Practice: Cramer's rule
Self-check
Section titled “Self-check”Six short questions. Answer each one in your head (or on paper) before opening the collapsible. Trying to retrieve the answer is where the learning sticks; rereading feels productive but does much less.
1. What is Cramer’s rule for solving M · v = b?
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Each coordinate is a ratio of determinants: replace the matching column of M with the target b, take that determinant, and divide by det(M). For 2x2: x = det(M with first column replaced by b) / det(M) and y = det(M with second column replaced by b) / det(M).
2. What single geometric idea is the rule built on?
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That a coordinate of the unknown is a signed area. The parallelogram spanned by the unknown [x, y] and j-hat has signed area exactly x (base 1, height x); the one spanned by i-hat and [x, y] has area y. Coordinates are areas in disguise.
3. Why does applying M turn that area into something you can compute from b?
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Applying M scales every area by det(M) (the determinant lesson’s central fact), so the area x becomes det(M) · x. Meanwhile M sends the parallelogram’s sides [x, y] and j-hat to b and the second column of M, so the transformed area is det([b | second column]). Equating gives x = det([b | second column]) / det(M).
4. To solve for x, which column do you replace? For y?
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For x (the first coordinate), replace the first column of M with b. For y (the second coordinate), replace the second column. Match the coordinate to its column; flipping this is the most common mistake.
5. What does det(M) = 0 mean for Cramer’s rule?
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The rule divides by zero, which is the honest signal that there is no unique solution. A zero determinant is the collapsed case from the inverses lesson: the system has either no solution or infinitely many, so there is no single answer to compute.
6. Is Cramer’s rule the fast way to solve large systems?
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No. It is a conceptual tool and a clean method for tiny systems. For anything sizable, Gaussian elimination or an iterative solver is far faster; computing a determinant for every unknown scales badly. Cramer’s value here is intuition and closure, not speed.
Try it yourself, part 1: solve with Cramer’s rule
Section titled “Try it yourself, part 1: solve with Cramer’s rule”Solve each system with the column-replacement formula, then check your answer in the original equations. About 8 minutes, pen and paper.
- a)
2x + 3y = 8,x + 2y = 5 - b)
4x - y = 5,2x + 3y = 13
Check your work
- a)
M = [[2, 3], [1, 2]],b = [8, 5],det(M) = (2)(2) - (3)(1) = 1.x = det([[8, 3], [5, 2]]) / 1 = (16 - 15)/1 = 1.y = det([[2, 8], [1, 5]]) / 1 = (10 - 8)/1 = 2. So[x, y] = [1, 2]. Check:2(1) + 3(2) = 8,1 + 2(2) = 5. Both hold. - b)
M = [[4, -1], [2, 3]],b = [5, 13],det(M) = (4)(3) - (-1)(2) = 12 + 2 = 14.x = det([[5, -1], [13, 3]]) / 14 = (15 + 13)/14 = 28/14 = 2.y = det([[4, 5], [2, 13]]) / 14 = (52 - 10)/14 = 42/14 = 3. So[x, y] = [2, 3]. Check:4(2) - 3 = 5,2(2) + 3(3) = 13. Both hold.
Try it yourself, part 2: the geometric why, and the collapsed case
Section titled “Try it yourself, part 2: the geometric why, and the collapsed case”About 7 minutes.
Step 1. The system in part 1a had solution [1, 2]. Confirm the geometric claim: compute the signed area of the parallelogram spanned by [1, 2] and j-hat = [0, 1] as the determinant det([[1, 0], [2, 1]]). Does it equal the x-coordinate?
Step 2. Try Cramer’s rule on x + 2y = 1, 2x + 4y = 3. What is det(M), and what does the rule report?
Check your work
Step 1. det([[1, 0], [2, 1]]) = (1)(1) - (0)(2) = 1, which equals x = 1. The coordinate really is the signed area of that parallelogram, exactly as the derivation claimed.
Step 2. M = [[1, 2], [2, 4]], det(M) = (1)(4) - (2)(2) = 0. Cramer’s rule would divide by zero, so it reports no unique solution. This is correct: the column space is the line through [1, 2] (points where the second coordinate is twice the first), and b = [1, 3] is off that line (3 ≠ 2·1), so the system has no solution at all. The division by zero is the rule being honest, not breaking.
Flashcards
Section titled “Flashcards”Ten cards. Click any card to reveal the answer. Use the Print flashcards button to lay out the full set as one card per page, ready to print or save as a PDF for offline review.
Q. What is Cramer's rule for solving M·v = b?
Each coordinate is det(M with its column replaced by b) / det(M). For 2x2: x = det([b | second column]) / det(M), y = det([first column | b]) / det(M).
Q. What single geometric idea is Cramer's rule built on?
A coordinate of the unknown is a signed area. The parallelogram spanned by [x, y] and j-hat has signed area x (base 1, height x); the one spanned by i-hat and [x, y] has area y.
Q. Why does applying M let you compute the coordinate from b?
M scales every area by det(M), so the area x becomes det(M)·x. M also sends the parallelogram’s sides to b and a column of M, so the new area is det([b | column]). Equate and divide.
Q. In Cramer's rule, which column do you replace for x? For y?
For x (first coordinate), replace the first column of M with b. For y (second coordinate), replace the second column. Match each coordinate to its own column.
Q. What does det(M) = 0 mean for Cramer's rule?
The rule divides by zero, honestly signaling no unique solution. A zero determinant is the collapsed case: the system has no solution or infinitely many, so there is no single answer to compute.
Q. Is Cramer's rule the fast way to solve large systems?
No. It is a conceptual tool and fine for tiny systems, but a determinant per unknown scales badly. Gaussian elimination and iterative solvers are far faster for real systems.
Q. How does Cramer's rule extend beyond 2x2?
The same way: for an n by n invertible system, the i-th unknown is det(M with its i-th column replaced by b) / det(M). In 3D the areas become volumes, scaled by the 3x3 determinant.
Q. What is the denominator in every Cramer's-rule coordinate?
det(M), the determinant of the original, unmodified matrix. Only the numerator changes per coordinate (one column swapped for b); the denominator is always the same.
Q. Why is a coordinate like x equal to a signed area?
The parallelogram with sides [x, y] and j-hat has base 1 (along the y-axis) and height equal to the horizontal distance of [x, y] from the y-axis, which is x. So its area is 1·x = x, confirmable as det([[x, 0], [y, 1]]) = x.
Q. What earlier question does Cramer's rule close?
The inverses lesson said the solution to M·v = b is M^-1·b but did not compute it. Cramer’s rule is a concrete way to compute that solution directly from the matrix entries, derived purely from “area scales by the determinant.”