Miscellaneous Exercise — Determinants

1. 4.M.15 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Prove that the determinant |[x,sin theta,cos theta],[-sin theta,-x,1],[cos theta,1,x]| is independent of theta.

   Hint. Expand along row 1 and simplify using sin^2+cos^2=1; if theta disappears entirely, independence is proved.

   Expanding along row 1: x(-x(x)-1(1)) - sin(theta)(-sin(theta)(x)-1(cos theta)) + cos(theta)(-sin(theta)(1)-(-x)(cos theta)) = x(-x^2-1) - sin(theta)(-x sin theta - cos theta) + cos(theta)(-sin theta + x cos theta). Expanding fully: -x^3-x + x sin^2(theta) + sin(theta)cos(theta) - sin(theta)cos(theta) + x cos^2(theta) = -x^3-x+x(sin^2 theta+cos^2 theta) = -x^3-x+x = -x^3, since the sin(theta)cos(theta) terms cancel and sin^2+cos^2=1.

   ✦ The determinant simplifies to -x^3, which has no theta in it — proving it is independent of theta.

2. 4.M.26 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Evaluate |[cos alpha cos beta,cos alpha sin beta,-sin alpha],[-sin beta,cos beta,0],[sin alpha cos beta,sin alpha sin beta,cos alpha]|.

   Hint. Expand along row 2, which has a convenient zero entry, then simplify using sin^2+cos^2=1 for both angles.

   Expanding along row 2 skips one term entirely, since it has a zero entry there. Collecting all the sin/cos terms for both alpha and beta and applying the Pythagorean identity to each pair, the determinant simplifies down to exactly 1, because every remaining sin^2+cos^2 combination reduces to 1.

   ✦ 1

3. 4.M.35 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   If A^{-1}=[[3,-1,1],[-15,6,-5],[5,-2,2]] and B=[[1,2,-2],[-1,3,0],[0,-2,1]], find (AB)^{-1}.

   Hint. Use (AB)^{-1}=B^{-1}A^{-1} directly, without ever recovering A itself.

   (AB)^{-1} = B^{-1}A^{-1}. First compute B^{-1}: |B| = 1(3-0)-2(-1-0)+(-2)(2-0) = 3+2-4 = 1, and computing the adjoint of B gives B^{-1}. Then multiplying B^{-1} by the given A^{-1} directly gives the answer.

   ✦ (AB)^{-1} = [[9,-3,5],[-2,1,0],[1,0,2]]

4. 4.M.46 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Let A=[[1,2,1],[2,3,1],[1,1,5]]. Verify that (i) [adj A]^{-1} = adj(A^{-1}) (ii) (A^{-1})^{-1} = A.

   Hint. Compute A^{-1} first via the adjoint formula, then compute both sides of each identity independently.

   |A| = 1(15-1)-2(10-1)+1(2-3) = 14-18-1 = -5. Computing adj(A) and A^{-1}=adj(A)/(-5). (i) Computing [adj A]^{-1} directly, and separately computing adj(A^{-1}) from the already-found A^{-1}, gives the identical matrix in both cases. (ii) Inverting A^{-1} a second time recovers exactly the original matrix A.

   ✦ Both identities verified: [adj A]^{-1} = adj(A^{-1}) = [[-1/5,-2/5,-1/5],[-2/5,-3/5,-1/5],[-1/5,-1/5,-1]], and (A^{-1})^{-1} = A.

5. 4.M.55 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Evaluate |[x,y,x+y],[y,x+y,x],[x+y,x,y]|.

   Hint. Add all three columns together first (a column operation that doesn't change the determinant) to spot a common factor.

   Adding all three columns into column 1 (C1 -> C1+C2+C3): every row becomes x+y+(x+y) = 2(x+y) in the first column, since each row's three entries are some permutation of x, y, x+y, always summing to 2(x+y). Factoring out 2(x+y) from column 1 and simplifying the remaining determinant gives the final factored answer.

   ✦ -2(x+y)(x^2-xy+y^2), equivalently -2(x^3+y^3)

6. 4.M.65 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Evaluate |[1,x,y],[1,x+y,y],[1,x,x+y]|.

   Hint. Subtract row 1 from rows 2 and 3 to create zeros in the first column below the top entry.

   R2 -> R2-R1 gives [0,y,0]. R3 -> R3-R1 gives [0,0,x]. The determinant becomes |[1,x,y],[0,y,0],[0,0,x]|, which is upper-triangular-like after the row operations, so expanding along column 1 gives 1*(yx-00) = xy.

   ✦ xy

7. 4.M.76 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Solve the system of equations: 2/x+3/y+10/z=4, 4/x-6/y+5/z=1, 6/x+9/y-20/z=2.

   Hint. Substitute u=1/x, v=1/y, w=1/z to turn this into an ordinary linear system, solve using the matrix method, then invert back.

   Let u=1/x, v=1/y, w=1/z. The system becomes 2u+3v+10w=4, 4u-6v+5w=1, 6u+9v-20w=2, an ordinary linear system in u,v,w. Solving via the matrix method gives u=1/2, v=1/3, w=1/5. Converting back: x=1/u=2, y=1/v=3, z=1/w=5.

   ✦ x=2, y=3, z=5

8. 4.M.82 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   If x,y,z are nonzero real numbers, then the inverse of matrix A=[[x,0,0],[0,y,0],[0,0,z]] is (A) diag(x^{-1},y^{-1},z^{-1}) (B) xyz*diag(x^{-1},y^{-1},z^{-1}) (C) (1/xyz)*diag(x,y,z) (D) (1/xyz)*I.

   Hint. Recall that the inverse of a diagonal matrix is simply the diagonal matrix of reciprocals — verify by direct multiplication.

   For a diagonal matrix, A^{-1} is the diagonal matrix of reciprocals: diag(1/x,1/y,1/z), since a diagonal matrix's cofactors are all zero off the diagonal. Checking directly: A * diag(1/x,1/y,1/z) = diag(x/x,y/y,z/z) = diag(1,1,1) = I, which means this candidate genuinely is the inverse. This matches option (A) exactly; the other options represent different (incorrect) matrices.

   ✦ (A) diag(x^{-1},y^{-1},z^{-1})

9. 4.M.96 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Let A=[[1,sin theta,1],[-sin theta,1,sin theta],[-1,-sin theta,1]], where 0<=theta<=2pi. Then (A) Det(A)=0 (B) Det(A) is in (2,infinity) (C) Det(A) is in (2,4) (D) Det(A) is in [2,4].

   Hint. Expand the determinant symbolically in terms of sin(theta), then use that sin^2(theta) ranges over [0,1] to find the range of the result.

   Expanding the determinant gives 2sin^2(theta)+2 (verified by direct cofactor expansion and simplification). Since sin^2(theta) ranges over [0,1] for any real theta, 2sin^2(theta)+2 ranges over [2,4], hitting 2 when sin(theta)=0 and 4 when sin(theta)=+-1.

   ✦ (D) Det(A) is in [2,4]

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.