6  Determinants

In this chapter we will look at determinants. These are useful theoretically, with a long history in multivariable analysis. We will start with the general definition of determinants, which looks rather complex at a first glance. After this we will consider how elementary row operations change the determinant. From this it will be clear that the determinant says something about how a matrix transforms a volume in \mathbb{R}^n. We all have some intuition about volumes in \mathbb{R}^3 and areas in \mathbb{R}^2, but when it comes to volumes in \mathbb{R}^n for n>3, things are different. We shall define this, and see that, if A is an n\times n-matrix and X a set in \mathbb{R}^n, then the set AX=\{A\mathbf{x}:\mathbf{x}\in X\} has a volume, written V(AX), which depends linearly on the volume V(X) of X - regardless how the set X must look. We thus have that V(AX)=kV(X) for a constant k which only depends on the matrix A. k can be interpreted as a magnification factor for volumes for the matrix A, and we shall see that magnification factors change under elementary row operations in the same way as determinants do. From this we deduce that the absolute value of the determinant equals the magnification factor.

One of the things we will show is that the determinant of a product equals the product of the determinants. If we disregard the sign of the determinant, this is perhaps simpler to see when thinking in terms of magnification factors: If A_1 and A_2 have magnification factors k_1 and k_2, respectively, then A_2X has volume k_2V(X), so that A_1A_2X has volume k_1(k_2V(X))=k_1k_2V(X), so that A_1A_2 has magnification factor k_1k_2.

We round off this chapter with a useful geometric interpretation of the determinant.

6.1 The definition of the determinant

The general definition of determinant is as follows

Definition 6.1 (Determinant) The determinant of an n\times n-matrix A is defined recursively as follows:

  1. The determinant of a 1\times 1-matrix A is defined as a_{11}.
  2. The determinant of an n\times n-matrix A is defined in terms of determinants of (n-1)\times(n-1)-submatrices of A as

\begin{aligned} \left| \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nn} \end{array}\right| =& a_{11} \left| \begin{array}{cccc} a_{22} & a_{23} & \cdots & a_{2n} \\ a_{32} & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n2} & a_{n3} & \cdots & a_{nn} \end{array}\right| - a_{12} \left| \begin{array}{cccc} a_{21} & a_{23} & \cdots & a_{2n} \\ a_{31} & a_{33} & \cdots & a_{3n} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n3} & \cdots & a_{nn} \end{array}\right| \\ &+\cdots + (-1)^{n+1} a_{1n} \left| \begin{array}{cccc} a_{21} & a_{22} & \cdots & a_{2(n-1)} \\ a_{31} & a_{32} & \cdots & a_{3(n-1)} \\ \vdots & \vdots & \vdots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{n(n-1)} \end{array}\right| \end{aligned} \tag{6.1} Row 1 of A has been left out in all submatrices on the right side. In the first submatrix column 1 of A has also been left out, in the second column 2 of A has been left out, and so on.

Vertical lines are used as notation for determinants above, but it is also common to write \det(A) for the determinant of A. Let us also write A^{(ij)} for the (n-1)\times(n-1)-matrix we obtain by leaving out row i and column j of A. The quantities \det\left(A^{(ij)}\right) are also called minors. With this notation Equation 6.1 can be written as \det(A)=a_{11}\det(A^{(11)})-a_{12}\det(A^{(12)})+a_{13}\det(A^{(13)})-\cdots + a_{1n}(-1)^{n+1}\det(A^{(1n)}), or even more compactly as \det(A)=\sum_{j=1}^n (-1)^{j+1}a_{1j}\det(A^{(1j)}). The determinant of a 1\times 1-matrix is per definition the one component in the matrix, and for a 2\times 2-matrix it is \det(A)=\left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| = a_{11}a_{22}-a_{12}a_{21}, The determinant of a 3\times 3-matrix is \begin{align*} \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| = a_{11}\left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| -a_{12}\left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| +a_{13}\left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right| \end{align*}

Example 6.1 (A 3\times 3-determinant) Let us compute the determinant of the matrix \begin{pmatrix} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 1 & -1\end{pmatrix} using the definition. We get \begin{align*} \left|\begin{array}{ccc} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 1 & -1\end{array}\right| &= 2 \left|\begin{array}{ccc} 4 & 2 \\ 1 & -1\end{array}\right| - 1 \left|\begin{array}{ccc} 1 & 2 \\ 3 & -1\end{array}\right| +3 \left|\begin{array}{ccc} 1 & 4 \\ 3 & 1 \end{array}\right| \\ &= 2(-4-2) -(-1-6) +3(1-12) = -12 +7 -33 = -38. \end{align*} We can check the answer with the built-in function for computing the determinant:

\clubsuit

Let K(n) be the number of operations needed to compute the determinant recursively using Equation 6.1. It is clear that K(n)=nK(n-1)+n+n-1\geq nK(n-1). We see from this that we must have K(n)\geq n!. When n is large this leads to too much computation, even for the most powerful computers. Laster in this chapter we will see that one can compute the determinant more efficiently using Gaussian elimination. The definition above is more of theoretical interest - it does not give a good implementation even if it is easily programmed, as seen below.

One could guess that Equation 6.1 is not used to compute the determinant in Example 6.1: The answer there comes with four decimals, so the answer is not -38 exactly. Had Equation 6.1 been used, only integer arithmetic would have been needed, and the answer should be -38 exactly. It is straightforward to verify that this is the case:

6.2 Properties of the determinant

Let us prove some useful properties of determinants. We start with the following.

Proposition 6.1 We have that \det(A)=0 if A has a row with zeros only, or a column with zeros only.

Proof. This is simplest to see for rows with zeros only: If we use Equation 6.1 so many times that all sub-matrices start with a row of zeros, it is clear that the next application of Equation 6.1 will give 0. Suppose instead that we have a column of zeros. For 1\times 1-matrices with a column of zeros it is clear that the determinant is 0. Suppose we have shown that all (n-1)\times(n-1)-matrices with a column of zeros have determinant 0. When we use Equation 6.1 on an n\times n-matrix with a column of zeros, we get a contribution from n-1 (n-1)\times(n-1)-sub-matrices with columns of zeros. And the contribution from the remaining column is 0, since we have a zero on the corresponding place in the first row that is multiplied in. It follows that n\times n-matrices with a zero column also have determinant 0.

Recall that a square matrix is called upper triangular if everything below the diagonal is 0, and lower triangular if everything above the diagonal is 0.

Proposition 6.2 The determinant of an upper- or lower triangular matrix equals the product of the diagonal elements.

Proof. The result is obvious for n=1. If A is lower triangular Equation 6.1 says that \det(A)=a_{11}\det(A^{(11)}). Since A^{(11)} also is lower triangular, and with a_{22}, a_{33},…,a_{nn} on the diagonal, it follows by induction that \det(A)=a_{11}(a_{22}a_{33}\cdots a_{nn})=a_{11}a_{22}a_{33}\cdots a_{nn}. Suppose instead that A is upper triangular. The first column in the minors A^{(12)},A^{(13)},...,A^{(1n)} are then all zero, and then Proposition 6.1 says that \det(A^{(12)})=\det(A^{(13)}=\cdots=\det(A^{(1n)})=0. Only the first minor contributes in Equation 6.1, so that \det(A)=a_{11}\det(A^{(11)}), which again by induction gives that \det(A)=a_{11}a_{22}a_{33}\cdots a_{nn}

We have the following result on how elementary row operations change the determinant.

Proposition 6.3 Let A and B be square matrices of equal dimension. The following hold:

  1. If B is obtained from a simple row interchange in A, then \det(B)=-\det(A).
  2. If B is obtained by multiplying a row in A with s, then \det(B)=s\det(A).
  3. If B is obtained by adding s times row i to row j in A, then \det(B)=\det(A).

Proof. We prove 1.. Let D_{ij} be the determinant of the matrix obtained by removing row 1 and 2 in A, as well as column i and j in A. We have that \begin{align*} \det(A)&=\sum_{i<j} ( (-1)^{i+1}(-1)^j a_{1i}a_{2j} + (-1)^{j+1}(-1)^{i+1}a_{1j}a_{2i})D_{ij}\\ &=\sum_{i<j} (-1)^{i+j}(a_{1j}a_{2i}-a_{1i}a_{2j})D_{ij} \end{align*} If B is the matrix where row 1 and 2 in A er swapped, we have that b_{1i}=a_{2i} and b_{1j}=a_{2j} and so on, and we get \begin{align*} \det(B) &= \sum_{i<j} (-1)^{i+j}(b_{1j}b_{2i}-b_{1i}b_{2j})D_{ij} = \sum_{i<j} (-1)^{i+j}(a_{2j}a_{1i}-a_{2i}a_{1j})D_{ij} \\ &= -\sum_{i<j} (-1)^{i+j} (a_{2i}a_{1j}-a_{2j}a_{1i})D_{ij} = -\det(A) \end{align*} If two other neighbouring rows are swapped it follows by induction, and from the definition \det(A)=\sum_{j=1}^n(-1)^{j+1}a_{1j}A_{1j} that the determinant also changes sign. Further, if row i and j not are neighbouring rows, a row interchange of these can be expressed as j-i+(j-i-1)=2(j-i)-1 neighbouring row interchanges. Since this is an odd number the determinant also then will change sign. This proves 1..

The proofs for 2. and 3. are given as exercises.

If A has two equal rows, we end up with the same matrix if these are interchanged. As a consequence of 1. above we must have that \det(A)=-\det(A), so that \det(A)=0.

Corollary 6.1 \det(A)=0 if A has two identical rows.

This result is useful when it comes to proving 3. above. We can also expand the determinant along any given row (This is an exercise):

Proposition 6.4 (Determinants can be expanded along any row) For all 1\leq i\leq n we have that \det(A)=\sum_{j=1}^{n} (-1)^{i+j}a_{ij}\det(A^{(ij)}).

If we in particular set A=I_n (which has determinant 1) and B=E (where E is an elementary matrix) in Proposition 6.3 we get that \det(E)= \begin{cases} -1 \text{ if $E$ is a row interchange.} \\ s \text{ if $E$ multiplies a row with $s$.} \\ 1 \text{ if $E$ adds a multiple of one row to another.} \end{cases} \tag{6.2}

We can now prove the following.

Proposition 6.5 If A and B both are square we have that \det(AB)=\det(A)\det(B).

Proof. we can write \begin{align*} A&=E_1E_2\cdots E_k C & B&=F_1F_2\cdots F_l D \end{align*} where E_1,F_1,E_2,F_2,... are elementary matrices, and C and D are reduced echelon forms. If C=D=I_n then \begin{align*} \det(AB) &= \det(E_1E_2\cdots E_kF_1F_2\cdots F_l) = \det(E_1)\det(E_2)\cdots \det(E_k)\det(F_1F_2\cdots F_l) \\ &= \det(E_1E_2\cdots E_k)\det(F_1F_2\cdots F_l) = \det(A)\det(B). \end{align*} Suppose then that one of C,D is\neq I_n. It follows that one of \det(A) or \det(B) equals 0, so that also \det(A)\det(B)=0. It is therefore enough to show that also \det(AB)=0. It is clear that \det(AB)=0 if and only if \det(CF_1F_2\cdots F_l D)=0. We consider two possibilities:

  1. If C=I_n, D\neq I_n: Then \det(D)=0 (since D has a row of zeros). But then \det(CF_1F_2\cdots F_l D)=\det(F_1F_2\cdots F_l D)=0.
  2. If C\neq I_n then C has a row of zeros. But then also CF_1F_2\cdots F_l D has a row of zeros, and then \det(CF_1F_2\cdots F_l D)=0

In particular we can find the determinant with the help of Gaussian elimination. Let us show this with an example.

Example 6.2 (A 3\times 3-determinant using Gaussian elimination) Let us verify the determinant we found in Example 6.1, using Gaussian elimination. We get \begin{align*} \begin{pmatrix} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 1 & -1\end{pmatrix} &\sim \begin{pmatrix} 1 & 4 & 2 \\ 2 & 1 & 3 \\ 3 & 1 & -1\end{pmatrix} \sim \begin{pmatrix} 1 & 4 & 2 \\ 0 & -7 & -1 \\ 0 & -11 & -7\end{pmatrix} \sim \begin{pmatrix} 1 & 4 & 2 \\ 0 & 1 & 1/7 \\ 0 & -11 & -7\end{pmatrix} \\ &\sim \begin{pmatrix} 1 & 4 & 2 \\ 0 & 1 & 1/7 \\ 0 & 0 & -38/7\end{pmatrix} \end{align*} The first row operation was a row interchange. This is followed by two operations where we add a multiple of row 1 to the other rows. Then we scale row 2 with (-1/7), before adding a multiple of row 2 to row 3. We ended up with an upper triangular matrix, where we know that the determinant is the product of the diagonal elements. We then get \left|\begin{array}{ccc} 1 & 4 & 2 \\ 0 & 1 & 1/7 \\ 0 & 0 & -38/7 \end{array}\right| = 1\times(-1/7)\times(1) \times(-1) \left|\begin{array}{ccc} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 1 & -1\end{array}\right| som gir at (-38/7)=(1/7) \left|\begin{array}{ccc} 2 & 1 & 3 \\ 1 & 4 & 2 \\ 3 & 1 & -1\end{array}\right|, so that the determinant is -38. \clubsuit

For large matrices (which often is the case in applications) it is more efficient to compute determinants by Gaussian elimination as above, when compared with using the definition of the determinant as in Example 6.1: In Section 3.1 we showed that the number of operations needed by Gaussian elimination is about 2n^3/3, while using the definition needed at least n! operations.

Again we can expand the theorem for invertible matrices with an equivalence. This will be the final version of this theorem:

Proposition 6.6 (Invertible matrix theorem) The following are equivalent for an n\times n-matrix A:

  1. A is row equivalent with I_n.
  2. A\mathbf{x}=\mathbf{0} has the unique solution \mathbf{x}=\mathbf{0}.
  3. A\mathbf{x}=\mathbf{b} has a unique solution for all right hand sides \mathbf{b}.
  4. A can be written as a product of elementary matrices.
  5. A is invertible.
  6. The columns of A provide a basis for \mathbb{R}^n.
  7. \det(A)\neq 0

Proof. 4.\Rightarrow 7.: If A=E_k\cdots E_1 then \det(A)=\det(E_k\cdots E_1)=\det(E_k)\cdots\det(E_1)\neq 0, where we used Proposition 6.5, and that elementary matrices have a determinant different from 0.

7.\Rightarrow 1. Suppose that A is not row equivalent with I_n. Then we can write C=E_k\cdots E_1A, and A=F_1\cdots F_kC, where C\neq I_n is the reduced echelon form of A. But then \det(C)=0, and it follows from Proposition 6.5 that \det(A)=0.

Proposition 6.7 For all square matrices A we have that \det(A)=\det(A^T).

Proof. It is clear that for all elementary matrices E, we have that \det(E)=\det(E^T). Suppose A=E_1E_2\cdots E_k C, where C is the reduced echelon form of A. We have that \begin{align*} \det(A^T) &= \det(C^TE_k^TE_{k-1}^T\cdots E_1^T) = \det(C^T) \det(E_k^T)\det(E_{k-1}^T)\cdots \det(E_1^T) \\ &= \det(E_1)\det(E_2)\cdots \det(E_k)\det(C) = \det(E_1E_2\cdots E_kC) = \det(A). \end{align*} We used here that \det(C)=\det(C^T). It is clear that this holds if C=I_n. If C\neq I_n then C has a row of zeros, so that \det(C)=0. But then C^T has a column of zeros, so that also \det(C^T)=0.

From this we can deduce that we also can expand the determinant along any column (this is an exercise).

Proposition 6.8 (Determinants can be expanded along any column) For all 1\leq j\leq n we have that \det(A)=\sum_{i=1}^{n} (-1)^{i+j}a_{ij}\det(A^{(ij)}).

From Proposition 6.4 and Proposition 6.8 we see that the minor \det(A^{(ij)}) should have an alternating sign (-1)^{i+j}, regardless of which row or column we expand along. This can also be indicated with the following sign table: \begin{pmatrix} + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ + & - & + & - & \cdots \\ - & + & - & + & \cdots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}. For practical use we should expand the determinant along the row/column which has the most zeros.

Example 6.3 Let us consider the 4\times 4-matrix A=\begin{pmatrix} 2 & 0 & 2 & 3 \\ 1 & 0 & 0 & 2 \\ 3 & 1 & -1 & 3 \\ -1 & 0 & 2 & 3 \end{pmatrix}. We see quickly that the row/column with most zeros is column 2. Expanding along this we get \begin{align*} \left|\begin{array}{cccc} 2 & 0 & 2 & 3 \\ 1 & 0 & 0 & 2 \\ 3 & 1 & -1 & 3 \\ -1 & 0 & 2 & 3 \end{array}\right| &= (-1)^{3+2} \left|\begin{array}{ccc} 2 & 2 & 3 \\ 1 & 0 & 2 \\ -1 & 2 & 3 \end{array}\right| \\ &= -\left( (-1)^{2+1}\left|\begin{array}{cc} 2 & 3 \\ 2 & 3 \end{array}\right| +(-1)^{2+3}2\left|\begin{array}{cc} 2 & 2 \\ -1 & 2 \end{array}\right| \right) \\ &= -(-(6-6)-2(4+2))=12. \end{align*} Here we also expanded along row 2, where there also was a zero. \clubsuit

6.3 Magnification factors for matrices(*)

Let us now consider the geometrical interpretation of the determinant as a magnification factor. When we now explain what we mean by the volume V(X) of a set X in \mathbb{R}^n for n>3, let us list some properties V(X) intuitively should have:

  1. Unit boxes B=\{\mathbf{x}\in\mathbb{R}^n:0\leq x_i\leq 1\forall i\} should have volume 1.
  2. If X is translated with a vector \mathbf{r}, then the volume should stay the same. In formula: V(X+\mathbf{r})=V(X), where X+\mathbf{r} = \{\mathbf{x}+\mathbf{r}:\mathbf{x}\in X\}.
  3. If X is stretched with a factor r in direction \mathbf{e}_i, the volume should be multiplied with r. In formula: V(X_{i,r})=rV(X), where X_{i,r}=\{\mathbf{x}: (x_1,...,x_{i-1},x_i/r,x_{i+1},...,x_n)\in X\}.
  4. If X_1 and X_2 are disjoint sets then V(X_1\cup X_2)=V(X_1)+V(X_2).
  5. Volume should equal base times height, as we are used to in \mathbb{R}^2 and \mathbb{R}^3. But now we must think of the base as a volume in \mathbb{R}^{n-1}.

If we use 1. 2., and 3. repeatedly on a unit box, the boxes B_{\mathbf{r},\delta}=\{\mathbf{x}\in\mathbb{R}^n:r_i\delta\leq x_i\leq(r_i+1)\delta\forall i\} should have volume \delta^n for all choices of vectors \mathbf{r}. This gives us the partition \mathbb{R}^n=\cup_\mathbf{r} B_{\mathbf{r},\delta}. Most sets in \mathbb{R}^n can be approximates arbitrarily well with the help of enough, and small enough, such boxes (more generally volume should be defined as a limit where \delta\to 0). Let us therefore restrict ourselves to regions which can be written as a finite union of boxes. Due to 4. we must have V(\cup_i B_{\mathbf{r}_i,\delta})=\sum_i \delta^n. If X= \cup_i B_{\mathbf{r}_i,\delta}, and A is invertible, it is clear that AX=A(\cup_i B_{\mathbf{r}_i,\delta}) = \cup_i A(B_{\mathbf{r}_i,\delta}) is a partition of AX into disjoint sets.

Proposition 6.9 Let X=\cup_i B_{\mathbf{r}_i,\delta} be a disjoint union of boxes. Then V(AX)=V(AB)V(X).

This means that V(AB) works as a magnification factor for A for sets X on the form \cup_i B_{\mathbf{r}_i,\delta}. The ide is that, since V(AX)=V(AB)V(X) for such X, and since moste sets can be approximated arbitrarily well with such sets, the also V(AX)=V(AB)V(X) for more general sets X. Thus, V(AB) works as a magnification factor more generally sets as well!

Proof. Since B_{\mathbf{r},\delta}=\delta (B + \mathbf{r}) we have that AB_{\mathbf{r},\delta}=A(\delta (B + \mathbf{r}))=\delta (AB + A\mathbf{r}). From property 2. and 3. it follows that V(AB_{\mathbf{r},\delta})=\delta^n V(AB). If we use property 4. we now get \begin{align*} V(AX)&=V(A(\cup_i B_{\mathbf{r}_i,\delta})) = V(\cup_i AB_{\mathbf{r}_i,\delta}) = \sum_i V(AB_{\mathbf{r}_i,\delta}) \\ &= \sum_i \delta^n V(AB) = V(AB) \sum_i \delta^n = V(AB)V(X). \end{align*}

Since B=\{ \sum_{i=1}^n x_i\mathbf{e}_i : 0\leq x_i\leq 1\forall i \} then AB=\left\{ \sum_{i=1}^n x_iA\mathbf{e}_i: 0\leq x_i\leq 1\forall i\right\} =\left\{ \sum_{i=1}^n x_i\mathbf{a}_i: 0\leq x_i\leq 1\forall i\right\}, where \mathbf{a}_i is column i in A. More generally, if \mathbf{a}_1,...,\mathbf{a}_m are vectors in \mathbb{R}^n, the set F(\mathbf{a}_1,...,\mathbf{a}_m):=\left\{\sum_{i=1}^mx_i\mathbf{a}_i,0\leq x_i\leq 1\forall i\right\} is called the parallelepiped spanned by \{\mathbf{a}_i\}_{i=1}^m (parallelograms correspond to setting m=2). Our magnification factors are thus volumes of parallelepipeds. Figure 6.1 shows a parallelepiped spanned by three vectors \mathbf{a}, \mathbf{b}, and \mathbf{c}, in space.

Figure 6.1: A parallelepiped in space.

Since A can be written as E_k\cdots E_1, where the E_i are elementary, we shall look at how volumes are magnified for every such. There are 3 different types of elementary matrices we will look at.

Proposition 6.10 The magnification factor of an elementary matrix E is given by V(EB)=\begin{cases} 1 \text{ if $E$ is a row interchange.} \\ |s| \text{ if $E$ multiplies a row with $s$.} \\ 1 \text{ if $E$ adds a multiple of one row to another.}\end{cases}

Proof. We go through the three possibilities.

  1. Suppose E swaps row i and j. Then EB=B, so that V(EB)=1.
  2. If E multiplies row i with s then EB=\left\{ \mathbf{x}: 0\leq x_i\leq s, 0\leq x_j\leq 1,j\neq i\right\}, so that V(EB)=s due to point 3..
  3. Suppose E adds c times row n to row n-1 (other choices of rows are handled similarly), i.e., E(x_1,...,x_{n-1},x_n)=(x_1,...,x_{n-1}+cx_n,x_n). For the cross section where x_n=k we get E(x_1,...,x_{n-1},k)=(x_1,...,x_{n-1},k)+(0,...,ck,0), so that the cross section where x_n=k is translated by E with a constant vector (0,...,ck,0), so that all cross sections of EB on the form x_n=k have the same area as B. This means that both base and height (x_n=1) are equal for B and EB. It follows then from point 5. that V(EB) is 1.

Comparing with Equation 6.2 we now see that |\det(E)|=V(EB) for all elementary matrices E, and from Proposition 6.3 applied several times it follows that \det(E_k\cdots E_1) = \det(E_k)\cdots \det(E_1). Since also V(E_k\cdots E_1B) = V(E_kB)\cdots V(E_1B) it follows from |\det(E_i)|=V(E_iB) that |\det(A)| =V(AB), as long as A can be written as a product of elementary matrices (i.e., that A is invertible). Vi har dermed vist følgende:

Proposition 6.11 If A is invertible then the magnification factor V(AB) equals |\det(A)|.

For n=2 this says in particular that \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| equals the area of the parallelogram spanned by \mathbf{a}_1 and \mathbf{a}_2. If we set \mathbf{a}_1=|\mathbf{a}_1|(\cos\theta_1,\sin\theta_1), and \mathbf{a}_2=|\mathbf{a}_2|(\cos\theta_2,\sin\theta_2) then \begin{align*} \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| &= \left| \begin{array}{cc} |\mathbf{a}_1|\cos\theta_1 & |\mathbf{a}_1|\sin\theta_1 \\ |\mathbf{a}_2|\cos\theta_2 & |\mathbf{a}_2|\sin\theta_2 \end{array} \right| = |\mathbf{a}_1||\mathbf{a}_2|( \cos\theta_1\sin\theta_2 - \sin\theta_1\cos\theta_2) \\ &= |\mathbf{a}_1||\mathbf{a}_2| \sin(\theta_2-\theta_1) \end{align*} This was also deduced in Exercise 1.9.

The magnification vector A=E_k\cdots E_1 is now obtained by multiplying the magnification factors from the elementary matrices.

6.4 Geometrical interpretation of the determinant. The vector product.

If we define \mathbf{b}=\left( \det(A^{(11)}),-\det(A^{(12)}),\det(A^{(13)}),...,(-1)^{n+1}\det(A^{(1n)}) \right), \tag{6.3} then the definition of the determinant given by Equation 6.1 can be written as the scalar product of \mathbf{a}_1\cdot \mathbf{b} (\mathbf{a}_i is here row i in A). This gives us the following geometrical interpretation.

Proposition 6.12 We have that

  1. \mathbf{b} is orthogonal to all \mathbf{a}_2,...,\mathbf{a}_n,
  2. |\mathbf{b}|=V(F(\mathbf{a}_2,...,\mathbf{a}_n)),
  3. |\mathbf{a}_1\cdot\mathbf{b}|=V(AB)=|\det(A)|.

Shortly we will look at the proof of this, but let us first take a closer look at the case n=3. Then we have that \mathbf{b}=\left( a_{22}a_{33}-a_{23}a_{32}, - (a_{21}a_{33}-a_{23}a_{31}),a_{21}a_{32}-a_{22}a_{31}\right). \tag{6.4} This also has its own name:

Definition 6.2 Equation 6.4 is called the vector product of \mathbf{a}_2 and \mathbf{a}_3. We also write \mathbf{a}_2\times\mathbf{a}_3 for the vector product of \mathbf{a}_2 and \mathbf{a}_3.

It is common to write the vector product as \begin{align*} &\left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| \\ =& (a_{22}a_{33}-a_{23}a_{32})\mathbf{e}_1 - (a_{21}a_{33}-a_{23}a_{31})\mathbf{e}_2 + (a_{21}a_{32}-a_{22}a_{31})\mathbf{e}_3 \\ =& \left(a_{22}a_{33}-a_{23}a_{32},- (a_{21}a_{33}-a_{23}a_{31}),a_{21}a_{32}-a_{22}a_{31}\right), \end{align*} i.e., we have inserted vectors instead of scalars in the first row in the matrix.

In particular \mathbf{a}_2\times\mathbf{a}_3 is orthogonal to \mathbf{a}_2 and \mathbf{a}_3, with length equal to the area of the parallelogram spanned by the two vectors. These two properties do not characterise the vector product completely, since if \mathbf{b} satisfy these two properties, then -\mathbf{b} will as well! It can be shown, however, that \mathbf{b}=\mathbf{a}_2\times\mathbf{a}_3 also satisfies what is called the right hand rule, something -\mathbf{b} not does:

Definition 6.3 Let the right index finger point in the same direction as \mathbf{a}_2, and let the middle finger point in the same direction as \mathbf{a}_3, in such a way that there is less than 180 degrees between the two. The thumb will now point in the direction of the vector product \mathbf{a}_2\times\mathbf{a}_3, perpendicular to both \mathbf{a}_2 and \mathbf{a}_3.

Proposition 6.13 \mathbf{b}=\mathbf{a}_2\times\mathbf{a}_3 is the unique vector orthogonal to both \mathbf{a}_2 and \mathbf{a}_3, with length equal to the area of the parallelogram spanned by \mathbf{a}_2 and \mathbf{a}_3, and with direction given by the right hand rule.

We will not prove that the direction of the vector product abides to the right hand rule. A 3\times 3 determinant can also be written as \det(A) = \mathbf{a}_1\cdot (\mathbf{a}_2\times\mathbf{a}_3), which due to Proposition 6.11 also equals the volume volumet of the parallelepiped spanned by \mathbf{a}_1, \mathbf{a}_2, and \mathbf{a}_3.

Note that \mathbf{x}\times\mathbf{y}=\mathbf{0} if \mathbf{x} and \mathbf{y} are parallel. This can be seen by observing that the parallelogram the two span then has area 0. The length of the vector product is then also 0, and then \mathbf{x}\times\mathbf{y}=\mathbf{0}. One can also argue that the vector product then is a determinant where the second and third rows are equal (after a scaling), which is 0 due to Corollary 6.1.

Example 6.4 Let \mathbf{a}_1=(1,3,2), and \mathbf{a}_2=(2,-1,3). Then we have that \begin{align*} \mathbf{a}_1\times\mathbf{a}_2 &= \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ 1 & 3 & 2 \\ 2 & -1 & 3 \end{array} \right| \\ &= (3\cdot 3-2\cdot(-1))\mathbf{e}_1 - (1\cdot 3 - 2\cdot 2)\mathbf{e}_2 + (1\cdot(-1) - 3\cdot 2)\mathbf{e}_3 \\ &= \left(9+2,-(3-4),-1-6\right) = (11,1,-7). \end{align*} The vector product can also be computed with help of the command cross:

\clubsuit

Example 6.5 In Example 1.7 we found the expression for a plane through a given point, and with a given normal vector. If we instead have three points, we can use the vector product to find a normal vector, and then also an expression for the plane: If the three points are \mathbf{x}_0, \mathbf{x}_1, and \mathbf{x}_2, then clearly \mathbf{x}_1-\mathbf{x}_0 and \mathbf{x}_2-\mathbf{x}_0 will be vectors in this plane, and the vector product of these will then give a normal vector.

To illustrate this, consider the three points \mathbf{a}=(4,-2,-1), \mathbf{b}=(1,0,0), and \mathbf{c}=(2,-1,0). We first get that \mathbf{b}-\mathbf{a}=(-3,2,1), and \mathbf{c}-\mathbf{a}=(-2,1,1). The vector product of these is \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ -3 & 2 & 1 \\ -2 & 1 & 1 \end{array} \right| = (2-1)\mathbf{e}_1 -(-3+2)\mathbf{e}_2 + (-3+4)\mathbf{e}_3 = \mathbf{e}_1+\mathbf{e}_2+\mathbf{e}_3 = (1,1,1). From this it follows that the plane can be written on the form x+y+z=1.

By the properties of the vector product we also have that

  • The area of the parallellogram spanned by \mathbf{a} and \mathbf{b} is the length of the vector \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ 4 & -2 & -1 \\ 1 & 0 & 0 \end{array} \right| =(0,-1,2), which is \sqrt{5}.

  • The volume of the parallellepiped spanned by \mathbf{a}, \mathbf{b}, and \mathbf{c}, is \left| \begin{array}{ccc} 4 & -2 & -1 \\ 1 & 0 & 0 \\ 2 & -1 & 0 \end{array} \right| = -\left| \begin{array}{cc} -2 & -1 \\ -1 & 0 \end{array} \right| = 1. \clubsuit

Proof. (for Proposition 6.12) We first prove point 1.. Define the matrix A_0=\begin{pmatrix} \mathbf{a}_i \\ \mathbf{a}_2 \\ \mathbf{a}_3 \\ \vdots \\ \mathbf{a}_n\end{pmatrix}, where \mathbf{a}_i is inserted in both row 1 and i. Since a matrix with two equal rows has determinant 0 (Corollary 6.1), we have that 0=\det(A_0)=\mathbf{a}_i\cdot\mathbf{b} \text{ for } i=2,...,n, so that \mathbf{b} is orthogonal to all \{\mathbf{a}_i\}_{i=2}^n.

To see that |\mathbf{b}|=V(F(\mathbf{a}_2,...,\mathbf{a}_n)) (punkt 2.), observe first that the projection of \mathbf{a}_1 onto \mathbf{b} is \frac{\mathbf{a}_1\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}, which has length \frac{|\mathbf{a}_1\cdot\mathbf{b}|}{|\mathbf{b}|}. This equals the distance from \mathbf{a}_1 to the plane spanned by \mathbf{a}_2,...,\mathbf{a}_n. Since volume equals base times height we get that V(AB)=V(F(\mathbf{a}_1,...,\mathbf{a}_n)) = \frac{|\mathbf{a}_1\cdot\mathbf{b}|}{|\mathbf{b}|}V(F(\mathbf{a}_2,...,\mathbf{a}_n)). Since |\det(A)|=V(AB) (Proposition 6.11) it follows that |\mathbf{a}_1\cdot\mathbf{b}| = \frac{|\mathbf{a}_1\cdot\mathbf{b}|}{|\mathbf{b}|}V(F(\mathbf{a}_2,...,\mathbf{a}_n)) so that |\mathbf{b}|=V(F(\mathbf{a}_2,...,\mathbf{a}_n)).

Point 3. follows directly from the definition of the determinant given by Equation 6.1, and the connection we have proved between volume and determinant.

The result above is perhaps simplest to interpret on (the obviously true) form |\mathbf{a}_1\cdot \mathbf{b}|=\frac{|\mathbf{a}_1\cdot\mathbf{b}|}{|\mathbf{b}|}|\mathbf{b}|. On the left side here stands the volume of the parallelepiped spanned by \mathbf{a}_1,\mathbf{a}_2,...,\mathbf{a}_n (V(AB)). On the right side \frac{|\mathbf{a}_1\cdot\mathbf{b}|}{|\mathbf{b}|} is the height over the plane spanned by \mathbf{a}_2,\mathbf{a}_3,...,\mathbf{a}_n, while |\mathbf{b}| is the area of the parallelepiped spanned by \mathbf{a}_2,\mathbf{a}_3,...,\mathbf{a}_n (V(F(\mathbf{a}_2,...,\mathbf{a}_n))). In other words, volume equals base times height!

Quiz

--- primary_color: orange secondary_color: lightgray text_color: black shuffle_questions: false shuffle_answers: false --- ## The determinant of a $3\times 3$-matrix The determinant of the matrix $$\begin{pmatrix} 0 & 0 & 3 \\ 0 & 2 & 0 \\ 1 & 0 & 0 \end{pmatrix}$$ is 1. [ ] 6 2. [ ] 0 3. [x] -6 ## The determinant of a reduced echelon form We bring a square matrix to reduced echelon form, and denote the resulting matrix by $C$. Then the determinant of $C$ is > The reduced echelon form of a square matrix is either the identity matrix, or has zero row. 1. [x] either 0 or 1 2. [ ] always 0 > The identity matrix is on reduced echelon form, and has determinant one 3. [ ] always 1 > The reduced echelon form of a square matrix can have a zero row at the end. in which case the determinant must be 0. ## The determinant of a $4\times 4$-matrix We have that $$\left|\begin{array}{cccc} 0 & 0 & 1 & 0 \\ 2 & 3 & 0 & 4 \\ 1 & 0 & 2 & 0 \\ 4 & 1 & 2 & 3\end{array}\right|$$ is 1. [x] -5 2. [ ] 5 3. [ ] 7 ## The determinant of a $4\times 4$-matrix. The determinant of $$\begin{pmatrix} 2 & 3 & -2 & 1 \\ 0 & 1 & 3 & 0 \\ 1 & -1 & 2 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix} $$ is > Expand the determinant along the last column, and then along the last row we get. 1. [ ] $-1$ 2. [x] $1$ 3. [ ] $2$ 4. [ ] $0$ 5. [ ] $3$ ## A vector product The vector product of ${\bf a}=(1,0,2)$ and ${\bf b}=(2,2,1)$ is 1. [x] ${\bf a}\times{\bf b}=(-4,3,2)$ 1. [ ] ${\bf a}\times{\bf b}=(2,3,4)$ 1. [ ] ${\bf a}\times{\bf b}=(-4,-3,-2)$ 1. [ ] ${\bf a}\times{\bf b}=(4,-3,-2)$ 1. [ ] ${\bf a}\times{\bf b}=(-4,-3,2)$ ## Area of parallellogram The area of the parallellogram spanned by the two vectors $(-1,2,3)$ and $(2,1,2)$ 1. [ ] $2$ 1. [x] $3\sqrt{10}$ 1. [ ] $4$ 1. [ ] $11$ 1. [ ] $\sqrt{119}$ ## The image of a square The matrix $$A=\begin{pmatrix} 1 & 2 \\ 2 & 1\end{pmatrix}$$ sends the square consisting of all points so that $0\leq x,y\leq 1$ onto 1. [ ] The square consisting of all points so that $0\leq x\leq 1$, $0\leq y\leq 2$. 1. [x] The parallellogram with vertices $(0,0)$, $(1,2)$, $(2,1)$, and $(3,3)$. 1. [ ] The parallellogram with vertices $(0,0)$, $(1,0)$, $(1,1)$, and $(2,1)$. 1. [ ] The parallellogram with vertices $(1,0)$, $(1,0)$, $(0,2)$, and $(2,1)$. 1. [ ] The parallellogram with vertices $(1,1)$, $(3,1)$, $(2,3)$, and $(3,2)$. ## Finding a plane passing through 3 given points The plane passing through the three points $\mathbf{a}=(2,1,2)$, $\mathbf{b}=(-1,3,1)$, $\mathbf{c}=(3,1,-1)$ is > Compute $\mathbf{b}-\mathbf{a}$ and $\mathbf{c}-\mathbf{a}$, and take the vector product of these 1. [x] $-6x-10y-2z=-26$ 1. [ ] $-6x-10y-2z=-24$ 1. [ ] $-6x-10y-2z=-20$ 1. [ ] $-7x-10y-2z=-22$ 1. [ ] $-7x-11y-2z=-21$ ## The volume of a parallellepiped The volume of the parallellepiped spanned by the three vectors $(1,1,2)$, $(2, 4,3)$, and $(0,2,1)$ is > Compute the $3\times 3$-determinant having the three vectors as rows. 1. [ ] $11$ 1. [ ] $5$ 1. [ ] $3$ 1. [ ] $0$ 1. [x] $4$

Exercises

Exercise 6.1 Compute

  1. \left|\begin{array}{cc} 4 & -1 \\ 2 & 3 \end{array}\right|
  2. \left|\begin{array}{ccc} 2 & 0 & 1 \\ 3 & -1 & 7 \\ 0 & 2 & 5 \end{array}\right|
  3. (2,0,4)\times(-1,2,1).

Exercise 6.2 (Properties of the vector product) Let \mathbf{x}, \mathbf{y}, and \mathbf{z} be vectors in \mathbb{R}^3. Show that

  1. \mathbf{x}\times\mathbf{y}=-\mathbf{y}\times\mathbf{x}.
  2. For all scalars s, \mathbf{x}\times(s\mathbf{y}) = (s\mathbf{x})\times\mathbf{y}=s(\mathbf{x}\times\mathbf{y}).
  3. \mathbf{x}\times(\mathbf{y}+\mathbf{z})=\mathbf{x}\times\mathbf{y}+\mathbf{x}\times\mathbf{z}.

Exercise 6.3 Let \mathbf{x} be a given vector in \mathbb{R}^3. From 2. and 3. in the preceding exercise it follows that the mapping \mathbf{y}\to\mathbf{x}\times\mathbf{y} (which goes from \mathbb{R}^3 to \mathbb{R}^3) is linear. What is the matrix of this linear mapping?

Define the mapping T by T(\mathbf{y})=\mathbf{x}\times\mathbf{y}. We get that

  • T(\mathbf{e}_1) = \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ x_1 & x_2 & x_3 \\ 1 & 0 & 0 \end{array} \right| = \left| \begin{array}{cc} \mathbf{e}_2 & \mathbf{e}_3 \\ x_2 & x_3 \end{array} \right| = x_3\mathbf{e}_2-x_2\mathbf{e}_3=(0,x_3,-x_2)
  • T(\mathbf{e}_2) = \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ x_1 & x_2 & x_3 \\ 0 & 1 & 0 \end{array} \right| = -\left| \begin{array}{cc} \mathbf{e}_1 & \mathbf{e}_3 \\ x_1 & x_3 \end{array} \right| = -(x_3\mathbf{e}_1-x_1\mathbf{e}_3)=(-x_3,0,x_1)
  • T(\mathbf{e}_3) = \left| \begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ x_1 & x_2 & x_3 \\ 0 & 0 & 1 \end{array} \right| = \left| \begin{array}{cc} \mathbf{e}_1 & \mathbf{e}_2 \\ x_1 & x_2 \end{array} \right| = x_2\mathbf{e}_1-x_1\mathbf{e}_2=(x_2,-x_1,0)

The matrix of T is thus \begin{pmatrix} 0 & -x_3 & x_2 \\ x_3 & 0 & -x_1 \\ -x_2 & x_1 & 0 \end{pmatrix}.

Exercise 6.4 We are given the three points \mathbf{a}=(-1,2,-2), \mathbf{b}=(2,-1,4), and \mathbf{c}=(3,3,1).

  1. Find an expression for the plane going through the three points.

We find first \mathbf{b}-\mathbf{a}=(3,-3,6) and \mathbf{c}-\mathbf{a}=(4,1,3). A normal vector to the plane is (\mathbf{b}-\mathbf{a})\times(\mathbf{c}-\mathbf{a}) = \left|\begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ 3 & -3 & 6 \\ 4 & 1 & 3 \end{array} \right| = (-9-6)\mathbf{e}_1 - (9-24)\mathbf{e}_2 + (3-12)\mathbf{e}_3 = (-15,15,15) The plane can thus be written -15x+15y+15z=d for some constant d. Inserting the point \mathbf{a}=(-1,2,-2) gives that d=15+30-30=15, so that the plane can be written as -15x+15y+15z=15, which can be simplified to -x+y+z=1.

  1. Find the area of the parallellogram spanned by \mathbf{a} and \mathbf{b}.

We compute \mathbf{a}\times\mathbf{b} = \left|\begin{array}{ccc} \mathbf{e}_1 & \mathbf{e}_2 & \mathbf{e}_3 \\ -1 & 2 & -2 \\ 2 & -1 & 4 \end{array} \right| = = (8-2)\mathbf{e}_1 - (-4+4)\mathbf{e}_2 + (1-4)\mathbf{e}_3 = (6,0,-3). The length of this vector is \sqrt{45}=3\sqrt{5}, which also is the area of the parallellogram.

  1. Find the volume of the parallellepiped spanned by \mathbf{a}, \mathbf{b}, and \mathbf{c}.

We compute \begin{aligned} \left|\begin{array}{ccc} -1 & 2 & -2 \\ 2 & -1 & 4 \\ 3 & 3 & 1 \end{array} \right| &= - \left| \begin{array}{cc} -1 & 4 \\ 3 & 1\end{array}\right| -2 \left| \begin{array}{cc} 2 & 4 \\ 3 & 1\end{array}\right| -2 \left| \begin{array}{cc} 2 & -1 \\ 3 & 3\end{array}\right| \\ &= -(-1-12)-2(2-12)-2(6+3)=13+20-18=15, \end{aligned} which also is the volume of the parallellepiped.

Exercise 6.5 Compute the determinant of
A=\begin{pmatrix} 2 & -1 & 1 \\ 0 & 1 & 4 \\ 1 & -2 & -6\end{pmatrix}. in two different ways: By help of the steps in Gaussian elimination (those were written down in Exercise 4.1), and by expanding the determinant along the first column.

In the steps of Gaussian elimination there is only one row interchange which contributes to changing the determinant, which therefore is -1. If we expand the determinant along the first column we get \left|\begin{array}{ccc} 2 & -1 & 1 \\ 0 & 1 & 4 \\ 1 & -2 & -6\end{array} \right| = 2(-6+8) - 0 + (-4-1) = 4 -5 = -1.

Exercise 6.6 Prove that the determinant of a vandermonde matrix with generators x_0,...,x_n is \prod_{i>j} (x_i-x_j).

Write down all row operations which change the determinant in Example 4.3.

Exercise 6.7 Prove Proposition 6.4.

First apply i-1 row interchanges so that row i occurs first. After this apply Equation 6.1. The sign becomes (-1)^{i+j}.

Exercise 6.8 Prove point 2. and 3. in Proposition 6.3.

Use point 1. of Proposition 6.3, and Corollary 6.1.

We first prove point 2. If the first row is multiplied with s it follows from Equation 6.1 that \begin{align*} \det(B) &= sa_{11}\det(A^{(11)}) + sa_{12}\det(A^{(12)}) + \cdots + sa_{1n}\det(A^{(1n)}) \\ &= s\left(a_{11}\det(A^{(11)}) + a_{12}\det(A^{(12)}) + \cdots + a_{1n}\det(A^{(1n)})\right) = s\det(A). \end{align*} If one of the other rows is multiplied with s, we combine three operations: First a row interchange with row 1, followed by multiplication of row 1 with s, and finally the same row interchange one more time. Proposition 6.3 then gives \begin{align*} \det(B) &= -(s(-\det(A)))= s\det(A). \end{align*} Let us now prove point 3.. Let C be the matrix obtained by swapping row 1 and j in B. The first row in C then has components a_{jk}+sa_{ik}, so that Equation 6.1 gives \begin{align*} -\det(B) &= \det(C) = \sum_{k=1}^n (-1)^{k+1}(a_{jk}+sa_{ik})\det(C_{1k}) \\ &= \sum_{k=1}^n (-1)^{k+1} a_{jk}\det(C_{1k}) + s\sum_{k=1}^n (-1)^{k+1} a_{ik}\det(C_{1k}) \\ &= \sum_{k=1}^n (-1)^{k+1} a_{jk}\det(C_{1k}), \end{align*} where the other sum became zero since row 1 and i are equal (Corollary 6.1). The last expression is the determinant of the matrix obtained by swapping rows 1 and j in A, and this is -\det(A). It follows that \det(A)=\det(B).

Exercise 6.9 Suppose A is invertible. Show that \det(A^{-1})=1/\det(A).

Since 1=\det(I_n)=\det(A^{-1}A)=\det(A^{-1})\det(A), it follows that \det(A^{-1})=1/\det(A).

Exercise 6.10 Suppose U is orthogonal, i.e., U is real, and U^TU=I_n. Show that \det(U) equals 1 or -1.

Exercise 6.11 Prove Proposition 6.8.

We have that \begin{align*} \det(A) = \det(A^T) = \sum_{j=1}^n (-1)^{j+i} a_{ji}\det(A^{(ji)}) = \sum_{i=1}^n (-1)^{i+j} a_{ij}\det(A^{(ij)}). \end{align*} where we expanded the determinant of A^T along row i.

Exercise 6.12 Let A and B be square matrices, and set D=\left(\begin{array}{c|c} A & C\\ \hline \mathbf{0} & B \end{array}\right), where \mathbf{0} and C have dimensions so that D is square. Show that \det(D)=\det(A)\det(B).

Exercise 6.13 (Cramer’s rule) Suppose A is invertible, and that \mathbf{x} solves the system A\mathbf{x}=\mathbf{b}. Let also A_i(\mathbf{b}) be the matrix A with column i replaced with \mathbf{b}.

  1. Show that A \begin{pmatrix} 1 & 0 & \cdots & 0 & x_1 & 0 & \cdots & 0 & 0 \\ 0 & 1 & \cdots & 0 & x_2 & 0 & \cdots & 0 & 0 \\ \vdots & \vdots & \ddots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \cdots & 0 & x_{n-1} & 0 & \cdots & 1 & 0 \\ 0 & 0 & \cdots & 0 & x_n & 0 & \cdots & 0 & 1 \end{pmatrix} = A_i(\mathbf{b}) The vector \mathbf{x} her comes in column i.

  2. Explain that the determinant of the large matrix above is x_i. Deduce from this that x_i=\det(A_i(\mathbf{b}))/\det(A).

As a result of this, the solution to a linear system can be expressed in terms of determinants. This is also called Cramer’s rule.

  1. Let A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} be a general invertible 2\times 2-matrix. In a previous exercise it was proved that A^{-1}=\frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}. Prove this also by using Cramer’s rule.

The inverse matrix of A can be obtained by solving A\mathbf{x}=\mathbf{e}_1, and A\mathbf{x}=\mathbf{e}_2. Cramer’s rule gives that A\mathbf{x}=\mathbf{e}_1 has solution \left( \frac{\left| \begin{array}{cc} 1 & b \\ 0 & d \end{array} \right|}{\left| \begin{array}{cc} a & b \\ c & d \end{array} \right|} , \frac{\left| \begin{array}{cc} a & 1 \\ c & 0 \end{array} \right|}{\left| \begin{array}{cc} a & b \\ c & d \end{array} \right|}\right) = \frac{1}{ad-bc}\begin{pmatrix} d \\ -c \end{pmatrix}, and that A\mathbf{x}=\mathbf{e}_2 has solution \left( \frac{\left| \begin{array}{cc} 0 & b \\ 1 & d \end{array} \right|}{\left| \begin{array}{cc} a & b \\ c & d \end{array} \right|} , \frac{\left| \begin{array}{cc} a & 0 \\ c & 1 \end{array} \right|}{\left| \begin{array}{cc} a & b \\ c & d \end{array} \right|}\right) = \frac{1}{ad-bc}\begin{pmatrix} -b \\ a \end{pmatrix}. The inverse matrix is thus \frac{1}{ad-bc}\begin{pmatrix} d & -b \\ -c & a \end{pmatrix}.

  1. Sometimes we want to restrict ourselves to matrices with integer entries which also have an inverse with integer entries. Explain how Cramer’s rule can be used to produce such matrices.

If A has integer entries and \mathbf{b}=\mathbf{e}_i for some i, we have that \det(A_i\mathbf{b}) also is an integer. if \det(A)=\pm 1 we will then have that A^{-1} has integer entries by Cramer’s rule. Matrices with determinant 1 are easy to construct - we can take a diagonal matrix with ones on the diagonal. And by adding multiples of rows to other rows the determinant will be unchanged at 1.