Class 12 NCERT Maths – Chapter 4 (Determinants) – Exercise 4.3 Solutions

Question 1: Evaluate Minors and Cofactors of the elements of the following determinant:

(i) \left| \begin{array}{cc} 2 & -4 \\ 0 & 3 \end{array} \right|

Solution:

Element a₁₁ = 2
M_{11} = 3
C_{11} = (+1) \times 3 = 3

Element a₁₂ = -4
M_{12} = 0
C_{12} = (-1) \times 0 = 0

Element a₂₁ = 0
M_{21} = -4
C_{21} = (-1) \times (-4) = 4

Element a₂₂ = 3
M_{22} = 2
C_{22} = (+1) \times 2 = 2

Summary Table:

\begin{array}{|c|c|c|} \hline \text{Element} & \text{Minor } M_{ij} & \text{Cofactor } C_{ij} \\ \hline a_{11} = 2 & 3 & 3 \\ a_{12} = -4 & 0 & 0 \\ a_{21} = 0 & -4 & 4 \\ a_{22} = 3 & 2 & 2 \\ \hline \end{array}

(ii) \left| \begin{array}{cc} a & c \\ b & d \end{array} \right|

Solution:

Element a₁₁ = a
M_{11} = d
C_{11} = (+1) \times d = d

Element a₁₂ = c
M_{12} = b
C_{12} = (-1) \times b = -b

Element a₂₁ = b
M_{21} = c
C_{21} = (-1) \times c = -c

Element a₂₂ = d
M_{22} = a
C_{22} = (+1) \times a = a

Summary Table:

\begin{array}{|c|c|c|} \hline \text{Element} & \text{Minor } M_{ij} & \text{Cofactor } C_{ij} \\ \hline a_{11} = a & d & d \\ a_{12} = c & b & -b \\ a_{21} = b & c & -c \\ a_{22} = d & a & a \\ \hline \end{array}

Question 2: Find Minors M_{ij} and Cofactors C_{ij} for each element of the determinant:

(i): \left| \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array} \right|

Solution:

Element at a₁₁=1
M_{11} = 1
C_{11} = (-1)^{1+1} \cdot 1 = 1

Element at a₁₂=0
M_{12} = 0
C_{12} = (-1)^{1+2} \cdot 0 = 0

Element at a₁₃=0
M_{13} = 0
C_{13} = (-1)^{1+3} \cdot 0 = 0

Element at a₂₁=0
M_{21} = 0
C_{21} = (-1)^{2+1} \cdot 0 = 0

Element at a₂₂=1
M_{22} = 1
C_{22} = (-1)^{2+2} \cdot 1 = 1

Element at a₂₃=0
M_{23} = 0
C_{23} = (-1)^{2+3} \cdot 0 = 0

Element at a₃₁=0
M_{31} = 0
C_{31} = (-1)^{3+1} \cdot 0 = 0

Element at a₃₂=0
M_{32} = 0
C_{32} = (-1)^{3+2} \cdot 0 = 0

Element at a₃₃=1
M_{33} = 1
C_{33} = (-1)^{3+3} \cdot 1 = 1

(ii): \left| \begin{array}{ccc} 1 & 0 & 4 \\ 3 & 5 & -1 \\ 0 & 1 & 2 \end{array} \right|

Solutions:

Element a_{11} = 1
M_{11} = \left| \begin{array}{cc} 5 & -1 \\ 1 & 2 \end{array} \right| = 5 \cdot 2 - -1 \cdot 1 = 11
C_{11} = (-1)^{1+1} \cdot 11 = 11

Element a_{12} = 0
M_{12} = \left| \begin{array}{cc} 3 & -1 \\ 0 & 2 \end{array} \right| = 3 \cdot 2 - -1 \cdot 0 = 6
C_{12} = (-1)^{1+2} \cdot 6 = -6

Element a_{13} = 4
M_{13} = \left| \begin{array}{cc} 3 & 5 \\ 0 & 1 \end{array} \right| = 3 \cdot 1 - 5 \cdot 0 = 3
C_{13} = (-1)^{1+3} \cdot 3 = 3

Element a_{21} = 3
M_{21} = \left| \begin{array}{cc} 0 & 4 \\ 1 & 2 \end{array} \right| = 0 \cdot 2 - 4 \cdot 1 = -4
C_{21} = (-1)^{2+1} \cdot -4 = 4

Element a_{22} = 5
M_{22} = \left| \begin{array}{cc} 1 & 4 \\ 0 & 2 \end{array} \right| = 1 \cdot 2 - 4 \cdot 0 = 2
C_{22} = (-1)^{2+2} \cdot 2 = 2

Element a_{23} = -1
M_{23} = \left| \begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right| = 1 \cdot 1 - 0 \cdot 0 = 1
C_{23} = (-1)^{2+3} \cdot 1 = -1

Element a_{31} = 0
M_{31} = \left| \begin{array}{cc} 0 & 4 \\ 5 & -1 \end{array} \right| = 0 \cdot -1 - 4 \cdot 5 = -20
C_{31} = (-1)^{3+1} \cdot -20 = -20

Element a_{32} = 1
M_{32} = \left| \begin{array}{cc} 1 & 4 \\ 3 & -1 \end{array} \right| = 1 \cdot -1 - 4 \cdot 3 = -13
C_{32} = (-1)^{3+2} \cdot -13 = 13

Element a_{33} = 2
M_{33} = \left| \begin{array}{cc} 1 & 0 \\ 3 & 5 \end{array} \right| = 1 \cdot 5 - 0 \cdot 3 = 5
C_{33} = (-1)^{3+3} \cdot 5 = 5

Question 3: Using Cofactors of elements of second Row, evaluate Δ (determinant)

\Delta = \left| \begin{array}{ccc} 5 & 3 & 8 \\ 2 & 0 & 1 \\ 1 & 2 & 3 \end{array} \right|

Solution.
Step 1: Compute Cofactors of Elements in Second Row

Element a_{21} = 2
M_{21} = \left| \begin{array}{cc} 3 & 8 \\ 2 & 3 \end{array} \right| = 3 \cdot 3 - 8 \cdot 2 = 9 - 16 = -7
C_{21} = (-1)^{2+1} \cdot M_{21} = -1 \cdot (-7) = 7

Element a_{22} = 0
M_{22} = \left| \begin{array}{cc} 5 & 8 \\ 1 & 3 \end{array} \right| = 5 \cdot 3 - 8 \cdot 1 = 15 - 8 = 7
C_{22} = (-1)^{2+2} \cdot M_{22} = +1 \cdot 7 = 7

Element a_{23} = 1
M_{23} = \left| \begin{array}{cc} 5 & 3 \\ 1 & 2 \end{array} \right| = 5 \cdot 2 - 3 \cdot 1 = 10 - 3 = 7
C_{23} = (-1)^{2+3} \cdot M_{23} = -1 \cdot 7 = -7

Step 2: Expand Determinant Using Second Row

\Delta = a_{21}C_{21} + a_{22}C_{22} + a_{23}C_{23}
\Delta = 2 \cdot 7 + 0 \cdot 7 + 1 \cdot (-7)
\Delta = 14 + 0 - 7 = 7

Answer: \boxed{Determinant, \\ \Delta = 7}

Question 4: Using Cofactors of elements of third Column, evaluate Δ (determinant)

\Delta = \left| \begin{array}{ccc} 1 & x & yz \\ 1 & y & zx \\ 1 & z & xy \end{array} \right|

Solution:
Step 1: Compute Cofactors of Elements in Third Column

Element a_{13} = yz
Minor M_{13} = \left| \begin{array}{cc} 1 & x \\ 1 & y \end{array} \right| = 1 \cdot y - x \cdot 1 = y - x
C_{13} = (-1)^{1+3} \cdot (y - x) = +1 \cdot (y - x) = y - x

Element a_{23} = zx
Minor M_{23} = \left| \begin{array}{cc} 1 & x \\ 1 & z \end{array} \right| = 1 \cdot z - x \cdot 1 = z - x
C_{23} = (-1)^{2+3} \cdot (z - x) = -1 \cdot (z - x) = x - z

Element a_{33} = xy
Minor M_{33} = \left| \begin{array}{cc} 1 & x \\ 1 & y \end{array} \right| = 1 \cdot y - x \cdot 1 = y - x
C_{33} = (-1)^{3+3} \cdot (y - x) = +1 \cdot (y - x) = y - x

Step 2: Expand Determinant Using Third Column
\Delta = a_{13}C_{13} + a_{23}C_{23} + a_{33}C_{33}
\Delta = yz(y - x) + zx(x - z) + xy(y - x)

Answer: \boxed{Determinant, \\ \Delta = yz(y - x) + zx(x - z) + xy(y - x)}

Question 5: If

\Delta = \left| \begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array} \right| and A_{ij} is the cofactor of a_{ij}, then the value of \Delta is given by:

(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}

Solution:

The value of a determinant can be computed by expanding along any row or column.
If we expand along the first column, we get:
\Delta = a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}
This matches option (D).

Answer: \boxed{\text{(D)} \quad a_{11}A_{11} + a_{21}A_{21} + a_{31}A_{31}}