2.5.8}}) \\ \amp = \frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n(-1)^{i+j}[A]_{ji}\det (A_{ji})^T \amp ([A^T]_{ij}=A_{ji}, (A^T)_{ij}=(A_{ji})^T)\\ \amp = \frac{1}{n}\sum_{i=1}^n\sum_{j=1}^n(-1)^{i+j}[A]_{ji}\det A_{ji} \amp (\det (A_{ji})^T=\det A_{ji}, \text{ by induction})\\ \amp= \frac{1}{n}\sum_{j=1}^n\sum_{i=1}^n(-1)^{i+j}[A]_{ji}\det A_{ji} \amp (\text{props. of addition})\\ \amp =\frac{1}{n}\sum_{j=1}^n\det A \amp (\knowl{./knowl/th_expansion_row.html}{\text{Theorem 2.5.8}})\\ \amp = \det A\text{.} \end{align*}
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.)
For any \(1\leq j\leq n\text{,}\) we have
\begin{align*} \sum_{k=1}^n(-1)^{k+j}a_{kj}M_{kj}\amp =\sum_{k=1}^n(-1)^{k+j}(A^T)_{jk}\det A_{kj} \amp (\knowl{./knowl/d_transpose.html}{\text{Definition 2.1.29}})\\ \amp = \sum_{k=1}^n(-1)^{k+j}(A^T)_{jk}\det (A^T)_{jk} \amp ( (A^T)_{ij}= A_{ji}) \\ \amp= \det A^T \amp (\knowl{./knowl/d_det.html}{\text{Definition 2.5.3}})\\ \amp=\det A \amp (\knowl{./knowl/th_det_transpose.html}{\text{Theorem 2.5.9}}) \text{.} \end{align*}

Example 2.5.11.

Compute \(\det A\) for
\begin{equation*} A=\begin{bmatrix}2\amp 1\amp 0\amp 3\\ 0\amp 0\amp 0\amp 2\\ 0\amp 2\amp 0\amp 0\\ 4\amp 2\amp 1\amp 0 \end{bmatrix}\text{.} \end{equation*}
Solution.
First we compute \(A\) by expanding along the second row. The only nonzero term of this expansion is the last one, yielding
\begin{equation*} \det A=(-1)^{2+4}[A]_{24}M_{24}=2\det A_{24}\text{.} \end{equation*}
We have
\begin{equation*} A_{24}=\begin{bmatrix} 2\amp 1\amp 0\\ 0\amp 2\amp 0\\ 4\amp 2\amp 1 \end{bmatrix}\text{.} \end{equation*}
To compute its determinant we expand along its third column:
\begin{equation*} \det A_{24}=(-1)^{3+3}1\det \begin{bmatrix} 2\amp 1\\ 0\amp 2 \end{bmatrix}=4. \end{equation*}
We conclude that
\begin{equation*} \det A=2\det A_{24}=2(4)=8\text{.} \end{equation*}

Remark 2.5.12. 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:
\begin{equation*} \begin{bmatrix} +\amp -\amp +\amp -\\ -\amp +\amp -\amp +\\ +\amp -\amp +\amp -\\ -\amp +\amp -\amp + \end{bmatrix}\text{.} \end{equation*}

Video example: determinant.

Figure 2.5.13. Video: determinant
The freedom to compute the determinant by expanding along any row or column gives rise to the following intuitive property.
The first statement is obvious since according to Theorem 2.5.8 and Corollary 2.5.10 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 2.5.9. 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
\begin{align*} \det A \amp=\sum_{j=1}^{n+1}(-1)^{\ell+j}a_{\ell j}\det A_{\ell j} \\ \amp = \sum_{j=1}^{n+1}(-1)^{\ell+j}a_{\ell j}(-\det {A'}_{\ell j})\\ \amp =-\sum_{j=1}^{n+1}(-1)^{\ell+j}a_{\ell j}\det {A'}_{\ell j} \\ \amp=-\det A' \text{.} \end{align*}
As a further consequence of Theorem 2.5.8 and Corollary 2.5.10, we can derive the adjoint matrix formula.

Definition 2.5.15. Adjoint matrix.

Let \(A\) be an \(n\times n\) matrix. The adjoint matrix of \(A\text{,}\) denoted \(\adj A\text{,}\) is the \(n\times n\) matrix whose \(ij\)-th entry is defined as follows:
\begin{equation*} (\adj A)_{ij}=(-1)^{i+j}M_{ji}\text{.} \end{equation*}

Remark 2.5.16.

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
\begin{equation*} A=\abcdmatrix{a}{b}{c}{d} \end{equation*}
we have
\begin{equation*} \adj A=\begin{amatrix}[rr] M_{11}\amp -M_{21}\\ -M_{12}\amp M_{22}\end{amatrix}= \begin{amatrix}[rr]d\amp -b\\ -c\amp a \end{amatrix}\text{.} \end{equation*}
For
\begin{equation*} A=\begin{bmatrix} a\amp b\amp c\\ d\amp e\amp f\\ g\amp h\amp i \end{bmatrix} \end{equation*}
we have
\begin{equation*} \adj A=\begin{amatrix}[rrr] M_{11}\amp -M_{21}\amp M_{31}\\ -M_{12}\amp M_{22}\amp -M_{32}\\ M_{13}\amp -M_{23}\amp M_{33} \end{amatrix} =\begin{amatrix}[rrr] (ei-fh)\amp -(bi-ch)\amp (bf-ce)\\ -(di-fg)\amp (ai-cg)\amp -(af-cd)\\ (dh-eg)\amp -(ah-bg)\amp (ae-bd) \end{amatrix}\text{.} \end{equation*}
First observe that the second statement regarding invertibility follows directly from (2.5.7), since in this case setting \(B=\frac{1}{\det A}\, \adj A\) we have
\begin{equation*} AB=BA=\frac{1}{\det A}(A\adj A)=\frac{1}{\det A}((\det A)\, I)=I\text{.} \end{equation*}
Thus it suffices to prove (2.5.7). To do so, we must show that
\begin{equation*} [A\, \adj A]_{ij}=[\adj A\, A]_{ij}= \begin{cases} \det A\amp \text{if } i=j\\ 0 \amp \text{if } i\ne j \end{cases}\text{.} \end{equation*}
Case: \(i=j\).
In this case we have
\begin{align*} [A\, \adj A]_{ii} \amp =\sum_{k=1}^na_{ik}[\adj A]_{ki} \\ \amp =\sum_{k=1}^n a_{ik}((-1)^{i+k}M_{ik}) \amp (\knowl{./knowl/d_adjoint.html}{\text{Definition 2.5.15}} )\\ \amp =\sum_{k=1}^n(-1)^{i+k}a_{ik}M_{ik} \\ \amp =\det A \text{.} \end{align*}
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
\begin{align*} [A\, \adj A]_{ij} \amp =\sum_{k=1}^na_{ik}[\adj A]_{kj} \\ \amp =\sum_{k=1}^n a_{ik}((-1)^{k+j}M_{jk}) \amp (\knowl{./knowl/d_adjoint.html}{\text{Definition 2.5.15}} )\\ \amp =\sum_{k=1}^n(-1)^{k+j}a_{ik}M_{jk} \\ \amp =\det C \text{,} \end{align*}
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 2.5.14 implies
\begin{equation*} [A\, \adj A]_{ij}=\det C=0\text{,} \end{equation*}
as desired. Once again, a similar argument using expansion along a column shows that \([\adj A\, A]_{ij}=0\text{.}\)

Example 2.5.18.

Use the adjoint matrix formula to compute \(A^{-1}\text{,}\) where
\begin{equation*} A=\begin{amatrix}[rrr]1\amp 2\amp 1 \\ -1\amp 1\amp -1\\ 0\amp 2\amp 2 \end{amatrix}\text{.} \end{equation*}
Solution.
First compute \(\det A=\) by expanding along the third row:
\begin{equation*} \det A=-2(-1+1)+2(2+1)=6\text{.} \end{equation*}
Next, compute
\begin{equation*} \adj A=\begin{amatrix}[rrr] 4\amp -2 \amp -3\\ 2\amp 2\amp 0\\ -2\amp -2\amp 3 \end{amatrix}\text{.} \end{equation*}
Then we have
\begin{equation*} A^{-1}=\frac{1}{6}\,\adj A=\frac{1}{6}\, \begin{amatrix}[rrr] 4\amp -2 \amp -3\\ 2\amp 2\amp 0\\ -2\amp -2\amp 3 \end{amatrix}\text{.} \end{equation*}

Remark 2.5.19.

Before you get too excited about the adjoint matrix 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 2.5.3 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.
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 2.5.14. 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
\begin{equation*} (B)_{ik}=a_{ik}+ca_{jk}\text{.} \end{equation*}
It follows that
\begin{align*} \det B \amp=\sum_{k=1}^n(-1)^{i+k}(a_{ik}+ca_{jk})M_{ik} \\ \amp = \sum_{k=1}^n(-1)^{i+k}a_{ik}M_{ik}+c\sum_{k=1}^n(-1)^{i+k}a_{jk}M_{ik}\\ \amp =\det A+\det C \text{,} \end{align*}
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 2.5.14 we conclude \(\det C=0\text{,}\) and thus
\begin{equation*} \det(\underset{r_i+cr_j}{E}\cdot A)=\det B=\det A+\det C=\det A\text{,} \end{equation*}
as desired.

Remark 2.5.21.

In the language of row operations, Theorem 2.5.20 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 2.5.22. Column operations and the determinant.

As shown in Exercise 2.5.4.26 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.
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 2.5.20.
Corollary 2.5.23 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 (2.5.9) representing the row reduction; then solve the corresponding equation (2.5.10) for \(\det A\) in terms of \(\det B\) and the \(\det E_i\text{.}\)

Example 2.5.24. Determinant via row reduction.

Suppose the matrix \(A\) can be row reduced to
\begin{equation*} B=\begin{amatrix}[rrr]2\amp -7\amp -8\\ 0\amp -3\amp 4\\ 0\amp 0\amp 2 \end{amatrix} \end{equation*}
by perfomring the following sequence of row operations:
  1. First swap the second and third rows.
  2. Then scale the first row by \(3\)
  3. Then replace the second row with the second row plus the first row.
Compute \(\det A\text{.}\)
Solution.
In terms of elementary matrices we have
\begin{equation*} B=\underset{r_2+r_1}{E}\underset{3r_1}{E}\underset{r_2\leftrightarrow r_3}{E}A, \end{equation*}
and hence
\begin{equation*} \det B=\det \underset{r_2+r_1}{E}\, \det \underset{3r_1}{E}\, \det \underset{r_2\leftrightarrow r_3}{E}\, \det A=(1)(3)(-1)\det A=(-3)\det A. \end{equation*}
We conclude that
\begin{equation*} \det A=-\frac{1}{3}\det B=-(1/3)(-12)=4\text{.} \end{equation*}
On the theoretical side, Corollary 2.5.23 implies both Theorem 2.5.25 and Theorem 2.5.26.
The implication \(\det A\ne 0\implies A \text{ invertible}\) was proved in Theorem 2.5.17.
For the other direction, assume \(A\) is invertible. Then Theorem 2.4.5 implies \(A\) is a product of elementary matrices:
\begin{equation*} A=E_1E_2\cdots E_r\text{.} \end{equation*}
Then Corollary 2.5.23 implies
\begin{equation*} \det A=\det E_1\, \det E_2\cdots \det E_1\text{.} \end{equation*}
Since \(\det E_i\ne 0\) for all \(i\) (Theorem 2.5.20), we conclude \(\det A\ne 0\text{.}\)
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 2.4.15), and hence \(\det AB=0\) by Theorem 2.5.25. By the same reasoning we must have \(\det A=0\) or \(\det B=0\text{.}\) It follows that
\begin{equation*} \det AB=\det A\, \det B=0 \end{equation*}
in this case.
\(A\) and \(B\) both invertible.
In this case we can write
\begin{align*} A \amp E_1E_2\cdots E_r \amp B\amp=E'_1E'_2\cdots E'_s \end{align*}
for some elementary matrices \(E_i\) and \(E'_j\) (Theorem 2.4.5). Then
\begin{align*} \det AB \amp= \det(E_1E_2\cdots E_rE'_1E'_2\cdots E'_s) \\ \amp= \det E_1\det E_2\cdots \det E_r\det E'_1\det E'_2\cdots \det E'_s \amp (\knowl{./knowl/cor_det_prod_elem.html}{\text{Corollary 2.5.23}})\\ \amp=\det A\, \det B \amp (\knowl{./knowl/cor_det_prod_elem.html}{\text{Corollary 2.5.23}}) \text{.} \end{align*}
We end this section (and chapter) by adding the results of Theorem 2.5.25 and one of our homework exercises to our invertibility theorem.

Exercises 2.5.4 Exercises

WeBWork Exercises

1.
Use the appropriate property of determinants to find \(a.\) Do not evaluate the determinants.
\begin{equation*} \begin{vmatrix} 2 \amp 23 \amp 11 \\ -43 \amp -31 \amp -17 \\ 2 \amp -3 \amp 9 \\ \end{vmatrix} = a \cdot \begin{vmatrix} -2 \amp -23 \amp -11 \\ 43 \amp 31 \amp 17 \\ -2 \amp 3 \amp -9\\ \end{vmatrix} \end{equation*}
Answer: \(a =\)
Answer.
\(-1\)
2.
Consider the following Gauss elimination:
\begin{equation*} A \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 0\cr 0 \amp 1 \amp 0\cr 0 \amp 0 \amp 5 \end{array}\right]}_{E_1} A \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 0\cr 0 \amp -3 \amp 0\cr 0 \amp 0 \amp 1 \end{array}\right]}_{E_2} E_1 A \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp -5\cr 0 \amp 1 \amp 0\cr 0 \amp 0 \amp 1 \end{array}\right]}_{E_3} E_2E_1 A \to \underbrace{\left[\begin{array}{ccc} 1 \amp 0 \amp 0\cr 0 \amp 0 \amp 1\cr 0 \amp 1 \amp 0 \end{array}\right]}_{E_4} E_3E_2E_1 A=\left[\begin{array}{ccc} -7 \amp 4 \amp -5\cr 0 \amp -2 \amp 7\cr 0 \amp 0 \amp -5 \end{array}\right] \end{equation*}
What is the determinant of \(A\text{?}\)
\(\det(A) =\)
Answer.
\(-4.66667\)
3.
Find \(k\) such that the following matrix \(M\) is singular.
\begin{equation*} M = \left[\begin{array}{ccc} 1 \amp -1 \amp -4\cr -3 \amp 1 \amp 10\cr 13+k \amp -5 \amp -32\cr \end{array}\right] \end{equation*}
\(k=\)
Answer.
\(-4\)
4.
If \(A\) and \(B\) are \(4 \times 4\) matrices, \(\det (A)=-5\text{,}\) \(\det (B)=-8\text{,}\) then
\(\det (AB) =\) ,
\(\det (-3 A) =\) ,
\(\det (A^{T}) =\) ,
\(\det (B^{-1}) =\) ,
\(\det (B^4) =\) .
Answer 1.
\(40\)
Answer 2.
\(-405\)
Answer 3.
\(-5\)
Answer 4.
\(-0.125\)
Answer 5.
\(4096\)

Written Exercises

5.
Let
\begin{equation*} A = \begin{bmatrix}-1\amp 1\amp 2\\ 3\amp 0\amp -5\\ 1\amp 7\amp 2 \end{bmatrix} \end{equation*}
  1. Compute \(\det(A)\) by expanding along the second row.
  2. Compute \(\det(A)\) by expanding along the third column.
Row/column expansion.
Compute the determinant of the given matrix. Indicate which row or column you expand along.
6.
\begin{equation*} A=\begin{amatrix}[rrr]2\amp 0\amp 3\\ 0\amp 3\amp 2\\ -2\amp 0\amp -4 \end{amatrix} \end{equation*}
7.
\begin{equation*} A=\begin{amatrix}[rrrr] 2\amp 1\amp 0\amp 5\\ 1\amp 2\amp 0\amp -1\\ 1\amp 4\amp -3\amp 0\\ 2\amp 6\amp 3\amp 2 \end{amatrix} \end{equation*}
8.
\begin{equation*} A=\begin{amatrix}[rrrrr]4\amp 0\amp 0\amp 1\amp 0\\ 3\amp 3\amp 3\amp -1\amp 0\\ 1\amp 2\amp 4\amp 2\amp 3\\ 9\amp 4\amp 6\amp 2\amp 3\\ 2\amp 2\amp 4\amp 2\amp 3 \end{amatrix} \end{equation*}
Determinant and invertibility.
For each matrix, find all values of \(c\) (if any) making the matrix invertible. Use the determinant.
9.
\begin{equation*} A=\begin{amatrix}[ccc] c \amp 1 \amp 1 \\ 1 \amp 1 \amp c \\ 1 \amp 1 \amp 1 \\ \end{amatrix} \end{equation*}
10.
\begin{equation*} A=\begin{amatrix}[ccc] c \amp 1 \amp 1 \\ 1 \amp 1 \amp c \\ 1 \amp 3 \amp 1 \\ \end{amatrix} \end{equation*}
11.
\begin{equation*} A=\begin{amatrix}[ccc] c \amp 1 \amp 1 \\ 1 \amp c \amp 1 \\ c+1 \amp c+1 \amp 2 \end{amatrix} \end{equation*}
12.
\begin{equation*} A=\begin{amatrix}[ccc] c \amp 1 \amp 1 \\ 1 \amp c \amp 1 \\ 1 \amp 1 \amp c \\ \end{amatrix} \end{equation*}
Adjoint matrix formula.
Use the adjoint matrix formula to compute the inverse of each matrix.
13.
\begin{equation*} A=\begin{amatrix}[rr]-7\amp -5\\ 1\amp 4 \end{amatrix} \end{equation*}
14.
\begin{equation*} A=\begin{amatrix}[rrr] 0\amp 2\amp -3\\ 1\amp -2\amp -1 \\ 0\amp 0\amp 1 \end{amatrix} \end{equation*}
15.
Let
\begin{equation*} A=\begin{amatrix}[rrrr]1\amp -3\amp 1\amp 1\\ -3\amp 1\amp 1\amp 1\\ 1\amp 1\amp -3\amp 1\\ 1\amp 1\amp 1\amp -3 \end{amatrix}\text{.} \end{equation*}
Show \(\det A=0\) without computing the determinant directly. In other words, use a row reduction technique or Theorem 2.5.27.
16.
Let \(A\) be an \(n\times n\) matrix, and let \(c\) be a scalar. State and prove a formula relating \(\det(cA)\) with \(\det(A)\text{.}\)
17.
Assume \(A\) is a \(5\times 5\) matrix satisfying \(\det A=2\text{.}\) Compute the determinant of the given matrix.
  1. \(\displaystyle \det(-A)\)
  2. \(\displaystyle \det(A^{-1})\)
  3. \(\displaystyle \det(2A^T)\)
  4. \(\displaystyle \det(A^3)\)
18.
Let \(A\) and \(B\) be \(n\times n\) matrices, and suppose \(B\) is invertible. Prove the following:
  1. \(\ds \det(B^{-1})=\frac{1}{\det(B)}\text{.}\)
  2. \(\ds \det(B^{-1}AB)=\det(A)\text{.}\)
19.
Assume the square matrix \(A\) satisfies \(A^TA=I\text{.}\) Show that \(\det A=\pm 1\text{.}\)
20.
Prove that a square matrix \(A\) is invertible if and only if \(A^TA\) is invertible.
Adjoint matrix.
The following exercises explore the relationship between a square matrix \(A\) and its adjoint \(\adj A\text{.}\)
21.
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.
22.
Let \(A\) be an \(n\times n\) matrix. Prove:
\begin{equation*} \det \adj A=(\det A)^{n-1}\text{.} \end{equation*}
23.
Prove: \(A\) is invertible if and only if \(\adj A\) is invertible.
24.
Assume \(A\) is invertible. Prove:
\begin{equation*} \adj(\adj A)=\frac{1}{(\det A)^{n-2}}\, A\text{.} \end{equation*}
25.
In our proof of statement (2) of Theorem 2.5.14 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.
26.
State and prove an analogue to Theorem 2.5.20 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 2.5.22).
Hint.
Express each of these types of column operations as multiplication on the right by an elementary matrix.
27.
Let \(A\) be the \(n\times n\) matrix with \(a's\) along the diagonal and \(b\)’s elsewhere: i.e.,
\begin{equation*} A=\begin{bmatrix}a\amp b\amp b\dots \\ b\amp a\amp b\amp \dots\\ \vdots \amp \amp \vdots \\ b\amp b\amp \dots \amp a \end{bmatrix}\text{.} \end{equation*}
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 2.5.20.
28.
Given scalars \(r_1, r_2, \dots, r_n\in \R\) the Vandermonde matrix is defined as
\begin{equation*} A_{r_1,r_2,\dots, r_n}=\begin{amatrix}[llcl]1\amp 1\amp \cdots \amp 1\\ r_1\amp r_2\amp \cdots \amp r_n\\ r_1^2\amp r_2^2\amp \cdots \amp r_n^2\\ \vdots \amp \vdots \amp \amp \vdots\\ r_1^{n-1}\amp r_2^{n-1}\amp \cdots \amp r_n^{n-1} \end{amatrix}\text{.} \end{equation*}
In other words, we have
\begin{equation*} (A_{r_1,r_2,\dots, r_n})_{ij}=r_j^{i-1} \end{equation*}
Prove:
\begin{equation*} \ds\det (A_{r_1,r_2,\dots, r_n})=\prod_{1\leq i\lt j\leq n}(r_j-r_i)\text{.} \end{equation*}
linear.ups.edu/fcla
http://linear.ups.edu/fcla/section-DM.html#theorem-DT
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