The Determinant: Fundamentals

Section 4.1 The Determinant: Fundamentals

Recall from Definition Definition 2.4.1 that the dot product is a function which takes input from \(\R^n\times\R^n\) (ordered pairs of \(n\)-vectors) and outputs the sum of the products of corresponding components of \(\u\) and \(\v.\) The value of the dot product provides information about the relationship between the two input vectors: for example, if the dot product of nonzero \(\u,\,\v\) equals zero the vectors are orthogonal. This is evident from Equation (2.4.5): \(\dpr{\u}{\v}=\|\u\|\,\|\v\|\cos\th.\)

The determinant is a real-valued function of square matrices of any size: \(\det:\R^{n\times n}\rightarrow\R;\) the determinant of \(\A\in\R^{n\times n}\) is denoted \(\det(\A).\) That is, the determinant takes input from ordered \(n\)-tuples of \(n\)-vectors and outputs a real number computed as the sum of \(n\) (signed) products of the entries of a matrix \(\A,\) one entry from each column. The formula for \(\det(\A)\) is simple for \(n=2,\) a bit more complex for \(n=3,\) still more elaborate for \(n=4,\) and so on.

The determinant of \(\A\) is so named because its value (more specifically whether it is nonzero or not) determines whether or not a linear system associated with \(\A\) has a unique solution or not.

Ultimately, we will give a recursive formula for \(\det(\A),\) which will allow us to compute the determinant of a \(4\times 4\) matrix by computing four \(3\times 3\) determinants (and of course for \(n=5\) we’d need to compute five \(4\times 4\) determinants, each of which would require the computation of four \(3\times 3\) determinants, and so on). The determinant function will tell us a lot about the matrix, including whether or not it is invertible, and we will find that it satisfies many interesting properties.

Definition 4.1.1. The determinant of a matrix.

The determinant of a square matrix \(\A,\) denoted \(\det(\A)\) or \(|\A|,\) is the unique function \(\det:\R^{n\times n}\longrightarrow\R\) satisfying the following properties:

For any \(n\in\N,\det(\boldsymbol{I_n})=1.\)
If \(\B\) is produced from \(\A\) by a single row-swap, \(\det(\B)=-\det(\A).\)
The determinant is linear in each row.

The notation for the determinant of \(\A\) is any of:

\begin{equation*} \det(\A)=|\A|=\left|\begin{array}{cccc}A_{11}\amp A_{12}\amp \cdots\amp A_{1n}\\A_{21}\amp A_{22}\amp \cdots\amp A_{2n}\\\vdots\amp \amp \ddots\amp \vdots\\ A_{n1}\amp \cdots\amp \cdots\amp A_{nn}\end{array}\right|. \end{equation*}

There is a subtlety to Definition 4.1.1, namely that a function satisfying the properties therein is unique. We omit the general proof of this uniqueness. From here on we will take the uniqueness of such a function as a given.

Definitions are crucial but do not always lead to convenient computation. We provide formulas for calculating determinants of matrices of various sizes, and show that our formulas do indeed satisfy the properties in Definition 4.1.1.

Theorem 4.1.2. Determinant of a \(\boldsymbol{2\times2}\) matrix.

Let \(\A=\begin{pmatrix}a\amp\!b\\c\amp\!d\end{pmatrix}\in\R^{2\times2}.\) Then

\begin{equation*} \det(\A)=ad\!-\!bc. \end{equation*}

Proof.

We show that the proposed formula \(\det(\A)=ad\!-\!bc\) satisfies the properties in Definition 4.1.1.

By the proposed formula applied to the \(2\times2\) identity we have \(\left|\begin{array}{cc}1\amp 0\\0\amp 1\end{array}\right|=1\cdot1-0\cdot0=1\) so the first property in Definition 4.1.1 holds.

Swapping the rows of \(\A\) and using the proposed formula yields

\begin{equation*} \left|\begin{array}{cc}c\amp\!d\\a\amp\!b\end{array}\right|=bc-ad=-(ad-bc)=-\left|\begin{array}{cc}a\amp b\\c\amp\!d\end{array}\right| \end{equation*}

verifying the second property in Definition 4.1.1.

Finally, we show that \(\det\) is linear in the first row by showing the effects of scalar multiplication and addition separately. Multiplying the first row through by the constant \(r,\) the proposed formula gives

\begin{equation*} \left|\begin{array}{cc}ra\amp rb\\c\amp\!d\end{array}\right|=(ra)d-(rb)c=r(ad-bc)=r\left|\begin{array}{cc}a\amp b\\c\amp\!d\end{array}\right|. \end{equation*}

To check additivity we have

\begin{align*} \left|\begin{array}{cc}a+a'\amp b+b'\\c\amp d\end{array}\right|=\amp (a+a')d-(b+b')c=ad-bc+a'd-b'c\\ =\amp\left|\begin{array}{cc}a\amp b\\c\amp d\end{array}\right|+\left|\begin{array}{cc}a'\amp b'\\c\amp d\end{array}\right|. \end{align*}

Since scalar multiples of the first row factor out of the proposed formula, and the proposed formula is additive in the first row, the \(2\times2\) determinant formula satisfies linearity in the first row. Similar calculations demonstrate linearity in the second row.

Thus the proposed \(2\times2\) determinant formula \(\left|\begin{array}{cc}a\amp b\\c\amp d\end{array}\right|=ad\!-\!bc\) is valid by Definition 4.1.1.

Example 4.1.3. Determinant of a \(\boldsymbol{2\times2}\) matrix.

For \(\A=\lmatrix{rr}7\amp 2\\3\amp 4\rmatrix\) we have

\begin{equation*} \left|\begin{array}{rr} 7\amp 2\\3\amp 4 \end{array}\right|=(7)(4)-(2)(3)=22. \end{equation*}

Theorem 4.1.4. Determinant of a \(\boldsymbol{3\times3}\) matrix.

Let \(\A = \left(\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right)\in \R^{3 \times 3}.\) Then

\begin{align*} \det(\A)=\amp A_{11}(A_{22}A_{33}-A_{23}A_{32})+(-1)A_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\qquad\,\,+A_{13}(A_{21}A_{32}-A_{22}A_{31}). \end{align*}

Proof.

To show Property 1 for \(n=3,\) we have \(I_3 = \lmatrix{ccc} 1 \amp 0 \amp 0 \\ 0 \amp 1 \amp 0 \\ 0 \amp 0 \amp 1\rmatrix.\) Using Theorem 4.1.4 we see

\begin{equation*} \det(I_3)=1(1-0)-0(0-0)+0(0-0)=1. \end{equation*}

To verify Property 2, for \(n=3,\) let

\begin{equation*} \A=\left(\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right) \end{equation*}

and let \(\B\) be the matrix \(\A\) after rows 1 and 2 have been swapped:

\begin{equation*} \B=\left(\begin{array}{rrr} A_{21} \amp A_{22} \amp A_{23}\\ A_{11} \amp A_{12} \amp A_{13} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right) \end{equation*}

Using Theorem 4.1.4 we find

\begin{align*} \det(\B)\amp =A_{21}(A_{12}A_{33}-A_{13}A_{32})+(-1)A_{22}(A_{11}A_{33}-A_{13}A_{31})\\ \amp \qquad\qquad\qquad+A_{23}(A_{11}A_{32}-A_{12}A_{31})\\ \amp = -A_{11}(A_{22}A_{33}-A_{23}A_{32})+A_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp \qquad\qquad\qquad -A_{13}(A_{21}A_{32}-A_{22}A_{31})\\ \amp =-\det(\A). \end{align*}

The result for swapping any other two rows follows similarly.

Finally, to show Property 3 for \(n=3\) let \((B_{11}\,\,B_{12} \,\, B_{13})\) be a row vector and let \(r\in\R.\) We form two matrices, for the purpose of proving separately each of the effects of linearity:

\begin{align*} \amp \lmatrix{ccc} A_{11}+B_{11} \amp A_{12}+B_{12} \amp A_{13}+B_{13} \\ A_{21} \amp A_{22} \amp A_{23}\\ A_{31} \amp A_{32} \amp A_{33} \rmatrix\\ \amp \text{where the first row is the sum } (A_{11}\,\,A_{12}\,\, A_{13})+ (B_{11}\,\,B_{12} \,\, B_{13}),\text{ and}\\ { }^{ }\\ \amp \left(\begin{array}{rrr} rA_{11} \amp rA_{12} \amp rA_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right)\\ \amp \text{where the first row is } r(A_{11}\,\,A_{12} \,\, A_{13}). \end{align*}

To prove Property 3 we must show

\begin{equation*} \left|\begin{array}{ccc} A_{11}\!+\!B_{11}\! \amp \!A_{12}\!+\!B_{12}\! \amp \!A_{13}\!+\!B_{13} \\ A_{21} \amp A_{22} \amp A_{23}\\ A_{31} \amp A_{32} \amp A_{33} \end{array}\right|=\left|\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right| \!+\!\left|\begin{array}{ccc} B_{11} \amp B_{12} \amp B_{13} \\ A_{21} \amp A_{22} \amp A_{23}\\ A_{31} \amp A_{32} \amp A_{33} \end{array}\right| \end{equation*}

and

\begin{equation*} \left|\begin{array}{rrr} rA_{11} \amp rA_{12} \amp rA_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right|=r\cdot\left|\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right| \end{equation*}

we would also need to demonstrate these two effects when the second or third row is so altered; each the following proofs can be modified accordingly.

Proof of (a). We have

\begin{align*} \amp \left|\begin{array}{ccc} A_{11}\!+\!B_{11}\! \amp \!A_{12}\!+\!B_{12}\! \amp \!A_{13}\!+\!B_{13} \\ A_{21} \amp A_{22} \amp A_{23}\\ A_{31} \amp A_{32} \amp A_{33} \end{array}\right|\\ { }^{ }\\ \amp\qquad=(A_{11}\!+\!B_{11})(A_{22}A_{33}-A_{23}A_{32})\\ \amp\qquad\qquad +(-1)(A_{12}\!+\!B_{12})(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\qquad +(A_{13}\!+\!B_{13})(A_{21}A_{32}-A_{22}A_{31})\\ { }^{ }\\ \amp\qquad=A_{11}(A_{22}A_{33}-A_{23}A_{32})+(-1)A_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\quad +A_{13}(A_{21}A_{32}-A_{22}A_{31})\\ \amp\qquad\quad +B_{11}(A_{22}A_{33}-A_{23}A_{32})+(-1)B_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\quad +B_{13}(A_{21}A_{32}-A_{22}A_{31})\\ { }^{ }\\ \amp \qquad=\left|\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right|+\left|\begin{array}{ccc} B_{11} \amp B_{12} \amp B_{13} \\ A_{21} \amp A_{22} \amp A_{23}\\ A_{31} \amp A_{32} \amp A_{33} \end{array}\right|. \end{align*}

Proof of (b). Now

\begin{align*} \amp \left|\begin{array}{rrr} rA_{11} \amp rA_{12} \amp rA_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right|\\ { }^{ }\\ \amp\qquad = rA_{11}(A_{22}A_{33}-A_{23}A_{32})+(-1)rA_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\qquad +rA_{13}(A_{21}A_{32}-A_{22}A_{31})\\ { }^{ }\\ \amp \qquad =r(A_{11}(A_{22}A_{33}-A_{23}A_{32})+(-1)A_{12}(A_{21}A_{33}-A_{23}A_{31})\\ \amp\qquad\qquad +A_{13}(A_{21}A_{32}-A_{22}A_{31}))\\ { }^{ }\\ \amp \qquad =r\cdot\left|\begin{array}{rrr} A_{11}\amp A_{12}\amp A_{13}\\ A_{21}\amp A_{22}\amp A_{23}\\ A_{31}\amp A_{32}\amp A_{33}\end{array}\right|. \end{align*}

Example 4.1.5. Determinant of a \(\boldsymbol{3\times3}\) matrix.

Let

\begin{equation*} \A=\left(\begin{array}{rrr} 1 \amp 2 \amp 3 \\ -2 \amp 2 \amp 0 \\ 4 \amp -1 \amp 2 \end{array}\right) \end{equation*}

then the determinant of \(\A\) is

\begin{align*} \left|\begin{array}{rrr} \boldsymbol{1} \amp \boldsymbol{2} \amp \boldsymbol{3} \\ -2 \amp 2 \amp 0 \\ 4 \amp -1 \amp 2\end{array}\right|=\amp\boldsymbol{1}\cdot[(2)(2) - (0)(-1)]+(-1)\cdot \boldsymbol{2}[(-2) (2)-(0)(4)]\\ \amp \qquad\qquad\,\,+\boldsymbol{3}\cdot[(-2)(-1)-(2)(4)]\\ =\amp-6. \end{align*}

Example 4.1.6. Demonstration of the determinant properties for a \(\boldsymbol{3\times3}\) matrix.

Suppose \(\A=\lmatrix{rrr} 1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 0 \amp 1 \amp 4\rmatrix.\)

By Theorem 4.1.4 we get

\begin{align*} \det(\A)=\amp\,1\cdot[(0)(4)-(1)(1)]-(-1)\cdot[(-2)(4)-(1)(0)]\\ \amp\qquad\quad+1\cdot[(-2)(1)-(0)(0)]\\ =\amp-11. \end{align*}
If we switch rows 1 and 3 of \(\A\) to get matrix \(\B=\lmatrix{rrr} 0 \amp 1 \amp 4\\ -2 \amp 0 \amp 1\\1 \amp -1 \amp 1 \rmatrix\) then by Definition 4.1.1 part 2 we know \(\det(\B)=-\det(\A)=11.\) We check this directly:

\begin{align*} \det(\B)=\amp0\cdot[(0)(1)-(1)(-1)]-(1)\cdot[(-2)(1)-(1)(1)]\\ \amp\qquad\quad+4\cdot[(-2)(-1)-(0)(1)]\\ =\amp 11. \end{align*}
If we add row vector \((2\,\,2\,\,1)\) to row \(3\) of matrix \(\A\) to get \(\B=\left( \begin{array}{rrr}1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 2 \amp 3 \amp 5\end{array}\right)\) then by Definition 4.1.1

\begin{align*} \det(\B)=\amp \left| \begin{array}{rrr}1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 0+2 \amp 1+2 \amp 4+1\end{array}\right|=\left| \begin{array}{rrr}1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 0 \amp 1 \amp 4\end{array}\right| + \left| \begin{array}{rrr}1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 2 \amp 2 \amp 1\end{array}\right|\\ \amp =\det(\A)+ \left| \begin{array}{rrr}1 \amp -1 \amp 1 \\ -2 \amp 0 \amp 1\\ 2 \amp 2 \amp 1\end{array}\right|\\ \amp =-11 + 1\cdot[(0)(1)-(1)(2)]-(-1)\cdot[(-2)(1)-(1)(2)]\\ \amp\qquad\quad +1\cdot[(-2)(2)-(0)(2)]\\ \amp =-11-10\\ =\amp-21. \end{align*}

Checking directly we get

\begin{align*} \det(\B)=\amp 1\cdot[(0)(5)-(1)(3)]-(-1)\cdot[(-2)(5)-(1)(2)]\\ \amp\qquad\quad +1\cdot[(-2)(3)-(0)(2)]\\ =\amp-21. \end{align*}
If we multiply row one of \(\A\) by 2 and row two of \(\A\) by 3 we get \(\B=\lmatrix{rrr} 2 \amp -2 \amp 2 \\ -6 \amp 0 \amp 3\\ 0 \amp 1 \amp 4\rmatrix\) and by Definition 4.1.1 part 3 (applied twice, once for each row) we get

\begin{equation*} \det(\B)=2\cdot 3 \cdot \det(\A)=-66. \end{equation*}

Checking directly:

\begin{align*} \det(\B)=\amp 2\cdot[(0)(4)-(3)(1)]-(-2)\cdot[(-6)(4)-(3)(0)]\\ \amp\qquad\quad +2\cdot[(-6)(1)-(0)(0)]\\ =\amp-66. \end{align*}

A useful implication of Definition 4.1.1 is next.

Before we give the general formula for the determinant of an \(n \, \times \, n\) matrix let us observe some patterns in the formulas for the \(n=2\) and \(n=3\) cases. Recall

\begin{equation*} \left| \begin{array}{rr} a \amp b \\ c \amp d \end{array} \right| = ad-bc. \end{equation*}

\begin{align*} \left|\begin{array}{rrr} A_{11}\! \amp \! A_{12} \! \amp \! A_{13}\\ A_{21} \! \amp \! A_{22} \! \amp \! A_{23} \\ A_{31} \! \amp \! A_{32} \! \amp \! A_{33} \end{array} \right| = \amp A_{11}(A_{22}A_{33}\!-\!A_{23}A_{32})+(-1)A_{12}(A_{21}A_{33}\!-\!A_{23}A_{31})\\ \amp \qquad\quad\, +A_{13}(A_{21}A_{32}\!-\!A_{22}A_{31}). \end{align*}

Note the determinant of the \(3 \times 3\) matrix is given as a linear combination of the determinants of \(2 \times 2\) matrices that can be found within the larger matrix. Indeed if we let the dash \(-\) stand for a deleted entry the determinant of a \(3\times3\) matrix \(\A\) may be expressed a linear combination of \(2\times2\) determinants, as in for example

\begin{align} \!\!\!\!\!\!\left|\begin{array}{ccc} \boldsymbol{A_{11}} \!\!\amp \!\boldsymbol{A_{12}}\!\! \amp \!\boldsymbol{A_{13}}\\ A_{21}\!\! \amp \! A_{22} \!\! \amp \! A_{23} \\ A_{31} \!\! \amp \! A_{32}\! \amp \! A_{33} \end{array} \right|\! = \amp\boldsymbol{A_{11}}\!\left|\begin{array}{ccc} - \!\amp \! - \!\amp \! -\\ - \! \amp \!\! A_{22}\! \amp \! A_{23} \\ - \! \amp \! \!A_{32} \! \amp \! A_{33} \end{array} \right|\!+(\!-1)\boldsymbol{A_{12}}\!\left|\begin{array}{ccc}\! - \! \amp \!\! - \!\!\amp \! -\\ A_{21} \!\amp \!\! - \!\!\amp \! A_{23} \\ A_{31} \!\amp \!\! - \!\!\amp \! A_{33} \end{array} \right|\notag\\ \amp\qquad\,\,\,\, + \boldsymbol{A_{13}}\!\left| \begin{array}{ccc} - \!\amp \! - \!\amp \! -\\ A_{21} \!\amp \! A_{22} \!\amp \!- \\ A_{31} \!\amp \! A_{32} \!\amp \! - \end{array}\right|.\tag{4.1.1} \end{align}

Definition 4.1.1 tells us that if we swap two rows of a matrix then the determinant changes by a sign. This means that if we swap any other row of our \(3 \times 3\) matrix with the top row, then the determinant changes by a sign. So for example, if we swap the first and second rows we find

\begin{align*} \amp\left|\begin{array}{rrr} A_{11} \amp A_{12} \amp A_{13}\\ A_{21} \amp A_{22} \amp A_{23} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right|=-\left|\begin{array}{rrr} A_{21} \amp A_{22} \amp A_{23}\\ A_{11} \amp A_{12} \amp A_{13} \\ A_{31} \amp A_{32} \amp A_{33} \end{array} \right|\\ \amp\\ \amp=-\left[ A_{21}(A_{12}A_{33}-A_{13}A_{32})+(-1)A_{22}(A_{11}A_{33}-A_{13}A_{31})\right.\\ \amp\left.\qquad\qquad+A_{23}(A_{11}A_{32}-A_{12}A_{31}) \right]\\ \amp=(-1)\boldsymbol{A_{21}}(A_{12}A_{33}+A_{13}A_{32})+\boldsymbol{A_{22}}(A_{11}A_{33}-A_{13}A_{31})\\ \amp\qquad\qquad+(-1)\boldsymbol{A_{23}}(A_{11}A_{32}-A_{12}A_{31}) \end{align*}

This means an alternative way to picture the computation is as a linear combination of a different set of determinants:

\begin{align} \!\!\!\!\!\!\left|\begin{array}{ccc} A_{11}\! \amp \!\! A_{12}\!\! \amp \! A_{13}\\ \boldsymbol{A_{21}} \!\amp \!\!\boldsymbol{A_{22}}\!\! \amp \! \boldsymbol{A_{23}} \\ A_{31} \!\amp \!\!A_{32}\!\! \amp \! A_{33} \end{array} \right|\! = \amp(-1)\boldsymbol{A_{21}}\!\left|\begin{array}{ccc} -\! \amp \!\!A_{12}\!\! \amp \! A_{13}\\ -\! \amp \!\!-\!\! \amp \! - \\ -\! \amp \!\!A_{32}\!\! \amp \! A_{33} \end{array} \right| + \boldsymbol{A_{22}}\!\left|\begin{array}{ccc} A_{11}\! \amp \!\!-\!\! \amp \! A_{13}\\ -\! \amp \!\!-\!\! \amp \! - \\ A_{31}\! \amp \!\!-\!\! \amp \! A_{33} \end{array} \right|\notag\\ \amp\qquad\,\,\,\, + (-1)\boldsymbol{A_{23}}\!\left|\begin{array}{ccc} A_{11}\! \amp \! A_{12}\! \amp \! -\\ -\! \amp \!-\! \amp \! - \\ A_{31}\! \amp \! A_{32}\! \amp \!- \end{array} \right|.\tag{4.1.2} \end{align}

In each of the terms in (4.1.2) the \(2\times 2\) matrix formed by deleting the \(i^{\text{th}}\) row and \(j^{\text{th}}\) column of the \(3 \times 3\) plays a crucial role; these matrices are defined below.

Definition 4.1.7.

Let \(\A\in\R^{n \times n}\) and let \(1\le i,j\le n.\) The \((n-1)\times (n-1)\) matrix formed by deleting the \(i^{\text{th}}\) row and \(j^{\text{th}}\) column of \(\A\) is called the \(ij^{\text{th}}\,\) minor of \(\A\) and denoted by \(\boldsymbol{M_{ij}}.\)

Figure 4.1.8. \(\boldsymbol{M_{ij}}:\) the minor matrix formed by eliminaing row i and column j of the matrix.

This process is called Laplacian expansion by minors (or just expansion) and is valid for square matrices of any size. In the above example line (4.1.1) we have expanded the determinant along row \(1\) and in example line (4.1.2) we have expanded the determinant along row \(2.\) We similarly could have put row \(3\) in position \(1\) and adjusting for signs we’d get a similar formula.

We are now ready to give our general formula for a determinant of an \(n\, \times \, n\) matrix \(\A \in \R^{n \times n}.\) The formula is found recursively by taking linear combinations of the determinants of the \((n-1) \, \times \, (n-1)\) minors of the matrix. Similar to the \(3 \, \times \, 3\) case we can expand by any row that we like. This is reflected in the following formula.

Theorem 4.1.9. Determinant expansion by rows.

Let \(\A \in \R^{n \times n}.\) Then \(\det(\A)\) is given by

\begin{equation} \left|\A\right|=\sum_{j=1}^n (-1)^{i+j}A_{ij}|\boldsymbol{M_{ij}}|.\tag{4.1.3} \end{equation}

Proof.

The proof is quite tedious and is omitted.

By Theorem 4.4.11 \(\,\det(\A)=\det(\A^T)\) so instead of expanding by rows we could expand by a column of \(\A\) if that is more convenient. The formula is modified in the obvious way; the expansion about column \(j\) is given in the next theorem.

Theorem 4.1.10. Determinant expansion by columns.

Let \(\A\in\R^{n\times n}.\) Then \(\det(\A)\) is given by

\begin{equation} \left|\A\right|=\sum_{i=1}^n (-1)^{i+j}A_{ij}|\boldsymbol{M_{ij}}|.\tag{4.1.4} \end{equation}

Proof.

The proof is almost immediate by Theorem 4.1.9, once we have proven that for any \(\A\in\R^{n\times n},\,\det(\A^T)=\det(\A).\) We do this in the next section.

The signs of the coefficients in the linear combinations alternate. It can be useful to remember how to place the signs by visualizing the matrix of signs in an \(n\, \times\, n\) matrix starting with a \(+\) in the upper left and alternating. For example in the \(3\times3\) case:

\begin{equation*} \left( \begin{array}{ccc} + \amp - \amp + \\ - \amp + \amp - \\ + \amp - \amp + \end{array}\right) \end{equation*}

The benefit of being able to expand by any row or column we choose is that we can choose the row or column that most simplifies our calculation. For example for \(\A=\left(\begin{array}{rrr}-5 \amp {\bf 0} \amp 2 \\ 2 \amp {\bf 2} \amp 9\\ 5 \amp {\bf 0} \amp 11 \end{array}\right)\) we would choose to expand by column \(2\) since there are two zeros:

\begin{equation*} \begin{array}{rl} \det(\A)\amp =-(\boldsymbol{0})|\boldsymbol{M_{12}}|+(\boldsymbol{2})|\boldsymbol{M_{22}}|-(\boldsymbol{0})|\boldsymbol{M_{32}}|\\ \amp = 2[(-5)(11)-(2)(5)]=-130 \end{array} \end{equation*}

Example 4.1.11.

Let \(\displaystyle{\A=\left(\begin{array}{rrrr} 1 \amp 2\amp 3 \amp -1 \\0 \amp 4 \amp -1 \amp 0 \\ 3 \amp 2 \amp 1 \amp 4\\ -1 \amp 1 \amp -1 \amp -3 \end{array} \right)}.\) To find the determinant we will expand about row \(2\) since there are two zeros:

\begin{equation*} \det(\A)=(-1)^{2+1} \cdot 0 \left| \begin{array}{rrr} 2 \amp 3 \amp -1 \\ 2 \amp 1 \amp 4 \\ 1 \amp -1 \amp -3 \end{array} \right| + (-1)^{2+2}\cdot 4 \left| \begin{array}{rrr} 1 \amp 3 \amp -1 \\ 3 \amp 1 \amp 4 \\ -1 \amp -1 \amp -3 \end{array} \right| \end{equation*}

\begin{equation*} + (-1)^{2+3}\cdot (-1) \left|\begin{array}{rrr} 1 \amp 2 \amp -1 \\ 3 \amp 2 \amp 4 \\ -1 \amp 1 \amp -3 \end{array} \right| + (-1)^{2+4} \cdot 0 \left|\begin{array}{rrr} 1 \amp 2 \amp 3 \\ 3 \amp 2 \amp 1 \\ -1 \amp 1 \amp -1 \end{array} \right| \end{equation*}

(Now we apply the formula again to find the \(3 \times 3\) determinants, this time expanding each along row \(1\))

\begin{align*} =\amp\,0 +4 \left[ (-1)^{1+1}\cdot 1\left| \begin{array}{rr} 1\amp 4 \\ -1 \amp -3 \end{array} \right| +(-1)^{1+2}\cdot 3 \left| \begin{array}{rr} 3 \amp 4 \\ -1 \amp -3 \end{array} \right|\right.\\ \amp\qquad\qquad + \left.(-1)^{1+3}\cdot (-1) \left| \begin{array}{rr} 3 \amp 1 \\ -1 \amp -1 \end{array} \right|\,\,\right]\\ \amp+1\left[(-1)^{1+1}\cdot 1\left| \begin{array}{rr} 2\amp 4 \\ 1 \amp -3 \end{array} \right| +(-1)^{1+2}\cdot 2\left| \begin{array}{rr} 3 \amp 4 \\ -1 \amp -3 \end{array} \right|\right.\\ \amp\qquad\qquad \left. + (-1)^{1+3}\cdot (-1) \left| \begin{array}{rr} 3 \amp 2 \\ -1 \amp 1 \end{array} \right|\,\,\right]+0 \end{align*}

(Finally we can use Theorem 4.1.2 to find the determinant of each \(2 \times 2\) to complete the computation)

\begin{align*} =\amp\,\,4\Bigl[1\cdot\bigl[(1)(-3)-(-1)(4)\bigr]-3\cdot\bigl[(3)(-3)-(-1)(4)\bigr]\Bigr.\\ \amp\qquad\quad\Bigl. -1\cdot\bigl[(3)(-1)-(-1)(1)\bigr]\,\,\Bigr] \\ \amp \,\,\,+1\Bigl[1\!\cdot\!\bigl[(2)(-3)\!-\!(1)(4)\bigr]-2\!\cdot\!\bigl[(3)(-3)\!-\!(-1)(4)\bigr]\Bigr.\\ \amp \qquad\quad\Bigl.-1\!\cdot\!\bigl[(3)(1)\!-\!(-1)(2)\bigr]\Bigr]\\ \amp=\,67. \end{align*}

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