Exercise 4.3 — Determinants

1. 4.3.14 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Write Minors and Cofactors of the elements of following determinants: (i) |[2,-4],[0,3]| (ii) |[a,c],[b,d]|.

   Hint. For a 2x2 determinant, each minor is just the single opposite diagonal entry, and cofactors apply the sign pattern +,-,-,+.

   For a 2x2 matrix [[p,q],[r,s]], the minor of each entry is simply the single opposite-diagonal entry, since deleting that entry's row and column leaves a 1x1 determinant. (i) M11=3, M12=0, M21=-4, M22=2. Applying the sign pattern: A11=3, A12=-0=0, A21=-(-4)=4, A22=2. (ii) M11=d, M12=b, M21=c, M22=a. Applying signs: A11=d, A12=-b, A21=-c, A22=a.

   ✦ (i) M11=3,M12=0,M21=-4,M22=2; A11=3,A12=0,A21=4,A22=2. (ii) M11=d,M12=b,M21=c,M22=a; A11=d,A12=-b,A21=-c,A22=a.

2. 4.3.26 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Write Minors and Cofactors of the elements of following determinants: (i) |[1,0,0],[0,1,0],[0,0,1]| (ii) |[1,0,4],[3,5,-1],[0,1,2]|.

   Hint. Delete the relevant row and column for each minor, then apply the (-1)^(i+j) sign pattern for each cofactor.

   (i) For the identity matrix: M11=1,M12=0,M13=0,M21=0,M22=1,M23=0,M31=0,M32=0,M33=1, and since every off-diagonal minor is 0, the cofactors equal the minors here: A11=1,A12=0,A13=0,A21=0,A22=1,A23=0,A31=0,A32=0,A33=1. (ii) M11=det[[5,-1],[1,2]]=11, so A11=11. M12=det[[3,-1],[0,2]]=6, so A12=-6. M13=det[[3,5],[0,1]]=3, so A13=3. M21=det[[0,4],[1,2]]=-4, so A21=4. M22=det[[1,4],[0,2]]=2, so A22=2. M23=det[[1,0],[0,1]]=1, so A23=-1. M31=det[[0,4],[5,-1]]=-20, so A31=-20. M32=det[[1,4],[3,-1]]=-13, so A32=13. M33=det[[1,0],[3,5]]=5, so A33=5.

   ✦ (i) All minors/cofactors are 0 except the diagonal ones, which are 1. (ii) A11=11,A12=-6,A13=3,A21=4,A22=2,A23=-1,A31=-20,A32=13,A33=5.

3. 4.3.34 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Using Cofactors of elements of second row, evaluate Delta = |[5,3,8],[2,0,1],[1,2,3]|.

   Hint. Compute the three cofactors of row 2 first, then combine as a21*A21+a22*A22+a23*A23.

   A21 = -det[[3,8],[2,3]] = -(9-16) = 7. A22 = det[[5,8],[1,3]] = 15-8 = 7. A23 = -det[[5,3],[1,2]] = -(10-3) = -7. Delta = a21A21+a22A22+a23*A23 = 2(7)+0(7)+1(-7) = 14+0-7 = 7.

   ✦ Delta = 7

4. 4.3.45 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   Using Cofactors of elements of third column, evaluate Delta = |[1,x,yz],[1,y,zx],[1,z,xy]|.

   Hint. Compute the three cofactors of column 3 (each a 2x2 determinant in the remaining rows/columns), then combine and factor.

   Since the expansion is along column 3, only the three column-3 cofactors are needed: A13 = det[[1,y],[1,z]] = z-y. A23 = -det[[1,x],[1,z]] = -(z-x) = x-z. A33 = det[[1,x],[1,y]] = y-x. Delta = yzA13+zxA23+xy*A33 = yz(z-y)+zx(x-z)+xy(y-x). Expanding and grouping terms, this simplifies to the factored form -(x-y)(x-z)(y-z), equivalently (x-y)(y-z)(z-x), because the expanded polynomial has (x-y), (y-z), and (z-x) as its only linear factors.

   ✦ Delta = (x-y)(y-z)(z-x)

5. 4.3.52 marksNCERT Class 12 Mathematics, Determinants, Reprint 2026-27

   If Delta=|[a11,a12,a13],[a21,a22,a23],[a31,a32,a33]| and A_ij is Cofactor of a_ij, then value of Delta is given by (A) a11A31+a12A32+a13A33 (B) a11A11+a12A21+a13A31 (C) a21A11+a22A12+a23A13 (D) a11A11+a21A21+a31A31.

   Hint. Recall that a determinant equals the sum of a row's (or column's) entries times their OWN matching cofactors — check which option pairs entries with cofactors from the same row or column.

   Option (D) pairs a11,a21,a31 (the entries of column 1) with A11,A21,A31 (the cofactors of those exact same entries) — this is valid expansion along column 1. The other options mix entries from one row/column with cofactors from a different row/column, which by the 'sum is zero' property of determinants gives 0, not Delta.

   ✦ (D) a11A11+a21A21+a31*A31

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.