Previous year questions: Matrices

Matrices28 min readLesson 6 of 6

Q1 (CBSE Board 2023, 1 mark) If A is a skew symmetric matrix of order 3 and its determinant equals x, then what is the value of 2023 raised to the power x? Choose from 2023, 1/2023, 2023 squared and 1. Answer: For a skew symmetric matrix of odd order, every diagonal entry is zero and the determinant is zero. Here the order is 3, which is odd, so x = 0. Then 2023 raised to the power 0 equals 1, which is the fourth option. The determinant of an odd order skew symmetric matrix is zero because its transpose is its negative, and the determinant changes sign under transposition. The fallback working for this MCQ is that every diagonal entry of an odd order skew symmetric matrix is zero, so the determinant vanishes. This is also a standard fact for any odd order skew symmetric matrix, so the number in the question does not change the reasoning. Writing the two diagonal entries and the transpose relation is enough to justify the zero determinant, and the powers of 2023 then follow from the exponent rule.

Q2 (CBSE Board 2024, 1 mark) A matrix has 36 elements. How many possible orders can it have? Choose from 13, 3, 5 and 9. Answer: An order is a pair of positive integers whose product is 36. The factor pairs are (1, 36), (2, 18), (3, 12), (4, 9) and (6, 6), and each unequal pair can be written in two orders. That gives 4 times 2 plus 1, which is 9 possible orders. The answer is 9. Each factor pair counts as two orders unless the two factors are equal, and the square pair (6, 6) counts once. The ordered pair count is small, and listing the pairs shows why each unequal pair appears twice.

Q3 (CBSE Board 2024, 1 mark) If A and B are two non-zero square matrices of the same order such that (A + B) squared equals A squared plus B squared, then which relation holds: AB = O, AB = minus BA, BA = O, or AB = BA? Answer: Expand the left side: (A + B) squared is A squared plus AB plus BA plus B squared. Setting it equal to A squared plus B squared leaves AB plus BA = O, so AB = minus BA. The answer is the second option. This also shows that the cross terms do not vanish unless the products cancel. Expanding the square and cancelling the squared terms leaves the cross terms, and that relation is the answer. The cross terms remain on the left after cancelling the squared terms, which is the only algebra needed. The expansion is short enough to check by pairing the positive and negative terms. Expanding the square gives the squared terms plus the two cross terms, and cancelling the squared terms leaves the relation between AB and BA.

Q4 (CBSE Board 2024, 1 mark) Let A be a 3 by 3 matrix with entries aij = i minus 3j. Which statement is false: a11 is less than 0, a12 + a21 = minus 6, a13 is greater than a31, or a31 = 0? Answer: Compute the entries. a11 = 1 minus 3 = minus 2, a12 = 1 minus 6 = minus 5, a21 = 2 minus 3 = minus 1, a13 = 1 minus 9 = minus 8 and a31 = 3 minus 3 = 0. The first, second and fourth statements are true, while a13 greater than a31 says minus 8 greater than 0, which is false. The false statement is the third one. Reading entries from the formula row by row avoids mistakes, and one false statement is enough to answer the question. The entries come from a formula, so only the four needed positions have to be evaluated.

Q5 (CBSE Board 2024, 1 mark) If A and B are two skew symmetric matrices of the same order, then AB + BA is which type of matrix? Answer: Take the transpose of AB + BA. It equals (AB)' + (BA)', which is B'A' + A'B'. Both A and B are skew symmetric, so A' = minus A and B' = minus B. Substituting gives BA + AB, which is the original expression. The expression equals its own transpose, so it is symmetric. The transpose test is the quickest route, and it uses both skew symmetric conditions at once. The transpose of the sum splits into two transposes, which is the whole trick. The transpose of a sum is the sum of transposes and the transpose of a product reverses the order, so the whole proof takes two lines.

Q6 (CBSE Board 2026, 1 mark) A 2 by 3 matrix is filled by choosing each of its six entries from the two values 1 and 2. In how many ways can such a matrix be formed? Choose from 16, 81, 64 and 32. Answer: A 2 by 3 matrix has six entries. Each entry has two independent choices, so the total count is 2 raised to the power 6, which is 64. The answer is the third option. The independence of the six choices is the key sentence, because the count is a product, not a sum, of possibilities. Six independent choices make the count a power of two, which is a common objective question pattern. The objective pattern is to multiply the number of choices per entry, and the order of the matrix fixes the count.

Q7 (CBSE Board 2026, 2 marks) If A = [[1, 2, 2], [2, 1, 2], [2, 2, 1]], compute A squared minus 4A minus 5I, where I is the identity matrix of order 3. Answer: First compute A squared. Row 1 times column 1 gives 1 + 4 + 4 = 9, row 1 times column 2 gives 2 + 2 + 4 = 8, and the matrix is symmetric, so A squared = [[9, 8, 8], [8, 9, 8], [8, 8, 9]]. Next 4A = [[4, 8, 8], [8, 4, 8], [8, 8, 4]] and 5I = [[5, 0, 0], [0, 5, 0], [0, 0, 5]]. Subtracting, every entry becomes zero, so the result is the zero matrix. Writing A squared as a matrix and then subtracting term by term makes the zero matrix appear without guesswork. Each diagonal entry cancels individually, and the off-diagonal zeros remain zero.

Q8 (CBSE Board 2026, 2 marks) If A = [[3, 2, 0], [1, 4, 0], [0, 0, 5]], compute A squared minus 7A plus 10I. Answer: Square the matrix block by block. The top left 2 by 2 block gives [[11, 14], [7, 18]], the off-diagonal block stays zero, and the last diagonal entry becomes 25. Then 7A = [[21, 14, 0], [7, 28, 0], [0, 0, 35]] and 10I has 10 on the diagonal. Subtracting and adding, each diagonal entry becomes 11 minus 21 plus 10 = 0, 18 minus 28 plus 10 = 0 and 25 minus 35 plus 10 = 0, while the off-diagonal entries cancel. The result is the zero matrix. The zero off-diagonal panels are what make the block arithmetic short.

Q9 (CBSE SQP 2021-22, 1 mark) If [[2a + b, a minus 2b], [5c minus d, 4c + 3d]] equals [[4, minus 3], [11, 24]], find a + b minus c + 2d. Answer: Equality gives four equations: 2a + b = 4, a minus 2b = minus 3, 5c minus d = 11 and 4c + 3d = 24. Solving the first pair gives a = 1 and b = 2. Solving the second pair gives c = 3 and d = 4. Substituting, a + b minus c + 2d = 1 + 2 minus 3 + 8 = 8. Solving the two independent pairs separately is faster than solving four equations together. Both pairs of equations are independent, so solving them separately is the fastest clean route. The answer follows after finding the four unknowns, and each pair of equations gives two of them. Solving each pair in two steps keeps the arithmetic clean, and the final substitution gives the numeric answer.

Q10 (CBSE SQP 2023-24, 1 mark) If A and B are invertible square matrices of the same order, which statement is not correct: (AB) inverse equals B inverse times A inverse, the determinant of AB equals determinant of A times determinant of B, (A transpose) inverse equals (A inverse) transpose, or (A + B) inverse equals A inverse plus B inverse? Answer: The first three statements are standard identities for invertible matrices. The fourth is not a valid rule: the inverse of a sum is not the sum of the inverses. A quick counterexample with A = I and B = I shows that (A + B) inverse is half of I, while A inverse plus B inverse is 2I. The incorrect statement is the fourth one. The counterexample uses the simplest invertible matrices, so no calculation is needed beyond the inverse of twice the identity. The false statement is the only one that mixes addition with inversion, which is not allowed. This is also a standard fact for any odd order skew symmetric matrix, so the number in the question does not change the reasoning. Writing the two diagonal entries and the transpose relation is enough to justify the zero determinant, and the powers of 2023 then follow from the exponent rule.