[B.5 PARTIAL](#page-6-0) FRACTION EXPANSION

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B.5 PARTIAL FRACTION EXPANSION

In the analysis of linear time-invariant systems, we encounter functions that are ratios of two polynomials in a certain variable, say, x. Such functions are known as rational functions. A rational function F(x) can be expressed as

$$F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} = \frac{P(x)}{Q(x)}$$

(B.22)

The function F(x) is improper if m ≥ n and proper if m < n. † An improper function can always be separated into the sum of a polynomial in x and a proper function. Consider, for example, the function

$$F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3}$$

Because this is an improper function, we divide the numerator by the denominator until the remainder has a lower degree than the denominator.

$$x^{2} + 4x + 3 \quad \begin{array}{c} 2x + 1 \\ 2x^{3} + 9x^{2} + 11x + 2 \\ 2x^{3} + 8x^{2} + 6x \\ x^{2} + 5x + 2 \\ x^{2} + 4x + 3 \\ x - 1 \end{array}$$

^† Some sources classify ^F(x) as strictly proper if ^{m &}lt; ⁿ, proper if ^{m ≤ n}, and improper if ^{m &}gt; ⁿ.

26 CHAPTER B BACKGROUND

Therefore, F(x) can be expressed as

$$F(x) = \frac{2x^3 + 9x^2 + 11x + 2}{x^2 + 4x + 3} = \underbrace{2x + 1}_{\text{polynomial in } x} + \underbrace{\frac{x - 1}{x^2 + 4x + 3}}_{\text{proper function}}$$

A proper function can be further expanded into partial fractions. The remaining discussion in this section is concerned with various ways of doing this.

B.5-1 Method of Clearing Fractions

A rational function can be written as a sum of appropriate partial fractions with unknown coefficients, which are determined by clearing fractions and equating the coefficients of similar powers on the two sides. This procedure is demonstrated by the following example.

EXAMPLE B.8 Method of Clearing Fractions

Expand the following rational function F(x) into partial fractions:

$$F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2}$$

This function can be expressed as a sum of partial fractions with denominators (x + 1), (x +2),(x +3), and (x +3)2, as follows:

$$F(x) = \frac{x^3 + 3x^2 + 4x + 6}{(x+1)(x+2)(x+3)^2} = \frac{k_1}{x+1} + \frac{k_2}{x+2} + \frac{k_3}{x+3} + \frac{k_4}{(x+3)^2}$$

To determine the unknowns k1, k2, k3, and k4, we clear fractions by multiplying both sides by (x +1)(x +2)(x +3)² to obtain

$$x^{3} + 3x^{2} + 4x + 6 = k_{1}(x^{3} + 8x^{2} + 21x + 18) + k_{2}(x^{3} + 7x^{2} + 15x + 9)$$

• $k_{3}(x^{3} + 6x^{2} + 11x + 6) + k_{4}(x^{2} + 3x + 2)$
  = $x^{3}(k_{1} + k_{2} + k_{3}) + x^{2}(8k_{1} + 7k_{2} + 6k_{3} + k_{4})$
• $x(21k_{1} + 15k_{2} + 11k_{3} + 3k_{4}) + (18k_{1} + 9k_{2} + 6k_{3} + 2k_{4})$

Equating coefficients of similar powers on both sides yields

$$k_1 + k_2 + k_3 = 1$$

\n

$$8k_1 + 7k_2 + 6k_3 + k_4 = 3$$

\n

$$21k_1 + 15k_2 + 11k_3 + 3k_4 = 4$$

\n

$$18k_1 + 9k_2 + 6k_3 + 2k_4 = 6$$

Solution of these four simultaneous equations yields

$$k_1 = 1$$

, $k_2 = -2$ , $k_3 = 2$ , $k_4 = -3$

Therefore,

$$F(x) = \frac{1}{x+1} - \frac{2}{x+2} + \frac{2}{x+3} - \frac{3}{(x+3)^2}$$

Although this method is straightforward and applicable to all situations, it is not necessarily the most efficient. We now discuss other methods that can reduce numerical work considerably.

B.5-2 The Heaviside “Cover-Up” Method

DISTINCT FACTORS OF Q(x)

We shall first consider the partial fraction expansion of F(x) = P(x)/Q(x), in which all the factors of Q(x) are distinct (not repeated). Consider the proper function

$$F(x) = \frac{b_m x^m + b_{m-1} x^{m-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0} \qquad m < n$$ $$= \frac{P(x)}{(x - \lambda_1)(x - \lambda_2) \cdots (x - \lambda_n)}$$

As seen in Ex. B.8, F(x) can be expressed as the sum of partial fractions

$$F(x) = \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n}$$

(B.23)

To determine the coefficient k1, we multiply both sides of Eq. (B.23) by x−λ¹ and then let x = λ1. This yields

$$(x - \lambda_1)F(x)|_{x = \lambda_1} = k_1 + \frac{k_2(x - \lambda_1)}{(x - \lambda_2)} + \frac{k_3(x - \lambda_1)}{(x - \lambda_3)} + \dots + \frac{k_n(x - \lambda_1)}{(x - \lambda_n)}\bigg|_{x = \lambda_1}$$

On the right-hand side, all the terms except k¹ vanish. Therefore,

$$k_1 = (x - \lambda_1)F(x)|_{x = \lambda_1}$$

Similarly, we can show that

$$k_r = (x - \lambda_r)F(x)|_{x = \lambda_r} \qquad r = 1, 2, \dots, n$$

(B.24)

This procedure also goes under the name method of residues.

EXAMPLE B.9 Heaviside “Cover-Up” Method

Expand the following rational function F(x) into partial fractions:

$$F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{k_1}{x+1} + \frac{k_2}{x-2} + \frac{k_3}{x+3}$$

To determine k1, we let x = −1 in (x + 1)F(x). Note that (x + 1)F(x) is obtained from F(x) by omitting the term (x + 1) from its denominator. Therefore, to compute k¹ corresponding to the factor (x + 1), we cover up the term (x + 1) in the denominator of F(x) and then substitute x = −1 in the remaining expression. [Mentally conceal the term (x + 1) in F(x) with a finger and then let x = −1 in the remaining expression.] The steps in covering up the function

$$F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}$$

are as follows.

Step 1. Cover up (conceal) the factor (x +1) from F(x):

$$\frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}$$

Step 2. Substitute x = −1 in the remaining expression to obtain k1:

$$k_1 = \frac{2 - 9 - 11}{(-1 - 2)(-1 + 3)} = \frac{-18}{-6} = 3$$

Similarly, to compute k2, we cover up the factor (x − 2) in F(x) and let x = 2 in the remaining function, as follows:

$$k_2 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=2} = \frac{8+18-11}{(2+1)(2+3)} = \frac{15}{15} = 1$$

and

$$k_3 = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)}\bigg|_{x=-3} = \frac{18 - 27 - 11}{(-3+1)(-3-2)} = \frac{-20}{10} = -2$$

Therefore,

$$F(x) = \frac{2x^2 + 9x - 11}{(x+1)(x-2)(x+3)} = \frac{3}{x+1} + \frac{1}{x-2} - \frac{2}{x+3}$$

COMPLEX FACTORS OF Q(x)

The procedure just given works regardless of whether the factors of Q(x) are real or complex. Consider, for example,

$$F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)}$$

= $\frac{k_1}{x+1} + \frac{k_2}{x+2-j3} + \frac{k_3}{x+2+j3}$ (B.25)

where

$$k_1 = \left[ \frac{4x^2 + 2x + 18}{(x+1)\ (x^2 + 4x + 13)} \right]_{x=-1} = 2$$

Similarly,

$$k_2 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2+j3} = 1+j2 = \sqrt{5}e^{j63.43^{\circ}}$$ $$k_3 = \left[ \frac{4x^2 + 2x + 18}{(x+1)(x+2-j3)(x+2+j3)} \right]_{x=-2-j3} = 1-j2 = \sqrt{5}e^{-j63.43^{\circ}}$$

Therefore,

$$F(x) = \frac{2}{x+1} + \frac{\sqrt{5}e^{i63.43^{\circ}}}{x+2-j3} + \frac{\sqrt{5}e^{-i63.43^{\circ}}}{x+2+j3}$$

The coefficients k² and k³ corresponding to the complex-conjugate factors are also conjugates of each other. This is generally true when the coefficients of a rational function are real. In such a case, we need to compute only one of the coefficients.

QUADRATIC FACTORS

Often we are required to combine the two terms arising from complex-conjugate factors into one quadratic factor. For example, F(x) in Eq. (B.25) can be expressed as

$$F(x) = \frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{k_1}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13}$$

The coefficient k¹ is found by the Heaviside method to be 2. Therefore,

$$\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{c_1x + c_2}{x^2 + 4x + 13}$$

(B.26)

The values of c¹ and c² are determined by clearing fractions and equating the coefficients of similar powers of x on both sides of the resulting equation. Clearing fractions on both sides of Eq. (B.26) yields

$$4x2 + 2x + 18 = 2(x2 + 4x + 13) + (c1x + c2)(x + 1)$$

= (2+c₁)x² + (8+c₁+c₂)x + (26+c₂)

Equating terms of similar powers yields c¹ = 2, c² = −8, and

$$\frac{4x^2 + 2x + 18}{(x+1)(x^2 + 4x + 13)} = \frac{2}{x+1} + \frac{2x - 8}{x^2 + 4x + 13}$$

SHORTCUTS

The values of c¹ and c² in Eq. (B.26) can also be determined by using shortcuts. After computing k¹ = 2 by the Heaviside method as before, we let x = 0 on both sides of Eq. (B.26) to eliminate c1. This gives us

$$\frac{18}{13} = 2 + \frac{c_2}{13} \qquad \Rightarrow \qquad c_2 = -8$$

To determine c1, we multiply both sides of Eq. (B.26) by x and then let x → ∞. Remember that when x → ∞, only the terms of the highest power are significant. Therefore,

$$4 = 2 + c_1 \qquad \Rightarrow \qquad c_1 = 2$$

In the procedure discussed here, we let x = 0 to determine c² and then multiply both sides by x and let x → ∞ to determine c1. However, nothing is sacred about these values (x = 0 or x = ∞). We use them because they reduce the number of computations involved. We could just as well use other convenient values for x, such as x = 1. Consider the case

$$F(x) = \frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{k}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5}$$

We find k = 1 by the Heaviside method in the usual manner. As a result,

$$\frac{2x^2 + 4x + 5}{x(x^2 + 2x + 5)} = \frac{1}{x} + \frac{c_1x + c_2}{x^2 + 2x + 5}$$

(B.27)

If we try letting x = 0 to determine c¹ and c2, we obtain ∞ on both sides. So let us choose x = 1. This yields

$$\frac{11}{8} = 1 + \frac{c_1 + c_2}{8} \qquad \text{or} \qquad c_1 + c_2 = 3$$

We can now choose some other value for x, such as x = 2, to obtain one more relationship to use in determining c¹ and c2. In this case, however, a simple method is to multiply both sides of Eq. (B.27) by x and then let x → ∞. This yields

$$2 = 1 + c_1 \qquad \Rightarrow \qquad c_1 = 1$$

Since c¹ +c² = 3, we see that c² = 2 and therefore,

$$F(x) = \frac{1}{x} + \frac{x+2}{x^2 + 2x + 5}$$

B.5-3 Repeated Factors of Q(x)

If a function F(x) has a repeated factor in its denominator, it has the form

$$F(x) = \frac{P(x)}{(x - \lambda)^r (x - \alpha_1)(x - \alpha_2) \cdots (x - \alpha_j)}$$

Its partial fraction expansion is given by

$$F(x) = \frac{a_0}{(x - \lambda)^r} + \frac{a_1}{(x - \lambda)^{r-1}} + \dots + \frac{a_{r-1}}{(x - \lambda)}$$

$$\frac{k_1}{x - \alpha_1} + \frac{k_2}{x - \alpha_2} + \dots + \frac{k_j}{x - \alpha_j}$$

(B.28)

The coefficients k1, k2,…, kj corresponding to the unrepeated factors in this equation are determined by the Heaviside method, as before [Eq. (B.24)]. To find the coefficients a0,a1, a2,…,ar−1, we multiply both sides of Eq. (B.28) by (x −λ)^r . This gives us

$$(x - \lambda)^r F(x) = a_0 + a_1(x - \lambda) + a_2(x - \lambda)^2 + \dots + a_{r-1}(x - \lambda)^{r-1} + k_1 \frac{(x - \lambda)^r}{x - \alpha_1} + k_2 \frac{(x - \lambda)^r}{x - \alpha_2} + \dots + k_n \frac{(x - \lambda)^r}{x - \alpha_n}$$

(B.29)

If we let x = λ on both sides of Eq. (B.29), we obtain

$$(x - \lambda)^r F(x)|_{x = \lambda} = a_0$$

Therefore, a⁰ is obtained by concealing the factor (x−λ)^r in F(x) and letting x =λ in the remaining expression (the Heaviside “cover-up” method). If we take the derivative (with respect to x) of both sides of Eq. (B.29), the right-hand side is a1+ terms containing a factor (x−λ) in their numerators. Letting x = λ on both sides of this equation, we obtain

$$\frac{d}{dx}\left[ (x - \lambda)^r F(x) \right] \Big|_{x = \lambda} = a_1$$

Thus, a¹ is obtained by concealing the factor (x−λ)^r in F(x), taking the derivative of the remaining expression, and then letting x = λ. Continuing in this manner, we find

$$a_j = \frac{1}{j!} \left. \frac{d^j}{dx^j} \left[ (x - \lambda)^r F(x) \right] \right|_{x = \lambda}$$

(B.30)

Observe that (x − λ)^r F(x) is obtained from F(x) by omitting the factor (x − λ)^r from its denominator. Therefore, the coefficient aj is obtained by concealing the factor (x − λ)^r in F(x), taking the jth derivative of the remaining expression, and then letting x = λ (while dividing by j!).

EXAMPLE B.10 Partial Fraction Expansion with Repeated Factors

Expand F(x) into partial fractions if

$$F(x) = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}$$

The partial fractions are

$$F(x) = \frac{a_0}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{k}{x+2}$$

The coefficient k is obtained by concealing the factor (x + 2) in F(x) and then substituting x = −2 in the remaining expression:

$$k = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-2} = 1$$

To find a0, we conceal the factor (x +1)³ in F(x) and let x = −1 in the remaining expression:

$$a_0 = \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)}\bigg|_{x=-1} = 2$$

To find a1, we conceal the factor (x + 1)³ in F(x), take the derivative of the remaining expression, and then let x = −1:

$$a_1 = \frac{d}{dx} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 1$$

Similarly,

$$a_2 = \frac{1}{2!} \frac{d^2}{dx^2} \left[ \frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} \right] \Big|_{x=-1} = 3$$

Therefore,

$$F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}$$

B.5-4 A Combination of Heaviside “Cover-Up” and Clearing Fractions

For multiple roots, especially of higher order, the Heaviside expansion method, which requires repeated differentiation, can become cumbersome. For a function that contains several repeated and unrepeated roots, a hybrid of the two procedures proves to be the best. The simpler coefficients are determined by the Heaviside method, and the remaining coefficients are found by clearing fractions or shortcuts, thus incorporating the best of the two methods. We demonstrate this procedure by solving Ex. B.10 once again by this method.

In Ex. B.10, coefficients k and a⁰ are relatively simple to determine by the Heaviside expansion method. These values were found to be k¹ = 1 and a⁰ = 2. Therefore,

$$\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2}$$

We now multiply both sides of this equation by (x +1)3(x +2) to clear the fractions. This yields

$$4x3 + 16x2 + 23x + 13 = 2(x+2) + a1(x+1)(x+2) + a2(x+1)2(x+2) + (x+1)3$$

= (1+a₂)x³ + (a₁+4a₂+3)x² + (5+3a₁+5a₂)x + (4+2a₁+2a₂+1)

Equating coefficients of the third and second powers of x on both sides, we obtain

$$\begin{array}{ccc} 1 + a_2 = 4 \\ a_1 + 4a_2 + 3 = 16 \end{array} \implies \begin{array}{c} a_1 = 1 \\ a_2 = 3 \end{array}$$

We may stop here if we wish because the two desired coefficients, a¹ and a2, are now determined. However, equating the coefficients of the two remaining powers of x yields a convenient check on the answer. Equating the coefficients of the x¹ and x⁰ terms, we obtain

$$23 = 5 + 3a_1 + 5a_2$$ $$13 = 4 + 2a_1 + 2a_2 + 1$$

These equations are satisfied by the values a¹ = 1 and a² = 3, found earlier, providing an additional check for our answers. Therefore,

$$F(x) = \frac{2}{(x+1)^3} + \frac{1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}$$

which agrees with the earlier result.

A COMBINATION OF HEAVISIDE “COVER-UP” AND SHORTCUTS

In Ex. B.10, after determining the coefficients a⁰ = 2 and k = 1 by the Heaviside method as before, we have

$$\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{a_2}{x+1} + \frac{1}{x+2}$$

There are only two unknown coefficients, a¹ and a2. If we multiply both sides of this equation by x and then let x → ∞, we can eliminate a1. This yields

$$4 = a_2 + 1 \quad \Longrightarrow \quad a_2 = 3$$

Therefore,

$$\frac{4x^3 + 16x^2 + 23x + 13}{(x+1)^3(x+2)} = \frac{2}{(x+1)^3} + \frac{a_1}{(x+1)^2} + \frac{3}{x+1} + \frac{1}{x+2}$$

34 CHAPTER B BACKGROUND

There is now only one unknown a1, which can be readily found by setting x equal to any convenient value, say, x = 0. This yields

$$\frac{13}{2} = 2 + a_1 + 3 + \frac{1}{2} \implies a_1 = 1$$

which agrees with our earlier answer.

There are other possible shortcuts. For example, we can compute a⁰ (coefficient of the highest power of the repeated root), subtract this term from both sides, and then repeat the procedure.

B.5-5 Improper ^F(x) with ^{m = n}

A general method of handling an improper function is indicated in the beginning of this section. However, for the special case of when the numerator and denominator polynomials of F(x) have the same degree (m = n), the procedure is the same as that for a proper function. We can show that for

$$F(x) = \frac{b_n x^n + b_{n-1} x^{n-1} + \dots + b_1 x + b_0}{x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0}$$

= $b_n + \frac{k_1}{x - \lambda_1} + \frac{k_2}{x - \lambda_2} + \dots + \frac{k_n}{x - \lambda_n}$

the coefficients k1, k2,…, kn are computed as if F(x) were proper. Thus,

$$k_r = (x - \lambda_r)F(x)|_{x = \lambda_r}$$

For quadratic or repeated factors, the appropriate procedures discussed in Secs. B.5-2 or B.5-3 should be used as if F(x) were proper. In other words, when m = n, the only difference between the proper and improper case is the appearance of an extra constant bn in the latter. Otherwise, the procedure remains the same. The proof is left as an exercise for the reader.

EXAMPLE B.11 Partial Fraction Expansion of Improper Rational Function

Expand F(x) into partial fractions if

$$F(x) = \frac{3x^2 + 9x - 20}{x^2 + x - 6} = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)}$$

Here, m = n = 2 with bn = b² = 3. Therefore,

$$F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{k_1}{x - 2} + \frac{k_2}{x + 3}$$

in which

$$k_1 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} \bigg|_{x=2} = \frac{12 + 18 - 20}{(2 + 3)} = \frac{10}{5} = 2$$

and

$$k_2 = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)}\bigg|_{x=-3} = \frac{27 - 27 - 20}{(-3 - 2)} = \frac{-20}{-5} = 4$$

Therefore,

$$F(x) = \frac{3x^2 + 9x - 20}{(x - 2)(x + 3)} = 3 + \frac{2}{x - 2} + \frac{4}{x + 3}$$

B.5-6 Modified Partial Fractions

In finding the inverse z-transform (Ch. 5), we require partial fractions of the form kx/(x −λi)^r rather than k/(x −λi)^r . This can be achieved by expanding F(x)/x into partial fractions. Consider, for example,

$$F(x) = \frac{5x^2 + 20x + 18}{(x+2)(x+3)^2}$$

Dividing both sides by x yields

$$\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2}$$

Expansion of the right-hand side into partial fractions as usual yields

$$\frac{F(x)}{x} = \frac{5x^2 + 20x + 18}{x(x+2)(x+3)^2} = \frac{a_1}{x} + \frac{a_2}{x+2} + \frac{a_3}{(x+3)} + \frac{a_4}{(x+3)^2}$$

Using the procedure discussed earlier, we find a¹ = 1, a² = 1, a³ = −2, and a⁴ = 1. Therefore,

$$\frac{F(x)}{x} = \frac{1}{x} + \frac{1}{x+2} - \frac{2}{x+3} + \frac{1}{(x+3)^2}$$

Now multiplying both sides by x yields

$$F(x) = 1 + \frac{x}{x+2} - \frac{2x}{x+3} + \frac{x}{(x+3)^2}$$

This expresses F(x) as the sum of partial fractions having the form kx/(x −λi)^r .

B.6 VECTORS AND MATRICES

An entity specified by n numbers in a certain order (ordered n-tuple) is an n-dimensional vector. Thus, an ordered n-tuple (x1, x2, …, xn) represents an n-dimensional vector x. A vector may be represented as a row (row vector):

$$\mathbf{x} = [x_1 \quad x_2 \quad \cdots \quad x_n]$$

or as a column (column vector):

$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$$

Simultaneous linear equations can be viewed as the transformation of one vector into another. Consider, for example, the m simultaneous linear equations

$$y_1 = a_{11}x_1 + a_{12}x_2 + \dots + a_{1n}x_n$$

\n

$$y_2 = a_{21}x_1 + a_{22}x_2 + \dots + a_{2n}x_n$$

\n

$$\vdots$$

\n

$$y_m = a_{m1}x_1 + a_{m2}x_2 + \dots + a_{mn}x_n$$

\n(B.31)

If we define two column vectors x and y as

$$\mathbf{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} \quad \text{and} \quad \mathbf{y} = \begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix}$$

then Eq. (B.31) may be viewed as the relationship or the function that transforms vector x into vector y. Such a transformation is called a linear transformation of vectors. To perform a linear transformation, we need to define the array of coefficients aij appearing in Eq. (B.31). This array is called a matrix and is denoted by A for convenience:

$$\mathbf{A} = \left[ \begin{array}{cccc} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{array} \right]$$

A matrix with m rows and n columns is called a matrix of order (m,n) or an (m × n) matrix. For the special case of m = n, the matrix is called a square matrix of order n.

It should be stressed at this point that a matrix is not a number such as a determinant, but an array of numbers arranged in a particular order. It is convenient to abbreviate the representation of matrix A with the form (aij)^m×ⁿ, implying a matrix of order m × n with aij as its ijth element. In practice, when the order m × n is understood or need not be specified, the notation can be abbreviated to (aij). Note that the first index i of aij indicates the row and the second index j indicates the column of the element aij in matrix A.

Equation (B.31) may now be expressed in a matrix form as

$$\begin{bmatrix} y_1 \\ y_2 \\ \vdots \\ y_m \end{bmatrix} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} \\ a_{21} & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \cdots & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix}$$

or $y = Ax$ (B.32)

At this point, we have not defined the multiplication of a matrix by a vector. The quantity Ax is not meaningful until such an operation has been defined.