Higher Mathematics 1st Paper · Classes 11-12 · Chapter 1 · 1.9

Systems of Linear Equations: Cramer's Rule and the Inverse-Matrix Method.

Part (c) of a creative question very often looks the same: three unknowns, three equations, and at the end “solve by Cramer's rule” or “solve using the inverse matrix”. Both methods are built from everything earlier in this chapter: the value of a determinant, cofactors, the adjoint and the inverse. Here they all work together.

1. Systems of linear equations

a₁x₁ + a₂x₂ + … + aₙxₙ = b, with a₁, …, aₙ and b constants, is a linear equation in the variables x₁, …, xₙ. No variable is raised above the first power and no two are multiplied together. Several such equations taken together form a system of linear equations.

When there are as many equations as variables, the system can be solved with determinants. The Swiss mathematician Gabriel Cramer established the method in 1750, which is why it is called Cramer's Rule.

2. Cramer's rule

Two variables: a₁x + b₁y = c₁ and a₂x + b₂y = c₂. The coefficients give D = |a₁, b₁ ; a₂, b₂|. Put the constants c₁, c₂ in place of the x column and you get D_x = |c₁, b₁ ; c₂, b₂|; put them in place of the y column and you get D_y = |a₁, c₁ ; a₂, c₂|. Then x = D_x/D and y = D_y/D.

Three variables follow the same rule. For a₁x + b₁y + c₁z = d₁, a₂x + b₂y + c₂z = d₂, a₃x + b₃y + c₃z = d₃, D = |a₁, b₁, c₁ ; a₂, b₂, c₂ ; a₃, b₃, c₃|. To get D_x, D_y or D_z, take out the x, y or z column and put the d₁, d₂, d₃ column in its place. Then x = D_x/D, y = D_y/D, z = D_z/D.

Every division has D underneath. So there is one condition: D ≠ 0. When D ≠ 0 the system has exactly one solution, a unique solution, and only then does Cramer's rule apply.

Watch the column swap happen. Press x, y or z and that column steps aside while the constants column takes its place.

Exercise 1(B) 15(i): x + 2y = 5, 3x + 5y = 14. D = |1, 2 ; 3, 5| = 5 − 6 = −1. D_x = |5, 2 ; 14, 5| = 25 − 28 = −3. D_y = |1, 5 ; 3, 14| = 14 − 15 = −1. So x = −3/−1 = 3 and y = −1/−1 = 1.

An answer can be checked by putting it back. Move x and y; both equations balance together at only one pair.

Now do it yourself. Pick a system, pick one determinant and type its value. Once all of them are right, x, y and z follow by themselves and are put back into the equations as a check.

The three-variable examples come from the book's exercise. 15(ii): x + 2y + 3z = 14, 2x + 4y + 7z = 31, 3x + 5y + 10z = 43 has D = 1, D_x = 1, D_y = 2, D_z = 3, so (x, y, z) = (1, 2, 3).

15(iii): 4x − y + 4z = 12, 2x + 3y + 8z = 12, 6x + 5y + 12z = 24 has D = −72 and (x, y, z) = (2, 0, 1). Here D_y = 0, so y = 0: a zero on top is no trouble, only a zero underneath stops the rule. 15(iv): 3x + 2y + z = 3, 6x + 4y + 3z = 7, 9x + 8y + 4z = 11 has D = −6 and (x, y, z) = (1/3, 1/2, 1).

The book's Task 1: x + 2y − z = 5, 3x − y + 3z = 7, 2x + 3y + z = 11 has D = −15, D_x = −30, D_y = −30, D_z = −15, so (x, y, z) = (2, 2, 1).

3. In matrix form: AX = B

The same system can be written with matrices: AX = B. A is the coefficient matrix, X the column of unknowns [[x], [y], [z]], and B the column of constants. If A is non-singular, multiply both sides on the left by A⁻¹: X = A⁻¹B. Then read off x, y and z by equating entries.

The two methods differ in the work they take. Cramer's rule needs four determinants: D, D_x, D_y, D_z. The inverse method needs |A| and adj A, and then X = (1/|A|) adj A · B. The question decides which: “by Cramer's rule” means Cramer, “using the inverse matrix” means A⁻¹. Doing the other one costs marks.

Exercise 1(B) 16(i): 2x + 3y = 4, x − y = 7 has A = [[2, 3], [1, −1]], |A| = −2 − 3 = −5, and X = A⁻¹B gives (x, y) = (5, −2). 16(ii): 2x − y − z = 6, x + 3y + 2z = 1, 3x − y − 5z = 1 has |A| = −27 and (x, y, z) = (3, −2, 2).

The book's Task 2: solving 2x + y − z = 5, x + 3y + 2z = 10, −x + 2y + z = 1 with A = [[2, 1, −1], [1, 3, 2], [−1, 2, 1]] and X = A⁻¹B gives |A| = −10 and (x, y, z) = (3, 1, 2).

Walk Example 6(c) all the way. The system is given as BX = C, with B = [[2, 1, −1], [3, 2, 2], [5, 4, 3]] and C = [[10], [1], [4]]. Each step produces one matrix.

4. From a story to equations

Many creative questions make you write the equations yourself. Example 5(c): call the production costs of a jug, a mug and a bucket x, y and z taka, and the cost totals give 2x + 6y + 5z = 89, 5x + 7y + 3z = 94, 3x + 5y + 4z = 83. D = −20, D_x = −160, D_y = −60, D_z = −220, so x = 8, y = 3 and z = 11 taka.

The trick is reading the table the right way: rows are events, columns are items. In creative question 1(c), packets of three products are bought over three weeks. Each week's numbers make one row: D = [[4, 2, 1], [6, 2, 1], [5, 1, 1]]. The price column is R = [[x], [y], [z]] and the weekly totals are C = [[475], [575], [425]]. Then DR = C, |D| = −2, and (x, y, z) = (50, 100, 75) taka. Tap the cells of the table below to see where each one goes.

Two more of the same kind. In creative question 19(c), Rajib, Riaz and Prokash buy 4, 2, 4; 5, 9, 5; and 2, 7, 8 units for 190, 426 and 377 taka. Here |A| = 156 and (p, q, r) = (15, 29, 18). Creative question 15(c) works out that setting up a laptop, a desktop and a printer takes 15, 30 and 10 minutes.

Sometimes the system arrives as a matrix product. Creative question 8(c): A = [[2, 3, −4], [1, 0, −2], [1, −5, 6]], B = [[x], [y], [z]] and AB = [[1], [3], [2]]. Multiply it out and you have 2x + 3y − 4z = 1, x − 2z = 3, x − 5y + 6z = 2. D = −24, D_x = −16, D_y = 40, D_z = 28, so x = 2/3, y = −5/3, z = −7/6.

5. When does Cramer's rule work?

Before any solving, look at D. If D ≠ 0 the rule works and the solution is unique. If D = 0 none of the divisions can be done, and the rule stops right there.

Three systems. Look at each one's D and say whether Cramer's rule will work.

6. Check yourself

Three questions. The first two use the system of Example 6(c), so watch which way up the division goes: x = D_x/D.

Source: Higher Mathematics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Matrices and Determinants”, section 1.9.