Exercise 4.4 Solutions

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Exercise 4.4 Solutions
Write Minors and Cofactors of the elements of following determinants:
1. (i) \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix} (ii) \begin{vmatrix} a & c \\[5pt] b & d \end{vmatrix}
(i) To write Minors and Cofactors of the elements of the determinants \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix}
Minor of an element {a_{ij}} of a determinant is the determinant obtained by deleting its {i^{th}} row and {j^{th}} column in which element {a_{ij}} lies and is denoted by {\text{M}_{ij}}
Cofactor of an element {a_{ij},} denoted by {\text{A}_{ij}} is defined by {\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij},} where {\text{M}_{ij}} is minor of {a_{ij}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = 2}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{3}\end{vmatrix}}
3
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}(3)} = 3
{a_{12} = -4}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
0
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}(0)} = 0
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{2}} & \textcolor{green}{-4} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
-4
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}(-4)} = 4
{a_{22} = 3}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{2} & \textcolor{red}{\cancel{-4}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{3}}\end{vmatrix}}
2
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}(2)} = 2
∴ The Minors and Cofactors of the elements of the determinant \begin{vmatrix} 2 & -4 \\[5pt] 0 & 3 \end{vmatrix} are {\text{M}_{11} = 3,} {\text{M}_{12} = 0,} {\text{M}_{21} = -4,} {\text{M}_{22} = 2,} {\text{A}_{11} = 3,} {\text{A}_{12} = 0,} {\text{A}_{21} = 4,} {\text{A}_{22} = 2,}
πŸ’‘Tip: The minor of any element in a 2 Γ— 2 matrix is the element diagonally opposite to the given element.
(ii) To write Minors and Cofactors of the elements of the determinants \begin{vmatrix} a & b \\[5pt] c & d \end{vmatrix}
Minor of an element {a_{ij}} of a determinant is the determinant obtained by deleting its {i^{th}} row and {j^{th}} column in which element {a_{ij}} lies and is denoted by {\text{M}_{ij}}
Cofactor of an element {a_{ij},} denoted by {\text{A}_{ij}} is defined by {\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij},} where {\text{M}_{ij}} is minor of {a_{ij}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
Minor
Value
Cofactor
{\text{A}_{ij} = (-1)^{i + j}\text{M}_{ij}}
{a_{11} = a}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{green}{d}\end{vmatrix}}
d
{\text{A}_{11} = (-1)^{1 + 1}\text{M}_{11} = (-1)^{1 + 1}a} = d
{a_{12} = c}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{green}{b} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
b
{\text{A}_{12} = (-1)^{1 + 2}\text{M}_{12} = (-1)^{1 + 2}b} = -b
{a_{21} = b}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{a}} & \textcolor{green}{c} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
c
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)^{2 + 1}c} = -c
{a_{22} = d}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{a} & \textcolor{red}{\cancel{c}} \\[5pt] \textcolor{red}{\cancel{b}} & \textcolor{red}{\cancel{d}}\end{vmatrix}}
a
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = (-1)^{2 + 2}d} = a
∴ The Minors and Cofactors of the elements of the determinant \begin{vmatrix} a & b \\[5pt] c & d \end{vmatrix} are {\text{M}_{11} = d,} {\text{M}_{12} = b,} {\text{M}_{21} = c,} {\text{M}_{22} = a,} {\text{A}_{11} = d,} {\text{A}_{12} = -b,} {\text{A}_{21} = -c,} {\text{A}_{22} = a,}
πŸ’‘Tip: The minor of any element in a 2 Γ— 2 matrix is the element diagonally opposite to the given element.
Write Minors and Cofactors of the elements of following determinants:
2. (i) {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}} (ii) {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}}
(i) To find the Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 0(1) – 0(0)
= 0 – 0
= 0
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 1 \\[5pt] 0 & 0 \end{vmatrix}}
= 0(0) – 1(0)
= 0 – 0
= 0
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{0} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 0(1) – 0(0)
= 0 – 0
= 0
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{0}} & \textcolor{green}{1} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{0} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 0 \end{vmatrix}}
= 1(0) – 0(0)
= 0 – 0
= 0
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 0 \\[5pt] 1 & 0 \end{vmatrix}}
= 0(0) – 0(1)
= 0 – 0
= 0
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{0} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{0} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 0 \end{vmatrix}}
= 1(0) – 0(0)}
= 0 – 0
= 0
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{0}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 1}
{\text{A}_{11} = (-1)^{1 + 1} Γ— \text{M}_{11} = 1 Γ— 1 = \textcolor{green}{\textbf{1}}}
{a_{12} = 0}
{\text{M}_{12} = 0}
{\text{A}_{12} = (-1)^{1 + 2} Γ— \text{M}_{12} = -1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{13} = 0}
{\text{M}_{13} = 0}
{\text{A}_{13} = (-1)^{1 + 3} Γ— \text{M}_{13} = 1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{21} = 0}
{\text{M}_{21} = 0}
{\text{A}_{21} = (-1)^{2 + 1} Γ— \text{M}_{21} = -1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{22} = 1}
{\text{M}_{22} = 1}
{\text{A}_{22} = (-1)^{2 + 2} Γ— \text{M}_{22} = 1 Γ— 1 = \textcolor{green}{\textbf{1}}}
{a_{23} = 0}
{\text{M}_{23} = 0}
{\text{A}_{23} = (-1)^{2 + 3} Γ— \text{M}_{23} = -1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{31} = 0}
{\text{M}_{31} = 0}
{\text{A}_{31} = (-1)^{3 + 1} Γ— \text{M}_{31} = 1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{32} = 0}
{\text{M}_{32} = 0}
{\text{A}_{32} = (-1)^{3 + 2} Γ— \text{M}_{32} = -1 Γ— 0 = \textcolor{green}{\textbf{0}}}
{a_{33} = 1}
{\text{M}_{33} = 1}
{\text{A}_{33} = (-1)^{3 + 3} Γ— \text{M}_{33} = 1 Γ— 1 = \textcolor{green}{\textbf{1}}}
∴ The Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 0 \\[5pt] 0 & 1 & 0 \\[5pt] 0 & 0 & 1 \end{vmatrix}} are {\text{M}_{11} = 1,} {\text{M}_{12} = 0,} {\text{M}_{13} = 0,} {\text{M}_{21} = 0,} {\text{M}_{22} = 1,} {\text{M}_{23} = 0,} {\text{M}_{31} = 0,} {\text{M}_{32} = 0,} {\text{M}_{33} = 1,} {\text{A}_{11} = 1,} {\text{A}_{12} = 0,} {\text{A}_{13} = 0,} {\text{A}_{21} = 0,} {\text{A}_{22} = 1,} {\text{A}_{23} = 0,} {\text{A}_{31} = 0,} {\text{A}_{32} = 0,} {\text{A}_{33} = 1,}
πŸ’‘Tip: The minor of the elements of the determinant of an identity matrix is same as the element i.e {\text{M}_{ij} = a_{ij}}.
πŸ’‘Tip: The cofactor of the elements of the determinant of an identity matrix is same as the element i.e {\text{A}_{ij} = \text{M}_{ij} = a_{ij}}.
(ii) To find the Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}}
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{11} = 1}
{\text{M}_{11}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{5} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 5 & -1 \\[5pt] 1 & 2 \end{vmatrix}}
= 5(2) – (-1)1
= 10 + 1
= 11
{a_{12} = 0}
{\text{M}_{12}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{5}} & \textcolor{green}{-1} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 3 & -1 \\[5pt] 0 & 2 \end{vmatrix}}
= 3(2) – (-1)0
= 6 – 0
= 6
{a_{13} = 0}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{5} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 3 & 5 \\[5pt] 0 & 1 \end{vmatrix}}
= 3(1) – 5(0)
= 3 – 0
= 3
{a_{21} = 0}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{green}{1} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 0 & 4 \\[5pt] 1 & 2 \end{vmatrix}}
= 0(2) – 4(1)
= 0 – 4
= -4
{a_{22} = 1}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{red}{\cancel{1}} & \textcolor{green}{2} \end{vmatrix}}
{= \begin{vmatrix} 1 & 4 \\[5pt] 0 & 2 \end{vmatrix}}
= 1(2) – 4(0)
= 2 – 0
= 2
{a_{23} = 0}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{red}{\cancel{5}} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{green}{0} & \textcolor{green}{1} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 0 & 1 \end{vmatrix}}
= 1(1) – 0(0)
= 1 – 0
= 1
{a_{31} = 0}
{\text{M}_{31}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{green}{0} & \textcolor{green}{4} \\[5pt] \textcolor{red}{\cancel{3}} & \textcolor{green}{5} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 0 & 4 \\[5pt] 5 & -1 \end{vmatrix}}
= 0(-1) – 4(5)
= 0 – 20
= -20
{a_{32} = 0}
{\text{M}_{32}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{red}{\cancel{0}} & \textcolor{green}{4} \\[5pt] \textcolor{green}{3} & \textcolor{red}{\cancel{5}} & \textcolor{green}{-1} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{4}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 4 \\[5pt] 3 & -1 \end{vmatrix}}
= 1(-1) – 4(3)
= -1 – 12
= -13
{a_{33} = 1}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{0} & \textcolor{red}{\cancel{4}} \\[5pt] \textcolor{green}{3} & \textcolor{green}{5} & \textcolor{red}{\cancel{-1}} \\[5pt] \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{2}} \end{vmatrix}}
{= \begin{vmatrix} 1 & 0 \\[5pt] 3 & 5 \end{vmatrix}}
= 1(5) – 0(3)
= 5 – 0
= 5
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{11} = 1}
{\text{M}_{11} = 11}
{\text{A}_{11} = (-1)^{1 + 1} Γ— \text{M}_{11} = 1 Γ— 11 = \textcolor{green}{\textbf{11}}}
{a_{12} = 0}
{\text{M}_{12} = 6}
{\text{A}_{12} = (-1)^{1 + 2} Γ— \text{M}_{12} = -1 Γ— 6 = \textcolor{green}{\textbf{-6}}}
{a_{13} = 4}
{\text{M}_{13} = 3}
{\text{A}_{13} = (-1)^{1 + 3} Γ— \text{M}_{13} = 1 Γ— 3 = \textcolor{green}{\textbf{3}}}
{a_{21} = 3}
{\text{M}_{21} = -4}
{\text{A}_{21} = (-1)^{2 + 1} Γ— \text{M}_{21} = -1 Γ— -4 = \textcolor{green}{\textbf{4}}}
{a_{22} = 5}
{\text{M}_{22} = 2}
{\text{A}_{22} = (-1)^{2 + 2} Γ— \text{M}_{22} = 1 Γ— 2 = \textcolor{green}{\textbf{2}}}
{a_{23} = -1}
{\text{M}_{23} = 1}
{\text{A}_{23} = (-1)^{2 + 3} Γ— \text{M}_{23} = -1 Γ— 1 = \textcolor{green}{\textbf{-1}}}
{a_{31} = 0}
{\text{M}_{31} = -20}
{\text{A}_{31} = (-1)^{3 + 1} Γ— \text{M}_{31} = 1 Γ— -20 = \textcolor{green}{\textbf{-20}}}
{a_{32} = 1}
{\text{M}_{32} = -13}
{\text{A}_{32} = (-1)^{3 + 2} Γ— \text{M}_{32} = -1 Γ— -13 = \textcolor{green}{\textbf{13}}}
{a_{33} = 2}
{\text{M}_{33} = 5}
{\text{A}_{33} = (-1)^{3 + 3} Γ— \text{M}_{33} = 1 Γ— 5 = \textcolor{green}{\textbf{5}}}
∴ The Minors and Cofactors of the elements of the determinant {\begin{vmatrix} 1 & 0 & 4 \\[5pt] 3 & 5 & -1 \\[5pt] 0 & 1 & 2 \end{vmatrix}} are {\text{M}_{11} = 11,} {\text{M}_{12} = 6,} {\text{M}_{13} = 3,} {\text{M}_{21} = -4,} {\text{M}_{22} = 2,} {\text{M}_{23} = 1,} {\text{M}_{31} = -20,} {\text{M}_{32} = -13,} {\text{M}_{33} = 5,} {\text{A}_{11} = 11,} {\text{A}_{12} = -6,} {\text{A}_{13} = 3,} {\text{A}_{21} = 4,} {\text{A}_{22} = 2,} {\text{A}_{23} = -1,} {\text{A}_{31} = -20,} {\text{A}_{32} = 13,} {\text{A}_{33} = 5,}
3. Using Cofactors of elements of third column, evaluate {Ξ” = \begin{vmatrix} 5 & 3 & 8 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 2 & 3 \end{vmatrix}}
To find the Cofactors, we first find the Minors of the elements of the second row and then deduce the Cofactors from the Minors as given below.
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{21} = 2}
{\text{M}_{21}}
{\begin{vmatrix} \textcolor{red}{\cancel{5}} & \textcolor{green}{3} & \textcolor{green}{8} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{green}{2} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 3 & 8 \\[5pt] 2 & 3 \end{vmatrix}}
= 3(3) – 8(2) = 9 – 16 = -7
= 9 – 16
= -7
{a_{22} = 0}
{\text{M}_{22}}
{\begin{vmatrix} \textcolor{green}{5} & \textcolor{red}{\cancel{3}} & \textcolor{green}{8} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{red}{\cancel{2}} & \textcolor{green}{3} \end{vmatrix}}
{= \begin{vmatrix} 5 & 8 \\[5pt] 1 & 3 \end{vmatrix}}
= 5(3) – 8(1)
= 15 – 8
= 7
{a_{23} = 1}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{5} & \textcolor{green}{3} & \textcolor{red}{\cancel{8}} \\[5pt] \textcolor{red}{\cancel{2}} & \textcolor{red}{\cancel{0}} & \textcolor{red}{\cancel{1}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{2} & \textcolor{red}{\cancel{3}} \end{vmatrix}}
{= \begin{vmatrix} 5 & 3 \\[5pt] 1 & 2 \end{vmatrix}}
= 5(2) – 3(1)
= 10 – 3
= 7
Now, the Cofactors of the elements of the second row are found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{21} = 2}
{\text{M}_{21} = -7}
{\text{A}_{21} = (-1)^{2 + 1}\text{M}_{21} = (-1)(-7) = 7}
{a_{22} = 0}
{\text{M}_{22} = 7}
{\text{A}_{22} = (-1)^{2 + 2}\text{M}_{22} = 1(7) = 7}
{a_{23} = 1}
{\text{M}_{23} = 7}
{\text{A}_{23} = (-1)^{2 + 3}\text{M}_{23} = (-1)7 = -7}
Now, the determinants of the matrix can be calculated from the elements of the second row, using the formula,
Ξ”
{= a_{21}\text{A}_{21} + a_{22}\text{A}_{22} + a_{23}\text{A}_{22}}
= 2(7) + 0(7) + 1(-7)
= 14 + 0 – 7
= 7
∴ The determinant {Ξ” = \begin{vmatrix} 5 & 3 & 8 \\[5pt] 2 & 0 & 1 \\[5pt] 1 & 2 & 3 \end{vmatrix}} evaluated using Cofactors of elements of second row is 7
4. Using cofactors of elements of third column evaluate {Ξ” = \begin{vmatrix} 1 & x & yz \\[5pt] 1 & y & yz \\[5pt] 1 & z & xy \end{vmatrix}}
To find the Cofactors, we first find the Minors of the elements of the third column and then deduce the Cofactors from the Minors as given below.
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Minor
Calculation
{a_{13} = yz}
{\text{M}_{13}}
{\begin{vmatrix} \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{x}} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{y} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{z} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & y \\[5pt] 1 & z \end{vmatrix}}
{= 1(z) - (y)1}
{= z - y}
{a_{23} = zx}
{\text{M}_{23}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{x} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{y}} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{z} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & x \\[5pt] 1 & z \end{vmatrix}}
{= 1(z) - x(1)}
{= z - x}
{a_{33} = xy}
{\text{M}_{33}}
{\begin{vmatrix} \textcolor{green}{1} & \textcolor{green}{x} & \textcolor{red}{\cancel{yz}} \\[5pt] \textcolor{green}{1} & \textcolor{green}{y} & \textcolor{red}{\cancel{zx}} \\[5pt] \textcolor{red}{\cancel{1}} & \textcolor{red}{\cancel{z}} & \textcolor{red}{\cancel{xy}} \end{vmatrix}}
{= \begin{vmatrix} 1 & x \\[5pt] 1 & y \end{vmatrix}}
{= 1(y) - x(1)}
{= y - x}
Now, the Cofactors of the elements of the third column are found as below:
Element
{a_{ij}}
Minor
{\text{M}_{ij}}
Cofactor
{\text{A}_{ij}}
{a_{13} = 2}
{\text{M}_{13} = z - y}
{\text{A}_{13} = (-1)^{1 + 3} Γ— \text{M}_{21} = 1(z - y) = z - y}
{a_{23} = 0}
{\text{M}_{23} = z - x}
{\text{A}_{23} = (-1)^{2 + 3} Γ— \text{M}_{22} = (-1)(z - x) = x - z}
{a_{33} = 1}
{\text{M}_{33} = y - z}
{\text{A}_{33} = (-1)^{3 + 3} Γ— \text{M}_{23} = 1(y - z) = y - z}
Now, the determinants of the matrix can be calculated from the elements of the third column, using the formula,
Ξ”
{= a_{13}\text{A}_{13} + a_{23}\text{A}_{23} + a_{33}\text{A}_{33}}
{= yz(z - y) + zx(x - z) + xy(y - x)}
{= z\left[y(z - y) + x(x - z)\right] + xy(y - x)}
{= z(yz - y^2 + x^2 - xz) + xy(y - x)}
{= z(x^2 - y^2 -xz + yz) + xy(y - x)}
{= z\left[(x + y)(x - y) - z(x - y)\right] + xy(y - x)}
{= (x - y)\left[z(x + y - z) - xy\right]}
{= (x - y)(xz + yz - z^2 - xy)}
{= (x - y)(yz - z^2 - xy + xz)}
{= (x - y)\left[z(y - z) - x(y - z)\right]}
= {(x - y)(y - z)(x - y)}
∴ The determinant {Ξ” = \begin{vmatrix} 1 & x & yz \\[5pt] 1 & y & zx \\[5pt] 1 & z & xy \end{vmatrix}} evaluated using Cofactors of elements of third column is {(x - y)(y - z)(x - y)}
5. If {Ξ” = \begin{vmatrix} a_{11} & a_{12} & a_{13} \\[5pt] a_{21} & a_{22} & a_{23} \\[5pt] a_{31} & a_{32} & a_{33} \end{vmatrix}} and {\text{A}_{ij}} is Cofactors of {a_{ij},} then value of Ξ” is given by
(A) {a_{11}\text{A}_{31} + a_{12}\text{A}_{32} + a_{13}\text{A}_{33}}
(B) {a_{11}\text{A}_{11} + a_{12}\text{A}_{21} + a_{13}\text{A}_{31}}
(C) {a_{21}\text{A}_{11} + a_{22}\text{A}_{12} + a_{23}\text{A}_{13}}
(D) {a_{11}\text{A}_{11} + a_{21}\text{A}_{21} + a_{31}\text{A}_{31}} βœ”
We know that the determinant Ξ” is equal to the sum of the product of elements of any row (or column) with their corresponding Cofactors.
We also know that if elements of a row (or column) are multiplied with cofactors of any other row (or column), then their sum is zero.
option A is the sum of the product of elements of first row and the Cofactors of the elements of the third row. So, it will be zero. so, (option A is not correct ❌)
option B is the sum of the product of elements of first row and the Cofactors of the elements of the third column. So, (option B is not correct ❌)
option C is the sum of the product of elements of second row and the Cofactors of the elements of the first row. So, (option C is not correct ❌)
option D is the sum of the product of elements of first column and the Cofactors of the elements of the first column. So, (option D) is the correct answer.