The value of λ, for which the matrix product 2 0 70 1 01 −2 1−λ 14λ 7λ0 1 0λ −4λ −2λ is an identity matrix is

2070101−21−λ14λ7λ016λ−4λ−2λ=100010001 −2λ+0+7d28λ+0−28λ14λ+0−14λ0+0+00+1+00+0+0−λ+0+λ14λ−2−4λ7λ−0−2λ=1 0 00 1 00 0 1 5λ00010010λ−25λ=100010001 ∴ 5λ=1∴ λ=15

The value of λ, for which the matrix product 2 0 70 1 01 −2 1−λ 14λ 7λ0 1 0λ −4λ −2λ is an identity matrix is

2070101−21−λ14λ7λ016λ−4λ−2λ=100010001 −2λ+0+7d28λ+0−28λ14λ+0−14λ0+0+00+1+00+0+0−λ+0+λ14λ−2−4λ7λ−0−2λ=1 0 00 1 00 0 1 5λ00010010λ−25λ=100010001 ∴ 5λ=1∴ λ=15

AI Mentor

Check Your IQ

Free Expert Demo

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

$The value of λ, for which the matrix product$
$[\begin{array}{rrr} 2 0 7 \\ 0 1 0 \\ 1 - 2 1 \end{array}] [\begin{array}{rrr} - λ 14 λ 7 λ \\ 0 1 0 \\ λ - 4 λ - 2 λ \end{array}]$
$is an identity matrix is$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{1}{5}$

$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{4}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$[\begin{array}{ccc} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & - 2 & 1 \end{array}] [\begin{array}{ccc} - λ & 14 λ & 7 λ \\ 0 & 1 & 6 \\ λ & - 4 λ & - 2 λ \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] [\begin{array}{ccc} - 2 λ + 0 + 7 d & 28 λ + 0 - 28 λ & 14 λ + 0 - 14 λ \\ 0 + 0 + 0 & 0 + 1 + 0 & 0 + 0 + 0 \\ - λ + 0 + λ & 14 λ - 2 - 4 λ & 7 λ - 0 - 2 λ \end{array}] = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] [\begin{array}{ccc} 5 λ & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 10 λ - 2 & 5 λ \end{array}] = [\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] \begin{aligned} ∴ 5 λ = 1 \\ ∴ λ = \frac{1}{5} \end{aligned}$