Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (v) maths textbook solution

Please solve RD Sharma class 12 chapter Solution of Simultaneous Linear Equation exercise 7.1 question 8 subquestion (v) maths textbook solution

Answers (1)

Answer:

\begin{aligned} x=2, y=-1, z=4 \end{aligned}

Given:

\begin{aligned} &A=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \quad, \quad B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ &y+2 z=7 \\ &x-y=3 \\ &2 x+3 y+11 z=17 \end{aligned}

Hint:

X=A-1B is used to solve this problem. First we find the determinant and co-factor of matrix A, take it’s transpose, and that will be Adj A using Adj A calculate A-1.

Solution:

        \begin{aligned} A &=\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right], B=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right] \\ B A &=\left[\begin{array}{ccc} 1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{array}\right]\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &=\left[\begin{array}{ccc} 2+4+0 & 2-2+0 & -4+4+0 \\ 4-12+8 & 4+6-4 & -8-12+20 \\ 0-4+4 & 0+2-2 & 0-4+10 \end{array}\right] \end{aligned}

        \begin{aligned} &=\left[\begin{array}{lll} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{array}\right] \\ &B A=6\left[\begin{array}{lll} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] \\ &B A=6 I_{3} \\ &B\left[\frac{1}{6} A\right]=I_{3} \end{aligned}

        \begin{aligned} &{\left[\frac{1}{11} B\right] A=I_{3}} \\ &B^{-1}=\frac{1}{6} A \\ &\quad=\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right] \\ &B X=C \\ &X=B^{-1} C \end{aligned}

        \begin{gathered} =\frac{1}{6}\left[\begin{array}{ccc} 2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5 \end{array}\right]\left[\begin{array}{c} 3 \\ 17 \\ 7 \end{array}\right] \\ =\frac{1}{6}\left[\begin{array}{c} 6+34-28 \\ -12+34-28 \\ 6-17+35 \end{array}\right] \end{gathered}

        \begin{aligned} &{\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\frac{1}{6}\left[\begin{array}{l} 12 \\ -6 \\ 24 \end{array}\right]} \\ &x=2, y=-1, z=4 \end{aligned}

Posted by

Gurleen Kaur

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads

Similar Questions

Trending Articles/News

anam-khan

PM Modi's Pariksha Pe Charcha 2025 to be held next month, registrations open till January 14
anam-khan

CBSE Board Exams 2025: Online portal for availing exemptions by CWSN students opens
anam-khan

Board Exams: States agree to equivalence; no question paper ‘jumbling’ from next year, says PARAKH CEO

Latest Question