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  • Stuck here, help me understand: - Matrices and Determinants - JEE Main-6

Let A be a 3\times3 matrix such that

Then A-1  is :

  • Option 1)

  • Option 2)

  • Option 3)

  • Option 4)

 

Answers (2)

best_answer

As we have learned

Inverse of a matrix -

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

-

 

We have  \begin{bmatrix} 1 &2 &3 \\ 0& 2& 3\\ 0& 1 &1 \end{bmatrix}= A^{-1} \begin{bmatrix} 0 &0 &1 \\ 1 & 0& 0\\ 0& 1 & 0 \end{bmatrix}

Now  let     B= \begin{bmatrix} 0 &0 &1 \\ 1& 0& 0\\ 0&1 &0 \end{bmatrix} 

so , adj(B) = B= \begin{bmatrix} 0 &0 &1 \\ 1& 0& 0\\ 0&1 &0 \end{bmatrix}^{T}

= \begin{bmatrix} 0 &1 &0 \\ 0& 0& 1\\ 1&0 &0 \end{bmatrix}

|B| = 1 

\therefore B^{-1} = \begin{bmatrix} 0 &1 &0 \\ 0& 0& 1\\ 1&0 &0 \end{bmatrix} 

 Therefore \therefore A^{-1} = \begin{bmatrix} 1 &2 &3 \\ 0& 2& 3\\ 0&1 &1 \end{bmatrix}\begin{bmatrix} 0 &1 &0 \\ 0& 0& 1\\ 1& 0 & 0 \end{bmatrix}= \begin{bmatrix} 3 & 1 & 2\\ 3& 0& 2\\ 1&0 &1 \end{bmatrix}

 

 

 

 

 

 


Option 1)

Option 2)

Option 3)

Option 4)

Posted by

gaurav

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