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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.

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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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