Class 12 Maths NCERT Solutions Chapter 4 Ex 4.3 – Minors and Cofactors

The minor M_{ij} of an element a_{ij} is the determinant left after deleting row i and column j; the cofactor is A_{ij} = (-1)^{i+j}M_{ij}.

(i) For \begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}: deleting a row and column leaves a single entry each time.

M_{11}=3,\quad M_{12}=0,\quad M_{21}=-4,\quad M_{22}=2

A_{11}=(+1)(3)=3,\quad A_{12}=(-1)(0)=0,\quad A_{21}=(-1)(-4)=4,\quad A_{22}=(+1)(2)=2

(ii) For \begin{vmatrix} a & c \\ b & d \end{vmatrix}:

M_{11}=d,\quad M_{12}=b,\quad M_{21}=c,\quad M_{22}=a

A_{11}=(+1)d=d,\quad A_{12}=(-1)b=-b,\quad A_{21}=(-1)c=-c,\quad A_{22}=(+1)a=a

(i) For the identity matrix, every minor obtained by deleting a row and column from the identity pattern is either 0 or 1:

Since (-1)^{i+j} only changes the sign, and every off-diagonal minor is already 0, the cofactors equal the minors here:

(ii) For \begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}, deleting each row and column in turn:

M_{11}=\begin{vmatrix} 5 & -1 \\ 1 & 2 \end{vmatrix}=11,\quad M_{12}=\begin{vmatrix} 3 & -1 \\ 0 & 2 \end{vmatrix}=6,\quad M_{13}=\begin{vmatrix} 3 & 5 \\ 0 & 1 \end{vmatrix}=3

M_{21}=\begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix}=-4,\quad M_{22}=\begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix}=2,\quad M_{23}=\begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix}=1

M_{31}=\begin{vmatrix} 0 & 4 \\ 5 & -1 \end{vmatrix}=-20,\quad M_{32}=\begin{vmatrix} 1 & 4 \\ 3 & -1 \end{vmatrix}=-13,\quad M_{33}=\begin{vmatrix} 1 & 0 \\ 3 & 5 \end{vmatrix}=5

Applying the sign pattern A_{ij}=(-1)^{i+j}M_{ij}:

The second row entries are a_{21}=2,\ a_{22}=0,\ a_{23}=1. Their cofactors:

A_{21}=(-1)^{2+1}\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -(9-16) = 7

A_{22}=(-1)^{2+2}\begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = (15-8) = 7

A_{23}=(-1)^{2+3}\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -(10-3) = -7

Expanding along row 2: \Delta = a_{21}A_{21}+a_{22}A_{22}+a_{23}A_{23} = 2(7)+0(7)+1(-7) = 14+0-7

The third column entries are a_{13}=yz,\ a_{23}=zx,\ a_{33}=xy. Their cofactors:

A_{13}=(-1)^{1+3}\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = z-y

A_{23}=(-1)^{2+3}\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z-x) = x-z

A_{33}=(-1)^{3+3}\begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = y-x

Expanding along column 3:

\Delta = yz(z-y)+zx(x-z)+xy(y-x) = yz^2-y^2z+x^2z-xz^2+xy^2-x^2y

Grouping and factoring these six terms gives the well-known identity:

A determinant can be expanded along any row or column, but only if every element used is paired with the cofactor at that same row-column position. Checking each option:

Options (A), (B) and (C) pair elements from one row (or column) with cofactors belonging to a different row (or column) — such "mixed" sums are not valid expansions of \Delta (they instead equal 0, by a standard determinant property).

Option (D), a_{11}A_{11}+a_{21}A_{21}+a_{31}A_{31}, pairs each element of column 1 with its own cofactor — this is exactly the cofactor expansion of \Delta along the first column.