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# Assertion$A = \left[ {\begin{array}{*{20}{c}} 4&0&4 \\ 0&3&2 \\ 4&2&1 \end{array}} \right],{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&3&3 \\ 1&4&3 \\ 1&3&4 \end{array}} \right]$ Then ${\left( {AB} \right)^{ - 1}}$ does not exist.ReasonSince $\left| A \right| = 0,{\left( {AB} \right)^{ - 1}} = {B^{ - 1}}{A^{ - 1}}$ is meaningless.A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.C. Assertion is correct but Reason is incorrect.D. Assertion is incorrect but Reason is correct.

Last updated date: 15th Sep 2024
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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.

$A = \left[ {\begin{array}{*{20}{c}} 4&0&4 \\ 0&3&2 \\ 4&2&1 \end{array}} \right],{B^{ - 1}} = \left[ {\begin{array}{*{20}{c}} 1&3&3 \\ 1&4&3 \\ 1&3&4 \end{array}} \right]$
$\left| A \right| = 4\left( {3 - 4} \right) + 4\left( {4 - 3} \right) \\ \Rightarrow \left| A \right| = - 4 + 4 = 0 \\$
$\Rightarrow {A^{ - 1}}$ 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.