Unit 2, lesson 9

Linear Systems with Parameters

How to solve a family of linear systems that depends on a coefficient parameter, find the exceptional values where division fails, classify every case, and tell a parameter apart from a free variable.

  • Video 9
  • Describing All Solutions

Keep the line x+y=2x + y = 2 fixed and let the second line be x+ay=3x + ay = 3. For a=2a = 2 the lines cross at (1,1)(1, 1), for a=3a = 3 they cross at (32,12)(\tfrac{3}{2}, \tfrac{1}{2}), and as aa approaches 11 the crossing point moves farther and farther away. At a=1a = 1 the lines are parallel and there is no crossing at all. This lesson solves all of these systems with one calculation, finds the values of aa where that calculation breaks down, and settles each of those values separately. Throughout, a tuple such as (2,3)(2, 3) stands for a column vector.

Key terms

Family of linear systems

Definition Let SS be a set of real numbers. A family of linear systems assigns to each aa in SS a linear system A(a)x=b(a)A(a)\mathbf{x} = \mathbf{b}(a), where the entries of the m×nm \times n matrix A(a)A(a) and of the vector b(a)\mathbf{b}(a) are given expressions in aa, defined for every aa in SS. A letter such as aa that appears among the coefficients is a coefficient parameter. It is not an unknown. For each fixed value of aa the coefficients are ordinary numbers, and the system is an ordinary linear system in the unknowns with its own solution set. Unless a problem restricts it, SS contains every real number at which these expressions are defined, negative values included.

In plain words A coefficient parameter is a dial that selects which system you are solving. Turning the dial changes the equations, and the solutions may change with them.

Example The family from the opening is A(a)x=bA(a)\mathbf{x} = \mathbf{b} with x=(x,y)\mathbf{x} = (x, y) and

A(a)=[111a],b=[23].A(a) = \begin{bmatrix} 1 & 1 \\ 1 & a \end{bmatrix}, \qquad \mathbf{b} = \begin{bmatrix} 2 \\ 3 \end{bmatrix}.

The value a=0a = 0 selects the system x+y=2x + y = 2, x=3x = 3, whose only solution is (3,−1)(3, -1). The value a=2a = 2 selects a different system, x+y=2x + y = 2, x+2y=3x + 2y = 3, whose only solution is (1,1)(1, 1).

Nonzero condition for division

Definition Multiplying an equation by an expression c(a)c(a), or dividing it by c(a)c(a), gives an equivalent equation for every value of aa with c(a)≠0c(a) \ne 0. When c(a)=0c(a) = 0, division is undefined, and multiplication replaces the equation by 0=00 = 0 and loses its information. So a step that divides by c(a)c(a), scales a row by c(a)c(a), or cancels a common factor c(a)c(a) from both sides is valid only under the condition c(a)≠0c(a) \ne 0, and every result derived after that step carries the same condition. A row replacement Ri→Ri+c(a)RjR_i \to R_i + c(a)R_j with i≠ji \ne j needs no condition when the multiplier c(a)c(a) is defined for every value of aa, as a polynomial in aa is, because Ri→Ri−c(a)RjR_i \to R_i - c(a)R_j then undoes it for every value of aa. A multiplier such as 1/(a−1)1/(a - 1) hides a division and is defined only for a≠1a \ne 1.

In plain words Before dividing by something that contains the parameter, write down where it is nonzero. The formula you get is promised only for those values.

Example For the family above, R2→R2−R1R_2 \to R_2 - R_1 is valid for every aa and gives

[1120a−11].\left[\begin{array}{cc|c} 1 & 1 & 2 \\ 0 & a - 1 & 1 \end{array}\right].

The second row says (a−1)y=1(a - 1)y = 1. If a≠1a \ne 1, we may divide by a−1a - 1, and then

y=1a−1,x=2−y=2a−3a−1.\begin{aligned} y &= \frac{1}{a - 1}, \\[0.4em] x &= 2 - y = \frac{2a - 3}{a - 1}. \end{aligned}

Substituting back confirms this for every a≠1a \ne 1. With x=2−yx = 2 - y the first equation holds, and the left side of the second equation is 2+(a−1)y=2+1=32 + (a - 1)y = 2 + 1 = 3. The formulas say nothing about a=1a = 1, and nearby values do not settle it either. At a=1.1a = 1.1 the solution is (−8,10)(-8, 10) and at a=0.9a = 0.9 it is (12,−10)(12, -10). Both points lie on the line x+y=2x + y = 2, but they solve two different systems, and neither is an answer at a=1a = 1.

Exceptional parameter value

Definition In a family solved by elimination, an exceptional parameter value is a value at which an expression used as a divisor in deriving the general solution equals zero, even if the final formula shows no denominator. Usually that expression is the entry that serves as a pivot for all other values. At an exceptional value the general formula was never derived, so it does not apply, and the system has to be solved again from the equations obtained before the division or from the original equations. If the row containing that divisor has no other nonzero coefficient, the row reads 0=c0 = c at the exceptional value, for some number cc. When c≠0c \ne 0 the system has no solution. When c=0c = 0 the row places no condition, and the remaining rows decide the solution set.

In plain words An exceptional value is where the usual recipe stops working. It is a warning, not a verdict, so the system there has to be checked on its own.

Example At a=1a = 1 the row (a−1)y=1(a - 1)y = 1 becomes 0=10 = 1, so the system has no solution. The original equations say the same thing, since they ask for x+y=2x + y = 2 and x+y=3x + y = 3 at once. Now keep A(a)A(a) and change the right side to (2,2)(2, 2). The same row replacement gives (a−1)y=0(a - 1)y = 0. For a≠1a \ne 1 this forces y=0y = 0 and then x=2x = 2. At a=1a = 1 the row becomes 0=00 = 0, the only remaining equation is x+y=2x + y = 2, and every pair (2−t,t)(2 - t, t) with tt real is a solution. The same coefficient matrix gives no solution for one right side and infinitely many for another, so at a=1a = 1 the matrix alone cannot tell which of these two outcomes occurs.

Case analysis

Definition Case analysis splits the allowed parameter values into cases that together cover every value, chosen so that each step used within a case is valid for every value in that case. Typically one case holds every value at which all divisors are nonzero, and the other cases cover the values at which some divisor is zero. Each case is solved separately. The result is a classification of the family, which states for every parameter value whether the system has exactly one solution, no solution or infinitely many, and gives all the solutions.

In plain words Split before you divide. Handle the general values with the formula and the exceptional ones by going back to the equations, and make sure no value is left out.

Example For the coefficient matrix A(a)A(a) above and the two right sides used so far, the classification is as follows.

CaseRight side (2,3)(2, 3)Right side (2,2)(2, 2)
a≠1a \ne 1exactly one, (2a−3a−1,1a−1)\left(\tfrac{2a - 3}{a - 1}, \tfrac{1}{a - 1}\right)exactly one, (2,0)(2, 0)
a=1a = 1noneinfinitely many, (2−t,t)(2 - t, t)

Reporting only (2,0)(2, 0) in the bottom right cell would be wrong. That pair is one solution when a=1a = 1, but the solution set there is a whole line. A formula that stays finite at an exceptional value can still miss solutions.

Parameter versus free variable

Definition A coefficient parameter is fixed before the system is solved. Different values can give different systems, and the answer is stated for each value. A free variable belongs to one fixed, consistent system. It is an unknown whose column has no pivot in an echelon form, it may take any real value tt, and once values are chosen for all the free variables the pivot variables are determined. Changing tt moves between solutions of the same system without changing any equation. A letter in a proposed solution formula is free only if the formula satisfies all the equations for every value of that letter.

In plain words The parameter aa chooses the equations. The parameter tt chooses a point in the solution set of equations that have already been chosen.

Example With right side (2,2)(2, 2) and a=3a = 3 the system has the single solution (2,0)(2, 0) and no free variable. With a=1a = 1 the system is x+y=2x + y = 2 twice, the variable yy is free, and y=ty = t gives the line of solutions (2−t,t)(2 - t, t). Now return to the right side (2,3)(2, 3) with a=2a = 2. The pairs (2−t,t)(2 - t, t) satisfy the first equation, but that does not make tt free. Substituting into the second equation gives

(2−t)+2t=2+t=3,(2 - t) + 2t = 2 + t = 3,

which forces t=1t = 1, so only (1,1)(1, 1) solves the system.

Check yourself

  1. For the family x+y=2x + y = 2, x+ay=3x + ay = 3, find the solution when a=−1a = -1 and check it in both original equations.
Show the answer

The value a=−1a = -1 is not exceptional, so the general formula applies. It gives

y=1−1−1=−12,x=2(−1)−3−1−1=52.\begin{aligned} y &= \frac{1}{-1 - 1} = -\frac{1}{2}, \\[0.4em] x &= \frac{2(-1) - 3}{-1 - 1} = \frac{5}{2}. \end{aligned}

The check works, since 52−12=2\tfrac{5}{2} - \tfrac{1}{2} = 2 and 52+(−1)(−12)=3\tfrac{5}{2} + (-1)\left(-\tfrac{1}{2}\right) = 3.

  1. Classify the family x+y=2x + y = 2, x+ay=a+1x + ay = a + 1 for every real aa. A classmate cancels a−1a - 1 and reports y=1y = 1 and x=1x = 1 for every aa. What does that answer miss?
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The replacement R2→R2−R1R_2 \to R_2 - R_1 is valid for every aa and gives (a−1)y=a−1(a - 1)y = a - 1. If a≠1a \ne 1, dividing gives y=1y = 1, and then x=1x = 1, so (1,1)(1, 1) is the only solution. If a=1a = 1, the row reads 0=00 = 0, the system is just x+y=2x + y = 2, and the solutions are (2−t,t)(2 - t, t) for every real tt.

Canceling a−1a - 1 is a division, so it needs a≠1a \ne 1. The classmate keeps only (1,1)(1, 1) at a=1a = 1 and loses every other solution there, for example (0,2)(0, 2), which satisfies 0+2=20 + 2 = 2 and 0+1⋅2=1+10 + 1 \cdot 2 = 1 + 1.

  1. For which values of kk does the system x+ky=1x + ky = 1, kx+y=1kx + y = 1 have exactly one solution, no solution or infinitely many? Give all solutions in each case.
Show the answer

The replacement R2→R2−kR1R_2 \to R_2 - kR_1 is valid for every kk and gives the row

(1−k2)y=1−k.(1 - k^2)y = 1 - k.

Since 1−k2=(1−k)(1+k)1 - k^2 = (1 - k)(1 + k), the exceptional values are k=1k = 1 and k=−1k = -1.

If k≠1k \ne 1 and k≠−1k \ne -1, division is allowed and

y=1−k(1−k)(1+k)=11+k,x=1−ky=11+k.\begin{aligned} y &= \frac{1 - k}{(1 - k)(1 + k)} = \frac{1}{1 + k}, \\[0.4em] x &= 1 - ky = \frac{1}{1 + k}. \end{aligned}

So there is exactly one solution, (11+k,11+k)\left(\tfrac{1}{1 + k}, \tfrac{1}{1 + k}\right).

If k=1k = 1, the row reads 0=00 = 0 and both equations are x+y=1x + y = 1, so there are infinitely many solutions, (1−t,t)(1 - t, t) for every real tt.

If k=−1k = -1, the row reads 0=20 = 2, so there is no solution. Indeed, the equations x−y=1x - y = 1 and −x+y=1-x + y = 1 add up to 0=20 = 2.

Watch for this trap. Simplifying the fraction to 1/(1+k)1/(1 + k) before splitting into cases hides the value k=1k = 1, where the formula stays finite but the system has infinitely many solutions.

How this lesson connects