Question

(M) Let \(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{25}}}&{ - {\bf{9}}}&{ - {\bf{27}}}\\{{\bf{546}}}&{{\bf{180}}}&{{\bf{537}}}\\{{\bf{154}}}&{{\bf{50}}}&{{\bf{149}}}\end{aligned}} \right)\). Find the second and third columns of \({A^{ - {\bf{1}}}}\) without computing the first column.

Short answer

\(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\)

Step-by-step solution

Find the expression \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{2}}}}&{{e_{\bf{3}}}}\end{aligned}} \right)\)

The matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_2}}&{{e_3}}\end{aligned}} \right)\) can be expressed as

\(\left( {\begin{aligned}{*{20}{c}}A&{{e_2}}&{{e_3}}\end{aligned}} \right) = \left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\\{546}&{180}&{537}&1&0\\{154}&{50}&{149}&0&1\end{aligned}} \right)\).

Convert the matrix \(\left( {\begin{aligned}{*{20}{c}}A&{{e_{\bf{2}}}}&{{e_{\bf{3}}}}\end{aligned}} \right)\) into the row-reduced echelon form

Consider the matrix\(\left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\\{546}&{180}&{537}&1&0\\{154}&{50}&{149}&0&1\end{aligned}} \right)\).

Use the code in MATLAB to obtain the row-reduced echelon form as shown below:

\(\begin{aligned}{l} > > {\rm{ A }} = {\rm{ }}\left( {\begin{aligned}{*{20}{c}}{ - 25}&{ - 9}&{ - 27}&0&0\end{aligned};{\rm{ }}\begin{aligned}{*{20}{c}}{546}&{180}&{537}&1&0\end{aligned};{\rm{ }}\begin{aligned}{*{20}{c}}{154}&{50}&{149}&0&1\end{aligned}{\rm{ }}} \right);\\ > > {\rm{ U}} = {\rm{rref}}\left( {\rm{A}} \right)\end{aligned}\)

\(\left( {\begin{aligned}{*{20}{c}}1&0&0&{\frac{3}{2}}&{ - \frac{9}{2}}\\0&1&0&{ -

\frac{{433}}{6}}&{\frac{{439}}{2}}\\0&0&1&{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\)

Find the column of the inverse matrix

In the matrix \(\left( {\begin{aligned}{*{20}{c}}1&0&0&{\frac{3}{2}}&{ - \frac{9}{2}}\\0&1&0&{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\0&0&1&{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\), the second and third columns are \(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\).

So, the third column of the inverse matrix is \(\left( {\begin{aligned}{*{20}{c}}{\frac{3}{2}}&{ - \frac{9}{2}}\\{ - \frac{{433}}{6}}&{\frac{{439}}{2}}\\{\frac{{68}}{3}}&{ - 69}\end{aligned}} \right)\).