Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • School
  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 53 math

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 53 math

Answers (1)

Answer:

A^{2}-5 A-14 I=\left[\begin{array}{cc}0 & 0 \\ 0 & 0\end{array}\right]=0\ \ and \ \ A^{3}=\left[\begin{array}{cc}187 & -195 \\ -156 & 148\end{array}\right]

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Given:

A=\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]

First, we solve this   A^{2}-5 A-14 I

Where I is an identity matrix

Then,

\\\\ \left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-5\left[\begin{array}{cc}3 & -5 \\ -4 & 2\end{array}\right]-14\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\\\\ \\=\left[\begin{array}{cc}9+20 & -15-10 \\ -12-8 & 20+4\end{array}\right]-\left[\begin{array}{cc}15 & -25 \\ -20 & 10\end{array}\right]-\left[\begin{array}{cc}14 & 0 \\ 0 & 14\end{array}\right]

\\\\ =\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]-\left[\begin{array}{cc} 15 & -25 \\ -20 & 10 \end{array}\right]-\left[\begin{array}{cc} 14 & 0 \\ 0 & 14 \end{array}\right]\\\\\\ =\left[\begin{array}{cc} 29-15-14 & -25+25-0 \\ -20+20-0 & 24-10-14 \end{array}\right]\\\\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]=0\\\\\\ A^{2}-5 A-14 I=0\\\\

Now,

\\\\ A^{2}-5 A-14 I=0\\\\ A^{2}=5 A+14 I\\\\ A^{3}=A^{2} A=(5 A+14 I) A\\\\ A^{3}=5 A^{2}+14 A\\\\ A^{3}=A^{2} A=5 A^{2}+14 A

As we know AI=A

\\\\ A^{3}=5\left[\begin{array}{cc} 29 & -25 \\ -20 & 24 \end{array}\right]+14\left[\begin{array}{cc} 3 & -5 \\ -4 & 2 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145 & -125 \\ -100 & 120 \end{array}\right]+\left[\begin{array}{cc} 42 & -70 \\ -56 & 28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 145+42 & -125-70 \\ -100-56 & 120+28 \end{array}\right]\\\\\\ A^{3}=\left[\begin{array}{cc} 187 & -195 \\ -156 & 148 \end{array}\right]

Posted by

infoexpert22

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

CBSE Class 12 result verification 2024 application out at cbse.gov.in; last date, fees
anam-khan

CBSE results 2024 later for 102 students over fake certificates; board issues show cause notice
anam-khan

What after CBSE Class 12 results 2024? From CUET UG to JEE Advanced, all you need to know

Latest Question