Determinants of order 2 and 3
Prices of items in two shops written in a square table form a square arrangement of numbers. The rule that turns that arrangement into one single number is the determinant. Building on classes 9–10 : In classes 9–10 we solved two equations by elimination; a determinant packs that same elimination into one formula.
A square arrangement of numbers in rows (across) and columns (down), placed between two vertical bars, is a determinant. With 2 rows and 2 columns it has order 2; with 3 rows and 3 columns, order 3. The value of a determinant is always one single number. Here a determinant is written as [[a, b], [c, d]]: the first row is a, b and the second row is c, d.
Rule 1: Determinant of order 2 — D = ad − bc: product along the main diagonal is positive, product along the other diagonal is negative.
In the figure : The green arrow is the main diagonal (a × d, positive) and the orange arrow is the other diagonal (b × c, negative).
Rule 2: Determinant of order 3 (expansion along the first row) — D = a₁(b₂c₃ − b₃c₂) − b₁(a₂c₃ − a₃c₂) + c₁(a₂b₃ − a₃b₂): multiply each number of the first row by the small determinant left after striking out its row and column; the signs are + − +.
How is the formula obtained? : Take the equations a₁x + b₁y = k₁ and a₂x + b₂y = k₂. Multiply the first by b₂ and the second by b₁ and subtract; y disappears: (a₁b₂ − a₂b₁)x = k₁b₂ − k₂b₁. The multiplier of x, a₁b₂ − a₂b₁, is exactly the value of a determinant of order 2, hence D = ad − bc. Doing the same elimination twice for three equations gives the order 3 formula.
- Easy : [[4, 3], [2, 5]] = 4 × 5 − 3 × 2 = 20 − 6 = 14.
- Medium : [[2, 1, 3], [0, 4, −1], [5, 2, 1]]: D = 2(4×1 − (−1)×2) − 1(0×1 − (−1)×5) + 3(0×2 − 4×5) = 2(6) − 1(5) + 3(−20) = 12 − 5 − 60 = −53.
- Exam-type : If [[x, 4], [3, x]] = 13, find x: x² − 12 = 13, x² = 25, so x = 5 or x = −5.
A determinant is a single number and is always square (equal rows and columns). In order 3 the signs alternate + − +.
Writing ad + bc instead of ad − bc, or forgetting the minus sign of the middle term in order 3.
Sarrus rule for order 3: rewrite the first two columns on the right. Add the three downward diagonal products and subtract the three upward diagonal products. (Only for 3 × 3.) Memory clue : "Down is plus, up is minus" — rain falls down (plus), smoke goes up (minus).
CET often gives an equation with x²; if x² = 25 take both answers (+5 and −5). If the options show only one, verify the other by substitution.
CET-style example : The value of [[2, −1, 3], [1, 0, 4], [−2, 5, 1]] is? (A) 16 (B) −16 (C) 24 (D) −24
Answer : (B). Along the second row (1, 0, 4): D = 1×C₂₁ + 0 + 4×C₂₃ = 1×16 + 4×(−8) = 16 − 32 = −16. Time-saving method : Expand along the row or column containing a 0; one term disappears and slips become rarer.
Spot the difference
| Point | Order 2 | Order 3 |
|---|---|---|
| Shape | 2 rows, 2 columns | 3 rows, 3 columns |
| Total numbers | 4 | 9 |
| Formula | ad − bc | three terms from the first row |
| Sarrus rule | not applicable | applicable |
Practice questions
- 1Find the value of [[6, 2], [3, 5]].
- 2Find the value of [[−2, 4], [1, −3]].
- 3Find the value of the determinant of the 3 × 3 identity matrix (1 on the diagonal, 0 elsewhere).
- 4Find the value of [[3, −1, 2], [1, 0, 4], [2, 5, −2]].
- 5Find the value of [[2, 0, 1], [3, 1, 0], [0, 4, 2]].
- 6If [[x, 2], [3, x − 1]] = 0, find x.
- 7Find all values of x for which [[x, 1, 1], [1, x, 1], [1, 1, x]] = 0.
- 8If [[x, 2], [3, 4]] = [[1, x], [2, 3]], find x.
Show answers (try them yourself first)
- 16×5 − 2×3 = 24.
- 2(−2)(−3) − 4×1 = 2.
- 31.
- 4−60.
- 516.
- 6x(x − 1) − 6 = 0, x² − x − 6 = 0, so x = 3 or x = −2.
- 7Expanding gives x³ − 3x + 2 = 0, i.e. (x − 1)²(x + 2) = 0; x = 1 or x = −2.
- 84x − 6 = 3 − 2x, so x = 3/2.
Minors and cofactors
Think of the benches of a class in rows and columns. Leave out one pupil's row and column and the remaining benches form a smaller square; a minor is made in just this way. Building on classes 9–10 : In classes 9–10 we learnt sign rules; in a cofactor the sign is fixed by the same rule through a power of (−1).
Strike out the row and column of one number of a determinant; the smaller determinant left is the minor of that number. If the number is aᵢⱼ (in row i and column j) its minor is Mᵢⱼ and its cofactor is Cᵢⱼ = (−1)^(i + j) × Mᵢⱼ.
Rule 1: The chess-board of signs — If i + j is even the cofactor equals the minor; if i + j is odd the cofactor is the minor with its sign reversed. The signs are + − + / − + − / + − + as in the figure.
In the figure : Signs of the cofactors for a 3 × 3 determinant; blue = plus, orange = minus.
Rule 2: Expansion along any row or column — D = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ and likewise along any row or column; but if the numbers of one row are multiplied by the cofactors of a different row the sum is 0.
How is the formula obtained? : Take the order 3 formula D = a₁(b₂c₃ − b₃c₂) − b₁(a₂c₃ − a₃c₂) + c₁(a₂b₃ − a₃b₂). The bracket beside a₁ is the minor of a₁ with sign +; the bracket beside b₁ is the minor of b₁ with sign −. These signs are exactly (−1)^(i + j), so D = a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃.
- Easy : In [[3, 1], [2, 4]]: M₁₁ = 4, M₁₂ = 2, M₂₁ = 1, M₂₂ = 3; cofactors C₁₁ = 4, C₁₂ = −2, C₂₁ = −1, C₂₂ = 3.
- Medium : In [[2, −1, 3], [4, 0, 5], [1, 2, −2]] the minor of 5: strike row 2 and column 3 → [[2, −1], [1, 2]] = 4 + 1 = 5; i + j = 5 is odd so the cofactor = −5.
- Exam-type : Value of [[3, 0, 2], [1, 5, 0], [4, 0, 1]]: only 5 is non-zero in column 2. D = 5 × C₂₂ = 5 × (3×1 − 2×4) = 5 × (−5) = −25.
A minor carries no sign; a cofactor does. Cofactor = (−1)^(i + j) × minor.
Forgetting to flip the sign for an odd position, or striking out the wrong row while forming the minor.
To evaluate a determinant choose the row or column with the most zeros; then only one cofactor is needed. Memory clue : "Zero means holiday" — where there is a zero, there is no work.
CET asks for "the cofactor of a particular number". Fix the sign from i + j first, then find the minor; the other way round tends to land on the wrong-sign option.
CET-style example : In [[2, 3, 1], [0, 4, 5], [1, 0, 6]] the cofactor of 5 is? (A) −3 (B) 3 (C) 5 (D) −5
Answer : (B). 5 is in row 2, column 3; striking them gives [[2, 3], [1, 0]] = 0 − 3 = −3 (the minor); i + j = 5 is odd so the cofactor = +3. Time-saving method : If the minor comes out −3 do not pick −3 from the options; after the sign flip the answer is +3.
Spot the difference
| Point | Minor | Cofactor |
|---|---|---|
| Sign | none | (−1)^(i + j) |
| How obtained | by striking out row and column | minor with the sign applied |
| Use | to find the cofactor | expansion of the determinant |
Practice questions
- 1Find all four cofactors of [[1, 2], [3, 4]].
- 2In [[5, 1, 2], [0, 3, 4], [1, 0, 2]] find the minor of 4.
- 3In the same determinant find the cofactor of 4.
- 4Find the value of [[5, 1, 2], [0, 3, 4], [1, 0, 2]] by expanding along the first row.
- 5In [[1, 0, 2], [3, 1, 0], [0, 2, 1]] find the cofactor of the 0 in row 1, column 2.
- 6In [[2, 1, 0], [3, 0, 1], [1, 2, 4]] find the cofactor of the 2 in row 3, column 2.
- 7In [[2, 1, 0], [3, 0, 1], [1, 2, 4]] multiply the numbers of the first row by the cofactors of the third row and add.
- 8Find the value of the same determinant by expanding along the second row.
Show answers (try them yourself first)
- 1C₁₁ = 4, C₁₂ = −3, C₂₁ = −2, C₂₂ = 1.
- 2Strike row 2 and column 3 → [[5, 1], [1, 0]] = −1.
- 3i + j = 5 is odd so the cofactor is +1.
- 45×6 + 1×4 + 2×(−3) = 30 + 4 − 6 = 28.
- 5C₁₂ = −(3×1 − 0×0) = −3.
- 6C₃₂ = −(2×1 − 0×3) = −2.
- 70 (a row multiplied by the cofactors of a different row).
- 8D = −15; it matches the value from the first row.
Properties of determinants
If two lines in a queue swap places the order is reversed, and a determinant reverses its sign in the same way. Remember a few such small rules and big determinants become easy. Building on classes 9–10 : In classes 9–10 we learnt simplification rules in algebra; these properties apply the same technique to determinants.
A property is a rule that tells the value or the sign of a determinant without expanding it fully. The same rules apply to rows and to columns.
Rule 1: Main properties — (1) Interchanging rows and columns leaves the value unchanged. (2) Swapping two rows reverses the sign. (3) If two rows are identical the value is 0. (4) Multiplying one row by k multiplies the value by k. (5) Adding k times one row to another row leaves the value unchanged.
In the figure : Three main properties: swapping rows, equal rows, multiplying a row by k.
Rule 2: More useful rules — If a whole row is zeros the value is 0. If all numbers on one side of the diagonal are zero (triangular) the value is the product of the diagonal numbers. If every number of an order n determinant is multiplied by k the value becomes kⁿ times.
How is the formula obtained? : Property (2): for order 2, [[a, b], [c, d]] = ad − bc. After swapping the rows, [[c, d], [a, b]] = cb − da = −(ad − bc), so the sign reversed. Property (3): if two rows are identical, swapping them leaves the determinant the same yet also reverses its sign; D = −D means 2D = 0, so D = 0.
- Easy : [[1, 2], [3, 4]] = 4 − 6 = −2; its transpose [[1, 3], [2, 4]] = 4 − 6 = −2 (property 1). Rows swapped: [[3, 4], [1, 2]] = 6 − 4 = 2 (sign reversed).
- Medium : In [[3, 6], [2, 5]] take 3 out of the first row: 3 × [[1, 2], [2, 5]] = 3 × (5 − 4) = 3. Directly: 15 − 12 = 3. Both agree.
- Exam-type : Value of [[1, 2, 3], [4, 5, 6], [7, 8, 9]]: after R₂ → R₂ − R₁ and R₃ → R₃ − R₂ the rows are (1, 2, 3), (3, 3, 3), (3, 3, 3). Two rows are equal, so D = 0.
Multiplying the whole order 3 determinant by k makes the value k³ times; multiplying one row by k makes it only k times.
Taking the value as k times after multiplying every row by k (it is kⁿ times) or forgetting the sign change after swapping rows.
For a big determinant first try subtracting or adding rows or columns to make two rows equal or one row zero; an answer of 0 saves long working. Memory clue : "Two equal rows, answer zero" — twins weigh the same, so their difference vanishes.
CET shows determinants in a, b, c; adding columns exposes a common factor like (a + b + c). Take it out and check whether two rows of what is left are equal; the value is 0.
CET-style example : The value of [[1, a, b + c], [1, b, c + a], [1, c, a + b]] is? (A) 0 (B) a + b + c (C) abc (D) 1
Answer : (A). After C₃ → C₃ + C₂ the third column becomes (a + b + c) × (1, 1, 1), which is a multiple of the first column, so D = 0. Time-saving method : Instead of substituting values for a, b, c just add columns; the answer comes in 10 seconds.
Spot the difference
| Change | Effect on the value |
|---|---|
| Interchange rows and columns | value unchanged |
| Swap two rows | sign reverses |
| Multiply one row by k | value becomes k times |
| Multiply all numbers of order n by k | value becomes kⁿ times |
| Add k times one row to another | value unchanged |
Practice questions
- 1Find the value of [[7, 3], [2, 5]] and of its transpose.
- 2Find the value of [[2, 5], [7, 3]] (the rows of the above determinant swapped).
- 3What is the value of [[2, 3, 1], [4, 6, 2], [5, 1, 7]]?
- 4Find the value of [[3, 1, 2], [0, 4, 5], [0, 0, 2]].
- 5If [[3, 4], [5, 7]] = 1, find [[6, 8], [5, 7]].
- 6An order 3 determinant has value 5. If every number is doubled find the new value.
- 7Does the value of [[3, 1, 2], [5, 4, 1], [2, 6, 3]] change after R₂ → R₂ + 2R₁?
- 8Find the value of [[1, 1, 1], [1, 2, 3], [1, 4, 9]].
Show answers (try them yourself first)
- 1Both are 29.
- 2−29 (the sign reversed).
- 30 (the second row is twice the first).
- 43 × 4 × 2 = 24 (triangular).
- 5The first row was multiplied by 2: 2 × 1 = 2 (directly: 42 − 40 = 2).
- 62³ × 5 = 40.
- 7No; the value stays the same (property 5).
- 8(2 − 1)(3 − 1)(3 − 2) = 2.
Applications (Cramer's rule, area, collinearity)
Finding the rate of each item from the total prices, or checking whether three villages lie on one straight road, both become one formula with determinants. Building on classes 9–10 : In classes 9–10 we solved equations in two variables and used the distance formula; here we handle the same equations and points with determinants.
Cramer's rule finds the variables as ratios of determinants: D is the determinant of the coefficients, and Dx, Dy, Dz come from putting the constants in place of one column. It works only when D ≠ 0.
Rule 1: Equations in two and three variables — x = Dx/D, y = Dy/D, z = Dz/D, where D is the determinant of the coefficients; in Dx replace the x column by the constants, in Dy the y column, in Dz the z column.
Rule 2: Area of a triangle and collinearity — Area = ½ × D (taking the positive value of D), where D = [[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]]. If three points are on one line the area is 0, i.e. D = 0. E.g. A(1, 1), B(4, 2), C(2, 5): D = 1(2 − 5) − 1(4 − 2) + 1(20 − 4) = −3 − 2 + 16 = 11, so the area = 11/2 = 5.5.
In the figure : A, B, C are the vertices; area = ½ × the positive value of the 3 × 3 determinant.
How is the formula obtained? : Take a₁x + b₁y = k₁ and a₂x + b₂y = k₂. Eliminating y gives (a₁b₂ − a₂b₁)x = k₁b₂ − k₂b₁, i.e. D·x = Dx, so x = Dx/D. Eliminating x likewise gives D·y = Dy. For the area: writing the base and the height in co-ordinates turns ½ × base × height into the determinant form.
- Easy : 2x + 3y = 13 and 4x − y = 5: D = [[2, 3], [4, −1]] = −2 − 12 = −14; Dx = [[13, 3], [5, −1]] = −13 − 15 = −28; Dy = [[2, 13], [4, 5]] = 10 − 52 = −42. So x = −28/−14 = 2 and y = −42/−14 = 3.
- Medium : x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2: D = 7, Dx = 7, Dy = 14, Dz = 21 (each order 3 determinant expanded). So x = 1, y = 2, z = 3. Check: 1 + 2 + 3 = 6, 2 − 2 + 3 = 3, 1 + 4 − 3 = 2.
- Exam-type : If A(1, 2), B(3, 4), C(k, 8) are collinear find k: D = [[1, 2, 1], [3, 4, 1], [k, 8, 1]] = 1(4 − 8) − 2(3 − k) + 1(24 − 4k) = −4 − 6 + 2k + 24 − 4k = 14 − 2k. D = 0 so k = 7.
If D = 0 Cramer's rule does not apply; the equations have no unique solution (none, or many). An area is never negative.
Replacing the y column instead of the x column while forming Dx, or forgetting the ½ in the area.
For collinearity compare slopes instead of expanding: equal slopes of two pairs mean the points are on one line. For the area, if one side is horizontal use ½ × base × height. Memory clue : "Flat base, straight height" — no determinant needed.
CET gives the area and asks for k; take the determinant as ±2 × area and keep both values of k. Both usually appear in the options.
CET-style example : The area of the triangle with vertices (1, 2), (5, 2), (3, 7) is? (A) 5 (B) 10 (C) 20 (D) 15
Answer : (B). Two points share y = 2, so the base = 5 − 1 = 4 and the height = 7 − 2 = 5; area = ½ × 4 × 5 = 10. The determinant also gives 10. Time-saving method : When the base is horizontal skip the determinant; five seconds.
Spot the difference
| Determinant | How it is formed | Used for |
|---|---|---|
| D | determinant of the coefficients | denominator of every ratio |
| Dx | D with the x column replaced by the constants | x = Dx/D |
| Dy | D with the y column replaced by the constants | y = Dy/D |
| Dz | D with the z column replaced by the constants | z = Dz/D |
Practice questions
- 1Solve x + y = 5 and x − y = 1 by Cramer's rule.
- 2Find the area of the triangle with vertices (0, 0), (4, 0), (0, 6).
- 3Are (1, 1), (2, 2), (3, 3) collinear?
- 4Solve 3x + 2y = 7 and x − 4y = 7 by Cramer's rule.
- 5Find the area of the triangle with vertices (2, 1), (6, 3), (4, 7).
- 6If (−1, 3), (2, 9), (k, 15) are collinear, find k.
- 7Solve x + y + z = 9, 2x − y + z = 5, x + 2y − z = 4 by Cramer's rule.
- 8Find k if the equations x + 2y = 3 and 2x + ky = 9 do not have a unique solution.
Show answers (try them yourself first)
- 1D = −2, Dx = −6, Dy = −4; x = 3, y = 2.
- 2½ × 4 × 6 = 12.
- 3Yes; the determinant is 0 (both slopes are 1).
- 4D = −14, Dx = −42, Dy = 14; x = 3, y = −1.
- 5½ × |−8 − 2 + 30| = ½ × 20 = 10.
- 6Slope (9 − 3)/(2 + 1) = 2; (15 − 9)/(k − 2) = 2, so k = 5.
- 7D = 7, Dx = 14, Dy = 21, Dz = 28; x = 2, y = 3, z = 4.
- 8D = k − 4 = 0, so k = 4 (then Dx = −6 ≠ 0, so there is no solution at all).
Matrices: types and equality
Marks of three pupils in four subjects written in a rectangular table form a matrix. Unlike a determinant it has no single value; the table itself is the thing. Building on classes 9–10 : In classes 9–10 we read statistical tables and equation coefficients; a matrix treats such tables as mathematical objects.
A rectangular arrangement of numbers in rows and columns, enclosed in brackets, is a matrix. With m rows and n columns its order is m × n and it has m·n numbers. The number in row i and column j is written aᵢⱼ.
Rule 1: Types of matrices — Row matrix: only one row. Column matrix: only one column. Square: rows = columns. Null matrix: every number 0. Diagonal: square with 0 outside the diagonal. Scalar: diagonal with equal diagonal numbers. Identity matrix: 1 on the diagonal, 0 elsewhere. Upper triangular: all numbers below the diagonal are 0. Lower triangular: all numbers above the diagonal are 0.
In the figure : A 2 × 3 matrix; a₂₃ is the number in row 2 and column 3.
Rule 2: Equality of matrices — Two matrices are equal when they have the same order and all corresponding numbers are equal.
How is the formula obtained? : Variables from equality: in [[x + y, 2], [3, x − y]] = [[9, 2], [3, 1]] match corresponding places: x + y = 9 and x − y = 1. Adding gives 2x = 10, x = 5; then y = 4. Each place gives one equation, and such sets are easy to solve.
- Easy : The 2 × 3 matrix with aᵢⱼ = 2i + j: a₁₁ = 3, a₁₂ = 4, a₁₃ = 5, a₂₁ = 5, a₂₂ = 6, a₂₃ = 7. So the matrix is [[3, 4, 5], [5, 6, 7]].
- Medium : Identify: [[2, 0, 0], [0, 2, 0], [0, 0, 2]] scalar; [[1, 0, 0], [0, 3, 0], [0, 0, 5]] diagonal; [[1, 2], [0, 3]] upper triangular; [[1, 0], [0, 1]] identity.
- Exam-type : If [[x + y, 2], [3, x − y]] = [[9, 2], [3, 1]], find x and y: x + y = 9, x − y = 1; x = 5 and y = 4.
Order is always "rows × columns". Every identity matrix is scalar and every scalar matrix is diagonal; the reverse is not true.
Treating a matrix as having a value like a determinant. A matrix is a table in brackets; a determinant is one number between bars.
Write the order as row first, then column; 2 × 3 means 2 rows and 3 columns. The count of numbers is the product of the two. Memory clue : "Row then column" — read it as RC, like RC cola.
CET asks "how many matrices are possible?". Multiply the choices for every place: a 2 × 3 matrix with entries 0 or 1 gives 2⁶ = 64.
CET-style example : How many matrices of order 2 × 3 can be formed whose every number is 0 or 1? (A) 12 (B) 36 (C) 64 (D) 8
Answer : (C). There are 6 places with 2 choices each; 2⁶ = 64. Time-saving method : Count the places, take the choices: 2 to the power 6. No need to list.
Spot the difference
| Point | Matrix | Determinant |
|---|---|---|
| Symbol | brackets [ ] | two vertical bars |
| Shape | any (m × n) | square only |
| Value | no value; a table | one number |
| After swapping rows | a different matrix | sign reverses |
Practice questions
- 1State the order and the total number of entries of [[1, 2, 3], [4, 5, 6]].
- 2Write the 2 × 2 matrix in which aᵢⱼ = i × j.
- 3What type is [[5, 0], [0, 5]]?
- 4Write the 3 × 3 matrix in which aᵢⱼ = i − j.
- 5If [[2x, 3], [y, 5]] = [[8, 3], [−2, 5]], find x and y.
- 6What type is [[0, 0], [0, 0]]?
- 7How many matrices of order 3 × 2 can be formed whose every number is 0, 1 or 2?
- 8If [[a + b, b + c], [c + a, 5]] = [[7, 9], [8, 5]], find a, b, c.
Show answers (try them yourself first)
- 1Order 2 × 3, total 6 entries.
- 2[[1, 2], [2, 4]].
- 3Scalar (and diagonal).
- 4[[0, −1, −2], [1, 0, −1], [2, 1, 0]].
- 52x = 8 so x = 4; y = −2.
- 6Null matrix.
- 73⁶ = 729.
- 8a + b = 7, b + c = 9, c + a = 8; adding, 2(a + b + c) = 24, a + b + c = 12; so a = 3, b = 4, c = 5.
Operations on matrices
If the number of items bought by two customers is one table and the price of the items is another, the bill comes by multiplying each row of quantities by the column of prices and adding; matrix multiplication does exactly this. Building on classes 9–10 : In classes 9–10 we learnt the rules of addition, subtraction and multiplication; for matrices the same rules act on tables of numbers, but the order of multiplication matters.
Matrix addition and subtraction work on corresponding places of matrices of the same order. Scalar multiplication multiplies every number by that scalar. The transpose Aᵀ is the matrix obtained by turning the rows of A into columns.
Rule 1: Addition, subtraction and scalar multiplication — Addition or subtraction is possible only when both matrices have the same order. In kA every number of A is multiplied by k. A + B = B + A, but in general AB ≠ BA.
Rule 2: Matrix multiplication — AB exists only when the columns of A = the rows of B. If A is m × n and B is n × p then AB is m × p. The number in place (i, j) of AB is the sum of products of row i of A with column j of B.
In the figure : A (2 × 3) times B (3 × 2) gives a 2 × 2 result; first row × first column = 5.
How is the formula obtained? : One customer bought 2 pens and 3 notebooks, another 1 pen and 4 notebooks; a pen costs 5 rupees and a notebook 20. Bills: first = 2×5 + 3×20 = 70, second = 1×5 + 4×20 = 85. So [[2, 3], [1, 4]] × [[5], [20]] = [[70], [85]]. Each bill is a row multiplied by the column and added; this is the rule.
Rule 3: Transpose, symmetric and skew-symmetric — (A + B)ᵀ = Aᵀ + Bᵀ and (AB)ᵀ = BᵀAᵀ (the order reverses). Symmetric: Aᵀ = A. Skew-symmetric: Aᵀ = −A (all diagonal numbers are 0). Any square A = ½(A + Aᵀ) + ½(A − Aᵀ), i.e. symmetric + skew-symmetric.
- Easy : A = [[2, 1], [0, 3]], B = [[1, 4], [2, 0]]: A + B = [[3, 5], [2, 3]]; 2A − B = [[4, 2], [0, 6]] − [[1, 4], [2, 0]] = [[3, −2], [−2, 6]].
- Medium : A = [[1, 2], [3, 4]], B = [[2, 0], [1, 5]]: AB = [[1×2 + 2×1, 1×0 + 2×5], [3×2 + 4×1, 3×0 + 4×5]] = [[4, 10], [10, 20]]; BA = [[2, 4], [16, 22]]. So AB ≠ BA.
- Exam-type : Symmetric and skew-symmetric parts of A = [[1, 4], [2, 3]]: ½(A + Aᵀ) = [[1, 3], [3, 3]] and ½(A − Aᵀ) = [[0, 1], [−1, 0]]. Their sum = [[1, 4], [2, 3]] = A.
Even when AB exists BA may not. (AB)ᵀ = BᵀAᵀ — the order reverses; not AᵀBᵀ.
Assuming AB = BA, or multiplying corresponding places instead of row × column.
Check the orders before multiplying: (m × n)(n × p) → m × p; the two middle numbers must match and the answer takes the two outer numbers. Many questions are solved by orders alone. Memory clue : "Inside match, outside answer."
CET asks the order of the transpose of a product. The order of (AB)ᵀ is that of AB reversed; if A is 2 × 3 and B is 3 × 4 then AB is 2 × 4 and (AB)ᵀ is 4 × 2.
CET-style example : If A is of order 2 × 3 and B is of order 3 × 4 then the order of (AB)ᵀ is? (A) 2 × 4 (B) 4 × 2 (C) 3 × 3 (D) not possible
Answer : (B). AB has order 2 × 4 (the middle 3s match); transposing gives 4 × 2. Time-saving method : No numbers needed; just the order digits: 2 × 4 reversed is 4 × 2.
Spot the difference
| Point | Symmetric | Skew-symmetric |
|---|---|---|
| Condition | Aᵀ = A | Aᵀ = −A |
| Diagonal numbers | anything | all 0 |
| Example | [[1, 2], [2, 3]] | [[0, 2], [−2, 0]] |
Practice questions
- 1If P = [[4, −1], [2, 5]] and Q = [[3, 2], [−1, 0]], find P + Q.
- 2For the same P find 3P.
- 3Write the transpose of [[1, 2, 3], [4, 5, 6]].
- 4If A = [[1, 0], [2, 3]] and B = [[4, 1], [0, 2]], find AB.
- 5State the order of the product of a 3 × 2 and a 2 × 4 matrix.
- 6Is [[0, 2, −3], [−2, 0, 5], [3, −5, 0]] skew-symmetric?
- 7Write [[3, 5], [1, 7]] as the sum of a symmetric and a skew-symmetric matrix.
- 8If A = [[2, −1], [3, 0]], find x and y in A² = xA + yI.
Show answers (try them yourself first)
- 1[[7, 1], [1, 5]].
- 2[[12, −3], [6, 15]].
- 3[[1, 4], [2, 5], [3, 6]].
- 4[[4, 1], [8, 8]].
- 53 × 4.
- 6Yes; Aᵀ = −A (zeros on the diagonal, and a₁₂ = −a₂₁ and so on).
- 7Symmetric part [[3, 3], [3, 7]] and skew-symmetric part [[0, 2], [−2, 0]]; their sum = [[3, 5], [1, 7]].
- 8A² = [[1, −2], [6, −3]]; 1 = 2x + y, −2 = −x; so x = 2 and y = −3.
Lesson 4: Determinants and Matrices — quick revision
| Topic | Formula or rule |
|---|---|
| Determinant of order 2 | ad − bc |
| Cofactor | Cᵢⱼ = (−1)^(i + j) Mᵢⱼ |
| Expansion | along any row: a₁₁C₁₁ + a₁₂C₁₂ + a₁₃C₁₃ |
| Properties | swap rows → sign reverses; equal rows → 0 |
| Cramer's rule | x = Dx/D, y = Dy/D, z = Dz/D (D ≠ 0) |
| Area | ½ × the positive value of [[x₁, y₁, 1], [x₂, y₂, 1], [x₃, y₃, 1]] |
| Matrix product | (m × n)(n × p) = m × p |
| Transpose | (AB)ᵀ = BᵀAᵀ |
🎯 Most important points for the exam
- A determinant is one number; a matrix is a table.
- In a cofactor the sign flips when i + j is odd.
- Pick the row with most zeros and try adding or subtracting rows.
- If D = 0 Cramer's rule does not work.
- AB ≠ BA; check the orders before multiplying.
- In (AB)ᵀ the order reverses.
Check the sign, check the order, and pick the row with zeros first.