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A matrix of order m × n has m rows and n columns, and the total number of entries is the product m × n. Here m = 4 and n = 3, so the matrix contains 4 × 3 = 12 entries. The order is written as 4 × 3, but the count of elements is the product of the two.
Two matrices can be added or subtracted only when they are of the same order, because addition is performed element by element at corresponding positions. P is 2 × 3 while Q is 3 × 2, so the positions do not match and the sum P + Q has no meaning. Their product PQ, however, is perfectly defined and has order 2 × 2.
The symbol aᵢⱼ denotes the entry lying in the i-th row and the j-th column of the matrix. So a₂₃ is read from row 2 and column 3, where the entry is 7. The row index is always quoted first, and the column index second.
The product AB is defined because the three columns of A match the three rows of B, and the resulting product has order 2 × 2. Transposing a matrix never changes its order, so (AB)ᵀ is also of order 2 × 2. Equivalently, (AB)ᵀ = BᵀAᵀ, which is 2 × 3 multiplied by 3 × 2, again giving 2 × 2.
For a 2 × 2 determinant whose first row is a, b and second row is c, d, the value is ad − bc. Here ad = 3 × 5 = 15 and bc = (−2) × 4 = −8, so the determinant equals 15 − (−8) = 23. Subtracting a negative product increases the value.
A square matrix is singular exactly when its determinant is zero. The determinant here is 6x − 12, so setting 6x − 12 = 0 gives x = 2. Substituting x = 2 back yields 12 − 12 = 0, confirming that the matrix has no inverse.
The element 5 lies in row 1 and column 2, so its minor is the determinant formed by the remaining entry, namely 3. The cofactor equals the minor multiplied by (−1)^(1+2), that is −1 × 3 = −3. The sign factor depends only on the position of the element, not on its value.
For A = [[a, b], [c, d]] with ad − bc ≠ 0, the inverse is (1/(ad − bc)) multiplied by [[d, −b], [−c, a]], so the diagonal entries are interchanged and both off-diagonal entries change sign. Here ad − bc = 8 − 3 = 5, giving A⁻¹ = (1/5)[[4, −3], [−1, 2]]. Checking by multiplication, A A⁻¹ reduces to the 2 × 2 identity matrix.
The coefficient determinant is Δ = (2)(−1) − (3)(1) = −5, which is non-zero, so the system has a unique solution. Replacing the x-column gives Δx = (12)(−1) − (3)(1) = −15 and replacing the y-column gives Δy = (2)(1) − (12)(1) = −10. Hence x = Δx/Δ = 3 and y = Δy/Δ = 2, which satisfies both equations.
For a square matrix of order n, det(kA) = kⁿ det A, because multiplying the matrix by k multiplies each of its n rows by k and each row contributes one factor k. Here n = 3 and k = 2, so det(2A) = 2³ × 5 = 40. Only in the 1 × 1 case would the factor k appear just once.
A matrix with m rows and n columns has m × n entries. Here the order is 3 × 4, so the number of elements is 3 × 4 = 12. The order is written rows first and columns second, and the count is a product, not a sum or concatenation.
A symmetric matrix equals its own transpose, so every element satisfies aᵢⱼ = aⱼᵢ. The condition Aᵀ = A is therefore the defining property. The other conditions describe skew-symmetric, idempotent and orthogonal matrices respectively.
First multiply each element of A by 2 to get 2A = [[2, 4], [6, 8]]. Then subtract B element by element: [[2−2, 4−0], [6−1, 8−3]] = [[0, 4], [5, 5]]. Scalar multiplication and subtraction are both performed entrywise.
The element (AB)₂₁ is the product of the second row of A with the first column of B. That gives 3 × 1 + 4 × 2 = 3 + 8 = 11. The other listed values are different entries of the product matrix AB.
The element 5 lies in the second row and third column. Deleting that row and column leaves the 2 × 2 determinant |2 1; 1 2| = 2 × 2 − 1 × 1 = 3. A minor is always the determinant of the remaining submatrix, without any sign change.
The determinant is upper triangular because every entry below the main diagonal is zero. For a triangular determinant, the value is the product of the diagonal entries: 3 × 2 × (−1) = −6. No further expansion is needed.
For A = [[a, b], [c, d]], the adjoint is adj A = [[d, −b], [−c, a]]. Substituting a = 2, b = −3, c = 4 and d = 1 gives [[1, 3], [−4, 2]]. The diagonal entries are interchanged and the other two entries change sign.
The transpose is Aᵀ = [[1, 3], [2, 4]]. Adding corresponding entries gives [[1+1, 2+3], [3+2, 4+4]] = [[2, 5], [5, 8]]. The result is symmetric, as expected for any matrix added to its transpose.
For a square matrix of order n, the adjoint satisfies |adj A| = |A|ⁿ⁻¹. Here n = 3 and |A| = 3, so |adj A| = 3² = 9. The exponent is one less than the order because each cofactor is of order n − 1.
The coefficient determinant is 1 × 4 − 2 × 2 = 0, so the system is not uniquely solvable. Dividing the second equation by 2 gives x + 2y = 3.5, which contradicts x + 2y = 3. Hence the two equations are inconsistent and the system has no solution.
The denominator is already a product of distinct linear factors (x − 1) and (x + 2), and the numerator has degree 1, which is less than the denominator's degree 3, so the fraction is proper. For each distinct linear factor the standard rule assigns a single constant numerator, giving A/(x − 1) + B/(x + 2). No extra polynomial part or repeated-factor term is needed because neither situation occurs here.
All five letters T, A, M, I, L are distinct, so each ordering of the five letters produces a different arrangement. The number of orderings of n distinct objects is n!, hence 5! = 5 × 4 × 3 × 2 × 1 = 120.
For combinations, ⁿCᵣ = n!/(r!(n − r)!), so ⁸C₃ = 8!/(3! × 5!) = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56. The division by 3! removes the internal orderings of each chosen triple, which is exactly what distinguishes a combination from a permutation.
Multiplying throughout by (x + 1)(x + 2) gives 3x + 5 = A(x + 2) + B(x + 1). Substituting x = −1 makes the term containing B vanish, leaving 3(−1) + 5 = A(1), that is A = 2. Substituting x = −2 in the same identity would instead give B = 1.
The word BOOKS has five letters in which O appears twice and the remaining letters B, K, S are distinct. The number of distinct arrangements is therefore 5!/2! = 120/2 = 60, because dividing by 2! cancels the double counting caused by interchanging the two identical O's.
Exactly one woman means choosing 1 woman from the 4 available and 2 men from the 5 available. The choices are independent, so multiply: ⁴C₁ × ⁵C₂ = 4 × 10 = 40. Combination notation is used because the order in which members are picked does not matter.
The general term of (1 + 2x)⁶ is ⁶Cᵣ (2x)ʳ, so the term containing x³ arises when r = 3. Its coefficient is ⁶C₃ × 2³ = 20 × 8 = 160. The factor 2³ must be included because the term is (2x)³, not x³.
In the expansion of (a + b)ⁿ the general term is Tᵣ₊₁ = ⁿCᵣ aⁿ⁻ʳ bʳ, so the fifth term corresponds to r = 4. Hence T₅ = ⁷C₄ x³ 2⁴ = 35 × 16 × x³ = 560x³. The index of x falls to 7 − 4 = 3 because the power of 2 rises to 4.
Multiply both sides by (x − 1) and then put x = 1: the fraction with denominator x² + 1 vanishes because it still carries the factor (x − 1). This leaves A = (2(1)² + 3)/((1)² + 1) = 5/2. The irreducible quadratic x² + 1 correctly takes a linear numerator Bx + C.
The general term is ⁹Cᵣ x^(9−r) (2/x²)ʳ = ⁹Cᵣ 2ʳ x^(9−3r). A term independent of x requires the exponent 9 − 3r to be zero, giving r = 3. The constant term is therefore ⁹C₃ × 2³ = 84 × 8 = 672.
A repeated linear factor (x − 3)² in the denominator needs one partial fraction for each power up to the second: A/(x − 3) plus B/(x − 3)². A single term with denominator (x − 3)² cannot represent the numerator's linear remainder in general.
By definition, nPr = n!/(n − r)!. Here ⁶P₂ = 6!/4! = 6 × 5 = 30, because only the last two factors of 6! remain after cancellation.
The identity ⁿCᵣ = ⁿC₍ₙ₋ᵣ₎ means equal combinations with different r-values occur when the two r-values add to n. Thus 4 + 6 = 10, so n = 10.
Multiplying by (x + 1)(x − 4) gives 5 = A(x − 4) + B(x + 1). Putting x = −1 gives A = −1, and putting x = 4 gives B = 1, so A + B = 0.
The number must end in 2 or 4, giving 2 choices for the units digit. The hundreds digit then has 4 choices and the tens digit has 3, so the total is 2 × 4 × 3 = 24.
An octagon has 8 vertices, and every pair of vertices determines a segment. There are ⁸C₂ = 28 such segments, but 8 of them are sides, so the diagonals number 28 − 8 = 20.
The general term of (x + 3)⁸ is ⁸Cᵣ x⁽⁸⁻ᵣ⁾ 3ᵣ. For the x⁵ term, 8 − r = 5, so r = 3. Its coefficient is ⁸C₃ × 3³ = 56 × 27 = 1512.
The expansion of (x + 2)⁶ has 7 terms, so the middle term is the 4th term. It is ⁶C₃ x³ 2³ = 20 × 8x³ = 160x³.
Divisibility by 5 requires the last digit to be 0 or 5. With last digit 0, there are 5 × 4! = 120 numbers; with last digit 5, the first digit has 4 choices and the rest 4! arrangements, giving 96. The total is 120 + 96 = 216.
Multiplying by (x − 1)²(x + 2) gives x² + 5 = A(x − 1)(x + 2) + B(x + 2) + C(x − 1)². Setting x = −2 isolates C: 9 = C(9), so C = 1.
For an arithmetic progression the nth term is given by tₙ = a + (n − 1)d. Here a = 7, d = 3 and n = 12, so t₁₂ = 7 + 11 × 3 = 7 + 33 = 40.
In a geometric progression each term is obtained by multiplying the previous term by a fixed common ratio. Dividing the second term by the first gives 6 ÷ 3 = 2, and checking 12 ÷ 6 = 2 confirms the ratio is 2.
The sum of the first n natural numbers is the standard result S = n(n + 1)/2, obtained by pairing the first and last terms. For example, with n = 4 the sum 1 + 2 + 3 + 4 = 10 and 4 × 5 / 2 = 10.
The sum of n terms of an arithmetic progression is Sₙ = n/2 [2a + (n − 1)d]. Substituting a = 5, d = 3 and n = 20 gives S₂₀ = 10 [10 + 57] = 10 × 67 = 670.
For an arithmetic progression the number of terms is n = (last term − first term)/d + 1. Here d = 9 − 5 = 4, so n = (85 − 5)/4 + 1 = 80/4 + 1 = 20 + 1 = 21.
The geometric mean of two positive numbers a and b is √(ab). Therefore the geometric mean is √(4 × 9) = √36 = 6, which lies between 4 and 9 as expected.
When the common ratio satisfies |r| < 1, the sum to infinity of a geometric progression is S∞ = a/(1 − r). Here a = 8 and r = 1/2, so S∞ = 8/(1 − 1/2) = 8/(1/2) = 16.
The sum of the squares of the first n natural numbers is n(n + 1)(2n + 1)/6. Putting n = 10 gives 10 × 11 × 21 / 6 = 2310 / 6 = 385.
Using tₙ = a + (n − 1)d we get t₅ = a + 4d = 17 and t₉ = a + 8d = 33. Subtracting the first equation from the second gives 4d = 16, so d = 4, and then a = 17 − 16 = 1.
Take the three terms as a/r, a and ar; their product is a³ = 216, so a = 6. The sum then gives 6/r + 6 + 6r = 26, that is 3r² − 10r + 3 = 0, so r = 3 or r = 1/3 and the numbers are 2, 6, 18.