Ex 4.2 Class 12 Maths Ncert Solutions

L.H.S. \( =\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & c+d\end{array}\right| \)
Applying Operations \( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) column by addition of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) column)
\(\therefore \text { L.H.S. }=\left|\begin{array}{lll}
x+a & a & x+a \\
y+b & b & y+b \\
z+c & c & z+c
\end{array}\right|\)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \) (i.e. Replacing \( 1^{\text {st }} \) column by subtraction of \( 1^{\text {st }} \) and \( 3^{\text {rd }} \) column)
\(\therefore \text {L.H.S. }=\left|\begin{array}{lll}
0 & a & x+a \\
0 & b & y+b \\
0 & c & z+c
\end{array}\right|\)
If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0 .
\(\begin{array}{l}
\therefore \text { LHS }=0=\text { RHS } \\
\therefore\left|\begin{array}{lll}
x & a & x+a \\
y & b & y+b \\
z & c & z+c
\end{array}\right|=0
\end{array}\)

Applying Operation \( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) column by addition of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) column)
\(\therefore \text { L.H.S. }=\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{3} \) (i.e. Replacing \( 1^{\text {st }} \) column by addition of \( 1^{\text {st }} \) and \( 3^{\text {rd }} \) column)
\(\therefore \text { L.H.S. }=\left|\begin{array}{lll}
0 & b-c & c-a \\
0 & c-a & a-b \\
0 & a-b & b-c
\end{array}\right|\)
If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0 .
\(\begin{array}{l}
\therefore \mathrm{LHS}=0=\mathrm{RHS} \\
\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|=0
\end{array}\)

\(\therefore \text { L.H.S }=\left|\begin{array}{lll}
2 & 7 & 65 \\
3 & 8 & 75 \\
5 & 9 & 86
\end{array}\right|\)
Applying Operation \( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+9 \mathrm{C}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) column by addition of \( 1^{\text {st }} \) column and 9 times second column)
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}2+9 \times 7 & 7 & 65 \\ 3+9 \times 8 & 8 & 75 \\ 5+9 \times 9 & 9 & 86\end{array}\right| \)
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}65 & 7 & 65 \\ 75 & 8 & 75 \\ 86 & 9 & 86\end{array}\right| \)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \) (i.e. Replacing \( 1^{\text {st }} \) column by subtraction of \( 1^{\text {st }} \) and \( 3^{\text {rd }} \) column)
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}0 & 7 & 65 \\ 0 & 8 & 75 \\ 0 & 9 & 86\end{array}\right| \)
If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0.
\( \therefore \) LHS \( =0= \) RHS
\( \therefore\left|\begin{array}{lll}2 & 7 & 65 \\ 3 & 8 & 75 \\ 5 & 9 & 86\end{array}\right| \)

\(\therefore \text { L.H.S }=\left|\begin{array}{lll}
1 & b c & a b+a c \\
1 & c a & b c+a b \\
1 & a b & a c+b c
\end{array}\right|\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & b(c+a) \\
1 & a b & c(a+b)
\end{array}\right|=0\)
\( \mathrm{C}_{3} \rightarrow \mathrm{C}_{2}+\mathrm{C}_{3} \) (i.e. replace \( 3^{\text {rd }} \) column by addition of \( 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column)
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}1 & b c & a b+b c+a c \\ 1 & c a & a b+b c+a c \\ 1 & a b & a b+b c+a c\end{array}\right| \)
Taking \( \mathrm{ab}+\mathrm{bc}+\mathrm{ac} \) outside determinant
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}1 & b c & 1 \\ 1 & c a & 1 \\ 1 & a b & 1\end{array}\right| \)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{3} \) (i.e. replace \( 1^{\text {st }} \) column by subtraction of \( 1^{\text {st }} \) and \( 3^{\text {rd }} \) column)
\( \therefore \) L.H.S \( =\left|\begin{array}{lll}0 & b c & 1 \\ 0 & c a & 1 \\ 0 & a b & 1\end{array}\right| \)
If any one of the rows or columns of a determinant is 0 then the value of that determinant is 0 .
\( \therefore \) LHS \( =0= \) RHS
\( \therefore\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0 \)

When two determinants are added each of the corresponding elements gets added.
Here we can split the LHS determinant as
\(\therefore \text { LHS }=\left|\begin{array}{lll}
b & q & y \\
c & r & z \\
a & p & x
\end{array}\right|+\left|\begin{array}{lll}
c & r & z \\
a & p & x \\
b & q & y
\end{array}\right|=\mathrm{u}+\mathrm{v}\)
For the determinant, u perform the following transformation \( \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{3} \) (i.e. interchange \( 1^{\text {st }} \) row with \( 3^{\text {rd }} \) row)
When two particular rows/columns of a determinant are interchanged the value becomes negative 1 times the original value.
\( \therefore \) LHS \( =(-1)\left|\begin{array}{lll}a & p & x \\ c & r & z \\ b & q & y\end{array}\right|+\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right|=\mathrm{u}+\mathrm{v} \)
\( \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2} \) (i.e. interchange \( 1^{\text {st }} \) row with \( 2^{\text {nd }} \) row)
\( \therefore \) LHS \( =(-1)^{2}\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right|\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right|=\mathrm{u}+\mathrm{v} \)
\( \therefore \) LHS \( =2\left|\begin{array}{lll}c & r & z \\ a & p & x \\ b & q & y\end{array}\right| \)
\( R_{1} \leftrightarrow R_{3} \) (i.e. interchange \( 1^{\text {st }} \) row with \( 3^{\text {rd }} \) row)
\( \therefore \) LHS \( =(-1)^{2}\left|\begin{array}{lll}b & q & y \\ a & p & x \\ c & r & z\end{array}\right| \)
\( \mathrm{R}_{1} \leftrightarrow \mathrm{R}_{2} \) (i.e. interchange \( 1^{\text {st }} \) row with \( 2^{\text {nd }} \) row)
\( \therefore \) LHS \( =(-1)^{2} \times 2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right| \)
\( \therefore \) LHS \( = \) RHS
\(\therefore\left|\begin{array}{lll}
b+c & q+r & y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|=2\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & z
\end{array}\right|\)

To Prove:
\(\left|\begin{array}{ccc}
0 & a & -b \\
-a & 0 & -c \\
b & c & 0
\end{array}\right|=0\)
\( R_{1} \rightarrow c R_{1} \) (i.e. replace \( 1^{\text {st }} \) row by multiplying it with \( c \) )
As we are multiplying we should also divide c so that the original given determinant is not changed
\( \therefore \) LHS \( =\frac{1}{c}\left|\begin{array}{ccc}0 & a c & -b c \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \)
\( R_{1} \rightarrow R_{1}-b R_{2} \) (i.e. replace \( 1^{\text {st }} \) row by subtraction of \( 1^{\text {st }} \) row and \( b \) times \( 2^{\text {nd }} \) row)
\( \therefore \) LHS \( =\frac{1}{c}\left|\begin{array}{ccc}a b & a b & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \)
Taking a outside the determinant from \( 1^{\text {st }} \) row
\( \therefore \) LHS \( =\frac{a}{c}\left|\begin{array}{ccc}b & c & 0 \\ -a & 0 & -c \\ b & c & 0\end{array}\right| \)
If any two rows or columns of a determinant are identical then the value of that determinant is 0 because we get a row or column with all elements 0 when we when we subtract those particular rows/columns here the transformation is \( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{3} \)
\(\begin{array}{l}
\therefore \mathrm{LHS}=0=\mathrm{RHS} \\
\therefore\left|\begin{array}{ccc}
0 & a & -b \\
-a & 0 & -c \\
b & c & 0
\end{array}\right|=0
\end{array}\)

Taking '\( a \)', '\( b \)' and '\( c \)' outside the determinant from \( 1^{\text {st}}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column respectively
\(\therefore \text { LHS }=\text { abc }\left|\begin{array}{ccc}
-a & a & a \\
b & -b & b \\
c & c & -c
\end{array}\right|\)
Taking '\( a \)', '\( b \)' and '\( c \)' outside the determinant from \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3{ }^{\text {rd }} \) row respectively
\(\therefore \text { LHS }=\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}\left|\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\)
\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) row by addition of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row)
\(\therefore \text { LHS }=a^{2} b^{2} c^{2}\left|\begin{array}{ccc}
0 & 0 & 2 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right|\)
pending the determinant along 1st row
\(\begin{array}{l}
\therefore \text { LHS }=\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2} \times 2\left|\begin{array}{cc}
1 & -1 \\
1 & 1
\end{array}\right| \\
\therefore \text { LHS }=\mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2} \times 2(1-(-1))=4 \mathrm{a}^{2} \mathrm{b}^{2}\mathrm{c}^{2}=\text { RHS } \\
\therefore\left|\begin{array}{ccc}
-a^{2} & a b & a c \\
b a & -b^{2} & b c \\
c a & c b & -c^{2}
\end{array}\right|=4 \mathrm{a}^{2} \mathrm{~b}^{2} \mathrm{c}^{2}
\end{array}\)

\( R_{1} \rightarrow R_{1}-R_{2} \) (i.e. Replacing \( 1^{\text {st }} \) row by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row)
\( R_{2} \rightarrow R_{2}-R_{3} \) (i.e. Replacing \( 2^{\text {nd }} \) row by subtraction of \( 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
0 & a-b & a^{2}-b^{2} \\
0 & b-c & b^{2}-c^{2} \\
1 & c & c^{2}
\end{array}\right|\)
Since we know \( \mathrm{a}_{2}-\mathrm{b}_{2}=(\mathrm{a}+\mathrm{b})(\mathrm{a}-\mathrm{b}) \)
Therefore, taking \( (\mathrm{a}-\mathrm{b}) \) and \( (\mathrm{b}-\mathrm{c}) \) outside the determinant from \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row respectively
\(\therefore \text { LHS }=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 1 & a+b \\
0 & 1 & b+c \\
1 & c & c^{2}
\end{array}\right|\)
\( R_{1} \rightarrow R_{1}-R_{2} \) (i.e. Replacing \( 1^{\text {st }} \) row by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row)
\(\therefore \text { LHS }=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 0 & a-c \\
0 & 1 & b+c \\
1 & c & c^{2}
\end{array}\right|\)
Expanding the determinant along 1st column
\(\begin{array}{l}
\therefore \mathrm{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ll}
0 & a-c \\
1 & b+c
\end{array}\right| \\
\therefore \mathrm{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(0-(\mathrm{a}-\mathrm{c})) \\
\therefore \mathrm{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})=\mathrm{RHS} \\
\therefore\left|\begin{array}{lll}
1 & a & a^{2} \\
1 & b & b^{2} \\
1 & c & c^{2}
\end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})
\end{array}\)

\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) column by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) column)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} \) (i.e. Replacing \( 2^{\text {nd }} \) column by subtraction of \( 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column)
\(\therefore \mathrm{LHS}=\left|\begin{array}{ccc}
0 & 0 & 1 \\
a-b & b-c & c \\
a^{3}-b^{3} & b^{3}-c^{3} & c^{3}
\end{array}\right|\)
We have \( a^{3}-b^{3}=(a-b)\left(a^{2}+a b+b^{2}\right) \) and
\( b^{3}-c^{3}=(b-c)\left(b^{2}+b c+c^{2}\right) \)
Therefore, taking \( (\mathrm{a}-\mathrm{b}) \) and \( (\mathrm{b}-\mathrm{c}) \) outside the determinant from \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) column respectively
\(\therefore \mathrm{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 0 & 0 \\
1 & 1 & c \\
a^{2}+a b+b^{2} & b^{2}+b c+c^{2} & c^{3}
\end{array}\right|\)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2} \) (i.e. Replacing \( 1^{\text {st }} \) column by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) column)
\(\therefore \text { LHS }=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & c \\
\left(a^{2}-c^{2}+b[a-c]\right) & b^{2}+b c+c^{2} & c^{3}
\end{array}\right|\)
As \( \left(a^{2}-c^{2}\right)=(a+c)(a-c) \) therefore taking \( (a-c) \) outside the determinant from \( 1^{\text {st }} \) column we get
\(\therefore \text { LHS }=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{a}-\mathrm{c})\left|\begin{array}{ccc}
0 & 0 & 1 \\
0 & 1 & c \\
(a+c)+b & b^{2}+b c+c^{2} & c^{3}
\end{array}\right|\)
Expanding the determinant along 1st row
\(\therefore \operatorname{LHS}=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{a}-\mathrm{c})(-(\mathrm{a}+\mathrm{b}+\mathrm{c}))\)
Adjusting the minus sign with \( (\mathrm{a}-\mathrm{c}) \)
\(\begin{array}{l}
\therefore \text { LHS }=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})=\text { RHS } \\
\therefore\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|=(\mathrm{a}-\mathrm{b})(\mathrm{b}-\mathrm{c})(\mathrm{c}-\mathrm{a})(\mathrm{a}+\mathrm{b}+\mathrm{c})
\end{array}\)

\( R_{1} \rightarrow R_{1}-R_{2} \) (i.e. Replacing \( 1^{\text {st }} \) row by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row)
\( \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3} \) (i.e. Replacing \( 2^{\text {nd }} \) row by subtraction of \( 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
x-y & x^{2}-y^{2} & y z-z x \\
y-z & y^{2}-z^{2} & z x-x y \\
z & z^{2} & x y
\end{array}\right|\)
We know \( x^{2}-y^{2}=(x+y)(x-y) \) and
\( y^{2}-z^{2}=(y+z)(y-z) \)
Therefore, taking \( (x-y) \) and \( (y-z) \) outside the determinant from \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row respectively
\( \therefore \) LHS \( =(x-y)(y-z)=\left|\begin{array}{ccc}1 & x+y & -z \\ 1 & y+z & -x \\ z & z^{2} & x y\end{array}\right| \)
\( R_{1} \rightarrow R_{1}-R_{2} \) (i.e. Replacing \( 1^{\text {st }} \) row by subtraction of \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row)
\( \therefore \) LHS \( =(x-y)(y-z)=\left|\begin{array}{ccc}0 & x-z & x-z \\ 1 & y+z & -x \\ z & z^{2} & x y\end{array}\right| \)
Taking \( (x-z) \) outside determinant from \( 1^{\text {st }} \) row
\( \therefore \) LHS \( =(x-y)(y-z)(x-z)\left|\begin{array}{ccc}0 & 1 & 1 \\ 1 & y+z & -x \\ z & z^{2} & x y\end{array}\right| \)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{3} \) (i.e. Replacing \( 2^{\text {nd }} \) column by subtraction of \( 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column)
\( \therefore \) LHS \( =(x-y)(y-z)(x-z)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & y+z+x & -x \\ z & z^{2}-x y & x y\end{array}\right| \)
Expanding the determinant along \( 1^{\text {st }} \) row
\( \therefore \) LHS \( =(x-y)(y-z)(x-z)\left|\begin{array}{cc}1 & x+y+z \\ z & z^{2}-x y\end{array}\right| \)
\( \therefore \) LHS \( =(x-y)(y-z)(x-z)\left(z^{2}-xy-xz-yz-z^{2}\right) \)
Cancelling \( z^{2} \) and adjusting the negative sign wHith \( (x-z) \)
\( \therefore \) LHS \( =(x-y)(y-z)(z-x)(xy+yz+ zx)= \) RHS
\(\therefore\left|\begin{array}{lll}
x & x^{2} & y z \\
y & y^{2} & z x \\
z & z^{2} & x y
\end{array}\right|=(x-y)(y-z)(x-z)(xy+yz+zx)\)

\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \) (i.e. replace \(1^{\text {st }}\) row by addition of \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
5 x+4 & 5 x+4 & 5 x+4 \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\)
Taking \( 5 x+4 \) outside the determinant from \( 1^{\text {st }} \) row
\(\therefore \text { LHS }=(5 x+4)\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 x & x+4 & 2 x \\
2 x & 2 x & x+4
\end{array}\right|\)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) (i.e. replace \( 2^{\text {nd }} \) column by subtraction of \( 2^{\text {nd }} \) and \( 1^{\text {st }} \) column)
\( \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1} \) (i.e. replace \( 3^{\text {rd }} \) column by subtraction of \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) column)
\(\therefore \text { LHS }=(5 x+4)\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 x & 4-x & 0 \\
2 x & 0 & 4-x
\end{array}\right|\)
Expanding the determinant along 1st row
\(\therefore \text { LHS }=(5 x+4)\left|\begin{array}{cc}
4-x & 0 \\
0 & 4-x
\end{array}\right|\)
\( \therefore \) LHS \( =(5 x-4)(4-x)^{2}= \) RHS
\( \therefore\left|\begin{array}{ccc}x+4 & 2 x & 2 x \\ 2 x & x+4 & 2 x \\ 2 x & 2 x & x+4\end{array}\right|=(5 x+4)(4-x)^{2} \)

\( \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \) (i.e. replace \( 1^{\text {st }} \) row by addition of \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
3 y+k & 3 y+k & 3 y+k \\
y & y+k & y \\
y & y & y+k
\end{array}\right|\)
Taking \( 3 \mathrm{y}+\mathrm{k} \) outside the determinant from \( 1^{\text {st }} \) row
\(\therefore \text { LHS }=(3 \mathrm{y}+\mathrm{k})\left|\begin{array}{ccc}
1 & 1 & 1 \\
y & y+k & y \\
y & y & y+k
\end{array}\right|\)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) (i.e. replace \( 2^{\text {nd }} \) column by subtraction of \( 2^{\text {nd }} \) and \( 1^{\text {st }} \) column)
\( \mathrm{C}_{3} \rightarrow \mathrm{C}_{3}-\mathrm{C}_{1} \) (i.e. replace \( 3^{\text {rd }} \) column by subtraction of \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) column)
\(\therefore \text { LHS }=(3 \mathrm{y}+\mathrm{k})\left|\begin{array}{lll}
1 & 0 & 0 \\
y & k & 0 \\
y & 0 & k
\end{array}\right|\)
Expanding the determinant along \( 1^{\text {st }} \) row
\(\therefore \text { LHS }=(3 \mathrm{y}+\mathrm{k})\left|\begin{array}{cc}
k & 0 \\
0 & k
\end{array}\right|\)
\(\begin{array}{l}
\therefore \text { LHS }=(3 \mathrm{y}+\mathrm{k}) \mathrm{k}^{2}=\text { RHS } \\
\therefore\left|\begin{array}{ccc}
y+k & y & y \\
y & y+k & y \\
y & y & y+k
\end{array}\right|=\mathrm{k}^{2}(3 \mathrm{y}+\mathrm{k})
\end{array}\)

\( R_{1} \rightarrow R_{1}+R_{2}+R_{3} \) (i.e. replace \( 1^{\text {st }} \) row by addition of \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
a+b+c & a+b+c & a+b+c \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)
Taking \( \mathrm{a}+\mathrm{b}+\mathrm{c} \) outside the determinant from \( 1^{\text {st }} \) row
\(\therefore \text { LHS }=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & 1 & 1 \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|\)
\( \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-\mathrm{C}_{1} \) (i.e. replace \( 2^{\text {nd }} \) column by subtraction of \( 2^{\text {nd }} \) and \( 1^{\text {st }} \) column)
\( C_{3} \rightarrow C_{3}-C_{1} \) (i.e. replace \( 3^{\text {rd }} \) column by subtraction of \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) column)
\(\therefore \text { LHS }=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{ccc}
1 & 0 & 0 \\
2 b & -b-c-a & 0 \\
2 c & 0 & -c-a-b
\end{array}\right|\)
Expanding the determinant along \( 1^{\text {st }} \) row
\(\begin{array}{l}
\therefore \text { LHS }=(\mathrm{a}+\mathrm{b}+\mathrm{c})\left|\begin{array}{cc}
-(a+b+c) & 0 \\
0 & -(a+b+c)
\end{array}\right| \\
\therefore \text { LHS }=(\mathrm{a}+\mathrm{b}+\mathrm{c})^{3}=\text { RHS } \\
\therefore\left|\begin{array}{ccc}
a-b-c & 2 a & 2 a \\
2 b & b-c-a & 2 b \\
2 c & 2 c & c-a-b
\end{array}\right|=(\mathrm{a}+\mathrm{b}+\mathrm{c})^{3}
\end{array}\)

\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C} 3 \) (i.e. replace \( 1^{\text {st }} \) column by addition of \( 1^{\text {st, }} 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
2 x+2 y+2 z & x & y \\
2 z+2 y+2 z & y+z+2 x & y \\
2 x+2 y+2 z & x & z+x+2 y
\end{array}\right|\)
Taking \( 2(x+y+z\) outside the determinant from \( 1^{\text {st }} \) column
\(\therefore \text { LHS }=2(x+y+z)\left|\begin{array}{ccc}
1 & x & y \\
1 & y+z+2 x & y \\
1 & x & z+x+2 y
\end{array}\right|\)
\( R_{2} \rightarrow R_{2}-R_{1} \) (i.e. replace \( 2^{\text {nd }} \) row by subtraction of \( 2^{\text {nd }} \) and \( 1^{\text {st }} \) row)
\( R_{3} \rightarrow R_{3}-R_{1} \) (i.e. replace \( 3^{\text {rd }} \) row by subtraction of \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) row)
\(\therefore \text { LHS }=2(x+y+z)\left|\begin{array}{ccc}
1 & x & y \\
0 & x+y+z & 0 \\
0 & 0 & x+y+z
\end{array}\right|\)
Expanding the determinant along \( 1^{\text {st }} \) column
\(\begin{array}{l}
\therefore \text { LHS }=2(x+y+z)\left|\begin{array}{cc}
x+y+z & 0 \\
0 & x+y+z
\end{array}\right| \\
\therefore \text { LHS }=2(x+y+z)^{3}=\mathrm{RHS} \\
\therefore\left|\begin{array}{ccc}
x+y+2 z & x & y \\
z & y+z+2 x & y \\
z & x & z+x+2 y
\end{array}\right|=2(x+y+z)^{3}
\end{array}\)

\( R_{1} \rightarrow R_{1}-x R_{2} \) (i.e. replace \( 1{ }^{\text {st }} \) row by subtraction of \( 1{ }^{\text {st }} \) row and '\( x \)' times \( 2^{\text {nd }} \) row)
\(\therefore \text { LHS }=\left|\begin{array}{ccc}
1-x^{3} & 0 & 0 \\
x^{2} & 1 & x \\
x & x^{2} & 1
\end{array}\right|\)
Taking (\( 1-x^{3} \)) outside the determinant from \( 1^{\text {st }} \) row
\(\therefore \text { LHS }=\left(1-x^{3}\right)\left|\begin{array}{ccc}
1 & 0 & 0 \\
x^{2} & 1 & x \\
x & x^{2} & 1
\end{array}\right|\)
Expanding the determinant along \( 1^{\text {st }} \) row
\(\begin{array}{l}
\therefore \text { LHS }=\left(1-x^{3}\right)\left|\begin{array}{cc}
1 & x \\
x^{2} & 1
\end{array}\right| \\
\therefore \text { LHS }=\left(1-x^{3}\right)^{2}=\mathrm{RHS} \\
\therefore\left|\begin{array}{ccc}
1 & x & x^{2} \\
x^{2} & 1 & x \\
x & x^{2} & 1
\end{array}\right|=\left(1-x^{3}\right)^{2}
\end{array}\)

\( R_{1} \rightarrow R_{1}+b R_{3} \) (i.e. replace \( 1^{\text {st }} \) row by addition of \( 1^{\text {st }} \) row and \( b \) times \( 3^{\text {rd }} \) row)
\( R_{2} \rightarrow R_{2}+a R_{3} \) (i.e. replace \( 2^{\text {nd }} \) row by subtraction of \( 2^{\text {nd }} \) row and a times \( 3^{\text {rd }} \) row)
\( \therefore \) LHS \( =\left|\begin{array}{ccc}1+a^{2}+b^{2} & 0 & -b\left(1+a^{2}+b^{2}\right) \\ 0 & 1+a^{2}+b^{2} & a\left(1+a^{2}+b^{2}\right) \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right| \)
Taking both \( \left(1+a^{2}+b^{2}\right) \) outside the determinant from \( 1^{\text {st }} \) and \( 2^{\text {nd }} \) row
\(\therefore \text { LHS }=(1+\mathrm{a}^{2}+\mathrm{b}^ {2})^{2}\left|\begin{array}{ccc}
1 & 0 & -b \\
0 & 1 & a \\
2 b & -2 a & 1-a^{2}-b^{2}
\end{array}\right|\)
Expanding the determinant along \( 1^{\text {st }} \) row
\( \therefore \) LHS \( =\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}\right)^{2}\left[\left|\begin{array}{cc}1 & a \\ -2 a & 1-a^{2}-b^{2}\end{array}\right|-b\left| \begin{array}{cc}0 & 1 \\ 2 b & -2 a\end{array} \right\rvert\,\right] \)
\( \therefore \text { LHS }=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}\right)^{2}\left[1+\mathrm{a}^{2}-\mathrm{b}^{2}-\left(-2 \mathrm{~b}^{2}\right)\right]\)
\(\therefore \text { LHS }=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}\right)^{2}\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}\right)\)
\(\therefore \text { LHS }=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}\right)^{3}=\mathrm{RHS} \)
\( \therefore\left|\begin{array}{ccc}1+a^{2}-b^{2} & 2 a b & -2 b \\ 2 a b & 1-a^{2}+b^{2} & 2 a \\ 2 b & -2 a & 1-a^{2}-b^{2}\end{array}\right| \)

Taking out \( \mathrm{a}, \mathrm{b} \) and c from the determinant from \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row respectively.
\( \therefore \) LHS \( =\mathrm{abc}\left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ a & b+\frac{1}{b} & c \\ a & b & c+\frac{1}{c}\end{array}\right| \)
\( R_{2} \rightarrow R_{2}-R_{1} \) (i.e. replace \( 2^{\text {nd }} \) row by subtraction of \( 2^{\text {nd }} \) and \( 1^{\text {st }} \) row)
\( R_{3} \rightarrow R_{3}-R_{1} \) (i.e. replace \( 3^{\text {rd }} \) row by subtraction of \( 3^{\text {rd }} \) and \( 1^{\text {st }} \) row)
\( \therefore \) LHS \( = \) abc \( \left|\begin{array}{ccc}a+\frac{1}{a} & b & c \\ -\frac{1}{a} & \frac{1}{b} & 0 \\ -\frac{1}{a} & 0 & \frac{1}{c}\end{array}\right| \)
\( \mathrm{C}_{1} \rightarrow \mathrm{aC}_{1} \) (i.e. replace \( 1^{\text {st }} \) column by 'a' times \( 1^{\text {st }} \) column)
\( \mathrm{C}_{2} \rightarrow \mathrm{bC}_{2} \) (i.e. replace \( 2^{\text {nd }} \) column by 'b' times \( 2^{\text {nd }} \) column)
\( \rightarrow \mathrm{cC}_{3} \) (i.e. replace \( 3{ }^{\text {rd }} \) column by ' c ' times \( 3{ }^{\text {rd }} \) column)
As we are multiplying by \( \mathrm{a}, \mathrm{b} \) and c we should also divide by \( \mathrm{a}, \mathrm{b} \) and \( c \) to keep the original determinant value unchanged.
\(\therefore \text { LHS }=\frac{a b c}{a b c}\left|\begin{array}{ccc}
a^{2}+1 & b^{2} & c^{2} \\
-1 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right|\)
\( \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3} \) (i.e. replace \( 1^{\text {st }} \) column by addition of \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) column)
\( \therefore \) LHS \( =\left|\begin{array}{ccc}1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right| \)
Expanding determinant along 1st column
\(\begin{array}{l}
\therefore \text { LHS }=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)\left|\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right| \\
\therefore \text { LHS }=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)=\text { RHS } \\
\therefore\left|\begin{array}{ccc}
a^{2}+1 & a b & a c \\
a b & b^{2}+1 & b c \\
c a & c b & c^{2}+1
\end{array}\right|=\left(1+\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}\right)
\end{array}\)

Let \( A \) be any \( 3 \times 3 \) matrix
\(\begin{array}{l}
\mathrm{A}=\left[\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right] \\
\therefore|\mathrm{A}|=\left|\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right|
\end{array}\)
\(\begin{array}{l}
\therefore \mathrm{KA}=\left[\begin{array}{lll}
k a & k b & k c \\
k d & k e & k f \\
k g & k h & k i
\end{array}\right] \\
\therefore|\mathrm{KA}|=\left|\begin{array}{lll}
k a & k b & k c \\
k d & k e & k f \\
k g & k h & k i
\end{array}\right|
\end{array}\)
Taking out k from the determinant from \( 1^{\text {st }}, 2^{\text {nd }} \) and \( 3^{\text {rd }} \) row
\(\begin{array}{l}
\therefore|\mathrm{kA}|=\mathrm{k}^{3}\left|\begin{array}{lll}
a & b & c \\
d & e & f \\
g & h & i
\end{array}\right| \\
\therefore|\mathrm{kA}|=\mathrm{k}^{3}|\mathrm{~A}|
\end{array}\)
Therefore, answer is option (C) \( \mathrm{k}^{3}|\mathrm{~A}| \)

Determinant is an operation which we perform on arranged numbers. A square matrix is set of arranged numbers. We perform some operations on a matrix and we get a value that value is called as determinant of that matrix hence determinant is a number associated to square matrix.