Class 12 Maths Chapter 4 Miscellaneous Exercise Solutions

Let \( \triangle=\left|\begin{array}{ccc}x & \sin \theta & \cos \theta \\ -\sin \theta & -x & 1 \\ \cos \theta & 1 & x\end{array}\right| \)
Expanding the above determinant along R1 i.e. Row 1
\( \triangle=x\left(-x^{2}-1\right)-\sin \theta(-x \times \sin \theta-\cos \theta \times 1)+\cos \theta\)\((-\sin \theta \times 1+x \cos \theta) \)
\( \triangle=-x^{3}-x+x \sin 2 \theta+\sin \theta \times \cos \theta-\sin \theta \times \cos \theta+x \cos 2 \theta \)
\( \triangle=-x^{3}-x+x(\sin 2 \theta+\cos 2 \theta) \)
Since, \( \sin 2 \theta+\cos 2 \theta=1 \)
\( \therefore \triangle=-x^{3}-x+x \)
Hence \( \triangle \) is independent of \( \theta \).

Let \( \triangle=\left|\begin{array}{ccc}\cos \alpha \cos \beta & \cos \alpha \sin \beta & -\sin \alpha \\ -\sin \beta & \cos \beta & 0 \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta & \cos \alpha\end{array}\right| \)
Expanding along \( \mathrm{C}_{3} \) we get
\( \triangle=-\sin \alpha(-\sin \beta \times \sin \alpha \sin \beta-\cos \beta \times \sin \alpha \cos \beta)-0(\sin \alpha \cos \beta \) \(\times \cos \alpha \sin \beta-\cos \alpha \cos \beta \times \sin \alpha \sin \beta)+\cos \alpha(\cos \alpha \cos \beta \times \cos \beta- \) \( \cos \alpha \sin \beta \times(-\sin \beta)) \)
\( \triangle=-\sin \alpha(-\sin \alpha \sin 2 \beta-\sin \alpha \cos 2 \beta)-0+\cos \alpha(\cos \alpha \cos 2 \beta \) \(+\cos \alpha \sin 2 \beta) \)
\( \triangle=\sin \alpha \times \sin \alpha(\sin 2 \beta+\cos 2 \beta)+\cos \alpha \times \cos \alpha(\cos 2 \beta+\sin 2 \beta) \)
[Taking \( -\sin \alpha \) and \( \cos \alpha \) common]
Since, \( \sin 2 \beta+\cos 2 \beta=1 \)
\( \therefore \triangle=\sin 2 \alpha+\cos 2 \alpha \)
\( \triangle=1[\sin 2 \alpha+\cos 2 \alpha=1] \)

Given, \( \triangle=\left|\begin{array}{lll}b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a\end{array}\right| \)
Applying Elementary transformations, we get
\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \) we have
\(\triangle=\left|\begin{array}{ccc}
2(a+b+c) & 2(a+b+c) & 2(a+b+c) \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)
\(\triangle=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & 1 & 1 \\
c+a & a+b & b+c \\
a+b & b+c & c+a
\end{array}\right|\)
Now applying \( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) and \( \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1} \)
\(\triangle=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & 0 & 0 \\
c+a & b-c & b-a \\
a+b & c-a & c-b
\end{array}\right|\)
Expanding along R1
\(\begin{array}{l}
\triangle=2(\mathrm{a}+\mathrm{b}+\mathrm{c})[1\{(\mathrm{~b}-\mathrm{c})(\mathrm{c}-\mathrm{b})-(\mathrm{b}-\mathrm{a})(\mathrm{c}-\mathrm{a})\}-0+0] \\
\triangle=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left[-\mathrm{b}^{2}-\mathrm{c}^{2}+2 \mathrm{bc}-\mathrm{bc}+\mathrm{ba}+\mathrm{ac}-\mathrm{a}^{2}\right] \\
\triangle=2(\mathrm{a}+\mathrm{b}+\mathrm{c})\left[\mathrm{ab}+\mathrm{bc}+\mathrm{ca}-\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}\right]
\end{array}\)
Given that \( \triangle=0 \)
\(\therefore 2(a+b+c)\left[a b+b c+c a-a^{2}-b^{2}-c^{2}\right]=0\)
\( \Rightarrow \) Either \( \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \), or \( \mathrm{ab}+\mathrm{bc}+\mathrm{ca}-\mathrm{a}^{2}-\mathrm{b}^{2}-\mathrm{c}^{2}=0 \)
Now
\(a b+b c+c a-a^{2}-b^{2}-c^{2}=0\)
Multiplying both sides by \(-2\)
\(\begin{array}{l}
\Rightarrow-2 a b-2 b c-2 c a+2 a^{2}+2 b^{2}+2 c^{2}=0 \\
\Rightarrow a^{2}-2 a b+b^{2}+b^{2}-2 b c+c^{2}+c^{2}-2 c a+a^{2}=0 \\
\Rightarrow(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0
\end{array}\)
Since, \( (a-b)^{2},(b-c)^{2},(c-a)^{2} \) are non-negative
\(\begin{array}{l}
\therefore(a-b)^{2}=(b-c)^{2}=(c-a)^{2} \\
\Rightarrow(a-b)=(b-c)=(c-a) \\
\Rightarrow a=b=c
\end{array}\)
Hence, if \( \triangle=0 \), then either \( \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \) or \( \mathrm{a}=\mathrm{b}=\mathrm{c} \)

Given, \( \left|\begin{array}{ccc}x+a & x & x \\ x & x+a & x \\ x & x & x+a\end{array}\right|=0 \)
Applying elementary transformations, we get,
\(\begin{array}{l}
\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\
\left|\begin{array}{ccc}
3 x+a & 3 x+a & 3 x+a \\
x & x+a & x \\
x & x & x+a
\end{array}\right|=0 \\
\Rightarrow(3 x+\mathrm{a})\left|\begin{array}{ccc}
1 & 1 & 1 \\
x & x+a & x \\
x & x & x+a
\end{array}\right|=0
\end{array}\)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) and \( \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1} \), we get
\((3 x+\mathrm{a})\left|\begin{array}{lll}
1 & 0 & 0 \\
x & a & 0 \\
x & 0 & a
\end{array}\right|=0\)
Expanding along R1 we have
\(\begin{array}{l}
(3 x+\mathrm{a})[1(\mathrm{a} \times \mathrm{a}-0 \times 0)-0(x \times \mathrm{a}-0 \times \mathrm{a})+0(x \times 0-\mathrm{a} \times x)]=0 \\
\Rightarrow(3 x+\mathrm{a})\left[1 \times \mathrm{a}^{2}\right]=0 \\
\Rightarrow \mathrm{a}^{2}(3 x+\mathrm{a})=0
\end{array}\)
Since, \( a \neq 0 \)
\(\begin{array}{l}
\therefore 3 x+\mathrm{a}=0 \\
\Rightarrow x=\frac{-\mathrm{a} }{ 3}
\end{array}\)

Given,
\(\begin{array}{l}
\text { LHS }=\left|\begin{array}{ccc}
a^{2} & b c & a c+c^{2} \\
a^{2}+a b & b^{2} & a c \\
a b & b^{2}+b c & c^{2}
\end{array}\right| \\
\text { RHS }=4 a^{2} b^{2} c^{2} \\
\text { LHS }=\triangle=\left|\begin{array}{ccc}
a^{2} & b c & a c+c^{2} \\
a^{2}+a b & b^{2} & a c \\
a b & b^{2}+b c & c^{2}
\end{array}\right|
\end{array}\)
Taking out common factors \( \mathrm{a}, \mathrm{b} \) and c from \( \mathrm{C}_{1}, \mathrm{C}_{2} \) and \( \mathrm{C}_{3} \), we have
\(\triangle=\operatorname{abc}\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
b & b+c & c
\end{array}\right|\)
Applying Elementary Transformations
\( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \) and \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \)
\(\begin{array}{l}
\triangle=\mathrm{abc}\left|\begin{array}{ccc}
a & c & a+c \\
b & b-c & -c \\
b-a & b & -a
\end{array}\right| \\
\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}+\mathrm{R}_{1} \\
\triangle=\mathrm{abc}\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
b-a & b & -a
\end{array}\right| \\
\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}+\mathrm{R}_{2}
\end{array}\)
\(\begin{array}{l}
\triangle=\mathrm{abc}\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
2 b & 2 b & 0
\end{array}\right| \\
\triangle=2 \mathrm{ab}^{2} \mathrm{c}\left|\begin{array}{ccc}
a & c & a+c \\
a+b & b & a \\
1 & 1 & 0
\end{array}\right| \\
\mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \\
\triangle=2 \mathrm{ab}^{2} \mathrm{c}\left|\begin{array}{ccc}
a & c-a & a+c \\
a+b & -a & a \\
1 & 0 & 0
\end{array}\right|
\end{array}\)
Expanding along R3 we get,
\(\begin{array}{l}
\triangle=2 \mathrm{ab}^{2} \mathrm{c}[\mathrm{a}(\mathrm{c}-\mathrm{a})+\mathrm{a}(\mathrm{a}+\mathrm{c})] \\
=2 \mathrm{ab}^{2} \mathrm{c}\left[\mathrm{ac}-\mathrm{a}^{2}+\mathrm{a}^{2}+\mathrm{ac}\right] \\
=2 \mathrm{ab}^{2} \mathrm{c}(2 \mathrm{ac}) \\
=4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2} \\
\triangle=\text { RHS } \\
\therefore \text { LHS = RHS }
\end{array}\)
Hence, Proved

\(A-1=\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
\( B=\left[\begin{array}{ccc}1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1\end{array}\right] \)
We need to find \(B^{-1}\)
To find the inverse of a matrix we need to find the Adjoint of that matrix For finding the adjoint of the matrix we need to find its cofactors
Let \( \mathrm{B}_{\mathrm{ij}} \) denote the cofactors of Matrix B
Minor of an element \( b_{i j}=M_{i j} \)
\( b_{11}=1 \), Minor of element
\( b_{11}=M_{11}=\left|\begin{array}{cc}3 & 0 \\ -2 & 1\end{array}\right|=(3 \times 1)-((-2) \times 0)=3 \)
\( b_{12}=2 \), Minor of element
\( b_{12}=M_{12}=\left|\begin{array}{cc}-1 & 0 \\ 0 & 1\end{array}\right|=((-1) \times 1)-(0 \times 0)=-1 \)
\( b_{13}=-2 \), Minor of element
\( b_{13}=M_{13}=\left|\begin{array}{cc}-1 & 3 \\ 0 & -2\end{array}\right|=((-1) \times(-2))-(3 \times \) \( 0)=2 \)
\( b_{21}=-1 \), Minor of element
\( b_{21}=M_{21}=\left|\begin{array}{cc}2 & -2 \\ -1 & 1\end{array}\right|=(2 \times 1)-((-2) \times(- \) 2)) \( =-2 \)
\( \mathrm{b}_{22}=3 \), Minor of element
\( \mathrm{b}_{22}=\mathrm{M}_{22}=\left|\begin{array}{cc}1 & -2 \\ 0 & 1\end{array}\right|=(1 \times 1)-((-2) \times 0)=1 \)
\( b_{23}=0 \), Minor of element
\( b_{23}=M_{23}=\left|\begin{array}{cc}1 & 2 \\ 0 & -2\end{array}\right|=(1 \times(-2))-(2 \times 0)=-2 \)
\( b_{31}=0 \), Minor of element
\( b_{31}=M_{31}=\left|\begin{array}{cc}2 & -2 \\ 3 & 0\end{array}\right|=(2 \times 0)-((-2) \times 3)=6 \)
\( b_{32}=-2 \), Minor of element
\( b_{32}=M_{32}=\left|\begin{array}{cc}1 & -2 \\ -1 & 0\end{array}\right|=(1 \times 0)-((-2) \times(- \) 1)) \( =-2 \)
\( b_{33}=1 \), Minor of element
\( b_{33}=M_{33}=\left|\begin{array}{cc}1 & 2 \\ -1 & 3\end{array}\right|=(1 \times 3)-(2 \times(-1))=5 \)
Cofactor of an element bij, \( \mathrm{bij},\mathrm{Bij}=(-1)^{\mathrm{i}+\mathrm{j}} \times \mathrm{M}_{\mathrm{ij}} \)
\(
\mathrm{B}_{11}=(-1)^{1+1} \times \mathrm{M}_{11}=1 \times 3=3\)
\(
\mathrm{B}_{12}=(-1)^{1+2} \times \mathrm{M}_{12}=(-1) \times(-1)=1\)
\(
\mathrm{B}_{13}=(-1)^{1+3} \times \mathrm{M}_{13}=1 \times 2=2\)
\(
\mathrm{B}_{21}=(-1)^{2+1} \times \mathrm{M}_{21}=(-1) \times(-2)=2\)
\(
B_{22}=(-1)^{2+2} \times \mathrm{M}_{22}=1 \times 1=1\)
\(
B_{23}=(-1)^{2+3} \times \mathrm{M}_{23}=(-1) \times(-2)=2\)
\(
B_{31}=(-1)^{3+1} \times \mathrm{M}_{31}=1 \times 6=6\)
\(
\mathrm{B}_{32}=(-1)^{3+2} \times \mathrm{M}_{32}=(-1) \times(-2)=2\)
\(
B_{33}=(-1)^{3+3} \times \mathrm{M}_{33}=1 \times 5=5
\)
\( \operatorname{Adj} B=\left[\begin{array}{lll}3 & 1 & 2 \\ 2 & 1 & 2 \\ 6 & 2 & 5\end{array}\right]=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right] \)
\(|\mathrm{B}|=1(3 \times 1-(-2) \times 0)-2 ((-1) \times 1-0 \times 0)+(-2)\)\(((-1) \times(-2)-3 \times 0)\)
\(|\mathrm{B}|=3-2(-1-0)-2(2-0)\)
\(|\mathrm{B}|=3+2-4=1\)
\( \therefore \mathrm{B}^{-1}=\frac{(\operatorname{Adj} \mathrm{B}) }{|\mathrm{B}|}=\frac{\left|\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right| }{ 1}=\left[\begin{array}{lll}3 & 2 & 6 \\ 1 & 1 & 2 \\ 2 & 2 & 5\end{array}\right] \)
We know \( (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{A}^{-1} \)
\((\mathrm{AB})^{-1}=\left[\begin{array}{lll}
3 & 2 & 6 \\
1 & 1 & 2 \\
2 & 2 & 5
\end{array}\right] \times\left[\begin{array}{ccc}
3 & -1 & 1 \\
-15 & 6 & -5 \\
5 & -2 & 2
\end{array}\right]\)
Solving the above matrix, we get
\( (\mathrm{AB})^{-1}=\left[\begin{array}{ccc}9 & -3 & 5 \\ -2 & 1 & 0 \\ 1 & 0 & 2\end{array}\right] \)

\(\begin{array}{l}
A=\left|\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right| \end{array}\) \(|A|=1(3 \times 5-1 \times 1)-(-2)((-2) \times 5-1 \times 1)+1\)\(((-2) \times 1-3 \times 1)\)
\(|A|=(15-1)+2(-10-1)+(-2-3) \)
\(|A|=14-22-5=-13\)
To find the inverse of a matrix we need to find the Adjoint of that matrix For finding the adjoint of the matrix we need to find its cofactors
Let Aij denote the cofactors of Matrix A
Minor of an element aij \( = \) Mij
\( a_{11}=1 \), Minor of element
\( a_{11}=M_{11}=\left|\begin{array}{ll}3 & 1 \\ 1 & 5\end{array}\right|=(3 \times 5)-(1 \times 1)=14 \)
\( a_{12}=-2 \), Minor of element
\( a_{12}=M_{12}=\left|\begin{array}{cc}-2 & 1 \\ 1 & 5\end{array}\right|=(-2 \times 5)-(1 \times 1)=- \) 11
\( a_{13}=1 \), Minor of element
\( a_{13}=M_{13}=\left|\begin{array}{cc}-2 & 3 \\ 1 & 1\end{array}\right|=(-2 \times 1)-(3 \times 1)=-5 \)
\( a_{21}=-2 \), Minor of element
\( a_{21}=M_{21}=\left|\begin{array}{cc}-2 & 1 \\ 1 & 5\end{array}\right|=((-2) \times 5)-(1 \times 1)=- 11\)
\( a_{22}=3 \), Minor of element
\( a_{22}=M_{22}=\left|\begin{array}{ll}1 & 1 \\ 1 & 5\end{array}\right|=(1 \times 5)-(1 \times 1)=4 \)
\( \mathrm{a}_{23}=1 \), Minor of element
\( \mathrm{a}_{23}=\mathrm{M}_{23}=\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=(1 \times 1)-((-2) \times 1)=3 \)
\( a_{31}=1 \), Minor of element
\( a_{31}=M_{31}=\left|\begin{array}{cc}-2 & 1 \\ 3 & 1\end{array}\right|=(-2 \times 1)-(3 \times 1)=-5 \)
\( a_{32}=1 \), Minor of element
\( a_{32}=M_{32}=\left|\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right|=(1 \times 1)-(1 \times(-2))=3 \)
\( a_{33}=5 \), Minor of element
\( a_{33}=M_{33}=\left|\begin{array}{cc}1 & -2 \\ -2 & 3\end{array}\right|=(1 \times 3)-((-2) \times(-2)) =-1 \)
Cofactor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{A}_{\mathrm{ij}} \)
\(
\mathrm{~A}_{11}=(-1)^{1+1} \times 14=1 \times 14=14\)
\(
\mathrm{~A}_{12}=(-1)^{1+2} \times(-11)=(-1) \times(-11)=11\)
\(
\mathrm{~A}_{13}=(-1)^{1+3} \times(-5)=1 \times(-5)=-5\)
\(
\mathrm{~A}_{21}=(-1)^{2+1} \times(-11)=(-1) \times(-11)=11\)
\(
\mathrm{~A}_{22}=(-1)^{2+2} \times 4=1 \times 4=4\)
\(
\mathrm{~A}_{23}=(-1)^{2+3} \times 3=(-1) \times 3=-3\)
\(
\mathrm{~A}_{31}=(-1)^{3+1} \times(-5)=1 \times(-5)=-5\)
\(
\mathrm{~A}_{32}=(-1)^{3+2} \times 3=(-1) \times 3=-3\)
\(
\mathrm{~A}_{33}=(-1)^{3+3} \times(-1)=1 \times(-1)=-1\)
\(\mathrm{Adj} {\mathrm{~A}}=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right] \)
\(\mathrm{A}^{-1}=\frac{(\mathrm{Adj} \mathrm{~A}) }{|\mathrm{A}|}\)
\(\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\
-\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right]\)
(i) \( |\operatorname{Adj} \mathrm{A}|=14(-4-9)-11(-11-15)-5(-33+20) \)
\(
=14 \times(-13)-11 \times(-26)-5(-13)\)
\(
=-182+286+65=169
\)
Similarly Finding the Adj (Adj A) as found above
\( \operatorname{Adj}(\operatorname{Adj} A)=\left[\begin{array}{ccc}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right] \)
\( [\operatorname{Adj} \mathrm{A}]^{-1}=\operatorname{Adj}\frac{(\operatorname{Adj} \mathrm{A}) }{|\operatorname{Adj} \mathrm{A}|} \)
\(\begin{array}{l}
=\frac{1}{169}\left[\begin{array}{ccc}
-13 & 26 & -13 \\
26 & -39 & -13 \\
-13 & -13 & -65
\end{array}\right] \\
=\frac{1}{13}\left[\begin{array}{ccc}
-1 & 2 & -1 \\
2 & -3 & -1 \\
-1 & -1 & -5
\end{array}\right]
\end{array}\)
\(\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\
-\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right]\)
Similarly Finding the \( \operatorname{Adj}\left(\mathrm{A}^{-1}\right) \) as found above
\( \operatorname{Adj}\left(\mathrm{A}^{-1}\right)=\frac{1}{169}\left[\begin{array}{ccc}-13 & 26 & -13 \\ 26 & -39 & -13 \\ -13 & -13 & -65\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right] \)
Hence, \( [\text {Adj } \mathrm{A}]^{-1}=\operatorname{Adj}\left(\mathrm{A}^{-1}\right) \)

\(\begin{array}{l}
A=\left[\begin{array}{ccc}
1 & -2 & 1 \\
-2 & 3 & 1 \\
1 & 1 & 5
\end{array}\right] \end{array}\)
\(|A|=1(3 \times 5-1 \times 1)-(-2)((-2) \times 5-1 \times 1)+1\)\(((-2) \times 1-3 \times 1)\)
\(|A|=(15-1)+2(-10-1)+(-2-3) \)
\(|A|=14-22-5=-13\)
To find the inverse of a matrix we need to find the Adjoint of that matrix For finding the adjoint of the matrix we need to find its cofactors
Let Aij denote the cofactors of Matrix A
Minor of an element \( \mathrm{aij}=\mathrm{M}_{\mathrm{ij}} \)
\( a_{11}=1 \), Minor of element
\( a_{11}=M_{11}=\left|\begin{array}{ll}3 & 1 \\ 1 & 5\end{array}\right|=(3 \times 5)-(1 \times 1)=14 \)
\( a_{12}=-2 \), Minor of element
\( a_{12}=M_{12}=\left|\begin{array}{cc}-2 & 1 \\ 1 & 5\end{array}\right|=(-2 \times 5)-(1 \times 1)=-11 \)
\( a_{13}=1 \), Minor of element
\( a_{13}=M_{13}=\left|\begin{array}{cc}-2 & 3 \\ 1 & 1\end{array}\right|=(-2 \times 1)-(3 \times 1)=-5 \)
\( a_{21}=-2 \), Minor of element
\( a_{21}=M_{21}=\left|\begin{array}{cc}-2 & 1 \\ 1 & 5\end{array}\right|=(-2 \times 5)-(1 \times 1)=- 11\)
\( \mathrm{a}_{22}=3 \), Minor of element
\( \mathrm{a}_{22}=\mathrm{M}_{22}=\left|\begin{array}{ll}1 & 1 \\ 1 & 5\end{array}\right|=(1 \times 5)-(1 \times 1)=4 \)
\( \mathrm{a}_{23}=1 \), Minor of element
\( \mathrm{a}_{23}=\mathrm{M}_{23}=\left|\begin{array}{cc}1 & -2 \\ 1 & 1\end{array}\right|=(1 \times 1)-((-2) \times 1)=3 \)
\( a_{31}=1 \), Minor of element
\( a_{31}=M_{31}=\left|\begin{array}{cc}-2 & 1 \\ 3 & 1\end{array}\right|=(-2 \times 1)-(3 \times 1)=-5 \)
\( \mathrm{a}_{32}=1 \), Minor of element
\( \mathrm{a}_{32}=\mathrm{M}_{32}=\left[\begin{array}{cc}1 & 1 \\ -2 & 1\end{array}\right]=(1 \times 1)-(1 \times-2)=3 \)
\( \mathrm{a}_{33}=5 \), Minor of element
\( \mathrm{a}_{33}=\mathrm{M}_{33}=\left|\begin{array}{cc}1 & -2 \\ -2 & 3\end{array}\right|=(1 \times 3)-(-2) \times(-2)= \) \( -1 \)
Cofactor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{A}_{\mathrm{ij}} \)
\(
\mathrm{~A}_{11}=(-1)^{1+1} \times 14=1 \times 14=14\)
\(
\mathrm{~A}_{12}=(-1)^{1+2} \times(-11)=(-1) \times(-11)=11\)
\(
\mathrm{~A}_{13}=(-1)^{1+3} \times(-5)=1 \times(-5)=-5\)
\(
\mathrm{~A}_{21}=(-1)^{2+1} \times(-11)=(-1) \times(-11)=11\)
\(
\mathrm{~A}_{22}=(-1)^{2+2} \times 4=1 \times 4=4\)
\(
\mathrm{~A}_{23}=(-1)^{2+3} \times 3=(-1) \times 3=-3\)
\(
\mathrm{~A}_{31}=(-1)^{3+1} \times(-5)=1 \times(-5)=-5\)
\(
\mathrm{~A}_{32}=(-1)^{3+2} \times 3=(-1) \times 3=-3\)
\(
\mathrm{~A}_{33}=(-1)^{3+3} \times(-1)=1 \times(-1)=-1
\)
\(\mathrm{Adj ~A}=\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right]=\left[\begin{array}{ccc}14 & 11 & -5 \\ 11 & 4 & -3 \\ -5 & -3 & -1\end{array}\right] \)
\(\begin{array}{l}
\mathrm{A}^{-1}=\frac{(\operatorname{Adj} \mathrm{A}) }{|\mathrm{A}|} \\
\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right]=\left[\begin{array}{ccc}
-\frac{14}{13} & -\frac{11}{13} & \frac{5}{13} \\
-\frac{11}{13} & -\frac{4}{13} & \frac{3}{13} \\
\frac{5}{13} & \frac{3}{13} & \frac{1}{13}
\end{array}\right]
\end{array}\)
(ii) To find \( \left(\mathrm{A}^{-1}\right)^{-1} \) we have to find out \( \operatorname{Adj}\left(\mathrm{A}^{-1}\right) \)
\(\begin{array}{l}
\mathrm{A}^{-1}=-\frac{1}{13}\left[\begin{array}{ccc}
14 & 11 & -5 \\
11 & 4 & -3 \\
-5 & -3 & -1
\end{array}\right] \end{array}\)
\(\left|\mathrm{A}^{-1}\right|=(-\frac{ 1 }{ 13 })^{3}[14(4 \times(-1)-(-3) \times(-3))-11(11 \times(-1)\) \(-(-3) \times(-5))+ (-5)(11 \times(-3)-4 \times(-5))]\)
\(|\mathrm{A}|=(-\frac{ 1 }{ 13 })^{3}[14(-4-9)-11(-11-15)-5(-33+20)]\)
\(|\mathrm{A}|=(-\frac{ 1 }{ 13 })^{3}[14 \times(-13)-11 \times(-26)-5 \times(-13)]\)
\(|\mathrm{A}|=(-\frac{ 1 }{ 13 })^{3} \times 169=-\frac{ 1 }{ 13 }\)
Cofactor of an element \( \mathrm{a}_{\mathrm{ij}}=\mathrm{A}_{\mathrm{ij}} \)
\( A_{11}=(-1)^{1+1} \times(\frac{-1 }{ 13})=1 \times(\frac{-1 }{ 13})=\frac{-1 }{ 13}\)
\(A_{12}=(-1)^{1+2} \times(\frac{-2 }{ 13})=(-1) \times(\frac{-2 }{ 13})=\frac{2 }{ 13}\)
\(A_{13}=(-1)^{1+3} \times(\frac{-1 }{ 13})=1 \times(\frac{-1 }{ 13})=\frac{-1 }{ 13}\)
\(A_{21}=(-1)^{2+1} \times(\frac{-2 }{ 13})=(-1) \times(\frac{-2 }{ 13})=\frac{2 }{ 13}\)
\(A_{22}=(-1)^{2+2} \times(\frac{3 }{ 13})=1 \times(\frac{-3 }{ 13})=\frac{-3 }{ 13}\)
\(A_{23}=(-1)^{2+3} \times(\frac{1 }{ 13})=(-1) \times(\frac{1 }{ 13})=\frac{-1 }{ 13}\)
\(A_{31}=(-1)^{3+1} \times(\frac{-1 }{ 13})=1 \times(\frac{-1 }{ 13})=\frac{-1 }{ 13}\)
\(A_{32}=(-1)^{3+2} \times(\frac{1 }{ 13})=(-1) \times \frac{1 }{ 13}=\frac{-1 }{ 13}\)
\(A_{33}=(-1)^{3+3} \times(\frac{-5 }{ 13})=1 \times(\frac{-5 }{ 13})=\frac{-5 }{ 13} \)
\( \operatorname{Adj}\left(\mathrm{A}^{-1}\right)=\left[\begin{array}{ccc}-\frac{1}{13} & \frac{2}{13} & -\frac{1}{13} \\ \frac{2}{13} & -\frac{3}{13} & -\frac{1}{13} \\ -\frac{1}{13} & -\frac{1}{13} & -\frac{5}{13}\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right] \)
\( \left(\mathrm{A}^{-1}\right)^{-1}=\operatorname{Adj}\frac{\left(\mathrm{A}^{-1}\right) }{\left|\mathrm{A}^{-1}\right|} \)
\( \left(A^{-1}\right)^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-1 & 2 & -1 \\ 2 & -3 & -1 \\ -1 & -1 & -5\end{array}\right]=\frac{-1}{13}\left[\begin{array}{ccc}1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5\end{array}\right]=A \)
\( \therefore\left(\mathrm{A}^{-1}\right)^{-1}=\mathrm{A} \)

Let \( \Delta=\left|\begin{array}{ccc}x & y & x+y \\ y & x+y & x \\ x+y & x & y\end{array}\right| \)
Applying Elementary Transformations.
Applying \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \), we have
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
2(x+y) & 2(x+y) & 2(x+y) \\
y & x+y & x \\
x+y & x & y
\end{array}\right| \\
\Delta & =2(x+y)\left|\begin{array}{ccc}
1 & 1 & 1 \\
y & x+y & x \\
x+y & x & y
\end{array}\right|
\end{aligned}\)
Applying \( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) and \( \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1} \), we have
\(\Delta=2(x+y)\left|\begin{array}{ccc}
1 & 0 & 0 \\
y & x & x-y \\
x+y & -y & -x
\end{array}\right|\)
Expanding along \( \mathrm{R}_{1} \), we have
\(\Delta=2(x+y)[1(x \times(-x)-(-y) \times(x-y))-0+0]\)
\(\Delta=2(x+y)[-x^{ 2}+y(x-y)]\)
\(\Delta=2(x+y)[-x^{2}+\mathrm{xy}-y^{2}]\)
\(\Delta=-2(x+y)\left[x^{2}-x y+y^{2}\right]\)
\(\Delta=-2\left(x^{3}+y^{3}\right)\)

Let \( \Delta=\left|\begin{array}{ccc}1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y\end{array}\right| \)
Applying Row transformations
\( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} ; \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \)
\(\begin{aligned}
\Delta & =\left|\begin{array}{ccc}
1 & x & y \\
1-1 & x+y-x & y-y \\
1-1 & x-x & x+y-y
\end{array}\right| \\
\Delta & =\left|\begin{array}{ccc}
1 & x & y \\
0 & y & 0 \\
0 & 0 & x
\end{array}\right|
\end{aligned}\)
Now, Expanding along \( \mathrm{C}_{1} \)
\(\Delta=1(x \times x-0 \times 0)-0(x \times x-0 \times x)+0(x \times 0-x \times x)\)
\(\Delta=1(xy)-0+0\)
\(\Delta=xy\)

Let \( \Delta=\left|\begin{array}{lll}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right| \)
Applying Row Transformations
\(\begin{array}{l}
\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \\
\Delta=\left|\begin{array}{ccc}
\alpha & \alpha^{2} & \beta+\gamma \\
\beta-\alpha & \beta^{2}-\alpha^{2} & \alpha-\beta \\
\gamma & \gamma^{2} & \alpha-\gamma
\end{array}\right| \\
\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \\
\Delta=\left|\begin{array}{ccc}
\alpha & \alpha^{2} & \beta+\gamma \\
\beta-a & \beta^{2}-\alpha^{2} & \alpha-\beta \\
\gamma-a & \gamma^{2}-\alpha^{2} & \alpha+\beta
\end{array}\right|
\end{array}\)
Taking \( (\beta-\alpha)(\gamma-\alpha) \) from \( \mathrm{R}_{2} \) and \( \mathrm{R}_{3} \) respectively
\(\Delta=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}
\alpha & \alpha^{2} & \beta+\gamma \\
1 & \beta+\gamma & -1 \\
1 & \gamma+\alpha & -1
\end{array}\right|\)
Applying \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2} \), we have
\(\Delta=(\beta-\alpha)(\gamma-\alpha)\left|\begin{array}{ccc}
\alpha & \alpha^{2} & \beta+\gamma \\
1 & \beta+\gamma & -1 \\
0 & \gamma-\beta & 0
\end{array}\right|\)
Expanding along \( \mathrm{R}_{3} \), we have
\(\Delta=(\beta-\alpha)(\gamma-\alpha)\left[0 \left(\alpha^{2} \times(-1)-(\beta+\gamma) \times(\beta+\gamma)-(\gamma-\beta)(-1) \right.\right.\)
\(\times \alpha-1 \times(\beta+\gamma)+0\left(\alpha \times(\beta+\gamma)-1 \times \alpha^{2}\right)\)
\(\Delta=(\beta-\alpha)(\gamma-\alpha)[0-(\gamma-\beta)(-\alpha-\beta-\gamma)+0]\)
\(\Delta=(\beta-\alpha)(\gamma-\alpha)(\gamma-\beta)(\alpha+\beta+\gamma)\)
\(\Delta=(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma)\)

Let \( \Delta=\left|\begin{array}{lll}x & x^{2} & 1+p x^{3} \\ y & y^{2} & 1+p y^{3} \\ x & z^{2} & 1+p z^{3}\end{array}\right| \)
Applying Elementary Row Transformations
\( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \) and \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \)
\(\Delta=\left|\begin{array}{ccc}
x & x^{2} & 1+p x^{3} \\
y-x & y^{2}-x^{2} & p\left(y^{3}-x^{3}\right) \\
z-x & z^{2}-x^{2} & p\left(z^{3}-x^{3}\right)
\end{array}\right|\)
Taking \( (y-x) \) and \( (z-x) \) common from \( \mathrm{R}_{2} \) and \( \mathrm{R}_{3} \) respectively
\(\Delta=(y-x)(z-x)\left|\begin{array}{ccc}
x & x^{2} & 1+p x^{3} \\
1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\
1 & z+x & p\left(z^{2}+x^{2}+x z\right)
\end{array}\right|\)
Applying \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2} \)
\(\Delta=(y-x)(z-x)\left|\begin{array}{ccc}
x & x^{2} & 1+p x^{3} \\
1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\
0 & z-y & p(z-y)(x+y+z)
\end{array}\right|\)
Taking \( (z-y) \) common from \( \mathrm{R}_{3} \)
\(\Delta=(y-x)(z-x)(z-y)\left|\begin{array}{ccc}
x & x^{2} & 1+p x^{3} \\
1 & y+x & p\left(y^{2}+x^{2}+x y\right) \\
0 & 1 & p(x+y+z)
\end{array}\right|\)
Expanding along \( \mathrm{R}_{3} \), we have
\(\Delta=(x-y)(y-z)(z-x)\left[0-1\left\{x \times \mathrm{p}\left(y^{2}+x^{2}+xy\right)-1 \times\left(1+px^{3}\right)\right\}\right.\)
\(+p(x+y+z)\left\{x \times(y+x)-1 \times x^{2}\right\}\)
\(\Delta=(x-y)(y-z)(z-x)\left(-px^{3}-pxy^{2}- px^{2} y+1+ px^{3}+px^{2} y\right.\)
\(pxy^{2}+pxyz)\)
\(\Delta=(x-y)(y-z)(z-x)(1+pxyz)\)
Hence, the given result is proved.

Let \( \Delta=\left|\begin{array}{ccc}3 a & -a+b & -a+c \\ -b+a & 3 b & -b+c \\ -c+a & -c+b & 3 c\end{array}\right| \)
Applying Elementary Transformations
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \), we have
\( \Delta=\left|\begin{array}{ccc}a+b+c & -a+b & -a+c \\ a+b+c & 3 b & -b+c \\ a+b+c & -c+b & 3 c\end{array}\right| \)
Taking \( (a+b+c) \) common from \( \mathrm{C}_{1} \), we get
\(\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
1 & 3 b & -b+c \\
1 & -c+b & 3 c
\end{array}\right|\)
Applying \( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1} \) and \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1} \)
\(\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & -a+b & -a+c \\
0 & 2 b+a & a-b \\
0 & a-c & 2 c+a
\end{array}\right|\)
Expanding along \( \mathrm{C}_{1} \)
\(\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})[1 \times\{(2 \mathrm{b}+\mathrm{a})(2 \mathrm{c}+\mathrm{a})-(\mathrm{a}-\mathrm{b})(\mathrm{a}-\mathrm{c})\}-0+0]\)
\(\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})[4 \mathrm{bc}+2 \mathrm{ab}+2 \mathrm{ac}+\mathrm{a} ^{2}-\mathrm{a}^{2}+\mathrm{ac}+\mathrm{ba}-\mathrm{bc}]\)
\(\Delta=(\mathrm{a}+\mathrm{b}+\mathrm{c})(3 \mathrm{ab}+3 \mathrm{bc}+3 \mathrm{ac})\)
\(\Delta=3(\mathrm{a}+\mathrm{b}+\mathrm{c})(\mathrm{ab}+\mathrm{bc}+\mathrm{ca})\)
Hence, the given result proved.

Let \( \Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right| \)
Applying Elementary Row Transformations
\(\begin{array}{l}
\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-2 \mathrm{R}_{1} \\
\Delta=\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
0 & 1 & 2+p \\
3 & 6+3 p & 10+6 p+3 q
\end{array}\right| \\
\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{1} \\
\Delta=\left|\begin{array}{ccc}
1 & 1+p & 1+p+q \\
0 & 1 & 2+p \\
0 & 3 & 7+3 p
\end{array}\right|
\end{array}\)
\( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-3 \mathrm{R}_{2} \)
\( \Delta=\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 0 & 1 & 2+p \\ 0 & 0 & 1\end{array}\right| \)
Expanding Along \(C_{1}\), we have
\( \Delta=1(1 \times 1-0 \times(2+\mathrm{p}))-0+0 \)
\( \Delta=1-0 \)
\( \Delta=1 \)
Hence, the given result is proved

Let \( \Delta=\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \cos (\alpha+\delta) \\ \sin \beta & \cos \beta & \cos (\beta+\delta) \\ \sin \gamma & \cos \gamma & \cos (\gamma+\delta)\end{array}\right| \)
\( \Delta=\left|\begin{array}{lll}\sin \alpha & \cos \alpha & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\ \sin \beta & \cos \beta & \cos \beta \cos \delta-\sin \beta \sin \delta \\ \sin \gamma & \cos \gamma & \cos \gamma \cos \delta-\sin \beta \sin \delta\end{array}\right| \)
\( \Delta=\frac{1}{\sin \delta \cos \delta}\left|\begin{array}{lll}\sin \alpha \sin \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\ \sin \beta \sin \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\ \sin \gamma \sin \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma \sin \delta\end{array}\right| \)
Applying Elementary Column Transformations
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3} \)
\( \Delta= \)
\(\frac{1}{\sin \delta \cos \delta}\left[\begin{array}{ccc}
\cos \alpha \cos \delta & \cos \alpha \cos \delta & \cos \alpha \cos \delta-\sin \alpha \sin \delta \\
\cos \beta \cos \delta & \cos \beta \cos \delta & \cos \beta \cos \delta-\sin \beta \sin \delta \\
\cos \gamma \cos \delta & \cos \gamma \cos \delta & \cos \gamma \cos \delta-\sin \gamma-\sin \gamma \sin \delta
\end{array}\right]\)
Since, the two columns are identical
[In a determinant if two columns are identical the value of determinant is 0 ]
So, the value of given determinant is 0
\(\therefore \Delta=0\)
Hence, the given result is proved.

Let \( \Delta=\left|\begin{array}{lll}x+2 & x+3 & x+2 a \\ x+3 & x+4 & x+2 b \\ x+4 & x+5 & x+2 c\end{array}\right| \)
Since, \( a, b, c \) is in A.P.
\(\begin{array}{l}
\therefore 2 \mathrm{b}=\mathrm{a}+\mathrm{c} \\
\Delta=\left|\begin{array}{ccc}
x+2 & x+3 & x+2 a \\
x+3 & x+4 & x+(a+c) \\
x+4 & x+5 & x+2 c
\end{array}\right|
\end{array}\)
Applying Elementary Row Transformations
\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2} \) and \( \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{2} \)
\( \Delta=\left|\begin{array}{ccc}-1 & -1 & a-c \\ x+3 & x+4 & x+(a+c) \\ 1 & 1 & c-a\end{array}\right| \)
\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{3} \), we have
\( \Delta=\left|\begin{array}{ccc}0 & 0 & 0 \\ x+3 & x+4 & x+a+c \\ 1 & 1 & c-a\end{array}\right| \)
[In a determinant if all elements of a row are 0 then the value of determinant is 0.]
So, here all the elements of first row \( \left(\mathrm{R}_{1}\right) \) are zero.
\( \therefore \Delta=0 \)

\( A=\left[\begin{array}{ccc}1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1\end{array}\right] \)
\( |A|=1(1 \times 1-\sin \theta \times(-\sin \theta))-\sin \theta((-\sin \theta) \times 1-(-1) \)
\( \times \sin \theta)+1((- \sin \theta) \times(-\sin \theta)-(-1) \times 1) \)
\( |A|=1+\sin 2 \theta+\sin 2 \theta-\sin 2 \theta+\sin 2 \theta+1\)
\(|A|=2+2 \sin 2 \theta\)
\(|A|=2(1+\sin 2 \theta) \)
Now, \( 0 \leq \theta \leq 2 \pi \)
\(\Rightarrow \sin 0 \leq \sin \theta \leq \sin 2 \pi\)
\(\Rightarrow 0 \leq \sin 2 \theta \leq 1\)
\(\Rightarrow 1+0 \leq 1+\sin 2 \theta \leq 1+1\)
\(\Rightarrow 2 \leq 2(1+\sin 2 \theta) \leq 4\)
\(\therefore \operatorname{Det} (A)\in[2,4]\)