RD Sharma Chapter 5 : Algebra of Matrices Exercise 5.3 Solutions
- Compute the indicated product : [[a, b], [-b, a]] [[a, -b], [b, a]] Watch Solution
- Compute the indicated product : [[1, -2], [2, 3]] [[1, 2, 3], [-3, 2, -1]] Watch Solution
- Compute the indicated product : [[2, 3, 4], [3, 4, 5], [4, 5, 6]] [[1, -3, 5], [0, 2, 4], [3, 0, 5]] Watch Solution
- Show that AB ≠ BA in the following : A = [[5, -1], [6, 7]] and B = [[2, 1], [3, 4]] Watch Solution
- Show that AB ≠ BA in the following A= [[-1, 1, 0], [0, -1, 1], [2, 3, 4]] and B = [[1, 2, 3], [0, 1, 0], [1, 1, 0]] Watch Solution
- Show that AB ≠ BA in the following A = [[1, 3, 0], [1, 1, 0], [4, 1, 0]] and B = [[0, 1, 0], [1, 0, 0], [0, 5, 1]] Watch Solution
- Compute the products AB and BA whichever exists in the case : A = [[1, -2], [2, 3]] and B = [[1, 2, 3], [2, 3, 1]] Watch Solution
- Compute the products AB and BA whichever exists in the case: A = [[3, 2], [-1, 0], [-1, 1]] and B = [[4, 5, 6], [0, 1, 2]] Watch Solution
- Compute the products AB and BA whichever exists in the case: A = [1, -1, 2, 3] and B = [[0], [1], [3], [2]] Watch Solution
- Compute the products AB and BA whichever exists in the case : [a, b] [[c], [d]] + [a, b, c, d] [[a], [b], [c], [d]] Watch Solution
- Show that AB ≠ BA in each of the case : A=[[1, 3, -1], [2, -1, -1], [3, 0, -1]] and B=[[-2, 3, -1], [-1, 2, -1], [-6, 9, -4]] Watch Solution
- Evaluate the following : ([[1, 3], [-1, -4]] + [[3, -2], [-1, 1]) [[1, 3, 5], [2, 4, 6]] Watch Solution
- Evaluate the following : [1, 2, 3] [[1, 0, 2], [2, 0, 1], [0, 1, 2]] [[2[, [4], [6]] Watch Solution
- Evaluate the following : [[1, -1], [0, 2], [2, 3]] ([[1, 0, 2], [2, 0, 1]] – [[0, 1, 2], [1, 0, 2]]) Watch Solution
- If A = [[1, 0], [0, 1]], B = [[1, 0], [0, -1]] and C = [[0, 1], [1, 0]], then show that A^2=B^2=C^2=I2. Watch Solution
- If A = [[2, -1], [3, 2]] and B = [[0, 4], [-1, 7]], find 3A^2 – 2B + I. Watch Solution
- If A = [[4, 2], [-1, 1]], prove that (A – 2I) (A – 3I) = O. Watch Solution
- If A=[[1, 1], [0, 1]], show that A^2 = [[1, 2], [0, 1]] and A^3 = [[1, 3], [0, 1]] Watch Solution
- If A=[[ab, b^2], [-a^2, -ab]], show tht A^2 = O. Watch Solution
- If A=[[cos2θ, sin2θ], [-sin2θ, cos2θ]], find A^2 Watch Solution
- If A = [[2, -3, -5], [-1, 4, 5], [1, -3, -4]] and B = [[-1, 3, 5], [1, -3, -5], [-1, 3, 5]], show that AB = BA = O3×3 Watch Solution
- If A = [[0, c, -b], [-c, 0, a], [b, -a, 0] and B = [[a^2, ab, ac], [ab, b^2, bc], [ac, bc, c^2]], show that AB = BA = O3×3 Watch Solution
- If A = [[2, -3, -5], [-1, 4 ,5], [1, -3, -4]] and B = [[2, -2, -4], [-1, 3, 4], [1, -2, -3]], show that AB = A and BA = B. Watch Solution
- Let A = [[-1, 1, -1], [3, -3, 3], [5, 5, 5]] and B = [[0, 4, 3], [1, -3, -3], [-1, 4, 4]], compare A^2 – B^2 Watch Solution
- For the matrix verify the associativity of matrix multiplication i.e. (AB)C = A(BC). A = [[1, 2, 0], [-1, 0, 1]], B=[[1, 0], [-1, 2], [0, 3]] and C=[[1], [-1]] Watch Solution
- For the matrix verify the associativity of matrix multiplication i.e. (AB)C = A(BC). A = [[4, 2, 3], [1, 1, 2], [3, 0, 1]], B = [[0, -1, 1], [0, 1, 2], [2, -1, 1]] and C = [[1, 2, -1], [3, 0, 1], [0, 0, 1]] Watch Solution
- For the following matrices verify the distributivity of matrix multiplication over matrix addition i.e. A(B+C) = AB + AC. A = [[1, -1], [0, 2]], B = [[-1, 0], [2, 1]] and C = [[0, 1], [1, -1]] Watch Solution
- For the matrix verify the distributivity of matrix multiplication over matrix addition i.e. A(B + C) = AB + AC. A = [[4, 2, 3], [1, 1, 2], [3, 0, 1]], B = [[1, -1, 1], [0, 1, 2], [2, -1, 1]] and C = [[1, 2, -1], [3, 0, 1], [0, 0, 1]] Watch Solution
- If A = [[2, 0, -2], [3, -1, 0], [-2, 1, 1]], B = [[0, 5, -4], [-2, 1, 3], [-1, 0, 2]] and C = [[1, 5, 2], [-1, 1, 0], [0, -1, 1]], verify that A(B – C) = AB – AC. Watch Solution
- Compute the elements a43 and a22 of the matrix: A = [[0, 1, 0], [2, 0, 2], [0, 3, 2], [4, 0, 4]] [[2, -1], [-3, 2], [4, 3]] [[0, 1, -1, 2, -2], [3, -3, 4, -4, 0]] Watch Solution
- If A = [[0, 1, 0], [0 0 1], [p, q, r]] and I is the identity matrix of order 3, show that A^3 = pI + qA + rA^2 Watch Solution
- If ω is a complex cube root of unity, show that ([[1, ω, ω^2], [ω, ω^2, 1], [ω^2, 1, ω]] + [[ω, ω^2, 1], [ω^2, 1, ω], [ω^2, 1, ω], [ω, ω^2, 1]]) [[1], [ω], [ω^2]] = [[0], [0], [0]] Watch Solution
- If A=[[2, -3, -5], [-1, 4, 5], [1, -3, -4]], show that A^2 = A. Watch Solution
- If A = [[4, -1, -4], [3, 0, -4], [3, -1, -3]]. show that A^2 = I3 Watch Solution
- If [1, 1, x] [[1, 0, 2], [0, 2, 1], [2, 1, 0]] [[1], [1], [1]] = 0, find x. Watch Solution
- If [[2, 3], [5, 7]] [[1, -3], [-2, 4]] = [[-4, 6], [-9, x]], find x Watch Solution
- If [x, 4 ,1] [[2, 1, 2], [1, 0, 2], [0, 2, -4]] [[x], [4], [-1]] = 0, find x. Watch Solution
- If [1, -1, x][[0, 1, -1], [2, 1, 3], [1, 1, 1]] [[0], [1], [1]] = 0 find x. Watch Solution
- If A = [[3, -2], [4, -2]] and I = [[1, 0], [0, 1]], then prove that A^2 – A + 2I = 0. Watch Solution
- If A = [[3, 1], [-1, 2]], and I = [[1, 0], [0, 1]], then find λ so that A^2 = 5A + λI. Watch Solution
- If A = [[3, 1], [-1, 2]], show that A^2 – 5A + 7I2 = O. Watch Solution
- If A = [[2, 3], [-1, 0]], show that A^2 – 2A + 3I2 = O. Watch Solution
- Show that the matrix A = [[2, 3], [1, 2]] satisfies the equation A^3 – 4A^2 + A = O. Watch Solution
- Show that the matrix A = [[5, 3], [12, 7]] is a root of the equation A^2 – 12A – I = O. Watch Solution
- If A = [[3, -5], [-4, 2]] , find A^2 – 5A + 14I. Watch Solution
- If A = [[3, 1], [-1, 2]], show that A^2 – 5A + 7I = O. Use this to find A^4. Watch Solution
- If A = [[3, -2], [4, -2]], find k such that A^2 = kA – 2I2 Watch Solution
- If A = [[1, 0],[-1, 7]], find k such that A^2 – 8A + kI = O. Watch Solution
- If A = [[1, 2], [2, 1]] and f(x) = x^2 – 2x – 3, show that f(A) = O. Watch Solution
- If A = [[2, 3], [1, 2]] and I = [[1, 0], [0, 1]], then find λ, μ so that A^2 = λA + μI. Watch Solution
- Find the value of x for which the matrix product. [[2, 0, 7], [0, 1, 0], [1, -2, 1]] [[-x, 14, 7x], [0, 1, 0], [x, -4x, -2x]] equal to an identity matrix. Watch Solution
- Solve the matrix equation [x, 1], [[1, 0], [-2, -3]] [[x], [5]] = 0 Watch Solution
- Solve the matrix equation [1, 2, 1] [[1, 2, 0], [2, 0, 1], [1, 0, 2]] [[0], [2], [x]] = 0 Watch Solution
- Solve the matrix equation [x, -5, -1] [[1, 0, 2], [0, 2, 1], [2, 0, 3]] [[x], [4], [1]] = 0 Watch Solution
- Solve the matrix equation [2x, 3] [[1, 2], [-3, 0], [[x], [8]] = 0 Watch Solution
- If A = [[1, 2, 0], [3, -4, 5], [0, -1, 3]], compare A^2 – 4A + 3I3 Watch Solution
- If f(x) = x^2 – 2x, find f(A), where A = [[0, 1, 2], [4, 5 ,0], [0, 2, 3]] Watch Solution
- If f(x) = x^3 + 4x^2 – x, find f(A), where A = [[0, 1, 2], [2, -3, 0], [1, -1, 0]] Watch Solution
- If A = [[1, 0, 2], [0, 2, 1], [2, 0, 3]], then show that A is a root of the polynomial f(x) = x^3 – 6x^2 + 7x + 2. Watch Solution
- If A = [[1, 2, 2], [2, 1, 2], [2, 2, 1]], then prove that A^2 – 4A + 5I = O. Watch Solution
- If A = [[3, 2, 0], [1, 4, 0], [0, 0, 5]], show that A^2 – 7A + 10I3 = O. Watch Solution
- Without using the concept of inverse of matrix, find the matrix [[x, y], [z, u]] such that [[5, -7], [-2, 3]] [[x, y], [z, u]] = [[-16, -6], [7, 2]] Watch Solution
- Find the matrix A such that [[1, 1], [0, 1]] A = [[3, 3, 5], [1, 0, 1]] Watch Solution
- Find the matrix A such that A[[1, 2, 3], [4, 5, 6]] = [[-7, -8, -9], [2, 4, 6]] Watch Solution
- Find the matrix A such that [[4], [1], [3]]A = [[-4, 8, 4], [-1, 2, 1], [-3, 6, 3]] Watch Solution
- Find the matrix such that [2, 1, 3] [[-1, 0, -1], [-1, 1, 0], [0, 1, 1]] [[1], [0], [-1]] = A Watch Solution
- Find the matrix A such that [[2, -1], [1, 0], [-3, 4]] A = [[-1, -8, -10], [1, -2, -5], [9, 22, 15]] Watch Solution
- Find the matrix A such that A [[1, 2, 3], [4, 5, 6]] = [[-7, -8, -9], [2, 4, 6], [11, 10, 9]] Watch Solution
- Find a 2×2 matrix A such that A[[1, -2], [1, 4]] = 6I2 Watch Solution
- If A = [[0, 0], [4, 0]], find A^16 Watch Solution
- If A=[[0, -x], [x, 0]], B = [[0, 1], [1, 0]] and x^2 = -1, then show that (A + B)^2 = A^2 + B^2 Watch Solution
- If A=[[1, 0, -3], [2, 1, 3], [0, 1, 1]], then verify that A^2 + A = A(A + I), where I is the identity matrix. Watch Solution
- If A=[[3, -5], [-4, 2]], then find A^2 – 5A – 14I. Hence, obtain A^2 – 5A + 14I. Hence, obtain A^3 Watch Solution
- If P(x) = [[cos x, sin x], [-sin x, cos x]], then show that P(x) P(y) = P(x + y) = P(y)P(x). Watch Solution
- If P = [[x, 0, 0], [0, y, 0], [0, 0, z]] and Q = [[a, 0, 0], [0, b, 0], [0, 0, c]], prove that PQ = [[xa, 0, 0], [0, yb, 0], [0, 0, zc]] = QP Watch Solution
- If A = [[2, 0, 1], [2, 1, 3], [1, -1, 0]], find A^2 – 5A + 4I and hence find a matrix X such that A^2 – 5A + 4I + X = 0. Watch Solution
- If A = [[1, 1], [0, 1]], prove that A^n = [[1, n], [0, 1]] for all positive integers n. Watch Solution
- If A = [[a, b], [0, 1]], prove that A^n = [[a^n, b((a^n-1)/(a-1))], [0, 1]] for every positive integer n. Watch Solution
- If A=[[cos θ, isin θ], [isin θ, cos θ]], then prove by principle of mathematical induction that A^n =[[cos nθ, isin nθ], [isin nθ, cos nθ]] for all n ∈ N. Watch Solution
- If A = [[cos α + sin α, √2sin α],[-√2sin α, cos α – sin α]], prove that If A^n = [[cos nα + sin nα, √2sin nα],[-√2sin nα, cos nα – sin nα]] for all n ∈ N Watch Solution
- If A=[[1, 1, 1], [0, 1, 1], [0, 0, 1]], then use the principle of mathematical induction to show that A^n = [[cos α + sin α, √2sin α], [-√sinn α, cos nα – sin nα]] for all n ∈ N. Watch Solution
- If B,C are n rowed square matrices and if A = B + C, BC = CB, C^2 = O. then show that for every n ∈ N, A^n+1 = B^n (B + (n + 1)C). Watch Solution
- If A = diag(a, b, c), show that A^n = diag(a^n, b^n, c^n) for all positive integer n. Watch Solution
- If A is a square matrix, using mathematical induction prove that (A^T)^n=(A^n)^T for all n ∈ N. Watch Solution
- A matrix X has a+b rows and a+2 column while the matrix Y has b+1 rows and a+3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal. Watch Solution
- Give example of matrix : A and B such that AB ≠ BA. Watch Solution
- Give example of matrix : A and B such that AB = O but A ≠ O, B ≠ O. Watch Solution
- Give example of matrix : A and B such that AB = O but BA ≠ O. Watch Solution
- Give example of matrix : A, B and C such that AB = AC but B ≠ C, A ≠ O. Watch Solution
- Let A and B be square matrix of the same order. Does (A + B)^2 = A^2 + 2AB + B^2 hold ? If not, why? Watch Solution
- If A and B are square matrices of the same order, explain, why in general (A + B)^2 ≠ A^2 + 2AB + B^2 Watch Solution
- If A and B are square matrices of the same order, explain, why in general (A – B)^2 ≠ A^2 – 2AB + B^2 Watch Solution
- If A and B are square matrices of the same order, explain, why in general (A + B)(A – B) ≠ A^2 – B^2 Watch Solution
- Let A and B be square matrices of the other 3×3. Is (AB)^2 = A^2 B^2 ? Give reasons. Watch Solution
- If A and B are square matrices of the same order such that AB = BA, then show that (A + B)^2 = A^2 + 2AB + B^2 Watch Solution
- Let A = [[1, 1, 1], [3, 3, 3]], B = [[3, 1], [5, 2], [-2, 4]] and C = [[4, 2], [-3, 5], [5, 0]] Verify that AB = AC though B ≠ C, A ≠ O. Watch Solution
- Three shopkeepers A, B and C go to a store by stationary. A purchases 12 dozen notebooks, 5 dozen pens and 6 dozen pencils. B purchases 10 dozen notebooks, 6 dozen pens and 7 dozen pencils. C purchases 11 dozen notebooks, 13 dozen pend and 8 dozen pencils. A notebook costs 40 paise, a pen costs ₹1.25 and a pencil costs 35 paise. Use matrix multiplication to calculate each individual’s bill. Watch Solution
- The cooperative stores of a particular school has 10 dozen physics books, 8 dozen chemistry books and 5 dozen mathematics books. Their selling prices are ₹8.30, ₹3.45 and ₹4.50 each respectively. Find the total amount the store will receive from selling all the items. Watch Solution
- In a legislative assembly election, a political group hired a public relations firm to promote its candidates in three ways: telephone, house calls and letters. The cost per contact (in paise) is given matrix A as A=Cost per contact [[40] – Telephone, [100]-House call, [50]-Letter] The number of contacts of each type made in two cities X and Y is given in matrix B as Find the total amount spend by the group in the two cities X and Y. Watch Solution
- A trust fund has ₹30000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using the trust fund must obtain an annual total interest of (i) ₹1800 (ii) ₹2000. Watch Solution
- To promote making of toilets for women, an organisation tried to generate awareness through (i) house calls (ii) letters, and (iii) announcements. The cost for each mode per attempt is given below: (i) ₹50 (ii) ₹20 (iii) ₹40 The number of attempts made in three village X, Y and Z are given below: Find the total cost incurred by the organisation for three villages separately, using matrices. Watch Solution
- There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women and 4 children in family B. The recommend daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grants of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrix. Using matrix multiplication, calculate the total requirement of calories and proteins for each of the two families. What awareness can you create among people about the planned diet from this question ? Watch Solution
- In a parliament election, a political party hired a public relations firm to promote its candidates in three ways – telephone, house calls and letters. The cost per contact (in paisa) is given in matrix A as The number of contacts of each type made in two cities X and Y is given in the matrix B as Find the total amount spent by the party in the two cities. What should one consider before casting his/her vote – party’s promotional activity or their social activities? Watch Solution
- The monthly incomes of Aryan and Babbar are in the ratio 3:4 and their monthly expenditures are in the ratio 5:7. If each saves ₹15000 per month, find their monthly incomes using matrix method. This problem reflects which value? Watch Solution
- A trust invested some money in two type of bonds. The first boys pays 10% interest and second bond pays 12% interest. The trust received ₹2800 as interest. However, if trust had interchanged money in bonds, they would have got ₹100 less as interest. Using matrix method, find the amount invested by the trust. Watch Solution
ALGEBRA OF MATRICES – R.D. Sharma Class 12th Math
- Algebra of Matrices Exercise 5.1 Video Solution
- Algebra of Matrices Exercise 5.2 Video Solution
- Algebra of Matrices Exercise 5.3 Video Solution
- Algebra of Matrices Exercise 5.4 Video Solution
- Algebra of Matrices Exercise 5.5 Video Solution
- Algebra of Matrices Very Short Answer Questions (VSAQs) Video Solution
- Algebra of Matrices Multiple Choice Questions (MCQs) Video Solution