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2.3 Additional Topics and Applications

In this section, we learn a method for computing the inverse of a nonsingular matrix A using determinants and we learn a method for solving linear systems using determinants. Both methods depend on Lemma 2.2.1. We also show how to use determinants to define the cross product of two vectors. The cross product is useful in physics applications involving the motion of a particle in 3-space.

The Adjoint of a Matrix

Let A be an $n \times n$ $n \times n$ matrix. We define a new matrix called the adjoint of A by

adj A = [\begin{matrix} A_{11} & A_{21} & \dots & A_{n 1} \\ A_{12} & A_{22} & \dots & A_{n 2} \\ ⋮ \\ A_{1 n} & A_{2 n} & \dots & A_{n n} \end{matrix}]

$adj A = [\begin{matrix} A_{11} & A_{21} & \dots & A_{n 1} \\ A_{12} & A_{22} & \dots & A_{n 2} \\ ⋮ \\ A_{1 n} & A_{2 n} & \dots & A_{n n} \end{matrix}]$

Thus, to form the adjoint, we must replace each term by its cofactor and then transpose the resulting matrix. By Lemma 2.2.1,

a_{i 1} A_{j 1} + a_{i 2} A_{j 2} + \dots + a_{i n} A_{j n} = {\begin{matrix} det (A) & if i = j \\ 0 & if i \neq j \end{matrix}

$a_{i 1} A_{j 1} + a_{i 2} A_{j 2} + \dots + a_{i n} A_{j n} = {\begin{matrix} det (A) & if i = j \\ 0 & if i \neq j \end{matrix}$

and it follows that

A (adj A) = det (A) I

$A (adj A) = det (A) I$

If A is nonsingular, det(A) is a nonzero scalar, and we may write

A (\frac{1}{det (A)} adj A) = I

$A (\frac{1}{det (A)} adj A) = I$

Thus,

Example 1

For a $2 \times 2$ $2 \times 2$ matrix,

adj A = [\begin{matrix} a_{22} & - a_{12} \\ - a_{21} & a_{11} \end{matrix}]

$adj A = [\begin{matrix} a_{22} & - a_{12} \\ - a_{21} & a_{11} \end{matrix}]$

If A is nonsingular, then

A^{- 1} = \frac{1}{a_{11} a_{22} - a_{12} a_{21}} [\begin{matrix} a_{22} & - a_{12} \\ - a_{21} & a_{11} \end{matrix}]

$A^{- 1} = \frac{1}{a_{11} a_{22} - a_{12} a_{21}} [\begin{matrix} a_{22} & - a_{12} \\ - a_{21} & a_{11} \end{matrix}]$

Example 2

Let

A = [\begin{matrix} 2 & 1 & 2 \\ 3 & 2 & 2 \\ 1 & 2 & 3 \end{matrix}]

$A = [\begin{matrix} 2 & 1 & 2 \\ 3 & 2 & 2 \\ 1 & 2 & 3 \end{matrix}]$

Compute adj A and $A^{- 1}$ $A^{- 1}$ .

SOLUTION

adj A = {[\begin{array}{r} | \begin{matrix} 2 & 2 \\ 2 & 3 \end{matrix} | & - | \begin{matrix} 3 & 2 \\ 1 & 3 \end{matrix} | & | \begin{matrix} 3 & 2 \\ 1 & 2 \end{matrix} | \\ - | \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix} | & | \begin{matrix} 2 & 2 \\ 1 & 3 \end{matrix} | & - | \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} | \\ | \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} | & - | \begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix} | & | \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} | \end{array}]}^{T} = [\begin{array}{r} 2 & 1 & - 2 \\ - 7 & 4 & 2 \\ 4 & - 3 & 1 \end{array}]

$adj A = {[\begin{array}{r} | \begin{matrix} 2 & 2 \\ 2 & 3 \end{matrix} | & - | \begin{matrix} 3 & 2 \\ 1 & 3 \end{matrix} | & | \begin{matrix} 3 & 2 \\ 1 & 2 \end{matrix} | \\ - | \begin{matrix} 1 & 2 \\ 2 & 3 \end{matrix} | & | \begin{matrix} 2 & 2 \\ 1 & 3 \end{matrix} | & - | \begin{matrix} 2 & 1 \\ 1 & 2 \end{matrix} | \\ | \begin{matrix} 1 & 2 \\ 2 & 2 \end{matrix} | & - | \begin{matrix} 2 & 2 \\ 3 & 2 \end{matrix} | & | \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} | \end{array}]}^{T} = [\begin{array}{r} 2 & 1 & - 2 \\ - 7 & 4 & 2 \\ 4 & - 3 & 1 \end{array}]$

A^{- 1} = \frac{1}{det (A)} adj A = \frac{1}{5} [\begin{array}{r} 2 & 1 & - 2 \\ - 7 & 4 & 2 \\ 4 & - 3 & 1 \end{array}]

$A^{- 1} = \frac{1}{det (A)} adj A = \frac{1}{5} [\begin{array}{r} 2 & 1 & - 2 \\ - 7 & 4 & 2 \\ 4 & - 3 & 1 \end{array}]$

Using the formula

A^{- 1} = \frac{1}{det (A)} adj A

$A^{- 1} = \frac{1}{det (A)} adj A$

we can derive a rule for representing the solution to the system $A x = b$ $A x = b$ in terms of determinants.

Cramer’s Rule

Theorem 2.3.1 Cramer’s Rule

Let A be a nonsingular $n \times n$ $n \times n$ matrix, and let $b \in ℝ^{n}$ $b \in ℝ^{n}$ . Let $A_{i}$ $A_{i}$ be the matrix obtained by replacing the ith column of A by replacing the ith column of A by b. If x is the unique solution of $A x = b$ $A x = b$ , then

\begin{matrix} x_{i} = \frac{det (A_{i})}{det (A)} & for i = 1, 2, \dots, n \end{matrix}

$\begin{matrix} x_{i} = \frac{det (A_{i})}{det (A)} & for i = 1, 2, \dots, n \end{matrix}$

Proof

Since

x = A^{- 1} b = \frac{1}{det (A)} (adj A) b

$x = A^{- 1} b = \frac{1}{det (A)} (adj A) b$

it follows that

\begin{matrix} x_{i} & = & \frac{b_{1} A_{1 i} + b_{2} A_{2 i} + \dots + b_{n} A_{n i}}{det (A)} \\ = & \frac{det (A_{i})}{det (A)} \end{matrix}

$\begin{matrix} x_{i} & = & \frac{b_{1} A_{1 i} + b_{2} A_{2 i} + \dots + b_{n} A_{n i}}{det (A)} \\ = & \frac{det (A_{i})}{det (A)} \end{matrix}$

∎

Example 3

Use Cramer’s rule to solve

\begin{matrix} x_{1} + 2 x_{2} + x_{3} & = & 5 \\ 2 x_{1} + 2 x_{2} + x_{3} & = & 6 \\ x_{1} + 2 x_{2} + 3 x_{3} & = & 9 \end{matrix}

$\begin{matrix} x_{1} + 2 x_{2} + x_{3} & = & 5 \\ 2 x_{1} + 2 x_{2} + x_{3} & = & 6 \\ x_{1} + 2 x_{2} + 3 x_{3} & = & 9 \end{matrix}$

SOLUTION

\begin{matrix} det (A) & = & | \begin{matrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} | & = & - 4 & det (A_{1}) & = & | \begin{matrix} 5 & 2 & 1 \\ 6 & 2 & 1 \\ 9 & 2 & 3 \end{matrix} | & = & - 4 \\ det (A_{2}) & = & | \begin{matrix} 1 & 5 & 1 \\ 2 & 6 & 1 \\ 1 & 9 & 3 \end{matrix} | & = & - 4 & det (A_{3}) & = & | \begin{matrix} 1 & 2 & 5 \\ 2 & 2 & 6 \\ 1 & 2 & 9 \end{matrix} | & = & - 8 \end{matrix}

$\begin{matrix} det (A) & = & | \begin{matrix} 1 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 2 & 3 \end{matrix} | & = & - 4 & det (A_{1}) & = & | \begin{matrix} 5 & 2 & 1 \\ 6 & 2 & 1 \\ 9 & 2 & 3 \end{matrix} | & = & - 4 \\ det (A_{2}) & = & | \begin{matrix} 1 & 5 & 1 \\ 2 & 6 & 1 \\ 1 & 9 & 3 \end{matrix} | & = & - 4 & det (A_{3}) & = & | \begin{matrix} 1 & 2 & 5 \\ 2 & 2 & 6 \\ 1 & 2 & 9 \end{matrix} | & = & - 8 \end{matrix}$

Therefore,

\begin{matrix} x_{1} = \frac{- 4}{- 4} = 1, & x_{2} = \frac{- 4}{- 4} = 1, & x_{3} = \frac{- 8}{- 4} = 2 \end{matrix}

$\begin{matrix} x_{1} = \frac{- 4}{- 4} = 1, & x_{2} = \frac{- 4}{- 4} = 1, & x_{3} = \frac{- 8}{- 4} = 2 \end{matrix}$

Cramer’s rule gives us a convenient method for writing the solution of an $n \times n$ $n \times n$ system of linear equations in terms of determinants. To compute the solution, however, we must evaluate $n + 1$ $n + 1$ determinants of order n. Evaluating even two of these determinants generally involves more computation than solving the system by Gaussian elimination.

Application 1

Coded Messages

A common way of sending a coded message is to assign an integer value to each letter of the alphabet and to send the message as a string of integers. For example, the message

SEND MONEY

$SEND MONEY$

might be coded as

5, 8, 10, 21, 7, 2, 10, 8, 3

$5, 8, 10, 21, 7, 2, 10, 8, 3$

Here, the S is represented by a 5, the E by an 8, and so on. Unfortunately, this type of code is generally easy to break. In a longer message, we might be able to guess which letter is represented by a number on the basis of the relative frequency of occurrence of that number. For example, if 8 is the most frequently occurring number in the coded message, then it is likely that it represents the letter E, the letter that occurs most frequently in the English language.

We can disguise the message further by using matrix multiplications. If A is a matrix whose entries are all integers and whose determinant is $\pm 1$ $\pm 1$ , then, since $A^{- 1} = \pm adj A$ $A^{- 1} = \pm adj A$ the entries of $A^{−1}$ $A^{−1}$ will be integers. We can use such a matrix to transform the message. The transformed message will be more difficult to decipher. To illustrate the technique, let

A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 2 & 3 & 2 \end{matrix}]

$A = [\begin{matrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 2 & 3 & 2 \end{matrix}]$

The coded message is put into the columns of a matrix B having three rows:

B = [\begin{array}{r} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{array}]

$B = [\begin{array}{r} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{array}]$

The product

A B = [\begin{matrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 2 & 3 & 2 \end{matrix}] [\begin{array}{r} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{array}] = [\begin{matrix} 31 & 37 & 29 \\ 80 & 83 & 69 \\ 54 & 67 & 50 \end{matrix}]

$A B = [\begin{matrix} 1 & 2 & 1 \\ 2 & 5 & 3 \\ 2 & 3 & 2 \end{matrix}] [\begin{array}{r} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{array}] = [\begin{matrix} 31 & 37 & 29 \\ 80 & 83 & 69 \\ 54 & 67 & 50 \end{matrix}]$

gives the coded message to be sent:

31, 80, 54, 37, 83, 67, 29, 69, 50

$31, 80, 54, 37, 83, 67, 29, 69, 50$

The person receiving the message can decode it by multiplying by $A^{−1}$ $A^{−1}$ :

[\begin{array}{r} 1 & - 1 & 1 \\ 2 & 0 & - 1 \\ - 4 & 1 & 1 \end{array}] [\begin{matrix} 31 & 37 & 29 \\ 80 & 83 & 69 \\ 54 & 67 & 50 \end{matrix}] = [\begin{matrix} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{matrix}]

$[\begin{array}{r} 1 & - 1 & 1 \\ 2 & 0 & - 1 \\ - 4 & 1 & 1 \end{array}] [\begin{matrix} 31 & 37 & 29 \\ 80 & 83 & 69 \\ 54 & 67 & 50 \end{matrix}] = [\begin{matrix} 5 & 21 & 10 \\ 8 & 7 & 8 \\ 10 & 2 & 3 \end{matrix}]$

To construct a coding matrix A, we can begin with the identity I and successively apply row operation III, being careful to add integer multiples of one row to another. Row operation I can also be used. The resulting matrix A will have integer entries, and since

det (A) = \pm det (I) = \pm 1

$det (A) = \pm det (I) = \pm 1$

$A^{- 1}$ $A^{- 1}$ will also have integer entries.

Reference

1. Hansen, Robert, “Integer Matrices Whose Inverses Contain Only Integers,” Two-Year College Mathematics Journal, 13(1), 1982.

The Cross Product

Given two vectors x and y in $ℝ^{3}$ $ℝ^{3}$ , one can define a third vector, the cross product, denoted $x \times y$ $x \times y$ , by

x \times y = [\begin{matrix} x_{2} y_{3} - y_{2} x_{3} \\ y_{1} x_{3} - x_{1} y_{3} \\ x_{1} y_{2} - y_{1} x_{2} \end{matrix}]

$x \times y = [\begin{matrix} x_{2} y_{3} - y_{2} x_{3} \\ y_{1} x_{3} - x_{1} y_{3} \\ x_{1} y_{2} - y_{1} x_{2} \end{matrix}]$ (1)

If C is any matrix of the form

C = [\begin{matrix} w_{1} & w_{2} & w_{3} \\ x_{1} & x_{2} & x_{3} \\ {\begin{matrix} y \end{matrix}}_{1} & y_{2} & y_{3} \end{matrix}]

$C = [\begin{matrix} w_{1} & w_{2} & w_{3} \\ x_{1} & x_{2} & x_{3} \\ {\begin{matrix} y \end{matrix}}_{1} & y_{2} & y_{3} \end{matrix}]$

then

x \times y = C_{11} e_{1} + C_{12} e_{2} + C_{13} e_{3} = [\begin{matrix} C_{11} \\ C_{12} \\ C_{13} \end{matrix}]

$x \times y = C_{11} e_{1} + C_{12} e_{2} + C_{13} e_{3} = [\begin{matrix} C_{11} \\ C_{12} \\ C_{13} \end{matrix}]$

Expanding det(C) by cofactors along the first row, we see that

det (C) = w_{1} C_{11} + w_{2} C_{12} + w_{2} C_{13} = w^{T} (x \times y)

$det (C) = w_{1} C_{11} + w_{2} C_{12} + w_{2} C_{13} = w^{T} (x \times y)$

In particular, if we choose $w = x$ $w = x$ or $w = y$ $w = y$ , then the matrix C will have two identical rows, and hence its determinant will be 0. We then have

x^{T} (x \times y) = y^{T} (x \times y) = 0

$x^{T} (x \times y) = y^{T} (x \times y) = 0$ (2)

In calculus books, it is standard to use row vectors

\begin{matrix} x = (x_{1}, x_{2}, x_{3}) & and & y \end{matrix} = (y_{1}, y_{2}, y_{3})

$\begin{matrix} x = (x_{1}, x_{2}, x_{3}) & and & y \end{matrix} = (y_{1}, y_{2}, y_{3})$

and to define the cross product to be the row vector:

x \times y = (x_{2} y_{3} - y_{2} x_{3}) i - (x_{1} y_{3} - y_{1} x_{3}) j + (x_{1} y_{2} - y_{1} x_{2}) k

$x \times y = (x_{2} y_{3} - y_{2} x_{3}) i - (x_{1} y_{3} - y_{1} x_{3}) j + (x_{1} y_{2} - y_{1} x_{2}) k$

where i, j, and k are the row vectors of the $3 \times 3$ $3 \times 3$ identity matrix. If one uses i, j, and k in place of $w_{1}$ $w_{1}$ , $w_{2}$ $w_{2}$ , and $w_{3}$ $w_{3}$ , respectively, in the first row of the matrix M, then the cross product can be written as a determinant.

x \times y = | \begin{matrix} i & j & k \\ x_{1} & x_{2} & x_{3} \\ y_{1} & {\begin{matrix} y \end{matrix}}_{2} & y_{3} \end{matrix} |

$x \times y = | \begin{matrix} i & j & k \\ x_{1} & x_{2} & x_{3} \\ y_{1} & {\begin{matrix} y \end{matrix}}_{2} & y_{3} \end{matrix} |$

In linear algebra courses, it is generally more standard to view x, y and $x \times y$ $x \times y$ as column vectors. In this case, we can represent the cross product in terms of the determinant of a matrix whose entries in the first row are $e_{1}$ $e_{1}$ , $e_{2}$ $e_{2}$ , $e_{3}$ $e_{3}$ , the column vectors of the $3 \times 3$ $3 \times 3$ identity matrix:

x \times y = | \begin{matrix} e_{1} & e_{2} & e_{3} \\ x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \end{matrix} |

$x \times y = | \begin{matrix} e_{1} & e_{2} & e_{3} \\ x_{1} & x_{2} & x_{3} \\ y_{1} & y_{2} & y_{3} \end{matrix} |$

The relation given in equation (2) has applications in Newtonian mechanics. In particular, the cross product can be used to define a binormal direction, which Newton used to derive the laws of motion for a particle in 3-space.

Application 2

Newtonian Mechanics

If x is a vector in either $ℝ^{2}$ $ℝ^{2}$ or $ℝ^{3}$ $ℝ^{3}$ then, we can define the length of x, denoted ‖x‖, by

‖ x ‖ = {(x^{T} x)}^{\frac{1}{2}}

$‖ x ‖ = {(x^{T} x)}^{\frac{1}{2}}$

A vector x is said to be a unit vector if $‖x‖ = 1$ $‖x‖ = 1$ . Unit vectors were used by Newton to derive the laws of motion for a particle in either the plane or 3-space. If x and y are nonzero vectors in $ℝ^{2}$ $ℝ^{2}$ , then the angle $θ$ $θ$ between the vectors is the smallest angle of rotation necessary to rotate one of the two vectors clockwise so that it ends up in the same direction as the other vector (see Figure 2.3.1).

Figure 2.3.1.

Figure 2.3.1. Full Alternative Text

A particle moving in a plane traces out a curve in the plane. The position of the particle at any time t can be represented by a vector $(x_{1} (t), x_{2} (t))$ $(x_{1} (t), x_{2} (t))$ . In describing the motion of a particle, Newton found it convenient to represent the position of vectors at time t as linear combinations of the vectors T(t) and N(t), where T(t) is a unit vector in the direction of the tangent line to curve at the point $(x_{1} (t), x_{2} (t))$ $(x_{1} (t), x_{2} (t))$ and N(t) is a unit vector in the direction of a normal line (a line perpendicular to the tangent line) to the curve at the given point (see Figure 2.3.2).

Figure 2.3.2.

Figure 2.3.2. Full Alternative Text

In Chapter 5, we will show that if x and y are nonzero vectors and $θ$ $θ$ is the angle between the vectors, then

x^{T} y = ‖ x ‖ ‖ y ‖ cos θ

$x^{T} y = ‖ x ‖ ‖ y ‖ cos θ$ (3)

This equation can also be used to define the angle between nonzero vectors in $ℝ^{3}$ $ℝ^{3}$ . It follows from (3) that the angle between the vectors is a right angle if and only if $x^{T} y = 0$ $x^{T} y = 0$ .

In this case, we say that the vectors x and y are orthogonal. In particular, since T(t) and N(t) are unit orthogonal vectors in $ℝ^{2}$ $ℝ^{2}$ , we have $‖ T (t) = ‖ N (t) ‖ = 1$ $‖ T (t) = ‖ N (t) ‖ = 1$ and the angle between the vectors is $\frac{π}{2}$ $\frac{π}{2}$ . It follows from (3) that

{T (t)}^{T} N (t) = 0

${T (t)}^{T} N (t) = 0$

In Chapter 5, we will also show that if x and y are vectors in $ℝ^{3}$ $ℝ^{3}$ and $θ$ $θ$ is the angle between the vectors, then

‖ x \times y ‖ = ‖ x ‖ ‖ y ‖ \sin θ

$‖ x \times y ‖ = ‖ x ‖ ‖ y ‖ \sin θ$ (4)

A particle moving in three dimensions will trace out a curve in 3-space. In this case, at time t the tangent and normal lines to the curve at the point $(x_{1} (t), x_{2} (t))$ $(x_{1} (t), x_{2} (t))$ determine a plane in 3-space. However, in 3-space the motion is not restricted to a plane. To derive laws describing the motion, Newton needed to use a third vector, a vector in a direction normal to the plane determined by T(t) and N(t). If z is any nonzero vector in the direction of the normal line to this plane, then the angle between the vectors z and T(t) and the angle between z and N(t) should both be right angles. If we set

B (t) = T (t) \times N (t)

$B (t) = T (t) \times N (t)$ (5)

then it follows from (2) that B(t) is orthogonal to both T(t) and N(t) and hence is in the direction of the normal line. Furthermore, B(t) is a unit vector since it follows from (4) that

‖ B (t) ‖ = ‖ T (t) \times N (t) ‖ = ‖ T (t) ‖ ‖ N (t) ‖ \sin \frac{π}{2} = 1

$‖ B (t) ‖ = ‖ T (t) \times N (t) ‖ = ‖ T (t) ‖ ‖ N (t) ‖ \sin \frac{π}{2} = 1$

The vector B(t) defined by (5) is called the binormal vector (see Figure 2.3.3).

Figure 2.3.3.

Figure 2.3.3. Full Alternative Text

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Table of Contents for 2.3 Additional Topics and Applications

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2.3 Additional Topics and Applications