The determinant of a 2 by 2 and a 3 by 3 matrix, using a determinant to find the area of a triangle, the coefficient matrix, Cramer's rule for solving a system, and what a zero determinant tells you.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 3 — Linear Systems and Matrices
Evaluate Determinants and Apply Cramer's Rule
Objectives
Five outcomes. The fifth is the one that makes the determinant worth computing before anything else.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-207 — the lesson these objectives are drawn from
Warm-up
Lesson 3.1 classified systems by comparing slopes. This lesson attaches a single number to a system that does the same job.
Discussion prompt
The systems 2x plus 3y equals 7 with 4x plus 6y equals 5, and 2x plus 3y equals 7 with 4x minus y equals 1, differ only in the second equation. Which has a unique solution, and what did you compare to decide?
Hint: Look at the ratio of the coefficients in each system.
Answer:
The second has a unique solution. In the first, the second equation's coefficients are exactly twice the first's, so the lines are parallel; in the second they are not proportional, so the lines cross.
\[ 2(6) - 4(3) = 0 \qquad 2(-1) - 4(3) = -14 \]
Those two numbers are the determinants of the coefficient matrices, and the fact that one is zero and the other is not is exactly what distinguished the two systems.
Concept
Every square matrix has a determinant: a single real number computed from its entries. It measures area, it decides whether a system has a unique solution, and it will decide in Lesson 3.8 whether the matrix has an inverse.
determinant — A real number associated with every square matrix, written det A or with vertical bars around the matrix.
\[ \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - cb \]
Only square matrices have determinants. A 2 by 3 matrix has none, which is one more reason square matrices are treated separately.
Figure (svg): A two by two matrix with its main diagonal and anti-diagonal marked, and the determinant computed as their difference
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-203 — The Determinant of a Matrix
Section
Section 1
Concept
Multiply along the main diagonal, multiply along the other diagonal, and subtract the second from the first. The order of the subtraction matters, so the main diagonal always comes first.
\[ \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix} = 5(1) - 3(4) = -7 \]
The notation with vertical bars means the determinant, a number, while square brackets mean the matrix itself. They are different objects and the notation keeps them apart.
Figure (svg): A two by two matrix with its main diagonal and anti-diagonal marked, and the determinant computed as their difference
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-203
Picture it
Example 1a: the determinant of the matrix with rows 5, 4 and 3, 1.
Figure (svg): A two by two matrix with its main diagonal and anti-diagonal marked, and the determinant computed as their difference
A determinant can be negative, as here, and that is not an error. Its sign carries information, which the area formula in Section 3 will have to handle deliberately.
Worked example
Example 1a. Two products and a subtraction.
\[ \text{Evaluate } \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix}. \]
Multiply along the main diagonal
Why: Top-left times bottom-right: five times one.
\[ 5 \]
Multiply along the other diagonal
Why: Bottom-left times top-right: three times four.
\[ 12 \]
Subtract, main diagonal first
Why: Five minus twelve.
\[ -7 \]
Note the sign
Why: A negative determinant is perfectly normal; nothing has gone wrong.
\[ \det = -7 \]
Figure (svg): The solution to Worked example a 2 by 2 determinant shown as a ladder of expressions, one row per algebraic move
\[ \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix} = -7 \]
Verify: swap the two rows and recompute
Why: Swapping gives the matrix with rows 3, 1 and 5, 4, whose determinant is 3 times 4 minus 5 times 1, which is 7. Swapping two rows reverses the sign of a determinant, and getting exactly the negative of the first answer confirms both computations.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-203
Fill the middle
A 2 by 2 with two negatives.
Fill in the blanks
\begin2 -3 & -4 \\ -1 & -2 \end___ = (-3)(-2) - (-1)(-4) = ___
Why: Negative three times negative two is six, and negative one times negative four is four, so the determinant is six minus four, which is two. Both diagonal products came out positive because each involved two negatives — a useful pattern to watch for, since it means the sign of the answer depends only on which product is larger.
Worked example
The same procedure when the entries carry signs.
\[ \text{Evaluate } \begin{vmatrix} 9 & 4 \\ 3 & -5 \end{vmatrix}. \]
Main diagonal product
Why: Nine times negative five.
\[ -45 \]
Other diagonal product
Why: Three times four.
\[ 12 \]
Subtract in order
Why: Negative forty-five minus twelve.
\[ -57 \]
State it
Why: The determinant is negative fifty-seven.
\[ \det = -57 \]
Figure (svg): The solution to Worked example a determinant with negatives shown as a ladder of expressions, one row per algebraic move
\[ \begin{vmatrix} 9 & 4 \\ 3 & -5 \end{vmatrix} = -57 \]
Verify: check the sign against the diagonal signs
Why: The main diagonal product is negative and the other is positive, so subtracting makes the result more negative still — the answer had to be a large negative number, which it is. Estimating the sign before computing catches a dropped minus.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 205-205
Trap
\[ \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix} \]
Subtract the main diagonal from the other
Why: The two products are computed correctly and then combined the wrong way round.
\[ 12 - 5 = 7 \quad \text{(wrong sign)} \]
Determinants are not symmetric in the two diagonals: reversing the subtraction always flips the sign.
\[ \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix} = 5(1) - 3(4) = -7 \]
Main diagonal first, always
Why: The formula is ad minus cb, reading the main diagonal from the top-left corner.
The sign matters. In Cramer's rule a flipped sign in the numerator and not the denominator would give the negative of the right answer, which passes no check.
Sorting
Compute mentally where you can.
Sort into buckets
Sort each matrix by whether its determinant is zero.
Proportional rows and a zero determinant are the same condition, which is exactly the parallel-lines condition from Lesson 3.1 wearing different clothes.
Prediction
Commit before computing.
Predict first
You double every entry of the first row of a 2 by 2 matrix. What happens to its determinant?
Correct: It doubles.
\[ \begin{vmatrix} 10 & 8 \\ 3 & 1 \end{vmatrix} = 10 - 24 = -14 = 2(-7) \]
Why: Each diagonal product contains exactly one entry from the first row, so doubling that row doubles both products and therefore doubles their difference. Doubling the WHOLE matrix would double both rows and quadruple the determinant, which is why the distinction matters. Testing on the matrix with rows 5, 4 and 3, 1 gives -7, and doubling the first row gives 10 times 1 minus 3 times 8, which is -14.
Explain it to yourself
Only square matrices have determinants.
\[ \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} \]
Discussion prompt
Explain why the diagonal recipe cannot be applied to this 2 by 3 matrix. What goes wrong when you try to follow the main diagonal?
Hint: Try to trace a diagonal from the top-left corner.
Answer:
Starting at the top-left and moving down and right, you reach 1 and then 5 and then run out of rows — the diagonal hits the bottom before it reaches the last column.
There is no way to pair every row with a distinct column when the counts differ, and every definition of a determinant depends on exactly that pairing. So the determinant of a non-square matrix is not merely hard to compute; it does not exist.
Section
Section 2
Concept
Copy the first two columns to the right of the matrix. Add the three products running down and to the right, subtract the three running up and to the right, and the difference is the determinant.
\[ \begin{vmatrix} a & b & c \\ d & e & f \\ g & h & i \end{vmatrix} = (aei + bfg + cdh) - (gec + hfa + idb) \]
The repeated columns are scaffolding, not part of the matrix. They exist only to make the six diagonals visible.
Figure (svg): A three by three determinant with the first two columns repeated, showing three downward and three upward products
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-203 — Determinant of a 3 by 3 Matrix
Picture it
Example 1b, with the first two columns repeated.
Figure (svg): A three by three determinant with the first two columns repeated, showing three downward and three upward products
The downward products add and the upward products subtract. Marking the two directions in different colours as you work is the simplest way to avoid mixing them up.
Worked example
Example 1b, all six products.
\[ \text{Evaluate } \begin{vmatrix} 2 & -1 & -3 \\ 4 & 1 & 0 \\ 3 & -4 & -2 \end{vmatrix}. \]
Repeat the first two columns to the right
Why: The array now has five columns, the last two copies of the first two.
\[ \text{columns } 2, -1\text{ and } 4, 1\text{ and } 3, -4\text{ repeated} \]
Take the three downward products
Why: Two times one times negative two is negative four; negative one times zero times three is zero; negative three times four times negative four is forty-eight.
\[ -4, 0, 48 \]
Take the three upward products
Why: Three times one times negative three is negative nine; negative four times zero times two is zero; negative two times four times negative one is eight.
\[ -9, 0, 8 \]
Subtract the second sum from the first
Why: Forty-four minus negative one.
\[ 44 + 1 = 45 \]
Figure (svg): The solution to Worked example a 3 by 3 determinant shown as a ladder of expressions, one row per algebraic move
\[ \det = 44 - (-1) = 45 \]
Verify: recount the number of products
Why: Six products in total, three added and three subtracted, and every one used exactly one entry from each row and each column. If any product repeats a row or a column, a diagonal was traced wrongly. Here the two zeros came from the single 0 entry, which correctly appears in exactly two of the six products.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-203
Fill the middle
Example 1b's largest downward product.
Fill in the blanks
(-3)(4)(-4) = 48
Why: Negative three times four is negative twelve, and negative twelve times negative four is positive forty-eight. This product uses the entry from row 1 column 3, row 2 column 1 and row 3 column 2 — one from each row and one from each column, as every valid product must.
Worked example
A case where the answer can be predicted before computing.
\[ \text{Evaluate } \begin{vmatrix} 2 & 5 & 1 \\ 0 & 0 & 0 \\ 3 & -4 & 7 \end{vmatrix}. \]
Notice the middle row
Why: Every entry is zero.
\[ \text{row } 2\text{ is all zeros} \]
Consider what each product contains
Why: Every one of the six diagonal products uses exactly one entry from each row, so every product includes a factor from the zero row.
Conclude
Why: Every product is zero, so both sums are zero and their difference is zero.
\[ \det = 0 \]
Compute it anyway to confirm
Why: All six products are zero, and zero minus zero is zero.
Figure (svg): The solution to Worked example a determinant with a row of zeros shown as a ladder of expressions, one row per algebraic move
\[ \det = 0 \]
Verify: say what a zero determinant will mean later
Why: A zero determinant will mean the matrix has no inverse and the corresponding system has no unique solution. A row of zeros corresponds to an equation reading 0 equals something, which is exactly the degenerate case from Lesson 3.4 — the algebra and the determinant agree.
Trap
\[ \begin{vmatrix} 2 & -1 & -3 \\ 4 & 1 & 0 \\ 3 & -4 & -2 \end{vmatrix} \text{ with two columns copied} \]
Count the copied columns as extra data and take more than six products
Why: The scaffolding is mistaken for the matrix itself.
A 3 by 3 determinant has exactly six products, three each way. Any more means a diagonal was traced twice or through the copies as if they were new.
\[ \text{six products: three downward, three upward} \]
Use the copies only to see the diagonals, and count exactly three in each direction
Why: Each diagonal starts in the top row of the original three columns and runs three entries long.
A quick check: every product must contain exactly one entry from each ROW and one from each COLUMN of the original matrix. Six such products exist and no more.
Prediction
Commit before doing any arithmetic.
Predict first
A 3 by 3 matrix has two identical rows. What is its determinant?
Correct: 0.
The same holds for two identical columns, and for a row that is any multiple of another. A quick scan for proportional rows or columns can settle a determinant instantly.
Why: Two identical rows means the rows are proportional in the strongest possible way, and proportional rows always give a zero determinant — the same fact that made the 2 by 2 with a doubled row come out zero. Geometrically it means the three rows do not span three dimensions, so the corresponding system's three planes cannot meet at a single point. Recognising this saves computing six products.
Ranking
Evaluating a 3 by 3 determinant.
Put in order
Why: The copies must exist before the diagonals can be traced. Keeping the two sets of products separate until the very end is what prevents the commonest error, which is subtracting one product too early and losing track of the signs. The final check is cheap and catches a mis-traced diagonal.
Comparison
Fill the blanks. The structure is the same; the counting is not.
Comparison matrix
| Feature | 2 by 2 | 3 by 3 |
|---|---|---|
| Number of products | 2 | 6 |
| Products added | 1 | 3 |
| Products subtracted | 1 | 3 |
| Extra columns needed | none | the first two, copied |
| Zero when rows are proportional | yes | yes |
The last row is the one that carries across every size, and it is the property this chapter actually uses.
Section
Section 3
Concept
The area of a triangle with given vertices is half the absolute value of the determinant of the matrix whose rows are those vertices, each with a 1 appended.
\[ \text{Area} = \pm\tfrac{1}{2}\begin{vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{vmatrix} \]
The plus-or-minus is there because a determinant can be negative while an area cannot. Choose whichever sign makes the answer positive.
Figure (svg): A triangle on a coordinate plane with its three vertices labelled and the determinant formula for its area beside it
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 204-204 — Area of a Triangle
Picture it
Example 2: the region bounded by three points, in miles.
Figure (svg): A triangle on a coordinate plane with its three vertices labelled and the determinant formula for its area beside it
The determinant came out as negative 1515, and half its absolute value is 757.5 — about 758 square miles. The negative sign records the order the vertices were listed, not anything about the triangle.
Worked example
Example 2. Three coordinates, one determinant.
\[ \text{Find the area of the triangle with vertices } (-1, 41), (38, -43) \text{ and } (0, 0). \]
Build the 3 by 3 matrix
Why: Each row is a vertex with a 1 appended.
\[ \text{rows } (-1, 41, 1), (38, -43, 1), (0, 0, 1) \]
Take the three downward products
Why: Negative one times negative forty-three times one is forty-three; the other two involve the zeros and vanish.
\[ 43, 0, 0 \]
Take the three upward products
Why: Two involve the zeros; the third is one times thirty-eight times forty-one, which is 1558.
\[ 0, 0, 1558 \]
Subtract and halve the absolute value
Why: Forty-three minus 1558 is negative 1515; half of 1515 is 757.5.
\[ 757.5 \]
Figure (svg): The solution to Worked example the sea lion triangle shown as a ladder of expressions, one row per algebraic move
\[ \text{Area} = \tfrac{1}{2}\lvert -1515 \rvert = 757.5 \text{ sq mi} \]
Verify: estimate the area from the picture
Why: The triangle spans about 39 miles horizontally and about 84 miles vertically, so a bounding rectangle would be roughly 3276 square miles. A triangle occupies less than half of a bounding box, and 757.5 is under a quarter of it — consistent with a long thin triangle, which is what the coordinates describe.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 204-204
Fill the middle
The sea lion triangle's largest upward product.
Fill in the blanks
1 \times 38 \times 41 = 1558
Why: Thirty-eight times forty-one is 1558. This is the only nonzero upward product, because the third vertex is the origin and contributes zeros to the other two. The whole determinant therefore reduces to 43 minus 1558, which is why placing one vertex at the origin makes these calculations so much shorter.
Worked example
The same formula on a triangle whose area can be checked by hand.
\[ \text{Find the area of the triangle with vertices } (0, 0), (6, 0) \text{ and } (0, 4). \]
Build the matrix
Why: Rows are (0,0,1), (6,0,1) and (0,4,1).
Take the downward products
Why: Zero times zero times one is zero; zero times one times zero is zero; one times six times four is 24.
\[ 0, 0, 24 \]
Take the upward products
Why: All three involve a zero factor.
\[ 0, 0, 0 \]
Subtract and halve
Why: Twenty-four minus zero is 24, and half of 24 is 12.
\[ \text{area } = 12 \]
Figure (svg): The solution to Worked example a triangle with easy coordinates shown as a ladder of expressions, one row per algebraic move
\[ \text{Area} = \tfrac{1}{2}(24) = 12 \]
Verify: compute the area the ordinary way
Why: This is a right triangle with legs of 6 and 4, so its area is half of 6 times 4, which is 12 — matching exactly. Testing the determinant formula against a triangle you can compute by hand is the best way to be confident you are applying it correctly.
Error analysis
A student computes the sea lion triangle's area and reports the determinant's sign.
Annotate
On: \( \text{Area} = \tfrac{1}{2}(-1515) = -757.5 \text{ square miles} \)
Whenever a formula produces a quantity that cannot be negative, the sign is telling you about the setup rather than the answer. Read it, then discard it.
Real world
A plot of land has corners at (0, 0), (120, 30) and (40, 90), measured in metres.
Discussion prompt
Use the determinant formula to find its area, then check your answer is plausible against a rough bounding rectangle.
Hint: One vertex at the origin will simplify the determinant considerably.
Answer:
\[ \begin{vmatrix} 0 & 0 & 1 \\ 120 & 30 & 1 \\ 40 & 90 & 1 \end{vmatrix} = (0 + 0 + 10800) - (0 + 0 + 1200) = 9600 \]
\[ \text{Area} = \tfrac{1}{2}(9600) = 4800 \text{ square metres} \]
A bounding rectangle 120 by 90 has area 10,800 square metres, and the triangle occupies 4800 of it — a little under half, which is what a triangle filling most of a rectangle looks like. The answer is plausible.
Prediction
Commit before computing.
Predict first
What determinant do you get for the vertices (0,0), (2,2) and (5,5)?
Correct: Zero — the three points lie on one line.
\[ \begin{vmatrix} 0 & 0 & 1 \\ 2 & 2 & 1 \\ 5 & 5 & 1 \end{vmatrix} = 0 \]
Why: All three points satisfy y equals x, so they are collinear and enclose no area at all. The formula correctly returns zero, which is a genuinely useful result: a zero determinant here is the test for whether three points lie on a straight line. That is the same zero-determinant signal that meant parallel lines in the warm-up and will mean no inverse in Lesson 3.8.
Explain it
A classmate is bothered by a formula that contains a choice.
Discussion prompt
In three sentences, explain why the area formula has a plus-or-minus rather than a fixed sign, and tell them how to decide which to use.
Hint: The sign depends on something that does not affect the triangle.
Answer:
The determinant's sign depends on the ORDER the three vertices are listed, and listing them the other way round flips it. But the triangle is the same triangle either way, so its area cannot depend on that choice.
The plus-or-minus lets the formula discard the sign: choose whichever gives a positive answer, which is the same as taking the absolute value.
Section
Section 4
Concept
For a two-variable system, form the coefficient matrix. Replace its first column with the constants to get the numerator for x, and its second column for y. Divide each by the determinant of the coefficient matrix.
coefficient matrix — The matrix formed from the coefficients of the variables in a linear system, with one row per equation.
\[ x = \frac{\begin{vmatrix} e & b \\ f & d \end{vmatrix}}{\det A}, \qquad y = \frac{\begin{vmatrix} a & e \\ c & f \end{vmatrix}}{\det A} \]
The rule requires the determinant of the coefficient matrix to be nonzero, because otherwise you would be dividing by zero.
Figure (svg): Cramer's rule shown as three determinants: the coefficient matrix and the two matrices with a column replaced
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 205-205 — Cramer's Rule for a 2 by 2 System
Picture it
Example 3: the system 9x plus 4y equals negative 6, and 3x minus 5y equals negative 21.
Figure (svg): Cramer's rule shown as three determinants: the coefficient matrix and the two matrices with a column replaced
The coefficient determinant appears in both denominators, so it is computed once. Each numerator differs from it in exactly one column.
Worked example
Example 3, both steps.
\[ \begin{cases} 9x + 4y = -6 \\ 3x - 5y = -21 \end{cases} \]
Evaluate the determinant of the coefficient matrix
Why: Nine times negative five minus three times four.
\[ -45 - 12 = -57 \]
Confirm it is not zero
Why: Negative fifty-seven is nonzero, so the rule applies and the system has exactly one solution.
Replace the x column with the constants and evaluate
Why: Negative six times negative five minus negative twenty-one times four is thirty plus eighty-four.
\[ 114 \]
Divide to get x
Why: One hundred and fourteen over negative fifty-seven.
\[ x = -2 \]
Replace the y column and divide
Why: Nine times negative twenty-one minus three times negative six is negative 189 plus 18, which is negative 171, over negative 57.
\[ y = 3 \]
Figure (svg): The solution to Worked example Cramer's rule on a 2 by 2 system shown as a ladder of expressions, one row per algebraic move
\[ (-2, 3) \]
Verify: substitute into both original equations
Why: First: 9 times -2 plus 4 times 3 is -18 plus 12, which is -6 — correct. Second: 3 times -2 minus 5 times 3 is -6 minus 15, which is -21 — correct. Both check, and the solution came out without a single elimination step.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 205-205
Fill the middle
Example 3's system.
Fill in the blanks
y = \frac-6 9 & ___ \\ 3 & -21 \end___}___
Why: To find y, the second column — the y coefficients 4 and negative 5 — is replaced by the constants negative 6 and negative 21. So the top-right entry becomes negative 6. The x column is left alone, which is what distinguishes this numerator from the one used for x.
Worked example
The same procedure on a system from Lesson 3.2.
\[ \begin{cases} 2x + 5y = -5 \\ x + 3y = 3 \end{cases} \]
Coefficient determinant
Why: Two times three minus one times five.
\[ 6 - 5 = 1 \]
Replace the x column
Why: Negative five times three minus three times five is negative fifteen minus fifteen.
\[ -30 \]
Divide for x
Why: Negative thirty over one.
\[ x = -30 \]
Replace the y column and divide
Why: Two times three minus one times negative five is six plus five, which is 11, over 1.
\[ y = 11 \]
Figure (svg): The solution to Worked example a second system by Cramer's rule shown as a ladder of expressions, one row per algebraic move
\[ (-30, 11) \]
Verify: compare with the Lesson 3.2 answer
Why: Lesson 3.2 solved this same system by substitution and got (-30, 11). Two entirely different methods reaching the same pair is strong confirmation, and it also shows that a determinant of 1 makes Cramer's rule especially quick — the divisions are trivial.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 205-205
Trap
\[ \begin{cases} 9x + 4y = -6 \\ 3x - 5y = -21 \end{cases} \]
Replace the second column with the constants to find x
Why: The constants are put in the column nearest to hand rather than in the one belonging to the variable being found.
\[ x = \frac{\begin{vmatrix} 9 & -6 \\ 3 & -21 \end{vmatrix}}{-57} = 3 \quad \text{(that is y)} \]
Substituting x equal to 3 into the first equation gives 27 plus 4y equals negative 6, so y would be negative 33 — nothing like the true solution.
\[ x = \frac{\begin{vmatrix} -6 & 4 \\ -21 & -5 \end{vmatrix}}{-57} = -2 \]
Replace the column belonging to the variable you are finding
Why: For x, the first column; for y, the second. The variable's own column is the one the constants take over.
A memorable phrasing: the constants move into the seat of the variable you want. Getting this backwards swaps the two answers, which a substitution check catches immediately.
Matching
Cramer's rule uses three determinants.
Match the pairs
Why: Only one determinant serves as a denominator, and it is computed first for a reason: if it is zero the rule cannot be used at all, and the other two determinants need not be computed. That ordering makes Cramer's rule an efficient way to test for a unique solution as well as to find one.
Comparison
Fill the blanks. Both give exact answers.
Comparison matrix
| Feature | Elimination (3.2) | Cramer's rule |
|---|---|---|
| What you compute | a sequence of equations | three determinants |
| Tells you about uniqueness | at the end | at the start |
| Finds one variable alone | no - you get one, then back-substitute | yes, each independently |
| Works for degenerate systems | yes, and identifies which kind | no - it stops at det = 0 |
Cramer's rule can find one variable without finding the other, which elimination cannot. Elimination can distinguish no solution from infinitely many, which Cramer's rule cannot.
Prediction
Commit before reasoning.
Predict first
For the system 2x + 5y = -5 and x + 3y = 3, is Cramer's rule quicker than substitution?
Correct: Yes, because the coefficient determinant is 1, making both divisions trivial.
\[ \det = 2(3) - 1(5) = 1 \;\Longrightarrow\; x = -30, \; y = 11 \text{ directly} \]
Why: A determinant of 1 means the numerators are the answers directly, with no division to do. When the determinant is an awkward number like negative 57, the two divisions are real work and substitution may well be faster — especially when a coefficient of one is available. Neither method is always quicker; computing the coefficient determinant first tells you which situation you are in.
Section
Section 5
Concept
Cramer's rule requires the coefficient determinant to be nonzero. When it is zero the rule cannot be applied — and that is not a limitation of the method but a fact about the system, which then has either no solution or infinitely many.
\[ \det A \neq 0 \;\Longleftrightarrow\; \text{exactly one solution} \]
A zero determinant means the coefficient rows are proportional, which means the lines are parallel or identical — exactly Lesson 3.1's two degenerate cases.
Figure (svg): Two columns contrasting a nonzero determinant with a zero one and what each says about the system
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 205-205
Picture it
Nonzero against zero, and what each permits.
Figure (svg): Two columns contrasting a nonzero determinant with a zero one and what each says about the system
Computing the determinant first is worth the twenty seconds. It tells you whether a unique solution exists before you invest any effort in finding one.
Worked example
A determinant of zero, and what to do next.
\[ \begin{cases} 2x + 3y = 7 \\ 4x + 6y = 5 \end{cases} \]
Evaluate the coefficient determinant
Why: Two times six minus four times three.
\[ 12 - 12 = 0 \]
Recognise that the rule does not apply
Why: Both denominators would be zero, and division by zero is undefined.
Fall back on Lesson 3.1's test
Why: The second equation's coefficients are twice the first's, but 5 is not twice 7, so the lines are parallel and distinct.
State the conclusion
Why: The system has no solution and is inconsistent.
Figure (svg): The solution to Worked example a system Cramer's rule cannot solve shown as a ladder of expressions, one row per algebraic move
\[ \det A = 0 \;\Longrightarrow\; \text{no unique solution} \]
Verify: confirm by elimination
Why: Multiplying the first equation by negative two and adding gives 0 equals negative 9, which is false — so no solution, exactly as the zero determinant warned. The determinant told you there was no unique solution; elimination told you which of the two degenerate cases it was.
Sorting
Compute the coefficient determinant mentally where you can.
Sort into buckets
Sort each system by whether Cramer's rule can be used.
Notice that two systems with identical coefficients share a determinant, whatever their constants. The determinant is a fact about the left-hand sides alone.
Worked example
The same determinant, a different outcome.
\[ \begin{cases} 2x + 3y = 7 \\ 4x + 6y = 14 \end{cases} \]
Evaluate the coefficient determinant
Why: Exactly as before: twelve minus twelve.
\[ \det = 0 \]
Note that the determinant cannot distinguish the two cases
Why: It depends only on the coefficients, and the coefficients here are identical to the previous system's.
Compare the constants
Why: Fourteen IS twice seven, so the second equation is exactly twice the first.
State the conclusion
Why: The graphs coincide, so there are infinitely many solutions.
Figure (svg): The solution to Worked example zero determinant, infinitely many solutions shown as a ladder of expressions, one row per algebraic move
\[ \det A = 0 \text{ and the equations are proportional} \;\Longrightarrow\; \text{infinitely many} \]
Verify: compare the two systems
Why: The two systems have identical coefficient matrices and therefore identical determinants, yet one has no solution and the other infinitely many. So a zero determinant rules out uniqueness without deciding which degenerate case applies — the constants settle that, and the determinant never looks at them.
Error analysis
A student applies Cramer's rule without checking the denominator.
Annotate
On: \( x = \frac{\begin{vmatrix} 7 & 3 \\ 5 & 6 \end{vmatrix}}{0} = \frac{27}{0} = 0 \)
Compute the coefficient determinant first and check it before computing anything else. That single check is why the rule is stated with the condition attached.
Two truths and a lie
All three are about zero determinants.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: C
Why: The survivor is the false one. A zero determinant rules out a UNIQUE solution but permits either degenerate case. The system with constants 7 and 14 has a zero determinant and infinitely many solutions, so no solution is only half the story.
Prediction
Commit before reasoning.
Predict first
Two systems have the same coefficient matrix but different constants. What must be true of their determinants?
Correct: They must be equal — the determinant depends only on the coefficients.
This also explains why Cramer's rule needs three determinants rather than one: the two numerators are where the constants finally enter.
Why: The coefficient matrix contains no constants at all, so changing them cannot change its determinant. That is why the determinant can tell you whether a unique solution exists without telling you what it is, and why it cannot distinguish an inconsistent system from a dependent one — both of those distinctions live entirely in the constants.
Socratic
One question, and nothing else on this slide.
\[ \det\begin{bmatrix} a & b \\ c & d \end{bmatrix} = ad - cb \]
Discussion prompt
The same number decides whether three points are collinear, whether two lines are parallel, and whether a system has a unique solution. What do those three situations have in common, and what does that suggest the determinant is measuring?
Hint: Think about what is missing in each degenerate case.
Answer:
In each case something has collapsed: three points that should outline a triangle fall on a line, two lines that should cross run parallel, a system that should pin down a point pins down a line instead. A dimension has been lost.
The determinant measures exactly that. For a 2 by 2 it is the area of the parallelogram the two rows span, and for a 3 by 3 it is a volume — so a zero determinant means the rows do not span the space they should, and the collapse follows.
Lesson 3.8 gives the same fact a fourth face: a matrix has an inverse if and only if its determinant is nonzero. One number, four consequences.
Comparison
Fill the blanks. One number, several jobs.
Comparison matrix
| Situation | Determinant is zero | Determinant is not zero |
|---|---|---|
| Three points as rows | collinear, no triangle | a genuine triangle |
| A system's coefficients | no unique solution | exactly one solution |
| Cramer's rule | does not apply | applies |
| The rows of the matrix | proportional | not proportional |
Every row says the same thing in different words: a zero determinant means a collapse, and a nonzero one means everything is in general position.
Pattern
One routine, whether you want an area or a solution.
Step four is what makes Cramer's rule safe. The determinant is both the denominator and the permission to divide.
OpenStax Algebra and Trigonometry 2e, §11.8 Solving Systems with Cramer's Rule §11.8
Check
A 2 by 2 determinant. Main diagonal first.
Check your understanding
Evaluate the determinant of [[9, 4], [3, -5]].
Answer: A
Why: The main diagonal product is 9 times -5, which is -45. The other diagonal product is 3 times 4, which is 12. Subtracting gives -45 minus 12, or -57.
Check
Cramer's rule. Replace the right column.
Check your understanding
For the system 9x + 4y = -6 and 3x - 5y = -21, what determinant is the numerator for x?
Answer: A
Why: To find x, the x column — the coefficients 9 and 3 — is replaced by the constants -6 and -21. The y column is left alone.
Check
The condition on the rule.
Check your understanding
The coefficient determinant of a system is zero. What can you conclude?
Answer: A
Why: A zero determinant means the coefficient rows are proportional, so the lines are parallel or identical. Which of the two depends on the constants, which the determinant never uses.
Real world
A surveyor has GPS coordinates for the three corners of a triangular field and needs its area without walking it.
Discussion prompt
Explain how the determinant formula gives the answer, why a sign might appear that has to be discarded, and what a determinant of zero would tell the surveyor about the three readings.
Hint: Consider what could make three GPS points give a zero determinant.
Answer:
Put the three coordinate pairs as rows with a 1 appended, take the determinant and halve the absolute value. That is the area, in whatever square units the coordinates use.
The sign depends on whether the three corners were listed clockwise or anticlockwise, which is a fact about the recording rather than the field, so it is discarded.
A determinant of zero would mean the three points are collinear — so either the field is degenerate, or, far more likely, two of the readings are the same point or one was mistyped. In surveying software that check runs automatically for exactly this reason.
Commit first
Answer, then rate your confidence honestly.
Predict first
Does swapping the two rows of a 2 by 2 matrix change its determinant?
Correct: Yes — the sign reverses.
\[ \begin{vmatrix} 5 & 4 \\ 3 & 1 \end{vmatrix} = -7 \qquad \begin{vmatrix} 3 & 1 \\ 5 & 4 \end{vmatrix} = 7 \]
Why: Swapping the rows of the matrix with rows a, b and c, d gives rows c, d and a, b, whose determinant is cb minus ad — exactly the negative of ad minus cb. This is why the area formula needs a plus-or-minus: listing the vertices in a different order swaps rows and flips the sign, without changing the triangle at all. The size of the determinant is unchanged, which is what makes the absolute value the meaningful quantity.
Explain it
They can solve systems by elimination and wonder why another method is needed.
Discussion prompt
In four sentences or fewer, tell them what a determinant is, what Cramer's rule does with it, and the one advantage the rule has over elimination.
Hint: The advantage is about finding one variable.
Answer:
A determinant is one number computed from a square matrix — for a 2 by 2 it is the main diagonal product minus the other diagonal product. Cramer's rule uses three of them: the coefficients alone give the denominator, and replacing a variable's column with the constants gives that variable's numerator.
The advantage is that each variable comes out independently, so you can find just y without ever finding x. It also tells you at the very first step, from the coefficient determinant, whether the system has a unique solution at all.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For the order, always start at the top-left corner and go down-right first. For the 3 by 3, copy the two columns and mark the two directions in different colours. For Cramer's rule, the constants take the seat of the variable you are finding. For a zero determinant, stop and use elimination to find out which degenerate case it is. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
At the top of a page write a 2 by 2 matrix of your own, draw both diagonals in different colours, and compute its determinant with the subtraction written out. Below it, write a 3 by 3 matrix, copy the first two columns to the right, and mark the three downward and three upward diagonals in your two colours, computing all six products. In the middle of the page, choose three vertices, build the area matrix with its column of ones, compute the determinant and the area, and sketch the triangle to check the answer is plausible. In the lower half, take a 2 by 2 system of your own, compute the coefficient determinant first and box it, then build both numerators by replacing the correct column each time. Finally, in a margin, write what a zero determinant would have meant and what you would have done instead.
The boxed determinant should be the first thing you computed. If it came last, reread Section 5 — it is both the denominator and the permission to divide.
Recap
Five things, and the last one is why the determinant is worth computing before anything else.
| If you see | Then |
|---|---|
| A 2 by 2 matrix | Main diagonal minus the other |
| A 3 by 3 matrix | Copy two columns; six products |
| Three vertices | Append ones; halve; drop the sign |
| A system to solve | Coefficient determinant first |
| A determinant of zero | Stop; no unique solution exists |
Lesson 3.8 uses the determinant one more way: a square matrix has an inverse exactly when its determinant is nonzero, and that inverse solves a whole system in a single multiplication.
McDougal Littell Algebra 2 (Texas Edition), Ch. 3 Linear Systems and Matrices — Lesson 3.7 Evaluate Determinants and Apply Cramer's Rule §3.7, pp. 203-207 — everything on these slides traces back here
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