Answer

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**Hint:**In order to solve the problem we will first find the modulus of the matrix further we will use the formula of the matrix to solve the given equation and we will substitute the value of modulus and further we will use the property of matrix.

**Complete step-by-step answer:**

Given that:

$A = \left[ {\begin{array}{*{20}{c}}

4&0&4 \\

0&3&2 \\

4&2&1

\end{array}} \right],{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}}

1&3&3 \\

1&4&3 \\

1&3&4

\end{array}} \right]$

$

\left| A \right| = 4\left( {3 - 4} \right) + 4\left( {4 - 3} \right) \\

\Rightarrow \left| A \right| = - 4 + 4 = 0 \\

$

$ \Rightarrow {A^{ - 1}}$ does not exist as in order to find the value of ${A^{ - 1}}$ we divide the term by modulus of A. but here modulus of A is 0 so the term ${A^{ - 1}}$ will not exist.

So, ${\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$ does not exist.

Therefore ${\left( {AB} \right)^{ - 1}}$ does not exist.

Hence, Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.

**So, the correct answer is “Option A”.**

**Note:**In order to solve such types of problems students must use the formula for the matrices and should not carry out the calculation in isolation. Students must remember different formulas of matrices such as the inverse formula and the formula to find out the modulus of the matrix.

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