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Matrices & Determinants: Let Pmatrix Pmatrix Value Equals

JEE Maths question with a full step-by-step solution.

Question
Let A=(ℓmnpqr111)A = \begin{pmatrix}\ell & m & n\\ p & q & r\\ 1 & 1 & 1\end{pmatrix} and B=A2B = A^2. If (ℓ−m)2+(p−q)2=9(\ell-m)^2+(p-q)^2 = 9, (m−n)2+(q−r)2=16(m-n)^2+(q-r)^2 = 16, (n−ℓ)2+(r−p)2=25(n-\ell)^2+(r-p)^2 = 25, then the value of det⁡(B)\det(B) equals:
Question diagram
A3636
B100100
C144144correct
D169169
Solution
Step 1: Interpret the given conditions geometrically The three quantities represent the squared distances between points P1(ℓ,p)P_1(\ell,p), P2(m,q)P_2(m,q), P3(n,r)P_3(n,r):
∣P1P2∣2=9,∣P2P3∣2=16,∣P3P1∣2=25|P_1P_2|^2 = 9, \quad |P_2P_3|^2 = 16, \quad |P_3P_1|^2 = 25
 Since 9+16=259+16=25, the triangle P1P2P3P_1P_2P_3 is a right triangle with legs 33 and 44. Step 2: Relate det⁡(A)\det(A) to the area of the triangle
Area=12×3×4=6\text{Area} = \frac{1}{2}\times3\times4 = 6
 Expanding det⁡(A)\det(A) along the third row gives det⁡(A)=ℓ(q−r)+m(r−p)+n(p−q)\det(A) = \ell(q-r)+m(r-p)+n(p-q), which equals twice the signed area of the triangle. Therefore ∣det⁡(A)∣=2×6=12|\det(A)| = 2\times 6 = 12. Step 3: Compute det⁡(B)\det(B)
det⁡(B)=det⁡(A2)=(det⁡A)2=122=144\det(B) = \det(A^2) = (\det A)^2 = 12^2 = 144
 Answer: (3)
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