Let Z and A are two matrices such that Z=112235410 and A=101210031 if Z=AB−1where B is a square matrix of order 3 and t=trace of (A(adj B)), then 28|t| is:

Given Z=AB−1⇒|Z|=|AB−1|⇒|Z|=|A|×|B|−1⇒−5=7(|B−1|)⇒|B−1|=−57Again Z=AB−1⇒Z=A1|B|adjB Z=A(1|B|adjB) ⇒Z|B|=A(adjB) trace (A(adjB)) = trace (Z|B|)=|B| trace (Z)=4(−75)=−285

Let Z and A are two matrices such that Z=112235410 and A=101210031 if Z=AB−1where B is a square matrix of order 3 and t=trace of (A(adj B)), then 28|t| is:

Given Z=AB−1⇒|Z|=|AB−1|⇒|Z|=|A|×|B|−1⇒−5=7(|B−1|)⇒|B−1|=−57Again Z=AB−1⇒Z=A1|B|adjB Z=A(1|B|adjB) ⇒Z|B|=A(adjB) trace (A(adjB)) = trace (Z|B|)=|B| trace (Z)=4(−75)=−285

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Let Z and A are two matrices such that $Z = [\begin{matrix} 1 & 1 & 2 \\ 2 & 3 & 5 \\ 4 & 1 & 0 \end{matrix}] and A = [\begin{matrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 0 & 3 & 1 \end{matrix}]$ if $Z = {AB}^{- 1}$ where B is a square matrix of order 3 and $t = trace of (A (adj B))$ , then $\frac{28}{| t |}$ is:

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Detailed Solution

Given $Z = {AB}^{- 1} \Rightarrow | Z | = | {AB}^{- 1} | \Rightarrow | Z | = | A | \times {| B |}^{- 1}$
$\Rightarrow - 5 = 7 (| B^{- 1} |) \Rightarrow | B^{- 1} | = - \frac{5}{7}$
Again
$Z = {AB}^{- 1} \Rightarrow Z = A \frac{1}{| B |} adjB Z = A (\frac{1}{| B |} adjB) \Rightarrow Z | B | = A (adjB) trace (A (adjB)) = trace (Z | B |) = | B | trace (Z) = 4 (- \frac{7}{5}) = - \frac{28}{5}$