Ex 4.4 Class 12 Maths Ncert Solutions

Minor: Minor of an element aij of a determinant is the determinant obtained by removing \( \text{ i}^{\text {th }} \) row and \( \text{j}^{\text {th }} \) column in which element \( a_{i j} \) lies. It is denoted by \( \mathrm{M}_{\mathrm{ij}} \).
Cofactor: Cofactor of an element aij, \( \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \mathrm{M}_{\mathrm{ij}} \).
Minor of element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{M}_{\mathrm{ij}} \)
\( \mathrm{a}_{11}=2 \), Minor of element \( \mathrm{a}_{11}=\mathrm{M}_{11}=3 \)
Here removing 1 st row and 1st column from the determinant we are left out with 3 so \( \mathrm{M}_{11}=3 \).
Similarly, finding other Minors of the determinant
\( a_{12}=-4 \), Minor of element \( a_{12}=M_{12}=0 \)
\( \mathrm{a}_{21}=0 \), Minor of element \( \mathrm{a}_{21}=\mathrm{M}_{21}=-4 \)
\( \mathrm{a}_{22}=3 \), Minor of element \( \mathrm{a}_{22}=\mathrm{M}_{22}=2 \)
Cofactor of an element aij, \( \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i+j}} \times \mathrm{M}_{\mathrm{ij}} \)
\(\begin{array}{l}
A_{11}=(-1)^{1+1} \times M_{11}=1 \times 3=3 \\
A_{12}=(-1)^{1+2} \times M_{12}=(-1) \times 0=0 \\
A_{21}=(-1)^{2+1} \times M_{11}=(-1) \times(-4)=4 \\
A_{22}=(-1)^{2+2} \times M_{22}=1 \times 2=2
\end{array}\)

\(\left|\begin{array}{ll}
a & c \\
b & d
\end{array}\right|\)
Minor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{M}_{\mathrm{ij}} \)
\( \mathrm{a}_{11}=\mathrm{a} \), Minor of element \( \mathrm{a}_{11}=\mathrm{M}_{11}=\mathrm{d} \)
Here removing 1st row and 1st column from the determinant we are left out with d so \( \mathrm{M}_{11}=\mathrm{d} \).
Similarly, finding other Minors of the determinant
\( \mathrm{a}_{12}=\mathrm{c} \), Minor of element \( \mathrm{a}_{12}=\mathrm{M}_{12}=\mathrm{b} \)
\( \mathrm{a}_{21}=\mathrm{b} \), Minor of element \( \mathrm{a}_{21}=\mathrm{M}_{21}=\mathrm{c} \)
\( \mathrm{a}_{22}=\mathrm{d} \), Minor of element \( \mathrm{a}_{22}=\mathrm{M}_{22}=\mathrm{a} \)
Cofactor of an element \( \mathrm{a}_{\mathrm{ij}}, \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \times \mathrm{M}_{\mathrm{ij}} \)
\(\begin{array}{l}
A_{11}=(-1)^{1+1} \times M_{11}=1 \times d=d \\
A_{12}=(-1)^{1+2} \times M_{12}=(-1) \times b=-b \\
A_{21}=(-1)^{2+1} \times M_{11}=(-1) \times c=-c \\
A_{22}=(-1)^{2+2} \times M_{22}=1 \times a=a
\end{array}\)

Minor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{M}_{\mathrm{ij}} \)
\( a_{11}=1 \), Minor of element
\( a_{11}=M_{11}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=(1 \times 1)-(0 \times 0)=1 \)
Here removing \( 1^{\text {st }} \) row and \( 1^{\text {st }} \) column from the determinant we are left out with the determinant \( \left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right| \). Solving this we get \( M_{11}=1 \)
Similarly, finding other Minors of the determinant
\( \mathrm{a}_{12}=0 \), Minor of element
\( \mathrm{a}_{12}=\mathrm{M}_{12}=\left|\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right|=(0 \times 1)-(0 \times 0)=0 \)
\( \mathrm{a}_{13}=0 \), Minor of element
\( \mathrm{a}_{13}=\mathrm{M}_{13}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=(0 \times 0)-(1 \times 0)=0 \)
\( \mathrm{a}_{21}=0 \), Minor of element
\( \mathrm{a}_{21}=\mathrm{M}_{21}=\left|\begin{array}{ll}0 & 1 \\ 0 & 0\end{array}\right|=(0 \times 1)-(0 \times 0)=0 \)
\( \mathrm{a}_{22}=1 \), Minor of element
\( \mathrm{a}_{22}=\mathrm{M}_{22}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=(1 \times 1)-(0 \times 0)=1 \)
\( \mathrm{a}_{23}=0 \), Minor of element
\( \mathrm{a}_{23}=\mathrm{M}_{23}=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=(1 \times 0)-(0 \times 0)=0 \)
\( \mathrm{a}_{31}=0 \), Minor of element
\( \mathrm{a}_{31}=\mathrm{M}_{31}=\left|\begin{array}{ll}0 & 0 \\ 1 & 0\end{array}\right|=(0 \times 0)-(0 \times 1)=0 \)
\( a_{32}=0 \), Minor of element
\( a_{32}=M_{32}=\left|\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right|=(1 \times 0)-(0 \times 0)=0 \)
\( \mathrm{a}_{33}=1 \), Minor of element
\( \mathrm{a}_{33}=\mathrm{M}_{33}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=(1 \times 1)-(0 \times 0)=1 \)
Cofactor of an element \( \mathrm{a}_{\mathrm{ij}}, \mathrm{A}_{\mathrm{ij}}=(-1)^{\mathrm{i}+\mathrm{j}} \times \mathrm{M}_{\mathrm{ij}} \)
\(A_{11}=(-1)^{1+1} \times M_{11}=1 \times 1=1\)
\(
A_{12}=(-1)^{1+2} \times M_{12}=(-1) \times 0=0\)
\(
A_{13}=(-1)^{1+3} \times M_{13}=1 \times 0=0\)
\(
A_{21}=(-1)^{2+1} \times M_{21}=(-1) \times 0=0\)
\(
A_{22}=(-1)^{2+2} \times M_{22}=1 \times 1=1\)
\(
A_{23}=(-1)^{2+3} \times M_{23}=(-1) \times 0=0\)
\(
A_{31}=(-1)^{3+1} \times M_{31}=1 \times 0=0\)
\(
A_{32}=(-1)^{3+2} \times M_{32}=(-1) \times 0=0\)
\(
A_{33}=(-1)^{3+3} \times M_{33}=1 \times 1=1
\)

\(\left|\begin{array}{ccc}
1 & 0 & 4 \\
3 & 5 & -1 \\
0 & 1 & 2
\end{array}\right|\)
Minor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{M}_{\mathrm{ij}} \)
\( \mathrm{a}_{11}=1 \), Minor of element
\( \mathrm{a}_{11}=\mathrm{M}_{11}=\left|\begin{array}{cc}5 & -1 \\ 1 & 2\end{array}\right|=(5 \times 2)-((-1) \times 1) \) \( =10+1=11 \)
Here removing \( 1^{\text {st }} \) row and 1 st column from the determinant we are left out with the determinant \( \left|\begin{array}{cc}5 & -1 \\ 1 & 2\end{array}\right| \). Solving this we get \( \mathrm{M}_{11}=11 \)
Similarly, finding other Minors of the determinant
\( \mathrm{a}_{12}=0 \), Minor of element
\( \mathrm{a}_{12}=\mathrm{M}_{12}=\left|\begin{array}{cc}3 & -1 \\ 0 & 2\end{array}\right|=(3 \times 2)-((-1) \times 0) \) \( =(6-0)=6 \)
\( \mathrm{a}_{13}=4 \), Minor of element
\( \mathrm{a}_{13}=\mathrm{M}_{13}=\left|\begin{array}{ll}3 & 5 \\ 0 & 1\end{array}\right|=(3 \times 1)-(5 \times 0)=3 \) \( -0=3 \)
\( \mathrm{a}_{21}=3 \), Minor of element
\( \mathrm{a}_{21}=\mathrm{M}_{21}=\left|\begin{array}{ll}0 & 4 \\ 1 & 2\end{array}\right|=(0 \times 2)-(4 \times 1)=0 \) \( -4=-4 \)
\( \mathrm{a}_{22}=5 \), Minor of element
\( \mathrm{a}_{22}=\mathrm{M}_{22}=\left|\begin{array}{ll}1 & 4 \\ 0 & 2\end{array}\right|=(1 \times 2)-(4 \times 0)=2 \) \( -0=2 \)
\( a_{23}=-1 \), Minor of element
\( a_{23}=M_{23}=\left|\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right|=(1 \times 1)-(0 \times 0)=1 \)
\( \mathrm{a}_{31}=0 \), Minor of element
\( \mathrm{a}_{31}=\mathrm{M}_{31}=\left|\begin{array}{cc}0 & 4 \\ 5 & -1\end{array}\right|=(0 \times(-1))-(4 \times 5) \) \( =0-20=-20 \)
\( \mathrm{a}_{32}=1 \), Minor of element
\( \mathrm{a}_{32}=\mathrm{M}_{32}=\left|\begin{array}{cc}1 & 4 \\ 3 & -1\end{array}\right|=(1 \times(-1))-(4 \times 3) \) \( =-1-12=-13 \)
\( \mathrm{a}_{33}=2 \), Minor of element
\( \mathrm{a}_{33}=\mathrm{M}_{33}=\left|\begin{array}{ll}1 & 0 \\ 3 & 5\end{array}\right|=(1 \times 5)-(0 \times 3)= \) \( (5-0)=5 \)
Cofactor of an element aij, \( \mathrm{Aij}=(-1)^{\mathrm{i}+\mathrm{j}} \times \mathrm{Mij} \)
\(\begin{array}{l}
A_{11}=(-1)^{1+1} \times M_{11}=1 \times 11=11 \\
A_{12}=(-1)^{1+2} \times M_{12}=(-1) \times 6=-6 \\
A_{13}=(-1)^{1+3} \times M_{13}=1 \times 3=3 \\
A_{21}=(-1)^{2+1} \times M_{21}=(-1) \times(-4)=4 \\
A_{22}=(-1)^{2+2} \times M_{22}=1 \times 2=2
\end{array}\)
\(
A_{23}=(-1)^{2+3} \times M_{23}=(-1) \times 1=-1\)
\(
A_{31}=(-1)^{3+1} \times M_{31}=1 \times(-20)=-20\)
\(
A_{32}=(-1)^{3+2} \times M_{32}=(-1) \times(-13)=13\)
\(
A_{33}=(-1)^{3+3} \times M_{33}=1 \times 5=5
\)

To evaluate a determinant using cofactors, Let
\(\mathrm{B}=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)
Expanding along Row 1
\( \mathrm{B}=(-1)^{1+1} \times \mathrm{a}_{11}\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|+(-1)^{1+2} \times \mathrm{a}_{12} \times\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{33}\end{array}\right| \) \( +(-1)^{1+3}\times a_{13} \times\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right| \)
\(B=a_{11} A_{11}+a_{12} A_{12}+a_{13} A_{13}\)
[Where Aij represents cofactors of aij of determinant B.]
\( B= \) Sum of product of elements of \( R_{1} \) with their corresponding cofactors
Similarly, the determinant can be solved by expanding along column
So, \( B= \) sum of product of elements of any row or column with their corresponding cofactors
\(\triangle=\left|\begin{array}{lll}
5 & 3 & 8 \\
2 & 0 & 1 \\
1 & 2 & 3
\end{array}\right|\)
Cofactors of second row
\( \mathrm{A}_{21}=(-1)^{2+1} \times \mathrm{M}_{21}=(-1) \times\left|\begin{array}{ll}3 & 8 \\ 2 & 3\end{array}\right|=(-1) \times(3 \times 3-8 \times 2) \) \( =(-1) \times(-7)=7 \)
\( A_{22}=(-1)^{2+2} \times M_{22}=1 \times\left|\begin{array}{ll}5 & 8 \\ 1 & 3\end{array}\right|=(5 \times 3-8 \times 1)=7 \) \( A_{23}=(-1)^{2+3} \times M_{23}=(-1) \times\left|\begin{array}{ll}5 & 3 \\ 1 & 2\end{array}\right|=(-1) \times(5 \times 2-3 \times 1) \) \( =(-1) \times 7=-7 \)
[Where Aij \( =(-1)^{i+j} \times M_{i j}, M_{i j}= \) Minor of ith row \(\&\) jth column]
Therefore,
\(\begin{array}{l}
\triangle=\mathrm{a}_{21} \mathrm{~A}_{21}+\mathrm{a}_{22} \mathrm{~A}_{22}+\mathrm{a}_{23} \mathrm{~A}_{23} \\
\triangle=2 \times 7+1 \times(-7)=14-7=7
\end{array}\)
Ans: \( \triangle=7 \)

To evaluate a determinant using cofactors, Let
\(\mathrm{B}=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)
Expanding along Row 1
\( \mathrm{B}=(-1)^{1+1} \times \mathrm{a}_{11} \times\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|+(-1)^{1+2} \times \mathrm{a}_{12} \times\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{33}\end{array}\right| \)
\(+(-1)^{1+3} \times a_{13} \times\left|\begin{array}{ll}a_{21} & a_{22} \\ a_{31} & a_{32}\end{array}\right| \)
\( B=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{12} \mathrm{~A}_{12}+\mathrm{a}_{13} \mathrm{~A}_{13} \)
[Where Aij represents cofactors of aij of determinant B.]
\( B= \) Sum of product of elements of \( R_{1} \) with their corresponding cofactors
Similarly, the determinant can be solved by expanding along column
So, \( B= \) sum of product of elements of any row or column with their corresponding cofactors
\( \triangle=\left|\begin{array}{lll}1 & x & y z \\ 1 & y & z x \\ 1 & z & x y\end{array}\right| \)
Cofactors of third column
\( \mathrm{A}_{13}=(-1)^{1+3} \times \mathrm{M}_{13}=1 \times\left|\begin{array}{ll}1 & y \\ 1 & z\end{array}\right|=1 \times(1 \times z-1 \times y)=(z-y) \)
\( A_{23}=(-1)^{2+3} \times M_{23}=(-1) \times\left|\begin{array}{ll}1 & x \\ 1 & z\end{array}\right|=(-1) \times(1 \times z-1 \times x) \)
\( =-(z-x)=(x-z) \)
\( \mathrm{A}_{33}=(-1)^{3+3} \times \mathrm{M}_{33}=1 \times\left|\begin{array}{ll}1 & x \\ 1 & y\end{array}\right|=1 \times(1 \times y-1 \times x)=(y-x) \)
[Where Aij \( =(-1)^{\mathrm{i}+\mathrm{j}} \times \mathrm{M}_{\mathrm{ij}}, \mathrm{M}_{\mathrm{ij}}= \) Minor of ith row \(\&\) jth column]
Therefore,
\(\triangle=\mathrm{a}_{13} \mathrm{~A}_{13}+\mathrm{a}_{23} \mathrm{~A}_{23}+\mathrm{a}_{33} \mathrm{~A}_{33}\)
\( \triangle=y z(z-y)+z x(x-z)+x y(y-x)=z[y(z-y)+x(x-z)]\) \(+x y(y - x)\)
\( \triangle=z\left(yz-y^{2}+x^{2}-xz\right)+xy(y-x)=z\left[(yz-xz)+\left(x^{2}-y^{2}\right)\right] \) \(+xy(y-x)\)
\( \triangle=z[z \times(y-x)+(x+y) \times(x-y)]+xy(y-x) \)
\( \triangle=z \times(y-x) \times(z-x-y)+xy(y-x) \)
\( \triangle=(y-x) \times\left(z^{2}-x z-y z+x y\right) \)
\( \triangle=(y-x) \times[z(z-x)-y(z-x)]=(y-x) \times(z-y) \times(z-x) \)
\( \triangle=(x-y)(y-z)(z-x) \)
Ans: \( \triangle=(x-y)(y-z)(z-x) \)

\(\triangle=\left|\begin{array}{lll}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}\right|\)
Expanding along Column 1
\( \triangle=\mathrm{B}=(-1)^{1+1} \times \mathrm{a}_{11} \times\left|\begin{array}{ll}a_{22} & a_{23} \\ a_{32} & a_{33}\end{array}\right|+(-1)^{2+1} \times \mathrm{a}_{21} \times \) \( \left|\begin{array}{ll}a_{12} & a_{13} \\ a_{32} & a_{33}\end{array}\right|+(-1)^{3+1} \times a_{31} \times\left|\begin{array}{ll}a_{12} & a_{13} \\ a_{22} & a_{23}\end{array}\right| \)
\(\triangle=\mathrm{a}_{11} \mathrm{~A}_{11}+\mathrm{a}_{21} \mathrm{~A}_{21}+\mathrm{a}_{31} \mathrm{~A}_{31} \)