#### Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 1 sub question 7 Maths textbook Solution.

Answer:$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|=512000$

Hint: First we will make a column with same elements

Given: $\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|$

Solution:$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|$

\begin{aligned} &\text { Applying } \mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}+\mathrm{C}_{3}+\mathrm{C}_{4}\\ &=\left|\begin{array}{cccc} 40 & 3 & 9 & 27 \\ 40 & 9 & 27 & 1 \\ 40 & 27 & 1 & 3 \\ 40 & 1 & 3 & 9 \end{array}\right|\\ &\text { Applying } \mathrm{R}_{1} \rightarrow R_{1}-R_{2} \text { and } \mathrm{R}_{2} \rightarrow R_{2}-R_{3} \text { and } \mathrm{R}_{3} \rightarrow R_{3}-R_{4}\\ &=\left|\begin{array}{cccc} 0 & -6 & -18 & 26 \\ 0 & -18 & 26 & -2 \\ 0 & 26 & -2 & -6 \\ 40 & 1 & 3 & 9 \end{array}\right| \end{aligned}

On changing all rows into columns

$=\left|\begin{array}{cccc} 0 & 0 & 0 & 40 \\ -6 & -18 & 26 & 1 \\ -18 & 26 & -2 & 3 \\ 26 & -2 & -6 & 9 \end{array}\right|$

If the rows and columns of a determinant are interchanged, the value of the determinant remains the same.

Expanding the determinant w.r.t $R_{1}$

\begin{aligned} &=0\left|\begin{array}{ccc} -18 & 26 & 1 \\ 26 & -2 & 3 \\ -2 & -6 & 9 \end{array}\right|-0\left|\begin{array}{ccc} -6 & 26 & 1 \\ -18 & -2 & 3 \\ 26 & -6 & 9 \end{array}\right|+0\left|\begin{array}{ccc} -6 & -18 & 1 \\ -18 & 26 & 3 \\ 26 & -2 & 9 \end{array}\right|-40\left|\begin{array}{ccc} -6 & -18 & 26 \\ -18 & 26 & -2 \\ 26 & -2 & -6 \end{array}\right| \\ &=0-0+0-40\left\{-6\left|\begin{array}{ll} 26 & -2 \\ -2 & -6 \end{array}\right|-(-18)\left|\begin{array}{cc} -18 & -2 \\ 26 & -6 \end{array}\right|+26\left|\begin{array}{cc} -18 & 26 \\ 26 & -2 \end{array}\right|\right\} \\ &=-40[-6\{26(-6)-(-2)(-2)\}+18\{(-18)(-6)-(-2)(26)\}+26\{(-18)(-2)-26 \times 26\}] \\ &=-40[-6(-156-4)+18(108+52)+26(36-676)] \\ &=-40[-6(-160)+18(160)+26(-640)] \\ &=-40(960+2880-16640) \\ &=-40(-12800) \\ &=512000 \end{aligned}

Hence,$\left|\begin{array}{cccc} 1 & 3 & 9 & 27 \\ 3 & 9 & 27 & 1 \\ 9 & 27 & 1 & 3 \\ 27 & 1 & 3 & 9 \end{array}\right|=512000$