lecture notes on determinants

1 Introduction.

The determinant operation is an algebraicformula involving addition and multiplication that combines the $n^2$entries of an $n\times n$ matrix of numbers into a single number.The determinant has many useful and surprising properties. Inparticular the determinant “determines” whether or not a matrix issingular. If the determinant is zero, the matrix is singular; if not,the matrix is invertible.

2 Notation.

We can regard an $n\times n$ matrix $A$ as an entity in and of itself,as a collection of $n^2$ numbers arranged in a table, or as a list ofcolumn vectors:

where

$${𝐚}_1=\left[\begin{array}{c}\hfill a_{11}\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill a_{n1}\hfill \end{array}\right]=a_{11}{𝐞}_1+\mathrm{\cdots }{a}_{n1}{𝐞}_n,{𝐚}_2=\left[\begin{array}{c}\hfill a_{12}\hfill \\ \hfill \mathrm{⋮}\hfill \\ \hfill a_{n2}\hfill \end{array}\right]=a_{12}{𝐞}_1+\mathrm{\cdots }{a}_{n2}{𝐞}_n,\text{etc.}$$

Correspondingly, we employ the following notation for the determinant:

3 Defining properties.

The determinant operation obeyscertain key properties. The correct application of these rules allowsus to evaluate the determinant of a given square matrix. For the sakeof simplicity, we describe these rules for the case of the $2\times 2$and $3\times 3$determinants. Determinants of larger sizes obey analogous rules.

1. Multi-linearity. The determinant operation is linearin each of its vector arguments.

   1. The determinant distributes over addition. Thus, for a$3\times 3$ matrix $A$ and a column vector $𝐛\in {ℝ}^3$, we have

      $\left|\begin{array}{c}\hfill {𝐚}_1+𝐛,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+\left|\begin{array}{c}\hfill 𝐛,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

      $\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2+𝐛,{𝐚}_3\hfill \end{array}\right|$$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐚}_1,𝐛,{𝐚}_3\hfill \end{array}\right|$

      $\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3+𝐛\hfill \end{array}\right|$$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,𝐛\hfill \end{array}\right|$

      Thus, if $A,B$ are two $2\times 2$ matrices the determinant of thesum $A+B$ will have a total of $4$ terms (think FOIL):

      $det(A+B)$$=\left|\begin{array}{c}\hfill {𝐚}_1+{𝐛}_1,{𝐚}_2+{𝐛}_2\hfill \end{array}\right|$

      $=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐚}_1,{𝐛}_2\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐛}_1,{𝐚}_2\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐛}_1,{𝐛}_2\hfill \end{array}\right|$

      $=det(A)+\left|\begin{array}{c}\hfill {𝐚}_1,{𝐛}_2\hfill \end{array}\right|+\left|\begin{array}{c}\hfill {𝐛}_1,{𝐚}_2\hfill \end{array}\right|+det(B)$

      Warning: the formula $det(A+B)=det(A)+det(B)$ is most certainlywrong; it has the F and the L terms from FOIL, but is missing theO and the I terms. Remember, the determinant is not linear; it’smulti-linear!

   2. Scaling one column of a matrix, scales the determinant by thesame amount. Thus, for a scalar $k\in ℝ$, we have

      $$\left|\begin{array}{c}\hfill k{𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=k\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=kdet(A).$$

      Similarly,

      $$\left|\begin{array}{c}\hfill {𝐚}_1,k{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,k{𝐚}_3\hfill \end{array}\right|=kdet(A).$$

      Let’s see what happens to the determinant if we scale the entirematrix:

      $det(kA)$$=\left|\begin{array}{c}\hfill k{𝐚}_1,k{𝐚}_2,k{𝐚}_3\hfill \end{array}\right|$

      $=k\left|\begin{array}{c}\hfill {𝐚}_1,k{𝐚}_2,k{𝐚}_3\hfill \end{array}\right|=k^2\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,k{𝐚}_3\hfill \end{array}\right|=k^3\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

      $=k^{3}det(A).$

   3. A matrix with a zero column has a zero determinant. Forexample,

2. Skew-symmetry. A multi-variable function or formula iscalled symmetric if it does not depend on the order of thearguments/variables. For example, ordinary addition andmultiplication are symmetric operations. A multi-variable functionis said to be skew-symmetric if changing the order of any twoarguments changes the sign of the operation. The 3-dimensionalcross-product is an example of a skew-symmetric operation:

   $$𝐮\times 𝐯=-𝐯\times 𝐮,𝐮,𝐯\in {ℝ}^3.$$

   Likewise, the $n\times n$ determinant is a skew-symmetric operation,albeit one with $n$ arguments.

   Thus, for a $2\times 2$ matrix $A$ we have

   $$det(A)=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2\hfill \end{array}\right|=-\left|\begin{array}{c}\hfill {𝐚}_2,{𝐚}_1\hfill \end{array}\right|.$$

   Thereare six possible ways to rearrange the columns of a $3\times 3$matrix. Correspondingly, for a $3\times 3$ matrix $A$ we have

   $det(A)$$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

   $=-\left|\begin{array}{c}\hfill {𝐚}_2,{𝐚}_1,{𝐚}_3\hfill \end{array}\right|=-\left|\begin{array}{c}\hfill {𝐚}_3,{𝐚}_2,{𝐚}_1\hfill \end{array}\right|=-\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_3,{𝐚}_2\hfill \end{array}\right|$

   $=+\left|\begin{array}{c}\hfill {𝐚}_2,{𝐚}_3,{𝐚}_1\hfill \end{array}\right|=+\left|\begin{array}{c}\hfill {𝐚}_3,{𝐚}_1,{𝐚}_2\hfill \end{array}\right|$

   The determinants in the 3rd line are equal to $det(A)$ because, ineach case, the matrices in question differ from $A$ by 2 columnexchanges.

   Skew-symmetry of the determinant operation has an important consequence: amatrix with two identical columns has zero determinant. Consider,for example a $3\times 3$ matrix $A$ with identical first and secondcolumns. By skew-symmetry, if we exchange the first two columns wenegate the value of the determinant:

   $$\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_1,{𝐚}_3\hfill \end{array}\right|=-\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_1,{𝐚}_3\hfill \end{array}\right|.$$

   Therefore, $det(A)$ is equal to its negation. This can only meanthat $det(A)=0$.

3. The identity rule. The determinant of the identitymatrix is equal to $1$. Written out in symbols for the$3\times 3$ case, this means that

   $$det(I_3)=\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2,{𝐞}_3\hfill \end{array}\right|=1.$$

4 Evaluation.

The above properties dictate the rules bywhich we evaluate a determinant. Overall, the process is veryreminiscent of binomial expansion — the algebra that goes intoexpanding an expression like ${(x+y)}^3$. The essential difference inevaluating determinants is that for scalar algebra the order of thevariables is unimportant: $x\cdot y=y\cdot x$. However, when weevaluate determinants, the choice of order can introduce a minus sign,or result in a zero answer, if some of the arguments are repeated.

Let’s see how evaluation works for $2\times 2$ determinants.

$=\left|\begin{array}{c}\hfill a{𝐞}_1+c{𝐞}_2,b{𝐞}_1+d{𝐞}_2\hfill \end{array}\right|$

$=ab\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_1\hfill \end{array}\right|+ad\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2\hfill \end{array}\right|+cb\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_1\hfill \end{array}\right|+cd\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_2\hfill \end{array}\right|$

$=0+ad\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_1\hfill \end{array}\right|-bc\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2\hfill \end{array}\right|+0$

$=(ad-bc)\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2\hfill \end{array}\right|$

$=ad-bc$

In the above evaluation we used skew-symmetry 3 times:

$$\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_1\hfill \end{array}\right|=0,\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_2\hfill \end{array}\right|=0,\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_1\hfill \end{array}\right|=-\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2\hfill \end{array}\right|.$$

At the very end, we also used the identity rule.It is useful to contrast the above manipulations with the expansion of

$$(ax+cy)(bx+dy)=ab{x}^2+(ac+bd)xy+cd{y}^2.$$

The distributive step is the same. The different answers arise becauseordinary multiplication is a symmetric operation.

Next, let’s see how to evaluate a $3\times 3$ determinant.

$=\left|\begin{array}{c}\hfill a_1{𝐞}_1+a_2{𝐞}_2+a_3{𝐞}_3,b_1{𝐞}_1+b_2{𝐞}_2+b_3{𝐞}_3,c_1{𝐞}_1+c_2{𝐞}_2+c_3{𝐞}_3\hfill \end{array}\right|$

The symmetric analogue would be an expansion of the form

$$(a_{1}x+a_{2}y+a_{3}z)(b_{1}x+b_{2}y+b_{3}z)(c_{1}x+c_{2}y+c_{3}z)=a_1{b}_1{c}_1{x}^3+\mathrm{\cdots }$$

In both cases, if we were to expandfully, we would get an expression involving $3\cdot 3\cdot 3=27$terms. However, skew-symmetry tells us that a determinant thatinvolves repeated standard vectors will equal to zero. Thus, from the27 terms we only keep the 6 terms that involve all three standardstandard vectors:

$=a_1{b}_2{c}_3\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2,{𝐞}_3\hfill \end{array}\right|+a_2{b}_3{c}_1\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_3,{𝐞}_1\hfill \end{array}\right|+a_3{b}_1{c}_2\left|\begin{array}{c}\hfill {𝐞}_3,{𝐞}_1,{𝐞}_2\hfill \end{array}\right|+$

$\mathrm{\hspace{1em}}+a_1{b}_3{c}_2\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_3,{𝐞}_2\hfill \end{array}\right|+a_2{b}_1{c}_3\left|\begin{array}{c}\hfill {𝐞}_2,{𝐞}_1,{𝐞}_3\hfill \end{array}\right|+a_3{b}_2{c}_1\left|\begin{array}{c}\hfill {𝐞}_3,{𝐞}_2,{𝐞}_1\hfill \end{array}\right|$

Using skew symmetry once again (one flip gives a minus sign; two flipsgive a plus sign), and the identity rule we obtain

$=a_1{b}_2{c}_3+a_2{b}_3{c}_1+a_3{b}_1{c}_2-a_1{b}_3{c}_2-a_2{b}_1{c}_3-a_3{b}_2{c}_1$

Notice that each of the 6 terms in the above expression involves anentry from all 3 rows and from all 3 columns; it’s a sudoku kind ofthing. The choice of sign depends on the number of column flipsthat takes the corresponding list of standard vectors to the identitymatrix; an even number of flips gives a $+$, an odd number a $-1$.

5 Row operations.

Above, we observed that the terms in the final expression treat rowsand columns on an equal footing.

Theorem 1.

Let $A$ be a square matrix. Then, $\det \mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}\det \mathit{}\mathrm{(}{A}^T\mathrm{)}$.

In other words, transposing a matrix does not alter the value of itsdeterminant. One consequence of the row-column symmetry is thateverything one can say about determinants in terms of columns, one cansay using rows. In particular, we can state some useful facts aboutthe effect of row operations on the value of a determinant.

Again, for the sake of simplicity we focus on the $3\times 3$ case.Let $A$ be a $3\times 3$ matrix, expressed as a column of row vectors.The hats above the symbols are there to remind us that we are dealingwith row rather than column vectors.

${\widehat{𝐚}}_1=\left[\begin{array}{c}\hfill a_{11},a_{12},a_{13}\hfill \end{array}\right],{\widehat{𝐚}}_2=\left[\begin{array}{c}\hfill a_{21},a_{22},a_{23}\hfill \end{array}\right],{\widehat{𝐚}}_3=\left[\begin{array}{c}\hfill a_{31},a_{32},a_{33}\hfill \end{array}\right].$

Below we list the effect of each of the 3 types of elementary rowoperations on the determinant of a matrix, and a provide an explanation.

1. Row replacements do not affect the valueof the determinant. Consider, for example the effect of a rowreplacement operation $R_2\to R_2+k{R}_1$. Here $k$ is a scalar and $E$ is an elementary matrixthat encodes the row operation in question:

2. Row scaling operations scale the determinant by the samefactor. Consider, for example, the operation $R_2\to k{R}_2$:

   $E$

3. A row exchange operation negates the determinant. Consider, forexample,the exchange of rows $1$ and $3$:

   $E$

Above, if we take $A$ to be the identity matrix, we obtain the valuefor the determinant of an elementary matrix. We summarize as follows.

Theorem 2.

Let $E$ be an elementary matrix. Then,

$$det(EA)=det(E)det(A)$$

where

Of course, a sequence of elementary row operations can be used totransform every matrix into reduced echelon form. This is a usefulobservation, because it gives us an alternative method for computingdeterminants; namely, we can compute the determinant of an $n\times n$square matrix $A$ by row reducing $A$ to row-echelon form. Let’ssummarize this process by writing

$E_k\mathrm{\cdots }{E}_{1}A$$=U,$

$det(E_k)\mathrm{\cdots }det(E_1)det(A)$$=det(U),$

where $E_1,\mathrm{\dots },E_k$ is a sequence of elementary row operations, andwhere $U$ is a matrix in reduced echelon form. If $A$ is singular,then the bottom row of $U$ willconsist of zeros, and hence $det(U)=0$. Since the determinant of anelementary matrix is never zero, this implies that $det(A)=0$,also.

If $A$ is invertible, then $U$ is the identity matrix, and hence

$$det(E_k)\mathrm{\cdots }det(E_1)det(A)=1.$$

Since the determinant of an elementary matrix is explicitly known (seeTheorem 2), thisgives us a way of calculating $det(A)$.We are also in a position to prove some important theorems.

Theorem 3.

Let $A$ be a square matrix. Then, $\det \mathit{}\mathrm{(}A\mathrm{)}\mathrm{=}\mathrm{0}$ ifand only if $A$ is singular.

Theorem 4.

Let $A\mathrm{,}B$ be $n\mathrm{\times }n$ matrices. Then,

$$det(AB)=det(A)det(B).$$

Proof. As above, Let $E_1,\mathrm{\dots },E_k$ be the elementarymatrices that row reduce $A$ to reduced echelon form;

$$E_k\mathrm{\cdots }{E}_{1}A=U.$$

Above, we showed that

$$det(UB)=det(E_k\mathrm{\cdots }{E}_{1}AB)=det(E_k)\mathrm{\cdots }det(E_1)det(AB).$$

If $A$ is singular, then the bottom row of $U$ will be zero, and sowill the bottom row of $UB$. Hence, in this case, $det(AB)=0=det(A)det(B)$, because $det(A)=0$, also. Suppose then that $A$ isinvertible. This means that $U$ is the identity matrix, and hence

$$det(B)=det(E_k)\mathrm{\cdots }det(E_1)det(AB).$$

However,

$$det(E_k)\mathrm{\cdots }det(E_1)=1/det(A),$$

and the desired conclusion follows.

Theorem 5.

Let $A$ be an invertible matrix. Then

$$det(A^{-1})=1/det(A).$$

Proof.By the above theorem,

$1$$=det(I)=det(A{A}^{-1})$

$=det(A)det(A^{-1}).$

6 Cofactor expansion.

Cofactor expansion is another method for evaluating determinants. Itorganizes the computation of larger determinants, and can be useful incalculating the determinants of matrices containing zero entries.At this point, we introduce some useful jargon. Given an $n\times n$matrix $A$, we call the $(n-1)\times (n-1)$ matrix obtained by deletingrow $i$ and column $j$ the $ij$ minor of $A$, and denote it by$A_{ij}$. We also set

$$C_{ij}={(-1)}^{i+j}det(A_{ij})$$

and call $C_{ij}$ the $ij$signed cofactor of $A$.We are going to prove the following.

Theorem 6.

Consider a $\mathrm{3}\mathrm{\times }\mathrm{3}$ matrix

The determinant of $A$ can beobtained by means of cofactor expansion along the first column:

or, along the first row:

More generally, the determinant of an $n\mathrm{\times }n$ matrix can beobtained by cofactor expansion along any column $j$:

$$det(A)=a_{1j}{C}_{1j}+a_{2j}{C}_{2j}+\mathrm{\cdots }+a_{nj}{C}_{nj},j=1,2,\mathrm{\dots },n;$$

or, along any row $i$:

$$det(A)=a_{i1}{C}_{i1}+a_{i2}{C}_{i2}+\mathrm{\cdots }+a_{in}{C}_{in},i=1,2,\mathrm{\dots },n.$$

The proof works by writing

$$A=\left[\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right],{𝐚}_1=a_{11}{𝐞}_1+a_{21}{𝐞}_2+a_{31}{𝐞}_3,$$

and then using multi-linearity:

$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=a_{11}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{21}\left|\begin{array}{c}\hfill {𝐞}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{31}\left|\begin{array}{c}\hfill {𝐞}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=a_{11}{C}_{11}+a_{21}{C}_{21}+a_{31}{C}_{31}.$

The intermediate steps, namely

$$\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=C_{11},\left|\begin{array}{c}\hfill {𝐞}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=C_{21},\left|\begin{array}{c}\hfill {𝐞}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=C_{31}.$$

need to be explained.

Theorem 7.

Let $A\mathrm{=}\mathrm{\left[}\begin{array}{c}\hfill {\mathrm{a}}_{\mathrm{1}}\mathrm{,}{\mathrm{a}}_{\mathrm{2}}\mathrm{,}{\mathrm{a}}_{\mathrm{3}}\mathrm{,}\mathrm{\dots }\mathrm{,}{\mathrm{a}}_n\hfill \end{array}\mathrm{\right]}$ be an $n\mathrm{\times }n$ matrix.Then,

$\left|\begin{array}{c}\hfill {𝐞}_i,{𝐚}_2,{𝐚}_3\mathrm{\dots },{𝐚}_n\hfill \end{array}\right|=C_{i1},$

$\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_i,{𝐚}_3,\mathrm{\dots },{𝐚}_n\hfill \end{array}\right|=C_{i2},$

$\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_i,\mathrm{\dots },{𝐚}_n\hfill \end{array}\right|=C_{i3},$

$\mathrm{\hspace{1em}\hspace{1em}}\text{𝑒𝑡𝑐}$

Proof. Expanding $\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$we obtain 9 terms:

$$\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=\left|\begin{array}{c}\hfill {𝐞}_1,a_{12}{𝐞}_1+a_{22}{𝐞}_2+a_{32}{𝐞}_3,a_{13}{𝐞}_1+a_{23}{𝐞}_2+a_{33}{𝐞}_3\hfill \end{array}\right|.$$

However, there can’t be any terms with a double occurrence of ${𝐞}_1$,and so we end up evaluating a $2\times 2$ determinant:

$\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=\left|\begin{array}{c}\hfill {𝐞}_1,a_{22}{𝐞}_2+a_{32}{𝐞}_3,a_{23}{𝐞}_2+a_{33}{𝐞}_3\hfill \end{array}\right|$

$=a_{22}{a}_{33}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2,{𝐞}_3\hfill \end{array}\right|+a_{22}{a}_{23}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_2,{𝐞}_2\hfill \end{array}\right|+a_{32}{a}_{23}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_3,{𝐞}_2\hfill \end{array}\right|+a_{32}{a}_{33}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐞}_3,{𝐞}_3\hfill \end{array}\right|$

$=C_{11}$

Similarly,

This argument generalizes to matrices of arbitrary size.

Next, by way of example, let’s consider the expansion of a $3\times 3$matrix along the 2nd column:

$=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=\left|\begin{array}{c}\hfill {𝐚}_1,a_{12}{𝐞}_1+a_{22}{𝐞}_2+a_{32}{𝐞}_3,{𝐚}_3\hfill \end{array}\right|$

$=a_{12}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_1,{𝐚}_3\hfill \end{array}\right|+a_{22}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_2,{𝐚}_3\hfill \end{array}\right|+a_{32}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_3,{𝐚}_3\hfill \end{array}\right|$

$=a_{12}{C}_{12}+a_{22}{C}_{22}+a_{32}{C}_{32}.$

The above argument generalizes to expansions along any column, andindeed to expansions of a matrix of arbitrary size.Forexample, for a $4\times 4$ matrix we write

$${𝐚}_1=a_{11}{𝐞}_1{a}_{21}{𝐞}_2+a_{31}{𝐞}_3+a_{41}{𝐞}_4$$

and use multi-linearity to obtain

$\mathrm{\hspace{1em}}=a_{11}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3,{𝐚}_4\hfill \end{array}\right|+a_{21}\left|\begin{array}{c}\hfill {𝐞}_2,{𝐚}_2,{𝐚}_3,{𝐚}_4\hfill \end{array}\right|+a_{31}\left|\begin{array}{c}\hfill {𝐞}_3,{𝐚}_2,{𝐚}_3,{𝐚}_4\hfill \end{array}\right|+a_{41}\left|\begin{array}{c}\hfill {𝐞}_4,{𝐚}_2,{𝐚}_3,{𝐚}_4\hfill \end{array}\right|$

$\mathrm{\hspace{1em}}=a_{11}{C}_{11}+a_{21}{C}_{21}+a_{31}{C}_{31}+a_{41}{C}_{41}.$

Working with row vectors, the same argument also establishes thevalidity of cofactor expansion along rows. For example, here is thederivation ofcofactor expansion along the 2nd row of a $3\times 3$ matrix:

$=\left|\begin{array}{c}\hfill {\widehat{𝐚}}_1\hfill \\ \hfill {\widehat{𝐚}}_2\hfill \\ \hfill {\widehat{𝐚}}_3\hfill \end{array}\right|=\left|\begin{array}{c}\hfill {\widehat{𝐚}}_1\hfill \\ \hfill a_{21}{\widehat{𝐞}}_1+a_{22}{\widehat{𝐞}}_2+a_{23}{\widehat{𝐞}}_3\hfill \\ \hfill {\widehat{𝐚}}_3\hfill \end{array}\right|$

$=a_{21}\left|\begin{array}{c}\hfill {\widehat{𝐚}}_1\hfill \\ \hfill {\widehat{𝐞}}_1\hfill \\ \hfill {\widehat{𝐚}}_3\hfill \end{array}\right|+a_{22}\left|\begin{array}{c}\hfill {\widehat{𝐚}}_1\hfill \\ \hfill {\widehat{𝐞}}_2\hfill \\ \hfill {\widehat{𝐚}}_3\hfill \end{array}\right|+a_{23}\left|\begin{array}{c}\hfill {\widehat{𝐚}}_1\hfill \\ \hfill {\widehat{𝐞}}_3\hfill \\ \hfill {\widehat{𝐚}}_3\hfill \end{array}\right|$

$=a_{21}{C}_{21}+a_{22}{C}_{22}+a_{23}{C}_{23}.$

Here ${\widehat{𝐞}}_i={𝐞}_i^T$ denote the elementary row vectors.

Here is a useful theorem about determinants that can be proved usingcofactor expansions.

Theorem 8.

The determinant of an uppertriangular matrix is the product of the diagonal entries.

Consider, for example the determinant of the following $4\times 4$upper triangular matrix; the stars indicate an arbitrary number. Werepeatedly use cofactor expansion along the first column. Themultiple zeros mean that, each time, the cofactor expansion has only one term.

$=abcd$

7 The adjugate matrix.

The cofactors $C_{ij}$ of an $n\times n$ matrix $A$ can be arrangedinto an $n\times n$ matrix, called $\mathrm{adj}A$, the adjugate of $A$. Inthe $3\times 3$ case we define

Note that the entries of $\mathrm{adj}A$ are indexed differently than theentries of $A$. For $A$, the entry in the $i$-th row and $j$-thcolumn is denoted by $a_{ij}$. However the cofactor $C_{ij}$ isplaced in row $j$ and column $i$. Remarkably, the matrix of cofactors$\mathrm{adj}A$ is closely related to the inverse of $A$.

Theorem 9.

Let $A$ be an $n\mathrm{\times }n$ matrix. Then, $A\mathit{}\mathrm{adj}\mathit{}A\mathrm{=}\det \mathit{}\mathrm{(}A\mathrm{)}\mathit{}I$.Furthermore, if $A$ is invertible, then $A^{\mathrm{-}\mathrm{1}}\mathrm{=}\mathrm{(}\mathrm{1}\mathrm{/}\det \mathit{}\mathrm{(}A\mathrm{)}\mathrm{)}\mathit{}\mathrm{adj}\mathit{}A$.

Let’s consider the proof for the $3\times 3$ case. We aimto show that

Writing $A=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$, using the properties of the determinant and Theorem7

$det(A)=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=a_{11}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{21}\left|\begin{array}{c}\hfill {𝐞}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{31}\left|\begin{array}{c}\hfill {𝐞}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=a_{11}{C}_{11}+a_{21}{C}_{21}+a_{31}{C}_{31}.$

$0=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_1,{𝐚}_3\hfill \end{array}\right|$$=a_{11}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_1,{𝐚}_3\hfill \end{array}\right|+a_{21}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_2,{𝐚}_3\hfill \end{array}\right|+a_{31}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_3,{𝐚}_3\hfill \end{array}\right|$

$=a_{11}{C}_{12}+a_{21}{C}_{22}+a_{31}{C}_{32}.$

$0=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_1\hfill \end{array}\right|$$=a_{11}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_1\hfill \end{array}\right|+a_{21}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_2\hfill \end{array}\right|+a_{31}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_3\hfill \end{array}\right|$

$=a_{11}{C}_{13}+a_{21}{C}_{23}+a_{31}{C}_{33}.$

This gives us the first column of the multiplication. To obtain thesecond column, we observe

$0=\left|\begin{array}{c}\hfill {𝐚}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=a_{12}\left|\begin{array}{c}\hfill {𝐞}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{22}\left|\begin{array}{c}\hfill {𝐞}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+a_{32}\left|\begin{array}{c}\hfill {𝐞}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=a_{12}{C}_{11}+a_{22}{C}_{21}+a_{32}{C}_{31}.$

$det(A)=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=a_{12}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_1,{𝐚}_3\hfill \end{array}\right|+a_{22}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_2,{𝐚}_3\hfill \end{array}\right|+a_{32}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐞}_3,{𝐚}_3\hfill \end{array}\right|$

$=a_{12}{C}_{12}+a_{22}{C}_{22}+a_{32}{C}_{32}.$

$0=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_2\hfill \end{array}\right|$$=a_{12}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_1\hfill \end{array}\right|+a_{22}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_2\hfill \end{array}\right|+a_{32}\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐞}_3\hfill \end{array}\right|$

$=a_{12}{C}_{13}+a_{22}{C}_{23}+a_{31}{C}_{33}.$

The values in the 3rd column are established in a similar fashion.

8 Cramer’s rule

We can also use determinants and cofactors to solve a linear system$A𝐱=𝐛$, where $A$ is an invertible, square matrix.Of course, if $A$ is invertible, then a solution exists and isunique; indeed, $𝐱=A^{-1}𝐛$. However, Cramer’s rule allows usto calculate $𝐱$ directly, without first calculating $A^{-1}$ andthen performing a matrix-vector multiplication.

Let’ssee how this works for the case of a $3\times 3$ matrix $A=\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$. Given a $𝐛\in {ℝ}^3$, we are searchingfor the $3$ numbers$x_1,x_2,x_3$ such that

$$𝐛=x_1{𝐚}_1+x_2{𝐚}_2+x_3{𝐚}_3.$$

Substituting this into thefollowing determinant and expanding produces a useful equation:

$\left|\begin{array}{c}\hfill 𝐛,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$$=\left|\begin{array}{c}\hfill x_1{𝐚}_1+x_2{𝐚}_2+x_3{𝐚}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=x_1\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+x_2\left|\begin{array}{c}\hfill {𝐚}_2,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+x_3\left|\begin{array}{c}\hfill {𝐚}_3,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|$

$=x_{1}det(A)+0+0.$

Similarly

$\left|\begin{array}{c}\hfill {𝐚}_1,𝐛,{𝐚}_3\hfill \end{array}\right|=x_1\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_1,{𝐚}_3\hfill \end{array}\right|+x_2\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|+x_3\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_3,{𝐚}_3\hfill \end{array}\right|=x_{2}det(A)$

$\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,𝐛\hfill \end{array}\right|=x_1\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_1\hfill \end{array}\right|+x_2\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_2\hfill \end{array}\right|+x_3\left|\begin{array}{c}\hfill {𝐚}_1,{𝐚}_2,{𝐚}_3\hfill \end{array}\right|=x_{3}det(A)$

Therefore, the desired solution can be obtained as follows:

$$x_1=\frac{|𝐛,{𝐚}_2,{𝐚}_3|}{|{𝐚}_1,{𝐚}_2,{𝐚}_3|}\mathit{\hspace{1em}}{x}_2=\frac{|{𝐚}_1,𝐛,{𝐚}_3|}{|{𝐚}_1,{𝐚}_2,{𝐚}_3|}\mathit{\hspace{1em}}{x}_3=\frac{|{𝐚}_1,{𝐚}_2,𝐛|}{|{𝐚}_1,{𝐚}_2,{𝐚}_3|}$$

We generalize and summarize as follows.

Theorem 10.

Let $A$ be an invertible $n\mathrm{\times }n$ matrix, and $\mathrm{b}\mathrm{\in }{\mathrm{R}}^n$ avector. Then the unique solution to the linear equation $A\mathit{}\mathrm{x}\mathrm{=}\mathrm{b}$is given by

$$x_i=\frac{det{A}_i(𝐛)}{detA},i=1,2,\mathrm{\dots },n,$$

where $A_i\mathit{}\mathrm{(}\mathrm{b}\mathrm{)}$ denotes the matrix obtained by replacing column $i$of $A$ with $\mathrm{b}$.