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Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants – Exercise 4.2 | Set 1

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  • Last Updated : 05 Apr, 2021
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Using the properties of determinants and without expanding in Exercise 1 to 7, prove that:

Question 1. \begin{vmatrix}x & a & x+a\\y & b & y+b\\z &c & z+c\end{vmatrix}=0

Solution:

L.H.S.=\begin{vmatrix}x & a & x+a\\y & b & y+b\\z &c & z+c\end{vmatrix}

                C1→C1+C2

=\begin{vmatrix}x+a & a & x+a\\y+b & b & y+b\\z+c &c & z+c\end{vmatrix}

According to Properties of Determinant   

=0                                [∵  C1 & C3 are identical]

Now, L.H.S.=R.H.S.

Hence Proved

Question 2. \begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}=0

Solution:

L.H.S.=\begin{vmatrix}a-b & b-c & c-a\\b-c & c-a & a-b\\c-a & a-b & b-c\end{vmatrix}

        \begin{array}{l} \text { } \qquad\mathrm{C}{1} \rightarrow \mathrm{C}{1}+\mathrm{C}{2}+\mathrm{C}{3} \\ =\left|\begin{array}{lll} a-b+b-c+c-a & b-c & c-a \\ b-c+c-a+a-b & c-a & a-b \\ c-a+a-b+b-c & a-b & b-c \end{array}\right| \\ =\left|\begin{array}{lll} 0 & b-c & c -a \\ 0 & c-a & a-b \\ 0 & a-b & b-c \end{array}\right|=0 \end{array}

=0                        [∵ Every element of C1 are 0]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 3. \begin{vmatrix}2 & 7 & 65\\3 & 8 & 75\\5 & 9 & 86\end{vmatrix}=0

Solution:

L.H.S.=\begin{vmatrix}2 & 7 & 65\\3 & 8 & 75\\5 & 9 & 86\end{vmatrix}

            C3→C3-C1 

        =\begin{vmatrix}2 & 7 & 63\\3 & 8 & 72\\5 & 9 & 81\end{vmatrix}

      =2\begin{vmatrix}2 & 7 & 7\\3 & 8 & 8\\5& 9 & 9\end{vmatrix}

=9 ×0=0          [∵C2 & C3  are identical]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 4. \begin{vmatrix}1 & bc & a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}=0

Solution:

L.H.S.=\begin{vmatrix}1 & bc & a(b+c)\\1 & ca & b(c+a)\\1 & ab & c(a+b)\end{vmatrix}

        =\text { } \mathrm{C}{3} \rightarrow \mathrm{C}{3}+\mathrm{C}_{2}\\\\ \text {= }\begin{vmatrix}1 & bc & ab+ac\\1 & ca & bc+ba\\1 & ab & ca+cb\end{vmatrix}\\\\ \text {= }\begin{vmatrix}1 & bc & ab+ac+bc\\1 & ca & bc+ba+ca\\1 & ab & ca+cb+ab\end{vmatrix}\\\\ \text {= }(ab+ac+bc)\begin{vmatrix}1 & bc & 1\\1 & ca &1\\1 & ab & 1\end{vmatrix}\\\\ \text {= }(ab+ac+bc)0\\\\ \text {= 0 }\qquad\qquad[∵Two \:columns\: are\: identical]

Now,  L.H.S.=R.H.S.

Hence Proved

Question 5. \begin{vmatrix}b+c & q+r & y+z\\c+a & r+p & z+x\\a+b & p+q & x+y\end{vmatrix}=2\begin{vmatrix}a& p & x\\b & q & y\\c & r & z\end{vmatrix}

Solution:

L.H.S.=\begin{vmatrix}b+c & q+r & y+z\\c+a & r+p & z+x\\a+b & p+q & x+y\end{vmatrix}

\begin{array}{l} \text { } \qquad\mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\ =\left|\begin{array}{lll} b+c+c+a+a+b & q+r+r+p+p+q & y+z+z+x+x+y \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|\\\\ =\left|\begin{array}{lll} 2(a+b+c) & 2(p+q+r) & 2(x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \\\\ =\left|\begin{array}{lll} (a+b+c) & (p+q+r) & (x+y+z) \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right| \end{array}

\begin{aligned} &\text { } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}-\mathrm{R}_{2}\\ &=2\left|\begin{array}{lcc} b & q & y \\ c+a & r+p & z+x \\ a+b & p+q & x+y \end{array}\right|\\ &\text { } \mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\\ &=2\left|\begin{array}{ccc} b & q & y \\ c+a & r+p & z+x \\ a & p & x \end{array}\right|\\ &\text { } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=2\left|\begin{array}{lll} b & q & y \\ c & r & z \\ a & p & x \end{array}\right|\\ &\text {Interchange } \mathrm{R}_{2} \text { and } \mathrm{R}_{3} \text { }\\ &=-2\left|\begin{array}{ccc} b & q & y \\ a & p & x \\ c & r & z \end{array}\right| \end{aligned}

=-(-2)\begin{vmatrix}a & p & q\\b & q & y\\c & r & z\end{vmatrix}\\\\ =2\begin{vmatrix}a & p & q\\b & q & y\\c & r & z\end{vmatrix}\\\\

Now, L.H.S.=R.H.S.

Hence Proved

Question 6. \begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}=0

Solution:

Let Δ=\begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}

Taking (-1) common from every row

  Δ=(-1)3 \begin{vmatrix}0& -a & b\\a & 0 & c\\-b & -c & 0\end{vmatrix}

Interchange rows and columns

Δ=-\begin{vmatrix}0& a & -b\\-a & 0 & -c\\b & c & 0\end{vmatrix}

Now, Δ=-Δ

       Î”+Δ=0

       2Δ=0

        Δ=0

Now,  L.H.S.=R.H.S.

Hence Proved

Question 7. \begin{vmatrix}-a^{2}& ab& ac\\ba & -b^{2} & bc\\ca & cb & -c^{2}\end{vmatrix}=4a^{2}b^{2}c^{2}

Solution:

L.H.S.=\begin{vmatrix}-a^{2}& ab& ac\\ba & -b^{2} & bc\\ca & cb & -c^{2}\end{vmatrix}

Taking common a from Row 1,

                          b from Row 2,

                          c from Row 3, we have

\begin{array}{l} =a b c\left|\begin{array}{ccc} -a & b & c \\ a & -b & b \\ a & b & -c \end{array}\right| \\ \\\text { } R_{1} \rightarrow R_{1}+R_{2} \\\\ =a b c\left|\begin{array}{ccc} 0 & 0 & 2 c \\ a & -b & c \\ a & b & -c \end{array}\right| \\\\ =a b c \cdot 2 c\left|\begin{array}{cc} a & -b \\ a & b \end{array}\right| \\\\ =a b c \cdot 2 c(a b+a b) \\\\ =a b c 2 c .2 a b=4 a^{2} b^{2} c^{2} \end{array}

Now,  L.H.S.=R.H.S.

Hence Proved

By using properties of determinants, in Exercises 8 to 14, show that:

Question 8(i). \begin{vmatrix}1 & a & a^{2}\\1 & b & b^{2}\\1 & c & c^{2}\end{vmatrix}=(a-b)(b-c)(c-a)

(ii)\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^{3} &b^{3} & c^{3}\end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)

Solution:

(i) L.H.S.=\begin{vmatrix}1 & a & a^{2}\\1 & b & b^{2}\\1 & c & c^{2}\end{vmatrix}

\begin{aligned} &R _{2} \rightarrow R _{2}- R _{1} \text { and } R _{3} \rightarrow R _{3}- R _{1}\\ &=\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & b^{2}-a^{2} \\ 1 & c-a & c^{2}-a^{2} \end{array}\right|\\ &\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & b-a & (b-a)(b+a) \\ 0 & c-a & (c-a)(c+a) \end{array}\right|\\ &\text { Takina (b-a) and (c-a) common from } R _{2} \text { and } R _{3} \text { respective }\\ &=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & 1 & (b+a) \\ 0 & 1 & (c+a) \end{array}\right|\\ &R_{2} \rightarrow R_{2}-R_{3}\\ &=(b-a)(c-a)\left|\begin{array}{ccc} 1 & a & a^{2} \\ 0 & 0 & (b-c) \\ 0 & 1 & (c+a) \end{array}\right| \end{aligned}

\begin{aligned} &\text { Expanding along } 1^{\text {st }} \text { column }\\ &=(b-a)(c-a) \begin{array}{ll} =\left|\begin{array}{lll} 0 & (b+a) \\ 1 & (c+a) \end{array}\right| \end{array}\\ &=(b-a)(c-a)(0-(b-c))\\ &=(b-a)(c-a)(c-b)\\ &=(a-b)(b-c)(c-a)\text { }\\ &=0 \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

(ii)  L.H.S.=\begin{vmatrix}1 & 1 & 1\\a & b & c\\a^{3} &b^{3} & c^{3}\end{vmatrix}

\begin{aligned} &\text { operating } C _{2} \rightarrow C _{2}- C _{1} \text { and } C _{3} \rightarrow C _{3}- C _{3}\\ &=\left|\begin{array}{ccc} 1 & 0 & 0 \\ a & b-a & c-a \\ a^{3} & b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right|\\ &=1\left|\begin{array}{cc} b-a & c-a \\ b^{3}-a^{3} & c^{3}-a^{3} \end{array}\right|\\ &=1\left|\begin{array}{cc} b-a & c-a \\ (b-a)\left(b^{2}+a^{2}+a b\right) & (c-a)\left(c^{2}+a^{2}+a c\right) \end{array}\right|\\ &=(b-a)(c-a) \mid\left(b^{2}+a^{2}+a b\right) \quad\left(c^{2}+a^{2}+a c\right)\\ &=(b-a)(c-a)\left(c^{2}+a^{2}+a c-b^{2}-a^{2}-a b\right)\\ &=(b-a)(c-a)\left(c^{2}-b^{2}+a c-a b\right)\\ &=(b-a)(c-a)[(c-b)(c+b)+a(c-b)]\\ &=(b-a)(c-a)(c-b)(c+b+a)\\ &=-(a-b)(c-a)[-(b-c)(c+b+a)]\\ &=(a-b)(b-c)(c-a)(a+b+c) \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

Question 9. \begin{vmatrix}x & x^{2} & yz\\y & y^{2} & zx\\z & z^{2} & xy\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)

Solution:

L.H.S.=\begin{vmatrix}x & x^{2} & yz\\y & y^{2} & zx\\z & z^{2} & xy\end{vmatrix}

R_{1} \rightarrow R_{1}-R_{2}

=\left|\begin{array}{ccc} (x-y) & (x+y)(x-y) & z(y-x) \\ y^{2} & y^{2} & z x \\ z &z^{2}& x y \end{array}\right|

\begin{aligned} &=(x-y)\left|\begin{array}{ccc} 1 & (x+y) & -2 \\ y^{2} & y^{2} x & 2 x \\ z & z^{2} & x y \end{array}\right|\\ &\begin{array}{l} \qquad R_{2} \rightarrow R_{2}-R_{3} \\ =(x-y)\left|\begin{array}{lll} 1 & (x+y) & -2 \\ y-2 & y^{2}-2^{2} & 2 x-x y \\ 2 & 2^{2} & x y \end{array}\right| \end{array}\\ &=(x-y)(y-2)\left|\begin{array}{ccc} 1 & (x+y) & -2 \\ 1 & (y+2) & 2 x-x y \\ 2 & z^{2} & x y \end{array}\right|\\\\ &=(x-y)(y-2)\left|\begin{array}{ccc} 1 & (x+y) & -z \\ 1 & (y+z) & -x \\ z & \left(z^{2}\right) & x y \end{array}\right|\\\\ &\begin{array}{l} R_{1} \rightarrow R_{1}-R_{2} \\ \end{array} \end{aligned}

=(x-y)(y-z)\left|\begin{array}{ccc} 0 & x-2 & -z+x \\ 1 & y+2 & -x \\ z & z^{2} & x y \end{array}\right|

=(x-y)(y-z)(z-x)\left|\begin{array}{ccc} 0 & -1 & -1 \\ 1 & y+z & -x \\ z & z^{2} & x y \end{array}\right|

=(x-y)(y-z)(z-x)\left|\begin{array}{ccc} 0 & 0 & -1 \\ 1 & x+y+z & -x \\ z & z^{2}-xy & x y \end{array}\right|\\\\ =(x-y)(y-z)(z-x)[-1(z^{2}-xy-zx-zy-z^{2}]\\\\ =(x-y)(y-z)(z-x)(xy+zx+zy)

Now, L.H.S.=R.H.S.

Hence Proved

Question 10.(i)\begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}=(5x+4)(4-x)^{2}

(ii)\begin{vmatrix}y+k & y & y\\y & y+k & y\\y & y & y+k\end{vmatrix}=k^{2}(3y+k)

Solution:

(i) L.H.S.=\begin{vmatrix}x+4& 2x & 2x\\2x & x+4 & 2x\\2x & 2x & x+4\end{vmatrix}

\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\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|\\ &\text { Taking } (5 x+4) \text { common from } R_{1}\\ &=(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|.\\ &\text {} C _{2} \rightarrow C _{2}- C _{1} \text { and } \left. C _{3} \rightarrow C _{3}- C _{1}\right]\\ &=(5 x+4)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 x & 4-x & 0 \\ 2 x & 0 & 4-x \end{array}\right|\\ &=(5 x+4) \cdot 1\left|\begin{array}{cc} 4-x & 0 \\ 0 & 4-x \end{array}\right|\\ &=(5 x+4)(4- x^{2}) \end{aligned}

Now,  L.H.S.=R.H.S.

Hence Proved

(ii) L.H.S.=\begin{vmatrix}y+k & y & y\\y & y+k & y\\y & y & y+k\end{vmatrix}

\begin{aligned} & C _{1} \rightarrow C _{1}+ C _{2}+ C _{3}\\ &=\left|\begin{array}{ccc} 3 y+k & y & y \\ 3 y+k & y+k & y \\ 3 y+k & y & y+k \end{array}\right|\\ &\text { Taking } 3 y+k \text { common from } C_{1}\\ &=(3 y+k)\left|\begin{array}{ccc} 1 & y & y \\ 1 & y+k & y \\ 1 & y & y+k \end{array}\right|\\ &\text { } C _{2} \rightarrow C _{2}- C _{1} \text { and } C _{3} \rightarrow C _{3}-C_{1} \end{aligned}

\begin{array}{l} =(3 y+k)\left|\begin{array}{lll} 1 & y & y \\ 0 & k & 0 \\ 0 & 0 & k \end{array}\right| \\ =(3 y+k) \cdot 1\left|\begin{array}{ll} k & 0 \\ 0 & k \end{array}\right| \\ =(3 y+k) k^{2}=k^{2}(3 y+k) \end{array}

Now,  L.H.S.=R.H.S.

Hence Proved

 Chapter 4 Determinants – Exercise 4.2 | Set 2 


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