Exercise 4.2 — Determinants

1. 4.2.16 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Find area of the triangle with vertices at the points given in each of the following: (i) (1,0), (6,0), (4,3) (ii) (2,7), (1,1), (10,8) (iii) (-2,-3), (3,2), (-1,-8).

   Hint. Apply the area formula (1/2)|det[[x1,y1,1],[x2,y2,1],[x3,y3,1]]| directly, remembering to take the absolute value.

   Since area must be non-negative, the absolute value of the determinant expression is taken in every part below. (i) Delta = (1/2)|1(0-3)-0(6-4)+1(18-0)| = (1/2)|(-3)-0+18| = (1/2)(15) = 15/2. (ii) Delta = (1/2)|2(1-8)-7(1-10)+1(8-10)| = (1/2)|(-14)+63+(-2)| = (1/2)(47) = 47/2. (iii) Delta = (1/2)|-2(2-(-8))-(-3)(3-(-1))+1(-24-(-2))| = (1/2)|-20+12-22| = (1/2)(30) = 15.

   ✦ (i) 15/2 square units (ii) 47/2 square units (iii) 15 square units

2. 4.2.24 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Show that points A(a,b+c), B(b,c+a), C(c,a+b) are collinear.

   Hint. Compute the determinant of the 3x3 vertex matrix symbolically and show it simplifies to zero.

   Delta = (1/2)det[[a,b+c,1],[b,c+a,1],[c,a+b,1]]. Adding column 1 to column 2 (a column operation that doesn't change the determinant): column 2 becomes a+b+c in every row, since (b+c)+a=(c+a)+b=(a+b)+c=a+b+c. With column 2 constant across all three rows, the determinant reduces to zero, because two proportional patterns across rows force a zero determinant.

   ✦ The determinant is 0, so the area of triangle ABC is 0, confirming A, B, C are collinear.

3. 4.2.36 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Find values of k if area of triangle is 4 sq. units and vertices are (i) (k,0), (4,0), (0,2) (ii) (-2,0), (0,4), (0,k).

   Hint. Set up the area expression as a function of k, then solve for both the +4 and -4 cases, since area is an absolute value.

   (i) Delta = (1/2)det[[k,0,1],[4,0,1],[0,2,1]] = (1/2)(k(0-2)-0(4-0)+1(8-0)) = (1/2)(-2k+8) = 4-k. Setting Delta=4: 4-k=4 gives k=0. Setting Delta=-4: 4-k=-4 gives k=8. (ii) Delta = (1/2)det[[-2,0,1],[0,4,1],[0,k,1]] = (1/2)(-2(4-k)-0+1(0-0)) = (1/2)(-8+2k) = k-4. Setting Delta=4: k-4=4 gives k=8. Setting Delta=-4: k-4=-4 gives k=0.

   ✦ (i) k=0 or k=8 (ii) k=0 or k=8

4. 4.2.44 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   (i) Find equation of line joining (1,2) and (3,6) using determinants. (ii) Find equation of line joining (3,1) and (9,3) using determinants.

   Hint. Set up det[[x,y,1],[x1,y1,1],[x2,y2,1]]=0 with the given two points, then expand and simplify.

   Since any point (x,y) on the line makes the three points collinear, its determinant with the two given points must be zero. (i) det[[x,y,1],[1,2,1],[3,6,1]] = 0. Expanding: x(2-6)-y(1-3)+1(6-6) = -4x+2y+0 = 0, so -4x+2y=0, which simplifies to y=2x. (ii) det[[x,y,1],[3,1,1],[9,3,1]] = 0. Expanding: x(1-3)-y(3-9)+1(9-9) = -2x+6y+0 = 0, so -2x+6y=0, which simplifies to x=3y.

   ✦ (i) y = 2x (ii) x = 3y

5. 4.2.53 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   If area of triangle is 35 sq units with vertices (2,-6), (5,4) and (k,4). Then k is (A) 12 (B) -2 (C) -12,-2 (D) 12,-2.

   Hint. Set up the area expression in terms of k and solve for both sign cases of the determinant.

   Delta = (1/2)det[[2,-6,1],[5,4,1],[k,4,1]] = (1/2)(2(4-4)-(-6)(5-k)+1(20-4k)) = (1/2)(0+30-6k+20-4k) = (1/2)(50-10k) = 25-5k. Setting Delta=35: 25-5k=35 gives k=-2. Setting Delta=-35: 25-5k=-35 gives k=12.

   ✦ (D) 12, -2

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.