NCERT Class 12 Mathematics — Determinants, Exercise 4.3. All 5 questions solved.

Exercise 4.3 names the pieces that Exercise 4.1 was already using without labelling them.

The minor $\mathrm{M}_{ij}$ of an element $a_{ij}$ is the determinant left after deleting row $i$ and column $j$. The cofactor $\mathrm{A}_{ij}$ is that minor with a sign attached:

$$\mathrm{A}_{ij} = (-1)^{i+j}\,\mathrm{M}_{ij}$$

so the signs follow the familiar chessboard pattern, $+$ wherever $i + j$ is even. With cofactors in hand, the determinant expands along any row or column:

$$\Delta = a_{i1}\mathrm{A}_{i1} + a_{i2}\mathrm{A}_{i2} + a_{i3}\mathrm{A}_{i3}$$

Key insight. Elements times the cofactors of their own row or column give $\Delta$. Elements times the cofactors of a different row or column give zero, every time. Question 5 is built entirely on that distinction, and it is also the reason $\mathrm{A}\,(\mathrm{adj}\,\mathrm{A}) = |\mathrm{A}|\,\mathrm{I}$ in the next exercise — the off-diagonal entries are those mismatched sums.

Question 1

Write Minors and Cofactors of the elements of the following determinants.

Question 1 (i)

$\begin{vmatrix} 2 & -4 \\ 0 & 3 \end{vmatrix}$

Solution. For a $2\times2$ determinant, deleting a row and a column leaves a single number, so each minor is just the entry diagonally opposite.

$$\mathrm{M}_{11} = 3, \qquad \mathrm{M}_{12} = 0, \qquad \mathrm{M}_{21} = -4, \qquad \mathrm{M}_{22} = 2$$

Now apply the signs $(-1)^{i+j}$ — positive on the main diagonal, negative off it:

$$\mathrm{A}_{11} = +3 = 3, \qquad \mathrm{A}_{12} = -0 = 0$$ $$\mathrm{A}_{21} = -(-4) = 4, \qquad \mathrm{A}_{22} = +2 = 2$$

Note that $\mathrm{M}_{21} = -4$ but $\mathrm{A}_{21} = 4$: the minor keeps the entry’s own sign, and the cofactor then flips it.

Question 1 (ii)

$\begin{vmatrix} a & c \\ b & d \end{vmatrix}$

Solution. The same pattern with letters. Watch the layout: $c$ sits in position $(1,2)$ and $b$ in position $(2,1)$, not the other way round.

$$\mathrm{M}_{11} = d, \qquad \mathrm{M}_{12} = b, \qquad \mathrm{M}_{21} = c, \qquad \mathrm{M}_{22} = a$$

$$\mathrm{A}_{11} = d, \qquad \mathrm{A}_{12} = -b, \qquad \mathrm{A}_{21} = -c, \qquad \mathrm{A}_{22} = a$$

Question 2

Write Minors and Cofactors of the elements of the following determinants.

Question 2 (i)

$\begin{vmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{vmatrix}$

Solution. This is the identity matrix. Deleting any row and column leaves a $2\times2$ block which is either the identity (determinant 1) or has a zero row (determinant 0).

$$\mathrm{M}_{11} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1, \qquad \mathrm{M}_{12} = \begin{vmatrix} 0 & 0 \\ 0 & 1 \end{vmatrix} = 0, \qquad \mathrm{M}_{13} = \begin{vmatrix} 0 & 1 \\ 0 & 0 \end{vmatrix} = 0$$

By the same reasoning the full set is

$$\mathrm{M}_{11} = \mathrm{M}_{22} = \mathrm{M}_{33} = 1, \qquad \text{all other } \mathrm{M}_{ij} = 0$$

Since every non-zero minor sits on the main diagonal, where the sign is positive, the cofactors are identical to the minors.

Question 2 (ii)

$\begin{vmatrix} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{vmatrix}$

Solution. Nine minors, each a $2\times2$ determinant. Taking them row by row:

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

$$\mathrm{M}_{21} = \begin{vmatrix} 0 & 4 \\ 1 & 2 \end{vmatrix} = -4, \quad \mathrm{M}_{22} = \begin{vmatrix} 1 & 4 \\ 0 & 2 \end{vmatrix} = 2, \quad \mathrm{M}_{23} = \begin{vmatrix} 1 & 0 \\ 0 & 1 \end{vmatrix} = 1$$

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

Now apply the sign pattern. The four positions with $i + j$ odd — $(1,2)$, $(2,1)$, $(2,3)$ and $(3,2)$ — change sign; the other five do not.

$$\mathrm{A}_{11} = 11, \quad \mathrm{A}_{12} = -6, \quad \mathrm{A}_{13} = 3$$ $$\mathrm{A}_{21} = 4, \quad \mathrm{A}_{22} = 2, \quad \mathrm{A}_{23} = -1$$ $$\mathrm{A}_{31} = -20, \quad \mathrm{A}_{32} = 13, \quad \mathrm{A}_{33} = 5$$

Question 3

Using Cofactors of elements of second row, evaluate $\Delta = \begin{vmatrix} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{vmatrix}$.

Solution. The question specifies the second row, so find the three cofactors of that row. Its signs run $-\,+\,-$.

$$\mathrm{A}_{21} = -\begin{vmatrix} 3 & 8 \\ 2 & 3 \end{vmatrix} = -(9 – 16) = 7$$

$$\mathrm{A}_{22} = +\begin{vmatrix} 5 & 8 \\ 1 & 3 \end{vmatrix} = 15 – 8 = 7$$

$$\mathrm{A}_{23} = -\begin{vmatrix} 5 & 3 \\ 1 & 2 \end{vmatrix} = -(10 – 3) = -7$$

Now multiply each by the element it belongs to, and add:

$$\Delta = a_{21}\mathrm{A}_{21} + a_{22}\mathrm{A}_{22} + a_{23}\mathrm{A}_{23} = 2(7) + 0(7) + 1(-7)$$

$$= 14 + 0 – 7 = 7$$

The zero in the middle does not remove the need to compute $\mathrm{A}_{22}$ if you are asked for the cofactors, but it does mean that term contributes nothing.

Question 4

Using Cofactors of elements of third column, evaluate $\Delta = \begin{vmatrix} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{vmatrix}$.

Solution. The third column has signs $+\,-\,+$, and each minor is a $2\times2$ determinant of the first two columns.

$$\mathrm{A}_{13} = +\begin{vmatrix} 1 & y \\ 1 & z \end{vmatrix} = z – y$$

$$\mathrm{A}_{23} = -\begin{vmatrix} 1 & x \\ 1 & z \end{vmatrix} = -(z – x) = x – z$$

$$\mathrm{A}_{33} = +\begin{vmatrix} 1 & x \\ 1 & y \end{vmatrix} = y – x$$

Expanding along the third column:

$$\Delta = yz(z – y) + zx(x – z) + xy(y – x)$$

$$= yz^2 – y^2z + x^2z – xz^2 + xy^2 – x^2y$$

This factorises. Grouping and taking out common factors gives

$$\Delta = (x – y)(y – z)(z – x)$$

which you can check by expanding the product back out. The factorised form is worth recognising: a determinant that vanishes whenever two of $x, y, z$ are equal must have each of those differences as a factor.

Question 5

If $\Delta = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{vmatrix}$ and $\mathrm{A}_{ij}$ is Cofactors of $a_{ij}$, then value of $\Delta$ is given by

Solution. A valid expansion pairs each element with the cofactor of its own position — all along one row, or all down one column. Check each option against that.

(A) takes elements from row 1 but cofactors from row 3. The indices do not match, so this is not an expansion at all; sums of this kind are always zero.

(B) pairs $a_{12}$ with $\mathrm{A}_{21}$ and $a_{13}$ with $\mathrm{A}_{31}$ — the subscripts are reversed, not matched.

(C) takes elements from row 2 but cofactors from row 1. Mismatched again, and therefore zero.

(D) pairs $a_{11}$, $a_{21}$, $a_{31}$ — the whole of the first column — each with its own cofactor. That is the expansion of $\Delta$ along the first column.

Common mistakes

• Confusing a minor with a cofactor. They differ only by the sign $(-1)^{i+j}$, and questions 1 and 2 ask for both, so the difference has to be written down explicitly. In question 1(i), $\mathrm{M}_{21} = -4$ while $\mathrm{A}_{21} = 4$.
• Applying the sign twice. The minor already carries whatever signs its own entries had. $\mathrm{M}_{32} = -13$ in question 2(ii); the cofactor is $-(-13) = 13$, not $-13$ again.
• Deleting the wrong row or column. $\mathrm{M}_{23}$ means delete row 2 and column 3. Reading the subscripts as (column, row) transposes every answer.
• Expanding along a row you were not asked to use. Questions 3 and 4 name a specific row and column, and part of the mark is for using it. The answer is the same either way — that is the point being demonstrated.
• Mixing elements from one row with cofactors from another. That sum is always zero, never $\Delta$. Three of the four options in question 5 are that mistake in different disguises.

Practise next

• Exercise 4.4 — the adjoint and the inverse, built directly from the cofactors computed here.
• Exercise 4.2 — the area determinant, if the $3\times3$ expansions still feel slow.