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  • Please Solve R.D. Sharma class 12 Chapter determinants Exercise 5.2 Question 17 Maths textbook Solution.

Please Solve R.D. Sharma class 12 Chapter determinants Exercise  5.2  Question  17 Maths textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}

Hint We will try to make some elements of the determinant zero

Given :\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}

Solution:

\text { L.H.S }\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow \mathrm{R}_{1}+\mathrm{R}_{2}+\mathrm{R}_{3} \\ &=\left|\begin{array}{ccc} a+a+2 b+a+b & a+b+a+a+2 b & a+2 b+a+b+a \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right| \\ &=\left|\begin{array}{ccc} 3 a+3 b & 3 a+3 b & 3 a+3 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right| \end{aligned}

\begin{aligned} &\text { Taking common }(3 a+3 b) \text { from } R_{1}\\ &=(3 a+3 b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|\\ &\text { On applying } \mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{3}\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b-(a+b) & a-(a+2 b) & a+b-a \\ a+b & a+2 b & a \end{array}\right|\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a+2 b-a-b & a-a-2 b & b \\ a+b & a+2 b & a \end{array}\right|\\ &=3(a+b)\left|\begin{array}{ccc} 1 & 1 & 1 \\ b & -2 b & b \\ a+b & a+2 b & a \end{array}\right| \end{aligned}

\begin{aligned} &\text { Taking common (b) from } R_{2}\\ &=3(a+b) b\left|\begin{array}{ccc} 1 & 1 & 1 \\ 1 & -2 & 1 \\ a+b & a+2 b & a \end{array}\right|\\ &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-\mathrm{C}_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-\mathrm{C}_{3}\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1-(-2) & -2-1 & 1 \\ a+b-(a+2 b) & a+2 b-a & a \end{array}\right|\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1+2 & -2-1 & 1 \\ a+b-a-2 b & a+2 b-a & a \end{array}\right|\\ &=3(a+b) b\left|\begin{array}{ccc} 0 & 0 & 1 \\ 3 & -3 & 1 \\ -b & 2 b & a \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } R_{1}\\ &=3(a+b) b\left[0\left|\begin{array}{cc} -3 & 1 \\ 2 b & a \end{array}\right|-0\left|\begin{array}{ll} 3 & 1 \\ -b & a \end{array}\right|+1\left|\begin{array}{cc} 3 & -3 \\ -b & 2 b \end{array}\right|\right]\\ &=3(a+b) b[0-0+1\{3 \times 2 b-(-3)(-b)\}]\\ &=3(a+b) b[1(6 b-3 b)]\\ &=3(a+b) b \times 3 b\\ &=9(a+b) b^{2} \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} a & a+b & a+2 b \\ a+2 b & a & a+b \\ a+b & a+2 b & a \end{array}\right|=9(a+b) b^{2}

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