Matrices & DeterminantshardFree

Matrices & Determinants — JEE Maths practice question

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

Question

Let

A = \begin{pmatrix}\ell & m & n\\ p & q & r\\ 1 & 1 & 1\end{pmatrix}

and

B = A^2

. If

(\ell-m)^2+(p-q)^2 = 9

(m-n)^2+(q-r)^2 = 16

(n-\ell)^2+(r-p)^2 = 25

, then the value of

\det(B)

equals:

36

100

144

correct

169

Solution

Step 1: Interpret the given conditions geometrically The three quantities represent the squared distances between points

P_1(\ell,p)

P_2(m,q)

P_3(n,r)

|P_1P_2|^2 = 9, \quad |P_2P_3|^2 = 16, \quad |P_3P_1|^2 = 25

Since

9+16=25

, the triangle

P_1P_2P_3

is a right triangle with legs

3

and

4

. Step 2: Relate

\det(A)

to the area of the triangle

\text{Area} = \frac{1}{2}\times3\times4 = 6

Expanding

\det(A)

along the third row gives

\det(A) = \ell(q-r)+m(r-p)+n(p-q)

, which equals twice the signed area of the triangle. Therefore

|\det(A)| = 2\times 6 = 12

. Step 3: Compute

\det(B)

\det(B) = \det(A^2) = (\det A)^2 = 12^2 = 144

Answer: (3)

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