Class 12th Determinants NCERT Solutions Chapter 4 Ex 4.3

Concept used. For a determinant of order 2, the minor M_ij of element a_ij is the (single-entry) determinant obtained by deleting row i and column j. Hence each minor is just the remaining diagonally-opposite entry. The cofactor is A_ij=(-1)^i+jM_ij.

Part (i).

1. Entries: a₁₁=2, a₁₂=-4, a₂₁=0, a₂₂=3.
2. Minors. Delete the relevant row and column of the 2× 2 matrix: M₁₁ = 3, M₁₂ = 0, M₂₁ = -4, M₂₂ = 2.
3. Cofactors using A_ij=(-1)^i+jM_ij: aligned A₁₁ &= (-1)²· 3 = 3,
   A₁₂ &= (-1)³· 0 = 0,
   A₂₁ &= (-1)³·(-4) = 4,
   A₂₂ &= (-1)⁴· 2 = 2. aligned

Part (ii).

1. Entries: a₁₁=a, a₁₂=c, a₂₁=b, a₂₂=d.
2. Minors: M₁₁=d, M₁₂=b, M₂₁=c, M₂₂=a.
3. Cofactors: aligned A₁₁ &= +d, A₁₂ = -b,
   A₂₁ &= -c, A₂₂ = +a. aligned

Concept used. For an order-3 determinant, the minor M_ij is the 2× 2 determinant left after deleting row i and column j. The cofactor is A_ij=(-1)^i+jM_ij.

Part (i): the identity I₃.

1. Each minor is the determinant of an order-2 sub-matrix. With the identity, deleting any row and column leaves either pmatrix 1 & 0 0 & 1 pmatrix (minor =1) when the deleted row and column are equal-indexed, or a matrix with a row of zeros (minor =0) otherwise.
2. Compute each minor explicitly: aligned M₁₁ &= vmatrix 1 & 0 0 & 1 vmatrix = 1, & M₁₂ &= vmatrix 0 & 0 0 & 1 vmatrix = 0, & M₁₃ &= vmatrix 0 & 1 0 & 0 vmatrix = 0,
   M₂₁ &= vmatrix 0 & 0 0 & 1 vmatrix = 0, & M₂₂ &= vmatrix 1 & 0 0 & 1 vmatrix = 1, & M₂₃ &= vmatrix 1 & 0 0 & 0 vmatrix = 0,
   M₃₁ &= vmatrix 0 & 0 1 & 0 vmatrix = 0, & M₃₂ &= vmatrix 1 & 0 0 & 0 vmatrix = 0, & M₃₃ &= vmatrix 1 & 0 0 & 1 vmatrix = 1. aligned
3. Cofactors. Each minor is multiplied by (-1)^i+j from the chessboard: A₁₁=1, A₂₂=1, A₃₃=1; all other A_ij=0. (Off-diagonal minors are already 0, so signs do not matter there.)

Part (ii). A=pmatrix 1 & 0 & 4 3 & 5 & -1 0 & 1 & 2 pmatrix.

1. Row 1 minors: aligned M₁₁ &= vmatrix 5 & -1 1 & 2 vmatrix = (5)(2) - (-1)(1) = 10 + 1 = 11,
   M₁₂ &= vmatrix 3 & -1 0 & 2 vmatrix = (3)(2) - (-1)(0) = 6 - 0 = 6,
   M₁₃ &= vmatrix 3 & 5 0 & 1 vmatrix = (3)(1) - (5)(0) = 3 - 0 = 3. aligned
2. Row 2 minors: aligned M₂₁ &= vmatrix 0 & 4 1 & 2 vmatrix = 0 - 4 = -4,
   M₂₂ &= vmatrix 1 & 4 0 & 2 vmatrix = 2 - 0 = 2,
   M₂₃ &= vmatrix 1 & 0 0 & 1 vmatrix = 1 - 0 = 1. aligned
3. Row 3 minors: aligned M₃₁ &= vmatrix 0 & 4 5 & -1 vmatrix = 0 - 20 = -20,
   M₃₂ &= vmatrix 1 & 4 3 & -1 vmatrix = -1 - 12 = -13,
   M₃₃ &= vmatrix 1 & 0 3 & 5 vmatrix = 5 - 0 = 5. aligned
4. Cofactors (apply chessboard signs): aligned A₁₁ &= +11, & A₁₂ &= -6, & A₁₃ &= +3,
   A₂₁ &= +4, & A₂₂ &= +2, & A₂₃ &= -1,
   A₃₁ &= -20, & A₃₂ &= +13, & A₃₃ &= +5. aligned (Notice A₁₂ flipped to -6, A₂₁ flipped to +4, A₂₃ to -1, A₃₁ stays -20, A₃₂ to +13.)

(ii) Minors 11,6,3,-4,2,1,-20,-13,5; Cofactors 11,-6,3,4,2,-1,-20,13,5.

Concept used. A determinant can be expanded along any row or column by the rule |A| = a_i1A_i1 + a_i2A_i2 + a_i3A_i3, where A_ij are the cofactors. Here we are told to expand along row 2.

1. Compute the row-2 cofactors. The minors are 2× 2 determinants obtained by deleting R₂ and the relevant column. aligned M₂₁ &= vmatrix 3 & 8 2 & 3 vmatrix = 9 - 16 = -7,
   M₂₂ &= vmatrix 5 & 8 1 & 3 vmatrix = 15 - 8 = 7,
   M₂₃ &= vmatrix 5 & 3 1 & 2 vmatrix = 10 - 3 = 7. aligned
2. Apply chessboard signs for row 2 (-,+,-): A₂₁ = -M₂₁ = 7, A₂₂ = +M₂₂ = 7, A₂₃ = -M₂₃ = -7.
3. Row-2 entries are a₂₁=2, a₂₂=0, a₂₃=1. Expand: Δ = 2· 7 + 0· 7 + 1·(-7) = 14 + 0 - 7 = 7.

Δ = 7.

Concept used. Expand along column 3: Δ = a₁₃A₁₃ + a₂₃A₂₃ + a₃₃A₃₃, with A_i3=(-1)ⁱ⁺³M_i3.

1. Column-3 entries: a₁₃=yz, a₂₃=zx, a₃₃=xy.
2. Minors (delete column 3 and the row of the entry): aligned M₁₃ &= vmatrix 1 & y 1 & z vmatrix = z - y,
   M₂₃ &= vmatrix 1 & x 1 & z vmatrix = z - x,
   M₃₃ &= vmatrix 1 & x 1 & y vmatrix = y - x. aligned
3. Apply signs (+,-,+) for column 3: A₁₃ = +(z-y), A₂₃ = -(z-x) = x-z, A₃₃ = +(y-x).
4. Expand: Δ = yz(z-y) + zx(x-z) + xy(y-x).
5. Expand each product: yz(z-y) = yz² - y²z, zx(x-z) = x²z - xz², xy(y-x) = xy² - x²y.
6. Sum: Δ = yz² - y²z + x²z - xz² + xy² - x²y.
7. Group and factor. Pair yz²-xz² = z²(y-x) and -x²y + x²z = x²(z-y) and xy²-y²z = y²(x-z). So Δ = z²(y-x) + x²(z-y) + y²(x-z). This is the well-known Vandermonde-style factor: (x-y)(y-z)(z-x) up to a sign.
8. Verify by checking a specific case. With x=y we expect Δ = 0 since columns 1 and 2 become identical after... actually with x=y the original determinant has R₁-R₂=(0,x-y,yz-zx)=(0,0,z(y-x))=(0,0,0), so Δ = 0. Our formula gives z²(y-x)+x²(z-y)+y²(x-z)|_x=y=z²· 0+x²(z-x)+x²(x-z)=0.

Δ = (x-y)(y-z)(z-x), equivalently -(x²y - x²z - xy² + xz² + y²z - yz²).

Concept used. Two parallel facts about cofactors.
(I) Same-row/column rule. The sum of products of elements of a row (or column) with their own cofactors equals |A|: |A| = a_i1A_i1 + a_i2A_i2 + a_i3A_i3 (row $i$), |A| = a_1jA_1j + a_2jA_2j + a_3jA_3j (column $j$). (II) Cross-row/column rule. The sum of products of elements of one row (or column) with cofactors of another row (or column) is zero. Symbolically, _j a_ijA_kj=0 when i≠ k.

1. Check option (A): a₁₁A₃₁+a₁₂A₃₂+a₁₃A₃₃. Row 1 elements times row 3 cofactors. By rule (II) this is 0, not Δ.  Reject.
2. Check option (B): a₁₁A₁₁+a₁₂A₂₁+a₁₃A₃₁. Row 1 elements times column 1 cofactors (mismatched indices). This is neither a row expansion nor a column expansion of Δ. Reject.
3. Check option (C): a₂₁A₁₁+a₂₂A₁₂+a₂₃A₁₃. Row 2 elements times row 1 cofactors. By rule (II), = 0. Reject.
4. Check option (D): a₁₁A₁₁+a₂₁A₂₁+a₃₁A₃₁. Column 1 elements times column 1 cofactors. This is exactly the column-1 expansion of |A|.  Accept.

Option (D): Δ = a₁₁A₁₁ + a₂₁A₂₁ + a₃₁A₃₁.