NCERT Solutions · Class 12 · Chapter 4: Determinants

NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.3

Exercise 4.3: Minors and cofactors. The minor \(M_{ij}\) is the determinant left after deleting row i and column j; the cofactor is \(A_{ij} = (-1)^{i+j}M_{ij}\). A determinant equals the sum of the elements of any one row (or column) times their own cofactors.

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Exercise 4.3 questions and solutions

Exercise 4.3, Question 1

Write the minors and cofactors of the elements.

(i) \(\begin{vmatrix}2 & -4 \\ 0 & 3\end{vmatrix}\)

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For a \(2 \times 2\), each minor is the single element left over.

Answer: \[\begin{aligned}M_{11} &= 3, M_{12} \\ &= 0, M_{21} \\ &= -4, M_{22} \\ &= 2\end{aligned}\]; \[\begin{aligned}A_{11} &= 3, A_{12} \\ &= 0, A_{21} \\ &= 4, A_{22} \\ &= 2\end{aligned}\]

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

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Delete row and column; the sign pattern is \(+, -, -, +\).

Answer: \[\begin{aligned}M_{11} &= d, M_{12} \\ &= b, M_{21} \\ &= c, M_{22} \\ &= a\end{aligned}\]; \[\begin{aligned}A_{11} &= d, A_{12} \\ &= -b, A_{21} \\ &= -c, A_{22} \\ &= a\end{aligned}\]

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Exercise 4.3, Question 2

Write the minors and cofactors of the elements.

(i) \(\begin{vmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{vmatrix}\)

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Deleting row i and column i leaves the \(2 \times 2\) identity (value 1); every other deletion leaves a determinant with a zero row or column.

Answer: \(M_{11} = M_{22} = M_{33} = 1\), all other minors 0; the cofactors are the same (\(A_{ii} = 1\), others 0)

(ii) \(\begin{vmatrix}1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2\end{vmatrix}\)

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E.g. \[\begin{aligned}M_{11} &= \begin{vmatrix}5 & -1 \\ 1 & 2\end{vmatrix} \\ &= 11\end{aligned}\], \[\begin{aligned}M_{12} &= \begin{vmatrix}3 & -1 \\ 0 & 2\end{vmatrix} \\ &= 6\end{aligned}\], and so on; then attach the signs \[\begin{smallmatrix}+ & - & + \\ - & + & - \\ + & - & +\end{smallmatrix}\]

Answer: Minors: \[\begin{aligned}M_{11} &= 11, M_{12} \\ &= 6, M_{13} \\ &= 3, M_{21} \\ &= -4, M_{22} \\ &= 2, M_{23} \\ &= 1, M_{31} \\ &= -20, M_{32} \\ &= -13, M_{33} \\ &= 5\end{aligned}\] Cofactors: \[\begin{aligned}A_{11} &= 11, A_{12} \\ &= -6, A_{13} \\ &= 3, A_{21} \\ &= 4, A_{22} \\ &= 2, A_{23} \\ &= -1, A_{31} \\ &= -20, A_{32} \\ &= 13, A_{33} \\ &= 5\end{aligned}\]

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Exercise 4.3, Question 3

Using the cofactors of the second row, evaluate \(\Delta = \begin{vmatrix}5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3\end{vmatrix}\).

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\[\begin{aligned}A_{21} &= -\begin{vmatrix}3 & 8 \\ 2 & 3\end{vmatrix} \\ &= 7\end{aligned}\], \[\begin{aligned}A_{22} &= \begin{vmatrix}5 & 8 \\ 1 & 3\end{vmatrix} \\ &= 7\end{aligned}\], \[\begin{aligned}A_{23} &= -\begin{vmatrix}5 & 3 \\ 1 & 2\end{vmatrix} \\ &= -7\end{aligned}\]
\(\Delta = 2(7) + 0(7) + 1(-7)\).

Answer: \(\Delta = 7\)

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Exercise 4.3, Question 4

Using the cofactors of the third column, evaluate \(\Delta = \begin{vmatrix}1 & x & y z \\ 1 & y & x z \\ 1 & z & x y\end{vmatrix}\).

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\[\begin{aligned}A_{13} &= \begin{vmatrix}1 & y \\ 1 & z\end{vmatrix} \\ &= z - y\end{aligned}\], \[\begin{aligned}A_{23} &= -\begin{vmatrix}1 & x \\ 1 & z\end{vmatrix} \\ &= x - z\end{aligned}\], \[\begin{aligned}A_{33} &= \begin{vmatrix}1 & x \\ 1 & y\end{vmatrix} \\ &= y - x\end{aligned}\]
\[\Delta = yz(z - y) + zx(x - z) + xy(y - x)\], which factorises.

Answer: \(\Delta = (x - y)(y - z)(z - x)\)

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Exercise 4.3, Question 5

\(\Delta = |a_{ij}|_{3 \times 3}\) with cofactors \(A_{ij}\). Then \(\Delta\) is: (A) \(a_{11}A_{31} + a_{12}A_{32} + a_{13}A_{33}\) (B) \(a_{11}A_{11} + a_{12}A_{21} + a_{13}A_{31}\) (C) \(a_{21}A_{11} + a_{22}A_{12} + a_{23}A_{13}\) (D) \(a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}\)

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\(\Delta\) is the elements of one column (or row) times their own cofactors: column 1 with \(A_{11}, A_{21}, A_{31}\). The other options mix a row with another row's cofactors (giving 0) or with the wrong cofactors.

Answer: (D)

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