Evaluate the determinant |[2,4],[-5,-1]|.
Hint. For a 2x2 determinant, compute the product of the main diagonal minus the product of the other diagonal.
For a 2x2 determinant |[a,b],[c,d]|, the value is ad-bc, since that is the defining formula. Here: 2(-1) - 4(-5) = -2+20 = 18.
✦ 18
Evaluate the determinants: (i) |[cos theta,-sin theta],[sin theta,cos theta]| (ii) |[x^2-x+1,x-1],[x+1,x+1]|.
Hint. For (i), expand and use sin^2+cos^2=1. For (ii), expand and simplify the resulting cubic in x.
(i) cos(theta)*cos(theta) - (-sin(theta))*sin(theta) = cos^2(theta)+sin^2(theta) = 1, since this is the Pythagorean identity. (ii) (x^2-x+1)(x+1) - (x-1)(x+1) = (x^3+x^2-x^2-x+x+1) - (x^2-1) = (x^3+1) - x^2+1 = x^3-x^2+2.
✦ (i) 1 (constant, independent of theta) (ii) x^3-x^2+2
If A=[[1,2],[4,2]], then show that |2A|=4|A|.
Hint. Compute |A| and |2A| separately and compare, remembering every entry gets scaled for |2A|.
|A| = 1(2)-2(4) = 2-8 = -6. 2A = [[2,4],[8,4]], so |2A| = 2(4)-4(8) = 8-32 = -24. Since 4|A| = 4(-6) = -24, this matches |2A|, because scaling a 2x2 matrix by k scales its determinant by k^2=4.
✦ |A|=-6, |2A|=-24=4|A|, verified.
If A=[[1,0,1],[0,1,2],[0,0,4]], then show that |3A|=27|A|.
Hint. Compute |A| directly, then note that scaling a 3x3 matrix by k scales its determinant by k^3.
Since A is upper triangular, |A| = 114 = 4 (the product of the diagonal entries). 3A = [[3,0,3],[0,3,6],[0,0,12]], also upper triangular, so |3A| = 3312 = 108. Since 27|A| = 27(4) = 108, this matches, because scaling a 3x3 matrix by k scales its determinant by k^3=27.
✦ |A|=4, |3A|=108=27|A|, verified.
Evaluate the determinants: (i) |[3,-1,-2],[0,0,-1],[3,-5,0]| (ii) |[3,-4,5],[1,1,-2],[2,3,1]| (iii) |[0,1,2],[-1,0,-3],[-2,3,0]| (iv) |[2,-1,-2],[0,2,-1],[3,-5,0]|.
Hint. Expand each along the row or column with the most zeros to minimise the arithmetic.
Since expanding along a row or column with zeros skips terms entirely, that row/column is chosen in each part below. (i) Expanding along row 2 (which has two zeros): -0*(...) + 0*(...) - (-1)det[[3,-1],[3,-5]] = 1(3(-5)-(-1)(3)) = 1*(-15+3) = -12. (ii) Expanding along row 1: 3(1(1)-(-2)(3)) - (-4)(1(1)-(-2)(2)) + 5(1(3)-1(2)) = 3(1+6)+4(1+4)+5(3-2) = 21+20+5 = 46. (iii) Expanding along row 1: 0*(...) - 1det[[-1,-3],[-2,0]] + 2det[[-1,0],[-2,3]] = -1(0-6) + 2(-3-0) = 6-6 = 0. (iv) Expanding along row 1: 2(2(0)-(-1)(-5)) - (-1)(0(0)-(-1)(3)) + (-2)(0(-5)-2(3)) = 2(0-5)+1(0+3)-2(0-6) = -10+3+12 = 5.
✦ (i) -12 (ii) 46 (iii) 0 (iv) 5
If A=[[1,1,-2],[2,1,-3],[5,4,-9]], find |A|.
Hint. Expand along any row; row 1 works fine here.
Expanding along row 1: 1(1(-9)-(-3)(4)) - 1(2(-9)-(-3)(5)) + (-2)(2(4)-1(5)) = 1(-9+12) - 1(-18+15) - 2(8-5) = 3 + 3 - 6 = 0.
✦ |A| = 0
Find values of x, if (i) |[2,4],[5,1]| = |[2x,4],[6,x]| (ii) |[2,3],[4,5]| = |[x,3],[2x,5]|.
Hint. Evaluate both determinants in terms of x and set them equal, then solve.
(i) Left side: 2(1)-4(5) = 2-20 = -18. Right side: 2x(x)-4(6) = 2x^2-24. Setting equal: 2x^2-24=-18, so 2x^2=6, x^2=3, x=+-sqrt3. (ii) Left side: 2(5)-3(4) = 10-12 = -2. Right side: x(5)-3(2x) = 5x-6x = -x. Setting equal: -x=-2, so x=2.
✦ (i) x = +-sqrt3 (ii) x = 2
If |[x,2],[18,x]| = |[6,2],[18,6]|, then x is equal to (A) 6 (B) +-6 (C) -6 (D) 0.
Hint. Evaluate both sides and solve the resulting equation for x, keeping both roots if there are two.
Left side: x^2 - 36. Right side: 6(6)-2(18) = 36-36 = 0. Setting equal: x^2-36=0, so x^2=36, giving x=6 or x=-6.
✦ (B) +-6
Solutions written by the tuition.in editorial team and checked against the NCERT Class 12 Mathematics textbook, Reprint 2026-27 (lemh104.pdf) — Exercise 4.1 (8 questions), Exercise 4.2 (5 questions), Exercise 4.3 (5 questions), Exercise 4.4 (18 questions), Exercise 4.5 (14 questions), plus the chapter's Miscellaneous Exercise (9 questions), 59 questions total, the largest exercise set in Class 12 Mathematics so far. Every exercise page was rendered as a 300dpi image to read the determinant bar notation and matrix brackets accurately. The old stub taught Cramer's Rule as a core technique, but the current book never mentions it anywhere — the only method taught for solving a linear system is X=A^{-1}B (the matrix method). The old stub also collapsed all six real exercises into one invented 35-question group with no solutions file behind it. A genuine book-level inconsistency was also found and preserved as a note rather than silently corrected: the Miscellaneous Exercise's own instruction lines reference 'Exercises 11 to 15' and 'Exercise 17 to 19', which no longer exist after rationalisation trimmed the exercise down to 9 questions — confirmed by reading straight through the exercise and finding no such numbered questions. Every determinant, adjoint, inverse, and system-solution computation in this file was independently verified with a Python script using sympy, catching zero discrepancies against the hand-derived working, including the two symbolic matrix-cubed identities (Q15, Q16 of Exercise 4.4) and the parametrised consistency question (Q4 of Exercise 4.5).. Questions are referenced from the NCERT textbook for identification.
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