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4.3 Properties of Determinants

In Theorem 3.1, we saw that performing an elementary row operation on a matrix can be accomplished by multiplying the matrix by an elementary matrix. This result is very useful in studying the effects on the determinant of applying a sequence of elementary row operations. Because the determinant of the $n \times n$ $n \times n$ identity matrix is 1 (see Example 4 in Section 4.2), we can interpret the statements on page 217 as the following facts about the determinants of elementary matrices.

(a) If E is an elementary matrix obtained by interchanging any two rows of I, then $det (E) = - 1$ $det (E) = - 1$ .
(b) If E is an elementary matrix obtained by multiplying some row of I by the nonzero scalar k, then $det (E) = k$ $det (E) = k$ .
(c) If E is an elementary matrix obtained by adding a multiple of some row of I to another row, then $det (E) = 1$ $det (E) = 1$ .

We now apply these facts about determinants of elementary matrices to prove that the determinant is a multiplicative function.

Theorem 4.7.

For any $A, B \in M_{n \times n} (F), det (A B) = det (A) \cdot det (B)$ $A, B \in M_{n \times n} (F), det (A B) = det (A) \cdot det (B)$ .

Proof.

We begin by establishing the result when A is an elementary matrix. If A is an elementary matrix obtained by interchanging two rows of I, then $det (A) = - 1$ $det (A) = - 1$ . But by Theorem 3.1 (p. 149), AB is a matrix obtained by interchanging two rows of B. Hence by Theorem 4.5 (p. 215), $det (A B) = - det (B) = det (A) \cdot det (B)$ $det (A B) = - det (B) = det (A) \cdot det (B)$ . Similar arguments establish the result when A is an elementary matrix of type 2 or type 3. (See Exercise 18.)

If A is an $n \times n$ $n \times n$ matrix with rank less than n, then $det (A) = 0$ $det (A) = 0$ by the corollary to Theorem 4.6 (p. 216). Since $rank (A B) \leq rank (A) < n$ $rank (A B) \leq rank (A) < n$ by Theorem 3.7 (p. 159), we have $det (A B) = 0$ $det (A B) = 0$ . Thus $det (A B) = det (A) \cdot det (B)$ $det (A B) = det (A) \cdot det (B)$ in this case.

On the other hand, if A has rank n, then A is invertible and hence is the product of elementary matrices (Corollary 3 to Theorem 3.6 p. 158), say, $A = E_{m} \dots E_{2} E_{1}$ $A = E_{m} \dots E_{2} E_{1}$ . The first paragraph of this proof shows that

\begin{array}{l} det (A B) = det (E_{m} \dots E_{2} E_{1} B) \\ = det (E_{m}) \cdot det (E_{m - 1} \dots E_{2} E_{1} B) \\ ⋮ \\ = det (E_{m}) \cdot \dots \cdot det (E_{2}) \cdot det (E_{1}) \cdot det (B) \\ = det (E_{m} \dots E_{2} E_{1}) \cdot det (B) \\ = det (A) \cdot det (B) . \end{array}

$\begin{array}{l} det (A B) = det (E_{m} \dots E_{2} E_{1} B) \\ = det (E_{m}) \cdot det (E_{m - 1} \dots E_{2} E_{1} B) \\ ⋮ \\ = det (E_{m}) \cdot \dots \cdot det (E_{2}) \cdot det (E_{1}) \cdot det (B) \\ = det (E_{m} \dots E_{2} E_{1}) \cdot det (B) \\ = det (A) \cdot det (B) . \end{array}$

Corollary.

A matrix $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ is invertible if and only if $det (A) \neq 0$ $det (A) \neq 0$ . Furthermore, if A is invertible, then $det (A^{- 1}) = \frac{1}{det (A)}$ $det (A^{- 1}) = \frac{1}{det (A)}$ .

Proof.

If $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ is not invertible, then the rank of A is less than n. So $det (A) = 0$ $det (A) = 0$ by the corollary to Theorem 4.6 (p. 217). On the other hand, if $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ is invertible, then

det (A) \cdot det (A^{- 1}) = det (A A^{- 1}) = det (I) = 1

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

by Theorem 4.7. Hence $det (A) \neq 0$ $det (A) \neq 0$ and $det (A^{- 1}) = \frac{1}{det (A)}$ $det (A^{- 1}) = \frac{1}{det (A)}$ .

In our discussion of determinants until now, we have used only the rows of a matrix. For example, the recursive definition of a determinant involved cofactor expansion along a row, and the more efficient method developed in Section 4.2 used elementary row operations. Our next result shows that the determinants of A and $A^{t}$ $A^{t}$ are always equal. Since the rows of A are the columns of $A^{t}$ $A^{t}$ this fact enables us to translate any statement about determinants that involves the rows of a matrix into a corresponding statement that involves its columns.

Theorem 4.8.

For any $A \in M_{n \times n} (F), det (A^{t}) = det (A)$ $A \in M_{n \times n} (F), det (A^{t}) = det (A)$ .

Proof.

If A is not invertible, then $rank (A) < n$ $rank (A) < n$ . But $rank (A^{t}) = rank (A)$ $rank (A^{t}) = rank (A)$ by Corollary 2 to Theorem 3.6 (p. 158), and so $A^{t}$ $A^{t}$ is not invertible. Thus $det (A^{t}) = 0 = det (A)$ $det (A^{t}) = 0 = det (A)$ in this case.

On the other hand, if A is invertible, then A is a product of elementary matrices, say $A = E_{m} \dots E_{2} E_{1}$ $A = E_{m} \dots E_{2} E_{1}$ . Since $det (E_{i}) = det (E_{i}^{t})$ $det (E_{i}) = det (E_{i}^{t})$ for every i by Exercise 29 of Section 4.2, by Theorem 4.7 we have

\begin{array}{l} \det (A^{t}) = det (E_{1}^{t} E_{2}^{t} \dots E_{m}^{t}) \\ = det (E_{1}^{t}) \cdot det (E_{2}^{t}) \cdot \dots \cdot det (E_{m}^{t}) \\ = det (E_{1}) \cdot det (E_{2}) \cdot \dots \cdot det (E_{m}) \\ = det (E_{m}) \cdot \dots \cdot det (E_{2}) \cdot det (E_{1}) \\ = det (E_{m} \dots E_{2} E_{1}) \\ = det (A) . \end{array}

$\begin{array}{l} \det (A^{t}) = det (E_{1}^{t} E_{2}^{t} \dots E_{m}^{t}) \\ = det (E_{1}^{t}) \cdot det (E_{2}^{t}) \cdot \dots \cdot det (E_{m}^{t}) \\ = det (E_{1}) \cdot det (E_{2}) \cdot \dots \cdot det (E_{m}) \\ = det (E_{m}) \cdot \dots \cdot det (E_{2}) \cdot det (E_{1}) \\ = det (E_{m} \dots E_{2} E_{1}) \\ = det (A) . \end{array}$

Thus, in either case, $det (A^{t}) = det (A)$ $det (A^{t}) = det (A)$ .

Among the many consequences of Theorem 4.8 are that determinants can be evaluated by cofactor expansion along a column, and that elementary column operations can be used as well as elementary row operations in evaluating a determinant. (The effect on the determinant of performing an elementary column operation is the same as the effect of performing the corresponding elementary row operation.) We conclude our discussion of determinant properties with a well-known result that relates determinants to the solutions of certain types of systems of linear equations.

Theorem 4.9

(Cramer’s Rule). Let $A x = b$ $A x = b$ be the matrix form of a system of n linear equations in n unknowns, where $x = {(x_{1}, x_{2}, \dots, x_{n})}^{t} .$ $x = {(x_{1}, x_{2}, \dots, x_{n})}^{t} .$ . If $det (A) \neq 0$ $det (A) \neq 0$ then this system has a unique solution, and for each k $(k = 1, 2, \dots, n)$ $(k = 1, 2, \dots, n)$ ,

x_{k} = \frac{det (M_{k})}{det (A)},

$x_{k} = \frac{det (M_{k})}{det (A)},$

where $M_{k}$ $M_{k}$ is the $n \times n$ $n \times n$ matrix obtained from A by replacing column k of A by b.

Proof.

If $det (A) \neq 0$ $det (A) \neq 0$ , then the system $A x = b$ $A x = b$ has a unique solution by the corollary to Theorem 4.7 and Theorem 3.10 (p. 173). For each integer k $(1 \leq k \leq n)$ $(1 \leq k \leq n)$ , let $a_{k}$ $a_{k}$ denote the kth column of A and $X_{k}$ $X_{k}$ denote the matrix obtained from the $n \times n$ $n \times n$ identity matrix by replacing column k by x. Then by Theorem 2.13 (p. 91), $A X_{k}$ $A X_{k}$ is the $n \times n$ $n \times n$ matrix whose ith column is

A e_{i} = a_{i} if i \neq k and A x = b if i = k .

$A e_{i} = a_{i} if i \neq k and A x = b if i = k .$

Thus $A X_{k} = M_{k}$ $A X_{k} = M_{k}$ . Evaluating $X_{k}$ $X_{k}$ by cofactor expansion along row k produces

det (X_{k}) = x_{k} \cdot det (I_{n - 1}) = x_{k} .

$det (X_{k}) = x_{k} \cdot det (I_{n - 1}) = x_{k} .$

Hence by Theorem 4.7,

det (M_{k}) = det (A X_{k}) = det (A) \cdot det (X_{k}) = det (A) \cdot x_{k} .

$det (M_{k}) = det (A X_{k}) = det (A) \cdot det (X_{k}) = det (A) \cdot x_{k} .$

Therefore

x_{k} = {[det (A)]}^{- 1} \cdot det (M_{k}) .

$x_{k} = {[det (A)]}^{- 1} \cdot det (M_{k}) .$

Example 1

We illustrate Theorem 4.9 by using Cramer’s rule to solve the following system of linear equations:

\begin{array}{r} x_{1} & + & 2 x_{2} & + & 3 x_{3} & = & 2 \\ x_{1} & + & x_{3} & = & 3 \\ x_{1} & + & x_{2} & - & x_{3} & = & 1. \end{array}

$\begin{array}{r} x_{1} & + & 2 x_{2} & + & 3 x_{3} & = & 2 \\ x_{1} & + & x_{3} & = & 3 \\ x_{1} & + & x_{2} & - & x_{3} & = & 1. \end{array}$

The matrix form of this system of linear equations is $A x = b$ $A x = b$ where

A = (\begin{array}{r} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & - 1 \end{array}) and b = (\begin{array}{r} 2 \\ 3 \\ 1 \end{array}) .

$A = (\begin{array}{r} 1 & 2 & 3 \\ 1 & 0 & 1 \\ 1 & 1 & - 1 \end{array}) and b = (\begin{array}{r} 2 \\ 3 \\ 1 \end{array}) .$

Because $det (A) = 6 \neq 0$ $det (A) = 6 \neq 0$ Cramer’s rule applies. Using the notation of Theorem 4.9, we have

x_{1} = \frac{det (M_{1})}{det (A)} = \frac{det (\begin{array}{r} 2 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 1 & - 1 \end{array})}{det (A)} = \frac{15}{6} = \frac{5}{2},

$x_{1} = \frac{det (M_{1})}{det (A)} = \frac{det (\begin{array}{r} 2 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 1 & - 1 \end{array})}{det (A)} = \frac{15}{6} = \frac{5}{2},$

x_{2} = \frac{det (M_{2})}{det (A)} = \frac{det (\begin{array}{r} 1 & 2 & 3 \\ 1 & 3 & 1 \\ 1 & 1 & - 1 \end{array})}{det (A)} = \frac{- 6}{6} = - 1,

$x_{2} = \frac{det (M_{2})}{det (A)} = \frac{det (\begin{array}{r} 1 & 2 & 3 \\ 1 & 3 & 1 \\ 1 & 1 & - 1 \end{array})}{det (A)} = \frac{- 6}{6} = - 1,$

and

x_{3} = \frac{det (M_{3})}{det (A)} = \frac{det (\begin{array}{r} 1 & 2 & 2 \\ 1 & 0 & 3 \\ 1 & 1 & 1 \end{array})}{det (A)} = \frac{3}{6} = \frac{1}{2} .

$x_{3} = \frac{det (M_{3})}{det (A)} = \frac{det (\begin{array}{r} 1 & 2 & 2 \\ 1 & 0 & 3 \\ 1 & 1 & 1 \end{array})}{det (A)} = \frac{3}{6} = \frac{1}{2} .$

Thus the unique solution to the given system of linear equations is

(x_{1}, x_{2}, x_{3}) = (\frac{5}{2}, - 1, \frac{1}{2}) .

$(x_{1}, x_{2}, x_{3}) = (\frac{5}{2}, - 1, \frac{1}{2}) .$

In applications involving systems of linear equations, we sometimes need to know that there is a solution in which the unknowns are integers. In this situation, Cramer’s rule can be useful because it implies that a system of linear equations with integral coefficients has an integral solution if the determinant of its coefficient matrix is $\pm 1$ $\pm 1$ . On the other hand, Cramer’s rule is not useful for computation because it requires evaluating $n + 1$ $n + 1$ determinants of $n \times n$ $n \times n$ matrices to solve a system of n linear equations in n unknowns. The amount of computation to do this is far greater than that required to solve the system by the method of Gaussian elimination, which was discussed in Section 3.4. Thus Cramer’s rule is primarily of theoretical and aesthetic interest, rather than of computational value.

As in Section 4.1, it is possible to interpret the determinant of a matrix $A \in M_{n \times n} (R)$ $A \in M_{n \times n} (R)$ geometrically. If the rows of A are $a_{1}, a_{2}, \dots, a_{n}$ $a_{1}, a_{2}, \dots, a_{n}$ respectively, then $| det (A) |$ $| det (A) |$ is the n-dimensional volume (the generalization of area in $R^{2}$ $R^{2}$ and volume in $R^{3}$ $R^{3}$ ) of the parallelepiped having the vectors $a_{1}, a_{2}, \dots, a_{n}$ $a_{1}, a_{2}, \dots, a_{n}$ as adjacent sides. (For a proof of a more generalized result, see Jerrold E. Marsden and Michael J. Hoffman, Elementary Classical Analysis, W.H. Freeman and Company, New York, 1993, p. 524.)

Example 2

The volume of the parallelepiped having the vectors $a_{1} = (1, - 2, 1), a_{2} = (1, 0, - 1),$ $a_{1} = (1, - 2, 1), a_{2} = (1, 0, - 1),$ and $a_{3} = (1, 1, 1)$ $a_{3} = (1, 1, 1)$ as adjacent sides is

| det (\begin{array}{r} 1 & - 2 & 1 \\ 1 & 0 & - 1 \\ 1 & 1 & 1 \end{array}) | = 6.

$| det (\begin{array}{r} 1 & - 2 & 1 \\ 1 & 0 & - 1 \\ 1 & 1 & 1 \end{array}) | = 6.$

Note that the object in question is a rectangular parallelepiped (see Figure 4.6) with sides of lengths $\sqrt{6}, \sqrt{2},$ $\sqrt{6}, \sqrt{2},$ and $\sqrt{3} .$ $\sqrt{3} .$ . Hence by the familiar formula for volume, its volume should be $\sqrt{6} \cdot \sqrt{2} \cdot \sqrt{3} = 6,$ $\sqrt{6} \cdot \sqrt{2} \cdot \sqrt{3} = 6,$ as the determinant calculation shows.

A parallelepiped determined by three vectors in R cubed on the x y plane. — Figure 4.6: Parallelepiped determined by three vectors in $R^{3}$ $R^{3}$ .

Figure 4.6: Full Alternative Text

In our earlier discussion of the geometric significance of the determinant formed from the vectors in an ordered basis for $R^{2}$ $R^{2}$ , we also saw that this determinant is positive if and only if the basis induces a right-handed coordinate system. A similar statement is true in $R^{n}$ $R^{n}$ . Specifically, if $γ$ $γ$ is any ordered basis for $R^{n}$ $R^{n}$ and $β$ $β$ is the standard ordered basis for $R^{n}$ $R^{n}$ , then $γ$ $γ$ induces a right-handed coordinate system if and only if $det (Q) > 0$ $det (Q) > 0$ , where Q is the change of coordinate matrix changing $γ$ $γ$ -coordinates into $β$ $β$ -coordinates. Thus, for instance,

γ = {(\begin{array}{r} 1 \\ 1 \\ 0 \end{array}), (\begin{array}{r} 1 \\ - 1 \\ 0 \end{array}), (\begin{array}{r} 0 \\ 0 \\ 1 \end{array})}

$γ = {(\begin{array}{r} 1 \\ 1 \\ 0 \end{array}), (\begin{array}{r} 1 \\ - 1 \\ 0 \end{array}), (\begin{array}{r} 0 \\ 0 \\ 1 \end{array})}$

induces a left-handed coordinate system in $R^{3}$ $R^{3}$ because

det (\begin{array}{r} 1 & 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & 1 \end{array}) = - 2 < 0,

$det (\begin{array}{r} 1 & 1 & 0 \\ 1 & - 1 & 0 \\ 0 & 0 & 1 \end{array}) = - 2 < 0,$

whereas

γ^{'} = {(\begin{array}{r} 1 \\ 2 \\ 0 \end{array}), (\begin{array}{r} - 2 \\ 1 \\ 0 \end{array}), (\begin{array}{r} 0 \\ 0 \\ 1 \end{array})}

$γ^{'} = {(\begin{array}{r} 1 \\ 2 \\ 0 \end{array}), (\begin{array}{r} - 2 \\ 1 \\ 0 \end{array}), (\begin{array}{r} 0 \\ 0 \\ 1 \end{array})}$

induces a right-handed coordinate system in $R^{3}$ $R^{3}$ because

det (\begin{array}{r} 1 & - 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}) = 5 > 0.

$det (\begin{array}{r} 1 & - 2 & 0 \\ 2 & 1 & 0 \\ 0 & 0 & 1 \end{array}) = 5 > 0.$

More generally, if $β$ $β$ and $γ$ $γ$ are two ordered bases for $R^{n}$ $R^{n}$ then the coordinate systems induced by $β$ $β$ and $γ$ $γ$ have the same orientation (either both are right-handed or both are left-handed) if and only if $det (Q) > 0,$ $det (Q) > 0,$ where Q is the change of coordinate matrix changing $γ$ $γ$ -coordinates into $β$ $β$ -coordinates.

Exercises

Label the following statements as true or false.
1. (a) If E is an elementary matrix, then $det (E) = \pm 1$ $det (E) = \pm 1$ .
2. (b) For any $A, B \in M_{n \times n} (F), det (A B) = det (A) \cdot det (B)$ $A, B \in M_{n \times n} (F), det (A B) = det (A) \cdot det (B)$ .
3. (c) A matrix $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ is invertible if and only if $det (M) = 0$ $det (M) = 0$ .
4. (d) A matrix $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ has rank n if and only if $det (M) \neq 0$ $det (M) \neq 0$ .
5. (e) For any $A \in M_{n \times n} (F), det (A^{t}) = - det (A)$ $A \in M_{n \times n} (F), det (A^{t}) = - det (A)$ .
6. (f) The determinant of a square matrix can be evaluated by cofactor expansion along any column.
7. (g) Every system of n linear equations in n unknowns can be solved by Cramer’s rule.
8. (h) Let $A x = b$ $A x = b$ be the matrix form of a system of n linear equations in n unknowns, where $x = {(x_{1}, x_{2}, \dots, x_{n})}^{t}$ $x = {(x_{1}, x_{2}, \dots, x_{n})}^{t}$ If $det (A) \neq 0$ $det (A) \neq 0$ and if $M_{k}$ $M_{k}$ is the $n \times n$ $n \times n$ matrix obtained from A by replacing row k of A by $b^{t}$ $b^{t}$ then the unique solution of $A x = b$ $A x = b$ is
  
  $x_{k} = \frac{det (M_{k})}{det (A)} for k = 1, 2, \dots, n .$ $x_{k} = \frac{det (M_{k})}{det (A)} for k = 1, 2, \dots, n .$

In Exercises 2-7, use Cramer’s rule to solve the given system of linear equations.

$\begin{array}{r} a_{11} x_{1} & + & a_{12} x_{2} & = & b_{1} \\ a_{21} x_{1} & + & a_{22} x_{2} & = & b_{2} \end{array}$ $\begin{array}{r} a_{11} x_{1} & + & a_{12} x_{2} & = & b_{1} \\ a_{21} x_{1} & + & a_{22} x_{2} & = & b_{2} \end{array}$

where $a_{11} a_{22} - a_{12} a_{21} \neq 0$ $a_{11} a_{22} - a_{12} a_{21} \neq 0$
$\begin{array}{r} 2 x_{1} & + & x_{2} & - & 3 x_{3} & = & 5 \\ x_{1} & - & 2 x_{2} & + & x_{3} & = & 10 \\ 3 x_{1} & + & 4 x_{2} & - & 2 x_{3} & = & 0 \end{array}$ $\begin{array}{r} 2 x_{1} & + & x_{2} & - & 3 x_{3} & = & 5 \\ x_{1} & - & 2 x_{2} & + & x_{3} & = & 10 \\ 3 x_{1} & + & 4 x_{2} & - & 2 x_{3} & = & 0 \end{array}$
$\begin{array}{r} 2 x_{1} & + & x_{2} & - & 3 x_{3} & = & 1 \\ x_{1} & - & 2 x_{2} & + & x_{3} & = & 0 \\ 3 x_{1} & + & 4 x_{2} & - & 2 x_{3} & = & - 5 \end{array}$ $\begin{array}{r} 2 x_{1} & + & x_{2} & - & 3 x_{3} & = & 1 \\ x_{1} & - & 2 x_{2} & + & x_{3} & = & 0 \\ 3 x_{1} & + & 4 x_{2} & - & 2 x_{3} & = & - 5 \end{array}$
$\begin{array}{r} x_{1} & - & x_{2} & + & 4 x_{3} & = & - 4 \\ - 8 x_{1} & + & 3 x_{2} & + & x_{3} & = & 8 \\ 2 x_{1} & - & x_{2} & + & x_{3} & = & 0 \end{array}$ $\begin{array}{r} x_{1} & - & x_{2} & + & 4 x_{3} & = & - 4 \\ - 8 x_{1} & + & 3 x_{2} & + & x_{3} & = & 8 \\ 2 x_{1} & - & x_{2} & + & x_{3} & = & 0 \end{array}$
$\begin{array}{r} x_{1} & - & x_{2} & + & 4 x_{3} & = & - 2 \\ - 8 x_{1} & + & 3 x_{2} & + & x_{3} & = & 0 \\ 2 x_{1} & - & x_{2} & + & x_{3} & = & 6 \end{array}$ $\begin{array}{r} x_{1} & - & x_{2} & + & 4 x_{3} & = & - 2 \\ - 8 x_{1} & + & 3 x_{2} & + & x_{3} & = & 0 \\ 2 x_{1} & - & x_{2} & + & x_{3} & = & 6 \end{array}$
$\begin{array}{r} 3 x_{1} & + & x_{2} & + & x_{3} & = & 4 \\ - 2 x_{1} & - & x_{2} & = & 12 \\ x_{1} & + & 2 x_{2} & + & x_{3} & = & - 8 \end{array}$ $\begin{array}{r} 3 x_{1} & + & x_{2} & + & x_{3} & = & 4 \\ - 2 x_{1} & - & x_{2} & = & 12 \\ x_{1} & + & 2 x_{2} & + & x_{3} & = & - 8 \end{array}$
Use Theorem 4.8 to prove a result analogous to Theorem 4.3 (p. 212), but for columns.
Prove that an upper triangular $n \times n$ $n \times n$ matrix is invertible if and only if all its diagonal entries are nonzero.
A matrix $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ is called nilpotent if, for some positive integer k, $M^{k} = O$ $M^{k} = O$ , where O is the $n \times n$ $n \times n$ zero matrix. Prove that if M is nilpotent, then $det (M) = 0$ $det (M) = 0$ .
A matrix $M \in M_{n \times n} (C)$ $M \in M_{n \times n} (C)$ is called skew-symmetric if $M^{t} = - M$ $M^{t} = - M$ . Prove that if M is skew-symmetric and n is odd, then M is not invertible. What happens if n is even?
A matrix $Q \in M_{n \times n} (R)$ $Q \in M_{n \times n} (R)$ is called orthogonal if $Q Q^{t} = I$ $Q Q^{t} = I$ . Prove that if Q is orthogonal, then $det (Q) = \pm 1$ $det (Q) = \pm 1$ .
For $M \in M_{n \times n} (C)$ $M \in M_{n \times n} (C)$ , let $\bar{M}$ $\bar{M}$ be the matrix such that ${(\bar{M})}_{i j} = \bar{M_{i j}}$ ${(\bar{M})}_{i j} = \bar{M_{i j}}$ for all i, j, where $\bar{M_{i j}}$ $\bar{M_{i j}}$ is the complex conjugate of $M_{i j}$ $M_{i j}$ .
1. (a) Prove that $det (\bar{M}) = \bar{det (M)}$ $det (\bar{M}) = \bar{det (M)}$ .
2. (b) A matrix $Q \in M_{n \times n} (C)$ $Q \in M_{n \times n} (C)$ is called unitary if $Q Q^{*} = I$ $Q Q^{*} = I$ , where $Q^{*} = \bar{Q^{t}}$ $Q^{*} = \bar{Q^{t}}$ . Prove that if Q is a unitary matrix, then $| det (Q) | = 1$ $| det (Q) | = 1$ .
Let $β = {u_{1}, u_{2}, \dots, u_{n}}$ $β = {u_{1}, u_{2}, \dots, u_{n}}$ be a subset of $F^{n}$ $F^{n}$ containing n distinct vectors, and let B be the matrix in $M_{n \times n} (F)$ $M_{n \times n} (F)$ having $u_{j}$ $u_{j}$ as column j. Prove that $β$ $β$ is a basis for $F^{n}$ $F^{n}$ if and only if $det (B) \neq 0$ $det (B) \neq 0$ .
^† Prove that if $A, B \in M_{n \times n} (F)$ $A, B \in M_{n \times n} (F)$ are similar, then $det (A) = det (B)$ $det (A) = det (B)$ .
Use determinants to prove that if $A, B \in M_{n \times n} (F)$ $A, B \in M_{n \times n} (F)$ are such that $A B = I$ $A B = I$ , then A is invertible (and hence $B = A^{- 1}$ $B = A^{- 1}$ ).
Let $A, B \in M_{n \times n} (F)$ $A, B \in M_{n \times n} (F)$ be such that $A B = - B A$ $A B = - B A$ . Prove that if n is odd and F is not a field of characteristic two, then A or B is not invertible.
Complete the proof of Theorem 4.7 by showing that if A is an elementary matrix of type 2 or type 3, then $(A B) = det (A) \cdot det (B)$ $(A B) = det (A) \cdot det (B)$ .
A matrix $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ is called lower triangular if $A_{i j} = 0$ $A_{i j} = 0$ for $1 \leq i < j \leq n$ $1 \leq i < j \leq n$ . Suppose that A is a lower triangular matrix. Describe det(A) in terms of the entries of A.
Suppose that $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ can be written in the form

$M = (\begin{array}{r} A & B \\ O & I \end{array}),$ $M = (\begin{array}{r} A & B \\ O & I \end{array}),$

where A is a square matrix. Prove that $det (M) = det (A)$ $det (M) = det (A)$ .
^† Prove that if $M \in M_{n \times n} (F)$ $M \in M_{n \times n} (F)$ can be written in the form

$M = (\begin{array}{r} A & B \\ O & C \end{array}),$ $M = (\begin{array}{r} A & B \\ O & C \end{array}),$

where A and C are square matrices, then $det (M) = det (A) \cdot det (C)$ $det (M) = det (A) \cdot det (C)$ . Visit goo.gl/4sG3iv for a solution.
Let T: $P_{n} (F) \to F^{n + 1}$ $P_{n} (F) \to F^{n + 1}$ be the linear transformation defined in Exercise 22 of Section 2.4 by $T (f) = (f (c_{0}), f (c_{1}), \dots, f (c_{n})),$ $T (f) = (f (c_{0}), f (c_{1}), \dots, f (c_{n})),$ where $c_{0}, c_{1}, \dots, c_{n}$ $c_{0}, c_{1}, \dots, c_{n}$ are distinct scalars in an infinite field F. Let $β$ $β$ be the standard ordered basis for $P_{n} (F)$ $P_{n} (F)$ and $γ$ $γ$ be the standard ordered basis for $F^{n + 1}$ $F^{n + 1}$ .
1. (a) Show that $M = {[T]}_{β}^{γ}$ $M = {[T]}_{β}^{γ}$ has the form
  
  $(\begin{array}{r} 1 & c_{0} & c_{0}^{2} & \dots & c_{0}^{n} \\ 1 & c_{1} & c_{1}^{2} & \dots & c_{1}^{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & c_{n} & c_{n}^{2} & \dots & c_{n}^{n} \end{array}) .$ $(\begin{array}{r} 1 & c_{0} & c_{0}^{2} & \dots & c_{0}^{n} \\ 1 & c_{1} & c_{1}^{2} & \dots & c_{1}^{n} \\ ⋮ & ⋮ & ⋮ & ⋮ \\ 1 & c_{n} & c_{n}^{2} & \dots & c_{n}^{n} \end{array}) .$
  
  A matrix with this form is called a Vandermonde matrix.
2. (b) Use Exercise 22 of Section 2.4 to prove that $det (M) \neq 0$ $det (M) \neq 0$ .
3. (c) Prove that
  
  $det (M) = \prod_{0 \leq i < j \leq n} (c_{j} - c_{i}),$ $det (M) = \prod_{0 \leq i < j \leq n} (c_{j} - c_{i}),$
  
  the product of all terms of the form $c_{j} - c_{i}$ $c_{j} - c_{i}$ for $0 \leq i \leq j \leq n$ $0 \leq i \leq j \leq n$ .
Let $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ be nonzero. For any $m (1 \leq m \leq n),$ $m (1 \leq m \leq n),$ an $m \times m$ $m \times m$ submatrix is obtained by deleting any $n - m$ $n - m$ rows and any $n - m$ $n - m$ columns of A.
1. (a) Let $k (1 \leq k \leq n)$ $k (1 \leq k \leq n)$ denote the largest integer such that some $k \times k$ $k \times k$ submatrix has a nonzero determinant. Prove that $rank (A) = k$ $rank (A) = k$ .
2. (b) Conversely, suppose that $rank (A) = k$ $rank (A) = k$ . Prove that there exists a $k \times k$ $k \times k$ submatrix with a nonzero determinant.
Let $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ have the form

$A = (\begin{array}{c} 0 & 0 & 0 & \dots & 0 & a_{0} \\ - 1 & 0 & 0 & \dots & 0 & a_{1} \\ 0 & - 1 & 0 & \dots & 0 & a_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & - 1 & a_{n - 1} \end{array}) .$ $A = (\begin{array}{c} 0 & 0 & 0 & \dots & 0 & a_{0} \\ - 1 & 0 & 0 & \dots & 0 & a_{1} \\ 0 & - 1 & 0 & \dots & 0 & a_{2} \\ ⋮ & ⋮ & ⋮ & ⋮ & ⋮ \\ 0 & 0 & 0 & \dots & - 1 & a_{n - 1} \end{array}) .$

Compute $det (A + t I)$ $det (A + t I)$ where I is the $n \times n$ $n \times n$ identity matrix.
Let $c_{j k}$ $c_{j k}$ denote the cofactor of the row j, column k entry of the matrix $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ .
1. (a) Prove that if B is the matrix obtained from A by replacing column k by $e_{j}$ $e_{j}$ then $det (B) = c_{j k}$ $det (B) = c_{j k}$ .
2. (b) Show that for $1 \leq j \leq n,$ $1 \leq j \leq n,$ we have
  
  $A (\begin{array}{r} c_{j 1} \\ c_{j 2} \\ ⋮ \\ c_{j n} \end{array}) = det (A) \cdot e_{j} .$ $A (\begin{array}{r} c_{j 1} \\ c_{j 2} \\ ⋮ \\ c_{j n} \end{array}) = det (A) \cdot e_{j} .$
  
  Hint: Apply Cramer’s rule to $A x = e_{j}$ $A x = e_{j}$ .
3. (c) Deduce that if C is the $n \times n$ $n \times n$ matrix such that $C_{i j} = c_{j i},$ $C_{i j} = c_{j i},$ then $A C = [det (A)] I$ $A C = [det (A)] I$ .
4. (d) Show that if $det (A) \neq 0,$ $det (A) \neq 0,$ then $A^{- 1} = {[det (A)]}^{- 1} C$ $A^{- 1} = {[det (A)]}^{- 1} C$ .

The following definition is used in Exercises 26-27.

Definition.

The classical adjoint of a square matrix A is the transpose of the matrix whose ij-entry is the ij-cofactor of A.

Find the classical adjoint of each of the following matrices.
1. $(\begin{array}{r} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array})$ $(\begin{array}{r} A_{11} & A_{12} \\ A_{21} & A_{22} \end{array})$
2. $(\begin{array}{r} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array})$ $(\begin{array}{r} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{array})$
3. $(\begin{array}{r} - 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{array})$ $(\begin{array}{r} - 4 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 5 \end{array})$
4. $(\begin{array}{r} 3 & 6 & 7 \\ 0 & 4 & 8 \\ 0 & 0 & 5 \end{array})$ $(\begin{array}{r} 3 & 6 & 7 \\ 0 & 4 & 8 \\ 0 & 0 & 5 \end{array})$
5. $(\begin{array}{c} 1 - i & 0 & 0 \\ 4 & 3 i & 0 \\ 2 i & 1 + 4 i & - 1 \end{array})$ $(\begin{array}{c} 1 - i & 0 & 0 \\ 4 & 3 i & 0 \\ 2 i & 1 + 4 i & - 1 \end{array})$
6. $(\begin{array}{r} 7 & 1 & 4 \\ 6 & - 3 & 0 \\ - 3 & 5 & - 2 \end{array})$ $(\begin{array}{r} 7 & 1 & 4 \\ 6 & - 3 & 0 \\ - 3 & 5 & - 2 \end{array})$
7. $(\begin{array}{r} - 1 & 2 & 5 \\ 8 & 0 & - 3 \\ 4 & 6 & 1 \end{array})$ $(\begin{array}{r} - 1 & 2 & 5 \\ 8 & 0 & - 3 \\ 4 & 6 & 1 \end{array})$
8. $(\begin{array}{c} 3 & 2 + i & 0 \\ - 1 + i & 0 & i \\ 0 & 1 & 3 - 2 i \end{array})$ $(\begin{array}{c} 3 & 2 + i & 0 \\ - 1 + i & 0 & i \\ 0 & 1 & 3 - 2 i \end{array})$
Let C be the classical adjoint of $A \in M_{n \times n} (F)$ $A \in M_{n \times n} (F)$ . Prove the following statements.
1. $det (C) = {[\det (A)]}^{n - 1}$ $det (C) = {[\det (A)]}^{n - 1}$ .
2. $C^{t}$ $C^{t}$ is the classical adjoint of $A^{t}$ $A^{t}$ .
3. If A is an invertible upper triangular matrix, then C and $A^{- 1}$ $A^{- 1}$ are both upper triangular matrices.
Let $y_{1}, y_{2}, \dots, y_{n}$ $y_{1}, y_{2}, \dots, y_{n}$ be linearly independent functions in $C^{\infty}$ $C^{\infty}$ . For each $y \in C^{\infty}$ $y \in C^{\infty}$ , define $T (y) \in C^{\infty}$ $T (y) \in C^{\infty}$ by

$[T (y)] (t) = det (\begin{array}{c} y (t) & y_{1} (t) & y_{2} (t) & \dots & y_{n} (t) \\ y^{'} (t) & {y^{'}}_{1} (t) & {y^{'}}_{2} (t) & \dots & {y^{'}}_{n} (t) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ y^{(n)} (t) & y_{1}^{(n)} (t) & y_{2}^{(n)} (t) & \dots & y_{n}^{(n)} (t) \end{array}) .$ $[T (y)] (t) = det (\begin{array}{c} y (t) & y_{1} (t) & y_{2} (t) & \dots & y_{n} (t) \\ y^{'} (t) & {y^{'}}_{1} (t) & {y^{'}}_{2} (t) & \dots & {y^{'}}_{n} (t) \\ ⋮ & ⋮ & ⋮ & ⋮ \\ y^{(n)} (t) & y_{1}^{(n)} (t) & y_{2}^{(n)} (t) & \dots & y_{n}^{(n)} (t) \end{array}) .$

The preceding determinant is called the Wronskian of $y, y_{1}, \dots, y_{n}$ $y, y_{1}, \dots, y_{n}$ .
1. Prove that T: $C^{\infty} \to C^{\infty}$ $C^{\infty} \to C^{\infty}$ is a linear transformation.
2. Prove that N(T) contains $span ({y_{1}, y_{2}, \dots, y_{n}})$ $span ({y_{1}, y_{2}, \dots, y_{n}})$ .

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Table of Contents for 4.3 Properties of Determinants

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Table of Contents for
4.3 Properties of Determinants