If beginbmatrix 1 & 1 \ 0 & 1 endbmatrix beginbmatrix 1 & 2 \ 0 & 1 endbmatrix beginbmatrix 1 & 3 \ 0 & 1 endbmatrix ·s beginbmatrix 1 & n-1 \ 0 & 1 endbmatrix = beginbmatrix 1 & 78 \ 0 & 1 endbmatrix, then the inverse of beginbmatrix 1 & n \ 0 & 1 endbmatrix is

beginbmatrix 1 & -13 \ 0 & 1 endbmatrix

beginbmatrix 1 & 0 \ 12 & 1 endbmatrix

beginbmatrix 1 & 0 \ 13 & 1 endbmatrix

beginbmatrix 1 & -12 \ 0 & 1 endbmatrix

If beginbmatrix 1 & 1 \ 0 & 1 endbmatrix beginbmatrix 1 & 2… — JEE Main 2019 Mathematics PYQ

JEE Main 2019 · 9 April · Shift I

\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & 3 \\ 0 & 1 \end{bmatrix} \cdots \begin{bmatrix} 1 & n-1 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 78 \\ 0 & 1 \end{bmatrix}

, then the inverse of

\begin{bmatrix} 1 & n \\ 0 & 1 \end{bmatrix}

A.
$\begin{bmatrix} 1 & 0 \\ 12 & 1 \end{bmatrix}$
B.
$\begin{bmatrix} 1 & -13 \\ 0 & 1 \end{bmatrix}$
C.
$\begin{bmatrix} 1 & 0 \\ 13 & 1 \end{bmatrix}$
D.
$\begin{bmatrix} 1 & -12 \\ 0 & 1 \end{bmatrix}$

Show answer and solution

Correct option: B

Solution

The product of upper triangular matrices [1,k;0,1] multiplies to give [1, 1+2+3+...+(n-1); 0, 1]. So 1+2+...+(n-1) = n(n-1)/2 = 78, giving n(n-1) = 156, so n = 13.

The inverse of [1, n; 0, 1] = [1, 13; 0, 1] is simply [1, -13; 0, 1], since for any upper triangular matrix of this form, [1,k;0,1]⁻¹ = [1,-k;0,1] (easily verified by multiplication giving the identity).

[JEE Main 2019 · 9 April · Shift I]

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