Question
Question: If matrix \(A = \left( {\begin{array}{*{20}{c}} 1&0&{ - 1} \\\ 3&4&5 \\\ 0&6&7 \end...
If matrix A = \left( {\begin{array}{*{20}{c}} 1&0&{ - 1} \\\ 3&4&5 \\\ 0&6&7 \end{array}} \right) and its inverse is denoted by {A^1} = \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right) , then the value of A23=
Solution
Here we will use some of the matrix rules and matrix formulas. Then we will substitute the values in the matrix. These values are denoted by the adjacent. Then that will be concluded using some rules. Finally, we will get the answer.
Complete step-by-step answer:
The question given,
A = \left( {\begin{array}{*{20}{c}}
1&0&{ - 1} \\\
3&4&5 \\\
0&6&7
\end{array}} \right)
A−1=detAadjA (here adjA means Adjacent A and det A means determinant A )
Here detA=1(28−30)−1(18)(formula of determinant a A = \left( {\begin{array}{*{20}{c}}
a&b;&c; \\\
d&e;&f; \\\
g&h;&i;
\end{array}} \right) is ∣A∣ = a(ei − fh) − b(di − fg) + c(dh − eg) )
Using above formula, we will get the detA
=−2−18 =−20
Then we will solve the adjA.
Here we will find adjacent of matrix
We have a method for finding adjacent matrices. Using that method find the adjacent matrix.
adjA = \left( {\begin{array}{*{20}{c}}
{ - 2}&{ - 6}&4 \\\
{ - 21}&7&{ - 8} \\\
{18}&{ - 6}&4
\end{array}} \right)