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  • Need Solution for R.D.Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 5 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter determinants  Exercise 5.2 Question 2 sub question 5 Maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|=\lambda^{2}(3 x+\lambda)

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

Given:\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|

Solution:\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|

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

\begin{aligned} &\text { On taking common }(3 x+\lambda) \text { from } \mathrm{R}_{1}\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow C_{1}-C_{2} \text { and } \mathrm{C}_{2} \rightarrow C_{2}-C_{3}\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-(x+\lambda) & (x+\lambda)-x & x \\ x-x & x-(x+\lambda) & x+\lambda \end{array}\right|\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-\lambda & x+\lambda-x & x \\ 0 & x-x-\lambda & x+\lambda \end{array}\right|\\ &=(3 x+\lambda)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -\lambda & \lambda & x \\ 0 & -\lambda & x+\lambda \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=(3 x+\lambda)\left[0\left|\begin{array}{cc} \lambda & x \\ -\lambda & x+\lambda \end{array}\right|-0\left|\begin{array}{cc} -\lambda & x \\ 0 & x+\lambda \end{array}\right|+1\left|\begin{array}{cc} -\lambda & \lambda \\ 0 & -\lambda \end{array}\right|\right]\\ &=(3 x+\lambda)[0-0+1[(-\lambda)(-\lambda)-0 \times \lambda]]\\ &=(3 x+\lambda)\left\{1\left(\lambda^{2}-0\right)\right\}\\ &=\lambda^{2}(3 x+\lambda) \end{aligned}

Hence\left|\begin{array}{ccc} x+\lambda & x & x \\ x & x+\lambda & x \\ x & x & x+\lambda \end{array}\right|=\lambda^{2}(3 x+\lambda)

 

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