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If A= \begin{bmatrix} 0 &-1&2 \\ 2& -2&0 \end{bmatrix}\, \, \,   ,         B= \begin{bmatrix} 0 &1 \\ 1&0 \\1&1\end{bmatrix}\, \, \, and\, \, \, n= AB,\, \, \ then\, \, \ n^{-1}\, \, \ is\, \, \ equal \, \, to

  • Option 1)

    \begin{bmatrix} 2 &-2 \\ 2& 1 \end{bmatrix}

  • Option 2)

    \begin{bmatrix}\frac{1}{3} &\frac{1}{3} \\ \frac{-1}{3} & \frac{1}{6} \end{bmatrix}

  • Option 3)

    \begin{bmatrix}\frac{1}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{bmatrix}

  • Option 4)

    \begin{bmatrix}\frac{1}{3} &\frac{-1}{3} \\ \frac{-1}{3} & \frac{1}{6} \end{bmatrix}

 

Answers (1)

best_answer

As we learnt in 

Multiplication of matrices -

-

 

 

Inverse of a matrix -

 A^{-1}=\frac{1}{\left | A \right |}\cdot adjA

-

 

 AB= \begin{bmatrix} 0&-1 &2 \\ 2& -2 &0 \end{bmatrix}\begin{bmatrix} 0 &1 \\ 1&0 \\ 1& 1 \end{bmatrix}

= \begin{bmatrix} 1 &2 \\ -2& 2 \end{bmatrix}

\left | AB \right |= 2+4=6

adj (AB)= \begin{bmatrix} 2 &2 \\ -2& 1 \end{bmatrix}

(AB)^{-1}= \frac{1}{6}\begin{bmatrix} 2 &2 \\ -2& 1 \end{bmatrix}

=\begin{bmatrix} \frac{1}{3} &\frac{1}{3} \\ \frac{-1}{3}& \frac{1}{6} \end{bmatrix}

 

 


Option 1)

\begin{bmatrix} 2 &-2 \\ 2& 1 \end{bmatrix}

Incorrect

Option 2)

\begin{bmatrix}\frac{1}{3} &\frac{1}{3} \\ \frac{-1}{3} & \frac{1}{6} \end{bmatrix}

Correct

Option 3)

\begin{bmatrix}\frac{1}{3} &\frac{-1}{3} \\ \frac{1}{3} & \frac{1}{6} \end{bmatrix}

Incorrect

Option 4)

\begin{bmatrix}\frac{1}{3} &\frac{-1}{3} \\ \frac{-1}{3} & \frac{1}{6} \end{bmatrix}

Incorrect

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Plabita

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