Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}, \mathrm{d}\) be in arithmetic progression with common difference \(\lambda\). If \(\left|\begin{array}{ccc}x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c\end{array}\right|=2\), then the value of \(\lambda^{2}\) is equal to........

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Determinants

Let

a, b, c, d

be in arithmetic progression with common difference

λ

. If

x + a - c x - 1 x - b + d x + b x + c x + d x + a x + b x + c = 2

, then the value of

λ^{2}

is equal to........

Answer: 1

With

b = a + λ

c = a + 2 λ

and

d = a + 3 λ

, the entries

a - c = - 2 λ

and

d - b = 2 λ

turn the determinant into

x - 2 λ x - 1 x + 2 λ x + a + λ x + a + 2 λ x + a + 3 λ x + a x + a + λ x + a + 2 λ

Down the second and third columns the entries are themselves in arithmetic progression, so the combination

R_{1} + R_{3} - 2 R_{2}

wipes both of them out and leaves

(2, 0, 0)

in the first column. Replacing

R_{1}

R_{1} + R_{3} - 2 R_{2}

does not change the value, and expanding along the new first row with

u = x + a + 2 λ

2 [u^{2} - (u - λ) (u + λ)] = 2 λ^{2}

Equating this to the given value

2

gives

λ^{2} = 1

JEE Main · 2021 · 20 Jul 2021 · Shift 1 · Integer answer