IPMPrepA = −2−131 satisfies A2 − pA + qI = O. Find p + q.
Solution
Every 2 × 2 matrix satisfies A2 − (trace A)A + (det A)I = O.
p = trace = −1, q = det = 1.
p + q = 0.
Answer: 0
Quantitative Ability · Algebra
Matrices and Determinants: IPMAT practice questions
24 questions, 16 of them short answer (type the number, as in IPMAT Indore). Each is marked with the exam level it suits. Answer, then open the step-by-step solution.
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For what value of k is the matrix k2−1004241 singular?
Solution
A matrix is singular when its determinant is 0.
The determinant works out to −16k + 16.
−16k + 16 = 0 gives k = 1.
Answer: 1
If A = 1252 and B = 1−2−1−1, find the sum of the diagonal entries (trace) of AB.
Solution
AB = −1−43−12.
Trace = −1 + (−12) = −13.
Answer: (D) −13
Find the value of the determinant −120−243−2−25.
Solution
Expand along the first row.
= −1(26) − (2)(−4) + (0)(12) = −18.
Answer: −18
Find the value of the determinant −140554525.
Solution
Expand along the first row.
= −1(17) − (4)(5) + (0)(−15) = −37.
Answer: −37
For what value of k is the matrix −225−11−1k−1−1 singular?
Solution
A matrix is singular when its determinant is 0.
The determinant works out to −7k + 7.
−7k + 7 = 0 gives k = 1.
Answer: 1
If A = 1604, find the entry in the first row and first column of A−1.
Solution
det A = 4.
For [[a, b], [c, d]] the inverse is (1/det) × [[d, −b], [−c, a]].
The required entry is 1.
Answer: (A) 1
If A = 2334, find the entry in the first row and second column of A−1.
Solution
det A = −1.
For [[a, b], [c, d]] the inverse is (1/det) × [[d, −b], [−c, a]].
The required entry is 3.
Answer: (A) 3
A is a 3 × 3 matrix with det A = −2. Find det(adj A).
Solution
For an n × n matrix, det(adj A) = (det A)n − 1.
= (−2)2 = 4.
Answer: 4
If A = 12−20 and B = −1245, find the entry in the first row and first column of AB.
Solution
That entry is row 1 of A times column 1 of B.
(1)(−1) + (2)(4) = 7.
Answer: 7
For what value of k is the matrix 2k14 singular?
Solution
A matrix is singular when its determinant is 0.
The determinant works out to −1k + 8.
−1k + 8 = 0 gives k = 8.
Answer: (D) 8
A and B are 3 × 3 matrices with det A = 5 and det B = 2. Find det(A2 B).
Solution
det(XY) = det X × det Y.
det(A2 B) = (5)2 × 2 = 50.
Answer: (A) 50
Find the value of the determinant 1−35−2.
Solution
For a 2 × 2 determinant: ad − bc = (1)(−2) − (−3)(5) = 13.
Answer: 13
If A = 2236, find the entry in the first row and first column of A−1.
Solution
det A = 6.
For [[a, b], [c, d]] the inverse is (1/det) × [[d, −b], [−c, a]].
The required entry is 1.
Answer: (B) 1
If A = −3−3−20 and B = 4−1−3−3, find the entry in the first row and first column of AB.
Solution
That entry is row 1 of A times column 1 of B.
(−3)(4) + (−3)(−3) = −3.
Answer: −3
A and B are 2 × 2 matrices with det A = 3 and det B = 2. Find det(A2 B).
Solution
det(XY) = det X × det Y.
det(A2 B) = (3)2 × 2 = 18.
Answer: 18
If A = 02−14, find the sum of all entries of A2.
Solution
A2 = −28−414.
Sum of entries = 16.
Answer: 16
A = −2−3−30 satisfies A2 − pA + qI = O. Find p + q.
Solution
Every 2 × 2 matrix satisfies A2 − (trace A)A + (det A)I = O.
p = trace = −2, q = det = −9.
p + q = −11.
Answer: −11
A = −3233 satisfies A2 − pA + qI = O. Find p + q.
Solution
Every 2 × 2 matrix satisfies A2 − (trace A)A + (det A)I = O.
p = trace = 0, q = det = −15.
p + q = −15.
Answer: −15
A and B are 3 × 3 matrices with det A = −2 and det B = 3. Find det(A2 B).
Solution
det(XY) = det X × det Y.
det(A2 B) = (−2)2 × 3 = 12.
Answer: 12
For what value of k is the matrix −1−3k−3 singular?
Solution
A matrix is singular when its determinant is 0.
The determinant works out to 3k + 3.
3k + 3 = 0 gives k = −1.
Answer: −1
If A = 4251 and B = 5−34−2, find the entry in the second row and first column of AB.
Solution
That entry is row 2 of A times column 1 of B.
(5)(5) + (1)(4) = 29.
Answer: 29
If A = 3221, find the entry in the first row and second column of A−1.
Solution
det A = −1.
For [[a, b], [c, d]] the inverse is (1/det) × [[d, −b], [−c, a]].
The required entry is 2.
Answer: (C) 2
Find the value of the determinant −20−2−1−21442.
Solution
Expand along the first row.
= −2(−8) − (0)(−6) + (−2)(4) = 8.
Answer: (C) 8
These are original practice questions written by Target Campus in the pattern, syllabus and difficulty of IPMAT Indore, IPMAT Rohtak, JIPMAT, NPAT and SET. They are not previous-year papers.