Question 4.6: Show that the matrices [1 1 -1 4] and [2 1 1 3] are similar ......
Show that the matrices \left[\begin{array}{c c}{1}&{1}\\ {-1}&{4}\end{array}\right] and \left[{\begin{array}{c c}{2}&{1}\\ {1}&{3}\end{array}}\right] are similar but that \left[\begin{array}{c c}{{3}}&{{1}}\\ {{-6}}&{{-2}}\end{array}\right] and \left[\begin{array}{c c}{-1}&{2}\\ {1}&{0}\end{array}\right] are not.
Let A_{1}={\left[\begin{array}{c c}{1}&{1}\\ {-1}&{4}\end{array}\right]},\ A_{2}={\left[\begin{array}{c c}{2}&{1}\\ {1}&{3}\end{array}\right]},\,A_{3}={\left[\begin{array}{c c}{3}&{1}\\ {-6}&{-2}\end{array}\right]},~{\mathrm{and~}}A_{4}={\left[\begin{array}{c c}{-1}&{2}\\ {1}&{0}\end{array}\right]}
\operatorname*{det}(A_{1})={\left|\begin{array}{c c}{1}&{1}\\ {-1}&{4}\end{array}\right|}=5,{\mathrm{~det}}(A_{2})={\left|\begin{array}{c c}{2}&{1}\\ {1}&{3}\end{array}\right|}=5
\\[1.5 em] \operatorname{det}(A_{3})={\left|\begin{array}{c c}{3}&{1}\\ {-6}&{-2}\end{array}\right|}=0,\;\operatorname{det}(A_{4})={\left|\begin{array}{c c}{-1}&{2}\\ {1}&{0}\end{array}\right|}=-2.
Therefore, if \operatorname*{det}(A_{1})={\mathrm{det}}(A_{2}), matrices A_{1} and A_{2} are similar.
If \operatorname*{det}(A_{3})\neq {\mathrm{det}}(A_{4}), matrices A_{1} and A_{2} are not similar.