#### Please Solve R. D. Sharma class 12 Chapter determinants Exercise 5.2 Question 1 sub question 5 Maths textbook Solution.

Answer:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|=-8$

Hint: We will expand it w.r.t $R_{1}$

Given:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$

Solution:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$

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

\begin{aligned} &=1\left|\begin{array}{cc} 9 & 16 \\ 16 & 25 \end{array}\right|-4\left|\begin{array}{ll} 4 & 16 \\ 9 & 25 \end{array}\right|+9\left|\begin{array}{cc} 4 & 9 \\ 9 & 16 \end{array}\right| \\ &=1(9 \times 25-16 \times 16)-4(4 \times 25-9 \times 16)+9(4 \times 16-9 \times 9) \\ &=(225-256)-4(100-144)+9(64-81) \\ &=-31+176-153 \\ &=176-184 \\ &=-8 \end{aligned}

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Answer:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|=-8$

Hint: We will expand it w.r.t $R_{1}$

Given:$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$

Solution: $\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|$

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

\begin{aligned} &=1\left|\begin{array}{cc} 9 & 16 \\ 16 & 25 \end{array}\right|-4\left|\begin{array}{ll} 4 & 16 \\ 9 & 25 \end{array}\right|+9\left|\begin{array}{cc} 4 & 9 \\ 9 & 16 \end{array}\right| \\ &=1(9 \times 25-16 \times 16)-4(4 \times 25-9 \times 16)+9(4 \times 16-9 \times 9) \\ &=(225-256)-4(100-144)+9(64-81) \\ &=-31+176-153 \\ &=176-184 \\ &=-8 \end{aligned}

Hence,$\left|\begin{array}{ccc} 1 & 4 & 9 \\ 4 & 9 & 16 \\ 9 & 16 & 25 \end{array}\right|=-8$