Matrices — Class 12 solved problems
22 problems from this chapter, each solved step by step.
- If AB = A and BA = B, then (B² + B) is equal to:
2B
- The number of all possible matrices of order 3 × 2 with each entry 1 or 2 is:
64
- Three schools A, B and C organised a mela for collecting funds for helping the rehabilitation of flood victims. They sold hand-made fans, mats and plates from r
(i) Price Matrix: P = 25 & 50 & 10 (ii) Sales Matrix: S = 50 & 30 & 35 60 & 35 & 40 40 & 50 & 25 (iii) (a) Funds collected by School B: ₹ 3000 (iii) (b) Total f
- Find the inverse of the matrix A = [[1, 2, 5], [2, 3, 1], [-1, 1, 1]] if it exists.
The inverse of the matrix A is A^(-1) = 2/21 & 3/21 & -13/21 -3/21 & 6/21 & 9/21 5/21 & -3/21 & -1/21.
- Let A=[a_ij] be a matrix of order 3 × 3, with a_ij=(√2)^(i+j). If the sum of all the elements in the third row of A² is α+β√2, α, β Z, then α+β is equal to:
224
- (b) Given that P = 2 & -1 3 & 4, Q = 5 & 2 7 & 4 and R = 2 & 5 3 & 8, find a matrix S such that PQ - RS is a null matrix. (matrices, matrix multiplication, null
The matrix S is -191 & -110 77 & 44.
- Let A=[ cc(1)/(√2) & -2 0 & 1 ] and P=[ cccos θ & -sin θ sin θ & cos θ ], θ>0. If B=PAP^(T), C=P^(T) B^(10) P and the sum of the diagonal elements of C is (m)/(
65
- Let A = [a_ij] be a 3 × 3 matrix such that A 0 1 0 = 0 0 1, A 4 1 3 = 0 1 0 and A 2 1 2 = 1 0 0, then a_23 equals:
-1
- If A, B, and (adj(A^(-1)) + adj(B^(-1))) are non-singular matrices of same order, then the inverse of A (adj(A^(-1)) + adj(B^(-1)))^(-1) B, is equal to
B^(-1) + A^(-1)
- (a) If P = 1 -1 0 2 3 4 0 1 2 and Q = 2 2 -4 -4 2 -4 2 -1 5, find (QP) and hence solve the following system of equations using matrices: x - y = 3, 2x + 3y + 4z
The product QP = 6 & 0 & 0 0 & 6 & 0 0 & 0 & 6. The solution to the system of equations is x = 2, y = -1, z = 4.
- The number of singular matrices of order 2, whose elements are from the set 2, 3, 6, 9, is _____
36
- (b) Given that P = 2 & -1 3 & 4, Q = 5 & 2 7 & 4 and R = 2 & 5 3 & 8, find a matrix S such that PQ - RS is a null matrix. (matrix, null matrix, matrix multiplic
S = -191 & -110 77 & 44
- Prove that for any square matrix A, the eigenvalues of AAᵀ and AᵀA are the same (excluding zero).
The non-zero eigenvalues of AA^T and A^TA are the same.
- (b) Given that P = [[2, -1], [3, 4]], Q = [[5, 2], [7, 4]] and R = [[2, 5], [3, 8]] find a matrix S such that PQ - RS is a null matrix.
The matrix S is -191 & -110 77 & 44.
- Let A be a 3 × 3 matrix such that X^T A X = 0 for all nonzero 3 × 1 matrices X = x y z. If A 1 1 1 = 1 4 -5, A 1 2 1 = 0 4 -8, and det(adj(2(A + I))) = 2^α 3^β
80
- Let A = [a_ij] be a square matrix of order 2 with entries either 0 or 1. Let E be the event that A is an invertible matrix. Then the probability P(E) is:
(3)/(8)
- Let S = m Z: A^(m² + A^m = 3 I - A^(-6), where A = [ cc 2 & -1 1 & 0 ]. Then n(S) is equal to
n(S) = 2
- For a 3 × 3 matrix M, let trace(M) denote the sum of all the diagonal elements of M. Let A be a 3 × 3 matrix such that |A|=(1)/(2) and trace(A)=3. If B=adj(adj(
280
- Let A= α-1 & -16 & β, α>0, such that det(A)=0 and α+β=1. If I denotes the 2 2 identity matrix, then the matrix (I+A)^8 is:
I + 8A
- Let M denote the set of all real matrices of order 3 × 3 and let S = -3, -2, -1, 1, 2. Let S_1 = A = [a_ij] M: A = A^T and a_ij S, i, j, S_2 = A = [a_ij] M: A =
1737
- Let A be a 3 × 3 real matrix such that A²(A - 2I) - 4(A - I) = O, where I and O are the identity and null matrices, respectively. If A^5 = α A² + β A + I, where
12
- Let A = cosθ & 0 & -sinθ 0 & 1 & 0 sinθ & 0 & cosθ. If for some θ (0,π), A² = A T, then the sum of the diagonal elements of the matrix (A + I)³ + (A - I)³ - 6A
6
More Class 12 chapters
- Relations and Functions39 solved
- Inverse Trigonometric Functions10 solved
- Determinants33 solved
- Continuity and Differentiability44 solved
- Application of Derivatives26 solved
- Integrals41 solved
- Application of Integrals28 solved
- Differential Equations29 solved
- Vector Algebra22 solved
- Three Dimensional Geometry37 solved
- Linear Programming5 solved
- Probability16 solved