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Class 12 NCERT Solutions- Mathematics Part I – Chapter 4 Determinants- Exercise 4.2 | Set 2
  • Last Updated : 05 Apr, 2021

Chapter 4 Determinants- Exercise 4.2 | Set 1

Question 11. (i)\begin{array}{ccc} \left|\begin{array}{lll} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{array} \mid=(a+b+c)^{3}

(ii)\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}

Solution:

(i) L.H.S.=\begin{array}{ccc} \left|\begin{array}{lll} a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b \end{array}\right| \end{array} \mid

\begin{aligned} &R_{1} \rightarrow R_{1}+R_{2}+R_{3} \\ &=\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|\\ &\text { Taking } a+b+c \text { common from } R_{1}\\ &=(a+b+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|\\ &[C_{2} \rightarrow C_{2}-C_{1} \text { and } C_{3} \rightarrow C_{3}-C_{1} \end{aligned}

\begin{array}{l} =(a+b+c)\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 b & -b-c-a & 0 \\ 2 c & 0 & -c-a-b \end{array}\right| \\ =(a+b+c) \cdot 1  \begin{array}{cc} =\left|\begin{array}{lll} -b-c-a & 0 \\ 0 & -c-a-b \mid \end{array}\right| \end{array} \\ =(a+b+c)\{-(b+c+a)\}\{-(c+a+b)\} \\ =  |a+b+c|^{3} \text {  } \end{array}

∵ L.H.S.=R.H.S



Hence proved

(ii) \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|

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

\begin{array}{l} =2(x+y+z)\left|\begin{array}{ccc} 1 & x & y \\ 0 & x+y+z & 0 \\ 0 & 0 & x+y+z \end{array}\right| \\ =2(x+y+z) \cdot 1 \cdot\left|\begin{array}{cc} x+y+z & 0 \\ 0 & x+y+z \end{array}\right| \\ =2(x+y+z)\left[(x+y+z)^{2}-0\right] \\ =2(x+y+z)^{3} \text {  } \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 12. \begin{aligned} &\text {  }\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ \end{aligned}

Solution:

\begin{aligned} &\text { L.H.S. }=\left|\begin{array}{lll} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ &R_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 1+x+x^{2} & 1+x+x^{2} & 1+x+x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right| \\ &\text { Taking } 1+x+x^{2} \text { common from } R_{1}\\ &=\left(1+x+x^{2}\right)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{2} & 1 \end{array}\right|\\ &\left[C_{2} \rightarrow C_{2}-C_{1}\right. \& C_{3} \rightarrow C_{3}-C_{1}\\ &=\left|1+x+x^{2}\right| \begin{array}{ccc} =\left|\begin{array}{lll} 1 & 0 & 0 \\ x^{2} & 1-x^{2} & x-x^{2} \\ x & x-x & 1-x \end{array}\right| \end{array} \mid \end{aligned}



\begin{array}{l} =\left(1+x+x^{2}\right) \cdot 1 \cdot\left|\begin{array}{ll} 1-x^{2} & x-x^{2} \\ x^{2}-x & 1-x \end{array}\right| \\ =\left(1+x+x^{2}\right) \cdot 1 \cdot \mid \begin{array}{ll} (1-x)(1+x) & x(1-x) \\ -x(1-x) & 1-x \end{array} \\ =\left(1+x+x^{2}\right)\left[(1-x)^{2}(1+x)+x^{2}(1-x)^{2}\right] \\ =\left(1+x+x^{2}\right)(1-x)^{2}\left(1+x+x^{2}\right) \\ =\left(1+x+x^{2}\right)^{2}(1-x)^{2} \\ =\left[\left(1+x+x^{2}\right)(1-x)\right]^{2} \\ =\left(1-x+x-x^{2}+x^{2}-x^{2}\right)^{2} \\ =\left(1-x^{3}\right)^{2} \quad \text {  } \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 13. \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|=\left(1+a^{2}+b^{2}\right)^{3}

Solution:

L.H.S.= \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|

\begin{array}{l} C _{1} \rightarrow C _{1}-b C _{3} \text { and } \left. C _{2} \rightarrow C _{2}+ aC _{3}\right] \\ =\begin{array}{ccc} \left|\begin{array}{lll} 1+a^{2}+b^{2} & 0 & -2 b \\ 0 & 1+a^{2}+b^{2} & 2 a \\ b\left(1+a^{2}+b^{2}\right) & -a\left(1+a^{2}+b^{2}\right) & 1-a^{2}-b^{2} \end{array}\right| \end{array}  \\ =\left(1+a^{2}+b^{2}\right)\left|\begin{array}{ccc} 1 & 0 & -2 b \\ 0 & 1 & 2 a \\ b & -a & 1-a^{2}-b^{2} \end{array}\right| \\ {\left[ R _{1} \rightarrow R _{1}-bR_{1} \right]} \end{array}

\begin{array}{l} =\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{ccc} 1 & 0 & -2 b \\ 0 & 1 & 2 a \\ 0 & -a & 1-a^{2}+b^{2} \end{array}\right| \\ =\left(1+a^{2}+b^{2}\right)^{2}\left|\begin{array}{cc} 1 & 2 a \\ -a & 1-a^{2}+b^{2} \end{array}\right| \\ =\left(1+a^{2}+b^{2}\right)^{2}\left(1-a^{2}+b^{2}+2 a^{2}\right) \\ =\left(1+a^{2}-b^{2}\right)^{3} \end{array}

∵ L.H.S.=R.H.S

Hence proved

Question 14. \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|=1+a^{2}+b^{2}+c^{2}

Solution:

L.H.S= \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|

Multiplying column1,column2,column3  by  a,b,c respectively and then dividing the determinant by  a,b,c .

\begin{aligned} &=\frac{1}{a b c}\left|\begin{array}{ccc} a\left(a^{2}+1\right) & a b^{2} & a c^{2} \\ a^{2} b & b\left(b^{2}+1\right) & b c^{2} \\ a^{2} c & b^{2} c & c\left(c^{2}+1\right) \end{array}\right|\\ &\text { Taking a } b \text { and } c \text { common from } R_{1}, R_{2} \text { and } R_{3} \text { respectively }\\ &=\frac{a b c}{a b c}  \begin{array}{ccc} =\left|\begin{array}{lll} a^{2}+1 & b^{2} & c^{2} \\ a^{2} & b^{2}+1 & c^{2} \\ a^{2} & b^{2} & c^{2}+1 \end{array}\right| \end{array}\\ &\left[ C _{1} \rightarrow C _{1}- C _{2}+ C _{3}\right]\\ &=\frac{a b c}{a b c}\left|\begin{array}{ccc} 1+a^{2}+b^{2}+c^{2} & b^{2} & c^{2} \\ 1+a^{2}+b^{2}+c^{2} & b^{2}-1 & c^{2} \\ 1-a^{2}+b^{2}-c^{2} & b^{2} & c^{2}+1 \end{array}\right| \end{aligned}

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

∵ L.H.S.=R.H.S

Hence proved

Choose the correct answer in Exercises 15 and 16.

Question 15. Let A be a square matrix of order 3 x 3, then|A| is equal to:

(A) k|A|

(B) k2|A|

(C) k3|A|

(D) 3k|A|

Solution:

 Let A =\begin{bmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{bmatrix}      be a square matrix of order 3 x 3   ….(1)

Now, kA=\begin{bmatrix}k1 & k2 & k3\\k4 & k5 & k6\\k7 & k8 & k9\end{bmatrix}

⇒|kA|=\begin{vmatrix}k1 & k2 & k3\\k4 & k5 & k6\\k7 & k8 & k9\end{vmatrix}

⇒|kA|=k3 \begin{vmatrix}1 & 2 & 3\\4 & 5 & 6\\7 & 8 & 9\end{vmatrix}

∴ |kA|= k3|A|            [ from eqn .(1) ]

∴ option (C) is correct.

Question 16.Which of the following is correct

(A) Determinant is a square matrix.

(B) Determinant is a number associated to a matrix.

(C) Determinant is a number associated to a square matrix.

(D) None of these

Solution:

Since, Determinant is a number which is always associated to a square matrix.

∴ option (C) is correct.

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