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  • Provide Solution For R. D. Sharma Maths Class 12 Chapter determinants Exercise 5.2 Question 25 Maths Textbook Solution.

Provide Solution For  R. D. Sharma Maths Class 12 Chapter  determinants Exercise 5.2 Question 25 Maths Textbook Solution.

Answers (1)

Answer:\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

Hint First we will try to convert some elements into zero

Given:\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

Solution:

\text { L.H.S }\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|

\begin{aligned} &\text { On applying } \mathrm{R}_{1} \rightarrow R_{1}+R_{2}+R_{3}\\ &=\left|\begin{array}{ccc} 3 x+4 & 3 x+4 & 3 x+4 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|\\ &\text { On taking } 3 x+4 \text { common from } \mathrm{R}_{1}\\ &=(3 x+4)\left|\begin{array}{ccc} 1 & 1 & 1 \\ x & x+4 & x \\ x & x & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}-\mathrm{C}_{2}, \mathrm{C}_{2} \rightarrow \mathrm{C}_{2}-C_{3} \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ x-x-4 & x+4-x & x \\ x-x & x-x-4 & x+4 \end{array}\right| \\ &=(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ -4 & 4 & x \\ 0 & -4 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On taking }(-4) \text { common from } \mathrm{C}_{1} \text { and }(-4) \text { from } \mathrm{C}_{2}\\ &=(3 x+4)(-4)(-4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right|\\ &=16(3 x+4)\left|\begin{array}{ccc} 0 & 0 & 1 \\ 1 & -1 & x \\ 0 & 1 & x+4 \end{array}\right| \end{aligned}

\begin{aligned} &\text { On expanding w.r.t } \mathrm{R}_{1}\\ &=16(3 x+4)\left[0\left|\begin{array}{cc} -1 & x \\ 1 & x+4 \end{array}\right|-0\left|\begin{array}{cc} 1 & x \\ 0 & x+4 \end{array}\right|+1\left|\begin{array}{cc} 1 & -1 \\ 0 & 1 \end{array}\right|\right]\\ &=16(3 x+4)[0-0+1(1-0)]\\ &=16(3 x+4) \times 1\\ &=16(3 x+4)\\ &=R \cdot H \cdot S \end{aligned}

Hence it is proved that

\left|\begin{array}{ccc} x+4 & x & x \\ x & x+4 & x \\ x & x & x+4 \end{array}\right|=16(3 x+4)

 

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