Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
0 votes
398 views
in Determinants by (40.7k points)
closed by

 आव्यूहों  \(\begin{bmatrix}1 & -1&1\\[0.3em]2&1&-3\\[0.3em]1&1&1 \\[0.3em] \end{bmatrix}\) का व्युत्क्रम आव्यूह ज्ञात कीजिए तथा इसकी सहायता से निम्नलिखित समीकरण निकाय को हल कीजिए।

\(\begin{bmatrix}1 &0&1\\[0.3em]2&1&0\\[0.3em]0&1&1 \\[0.3em] \end{bmatrix}\) \(\begin{bmatrix}x \\[0.3em]y\\[0.3em]z \\[0.3em] \end{bmatrix}\) = 1/2  \(\begin{bmatrix}2y \\[0.3em]6z\\[0.3em]-2x \\[0.3em] \end{bmatrix}\) +2 \(\begin{bmatrix}2 \\[0.3em]0\\[0.3em]1 \\[0.3em] \end{bmatrix}\) 

1 Answer

+1 vote
by (40.4k points)
selected by
 
Best answer

दिया गया आव्यह  \(\begin{bmatrix}1 & -1&1\\[0.3em]2&1&-3\\[0.3em]1&1&1 \\[0.3em] \end{bmatrix}\) 

A = \(\begin{bmatrix}1 & -1&1\\[0.3em]2&1&-3\\[0.3em]1&1&1 \\[0.3em] \end{bmatrix}\) 

 |A| = \(\begin{bmatrix}1 & -1&1\\[0.3em]2&1&-3\\[0.3em]1&1&1 \\[0.3em] \end{bmatrix}\) 

= 1(1 + 3) + 1(2 + 3) + 1(2 – 1)

= 4 + 5 + 1 = 10 ≠ 0

अतः A-1 का अस्तित्व है।

A के सहखण्ड ज्ञात करने पर,

Related questions

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

...