If A and B are symmetric matrices, prove that AB-BA is a skew symmetric matrix.
Hint. Use A'=A, B'=B and the reversed-order transpose-of-product rule.
(AB-BA)' = (AB)'-(BA)' = B'A'-A'B' = BA-AB (since A'=A, B'=B) = -(AB-BA). So (AB-BA)'=-(AB-BA), which is exactly the condition for AB-BA to be skew-symmetric.
✦ Proved: (AB-BA)' = -(AB-BA), so AB-BA is skew-symmetric.
Show that the matrix B'AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Hint. Compute (B'AB)' using the reversed-order rule repeatedly, then substitute A'=A or A'=-A.
(B'AB)' = B'A'(B')' = B'A'B (using (B')'=B). If A is symmetric (A'=A), this becomes B'AB, matching the original — so B'AB is symmetric. If A is skew-symmetric (A'=-A), this becomes B'(-A)B = -(B'AB), matching the skew-symmetric condition.
✦ Proved: B'AB is symmetric when A is symmetric, and skew-symmetric when A is skew-symmetric.
Find the values of x, y, z if the matrix A=[[0,2y,z],[x,y,-z],[x,-y,z]] satisfies the equation A'A=I.
Hint. Compute A'A symbolically and match every entry to the identity matrix, giving equations in x, y, z.
Computing A'A and setting it equal to I gives (after collecting terms): 2x^2=1, 6y^2=1, 3z^2=1, together with the off-diagonal entries automatically vanishing. Solving: x^2=1/2 so x=+-1/sqrt2; y^2=1/6 so y=+-1/sqrt6; z^2=1/3 so z=+-1/sqrt3, with all eight sign combinations satisfying the equation independently.
✦ x = +-1/sqrt2, y = +-1/sqrt6, z = +-1/sqrt3 (any combination of signs).
For what values of x: [1,2,1][[1,2,0],[2,0,1],[1,0,2]][[0],[2],[x]] = O?
Hint. Multiply the row vector by the matrix first, then take the dot product with the column vector and set it to zero.
[1,2,1] times the 3x3 matrix gives the row vector [11+22+11, 12+20+10, 10+21+12] = [6,2,4]. Then [6,2,4] dotted with [0,2,x] gives 60+22+4x = 4+4x. Setting this to 0: 4+4x=0, so x=-1.
✦ x = -1
If A=[[3,1],[-1,2]], show that A^2-5A+7I=0.
Hint. Compute A^2 via matrix multiplication, then combine with -5A and 7I entry by entry.
Since the identity involves A^2, compute that first: A^2 = [[33+1-1, 31+12],[-13+2-1, -11+22]] = [[8,5],[-5,3]]. 5A=[[15,5],[-5,10]]. 7I=[[7,0],[0,7]]. Combining, A^2-5A+7I = [[8-15+7,5-5+0],[-5+5+0,3-10+7]] = [[0,0],[0,0]].
✦ A^2-5A+7I is the zero matrix, as required.
Find x, if [x,-5,-1][[1,0,2],[0,2,1],[2,0,3]][[x],[4],[1]] = O.
Hint. Multiply the row vector by the matrix symbolically first, then take the dot product with the column vector.
[x,-5,-1] times the 3x3 matrix gives [x1+-50+-12, x0+-52+-10, x2+-51+-1*3] = [x-2, -10, 2x-8]. Dotting with [x,4,1]: (x-2)(x) + (-10)(4) + (2x-8)(1) = x^2-2x-40+2x-8 = x^2-48. Setting this to 0: x^2=48, so x=+-4sqrt3.
✦ x = 4sqrt3 or x = -4sqrt3
A manufacturer produces three products x, y, z which he sells in two markets. Annual sales are: Market I: 10,000 (x), 2,000 (y), 18,000 (z); Market II: 6,000 (x), 20,000 (y), 8,000 (z). (a) If unit sale prices of x, y, z are Rs 2.50, Rs 1.50 and Rs 1.00 respectively, find the total revenue in each market using matrix algebra. (b) If the unit costs of the three commodities are Rs 2.00, Rs 1.00 and 50 paise respectively, find the gross profit.
Hint. Set up sales as a 2x3 matrix, prices and costs as 3x1 column vectors, multiply for revenue and cost, then subtract.
Since sales and prices multiply market-by-product against product-by-price, set up Sales matrix S=[[10000,2000,18000],[6000,20000,8000]]. (a) Price vector P=[2.50,1.50,1.00]'. Revenue = S.P = [10000(2.5)+2000(1.5)+18000(1), 6000(2.5)+20000(1.5)+8000(1)] = [25000+3000+18000, 15000+30000+8000] = [46000,53000]. (b) Cost vector C=[2.00,1.00,0.50]'. Cost = S.C = [10000(2)+2000(1)+18000(0.5), 6000(2)+20000(1)+8000(0.5)] = [20000+2000+9000, 12000+20000+4000] = [31000,36000]. Since gross profit is revenue minus cost, Gross profit = Revenue-Cost = [46000-31000, 53000-36000] = [15000,17000].
✦ (a) Revenue: Rs 46,000 in Market I, Rs 53,000 in Market II. (b) Gross profit: Rs 15,000 in Market I, Rs 17,000 in Market II.
Find the matrix X so that X[[1,2,3],[4,5,6]] = [[-7,-8,-9],[2,4,6]].
Hint. X must be a 2x2 matrix for this product to be defined and give a 2x3 result; set up entry equations and solve.
Let X=[[a,b],[c,d]]. Then X times the given 2x3 matrix gives [[a+4b, 2a+5b, 3a+6b],[c+4d, 2c+5d, 3c+6d]]. Matching to [[-7,-8,-9],[2,4,6]]: from a+4b=-7 and 2a+5b=-8, solving gives a=1, b=-2. From c+4d=2 and 2c+5d=4, solving gives c=2, d=0. Checking the third column confirms both solutions.
✦ X = [[1,-2],[2,0]]
If A=[[alpha,beta],[gamma,-alpha]] is such that A^2=I, then (A) 1+alpha^2+betagamma=0 (B) 1-alpha^2+betagamma=0 (C) 1-alpha^2-betagamma=0 (D) 1+alpha^2-betagamma=0.
Hint. Compute A^2 directly and match its diagonal entries to those of the identity matrix.
Computing A^2 directly: [[alpha^2+betagamma, alphabeta-alphabeta],[gammaalpha-alphagamma, gammabeta+alpha^2]], which simplifies to [[alpha^2+betagamma, 0],[0, alpha^2+betagamma]] since the off-diagonal terms cancel. Setting this equal to I gives alpha^2+betagamma=1, which means 1-alpha^2-betagamma=0.
✦ (C) 1-alpha^2-beta*gamma=0
If the matrix A is both symmetric and skew symmetric, then (A) A is a diagonal matrix (B) A is a zero matrix (C) A is a square matrix (D) None of these.
Hint. Symmetric means A'=A; skew-symmetric means A'=-A. Combine both conditions.
If A is symmetric, A'=A. If A is also skew-symmetric, A'=-A. Combining: A=-A, so 2A=0, meaning every entry of A is 0.
✦ (B) A is a zero matrix
If A is square matrix such that A^2=A, then (I+A)^3-7A is equal to (A) A (B) I-A (C) I (D) 3A.
Hint. Expand (I+A)^3 using the binomial expansion, then repeatedly substitute A^2=A to simplify every power of A.
(I+A)^3 = I+3A+3A^2+A^3. Since A^2=A, also A^3=A^2.A=A.A=A^2=A. So (I+A)^3 = I+3A+3A+A = I+7A. Subtracting 7A: (I+A)^3-7A = I+7A-7A = I.
✦ (C) I
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh103.pdf) — Exercise 3.1 (10 questions), Exercise 3.2 (22 questions), Exercise 3.3 (12 questions), Exercise 3.4 (1 question), plus the chapter's Miscellaneous Exercise (11 questions), 56 questions total. Exercise pages were rendered as 300dpi images throughout to read matrix bracket notation accurately, since raw text extraction badly garbled every matrix in the chapter. The old stub taught 'finding the inverse using elementary row/column operations' as a core section, but that method has been fully removed from the current edition — Section 3.7 only proves existence and uniqueness of an inverse, and Exercise 3.4 has shrunk from what older editions had down to a single leftover MCQ, confirmed by reading straight through the section with nothing about row-reducing a matrix anywhere in it. The old stub also collapsed all four real exercises into one invented 30-question group with no solutions file behind it. Every matrix computation in this file (all 56 questions) was independently verified with a Python script using numpy for numeric matrices and sympy for the symbolic ones (F(x)F(y)=F(x+y), the A'A=I system, and the two symbolic quadratic-in-x miscellaneous questions), catching zero discrepancies against the hand-derived working.. Questions are referenced from the NCERT textbook for identification.
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