Math, asked by omkarchudhari2004, 3 months ago

11 0 0
Ex. (7) If A= 5 1 0
that BA = I.
1 3
1
then find matrix B such that AB = I. Verify​

Answers

Answered by yogitakarnani8401267
0

\textbf{Given:}Given:

\begin{gathered}\mathsf{A=\left(\begin{array}{ccc}1&0&0\\5&1&0\\1&3&1\end{array}\right)}\end{gathered}

A=

1

5

1

0

1

3

0

0

1

\textbf{To find:}To find:

\textsf{The matrix B}The matrix B

\textbf{Solution:}Solution:

\textsf{Consider,}Consider,

\mathsf{AB=I}AB=I

\implies\mathsf{B=A^{-1}}⟹B=A

−1

\begin{gathered}\mathsf{|A|=\left|\begin{array}{ccc}1&0&0\\5&1&0\\1&3&1\end{array}\right|}\end{gathered}

∣A∣=

1

5

1

0

1

3

0

0

1

\mathsf{=1(1-0)-0+0=1\neq\,0}=1(1−0)−0+0=1

=0

\mathsf{A^{-1}\;exists}A

−1

exists

\begin{gathered}\mathsf{adjA=\left(\begin{array}{ccc}(1-0)&-(5-0)&(15-1)\\-(0-0)&(1-0)&-(3-0)\\(0-0)&-(0-0)&(1-0)\end{array}\right)^T}\end{gathered}

adjA=

(1−0)

−(0−0)

(0−0)

−(5−0)

(1−0)

−(0−0)

(15−1)

−(3−0)

(1−0)

T

\begin{gathered}\mathsf{adjA=\left(\begin{array}{ccc}1&-5&14\\0&1&-3\\0&0&1\end{array}\right)^T}\end{gathered}

adjA=

1

0

0

−5

1

0

14

−3

1

T

\begin{gathered}\mathsf{adjA=\left(\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right)}\end{gathered}

adjA=

1

−5

14

0

1

−3

0

0

1

\mathsf{Now,\;A^{-1}=\dfrac{1}{|A|}adjA}Now,A

−1

=

∣A∣

1

adjA

\begin{gathered}\mathsf{A^{-1}=\left(\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right)}\end{gathered}

A

−1

=

1

−5

14

0

1

−3

0

0

1

\begin{gathered}\implies\boxed{\mathsf{B=\left(\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right)}}\end{gathered}

B=

1

−5

14

0

1

−3

0

0

1

\mathsf{BA}BA

\begin{gathered}\mathsf{=\left(\begin{array}{ccc}1&0&0\\-5&1&0\\14&-3&1\end{array}\right)\left(\begin{array}{ccc}1&0&0\\5&1&0\\1&3&1\end{array}\right)}\end{gathered}

=

1

−5

14

0

1

−3

0

0

1

1

5

1

0

1

3

0

0

1

\begin{gathered}\mathsf{=\left(\begin{array}{ccc}1+0+0&0+0+0&0+0+0\\-5+5+0&0+1+0&0+0+0\\14-15+1&0-3+3&0+0+1\end{array}\right)}\end{gathered}

=

1+0+0

−5+5+0

14−15+1

0+0+0

0+1+0

0−3+3

0+0+0

0+0+0

0+0+1

\begin{gathered}\mathsf{=\left(\begin{array}{ccc}1&0&0\\0&1&0\\0&0&1\end{array}\right)}\end{gathered}

=

1

0

0

0

1

0

0

0

1

\mathsf{=I}=I

\mathsf{Hence\;verified}Henceverified

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