Chapter 2: Q17SE (page 93)
Let A be a \(6 \times 4\) matrix and B a \(4 \times 6\) matrix. Show that the \(6 \times 6\) matrix \(AB\) cannot be invertible.
It is proved that the \(6 \times 6\) matrix \(AB\) cannot be invertible.
Over 22 million students worldwide already upgrade their learning with Vaia!
All the tools & learning materials you need for study success - in one app.
Get started for free
Suppose AB = AC, where Band Care \(n \times p\) matrices and A is invertible. Show that B = C. Is this true, in general, when A is not invertible.
Suppose Tand U are linear transformations from \({\mathbb{R}^n}\) to \({\mathbb{R}^n}\) such that \(T\left( {U{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\) . Is it true that \(U\left( {T{\mathop{\rm x}\nolimits} } \right) = {\mathop{\rm x}\nolimits} \) for all x in \({\mathbb{R}^n}\)? Why or why not?
Let \(S = \left( {\begin{aligned}{*{20}{c}}0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\\0&0&0&0&1\\0&0&0&0&0\end{aligned}} \right)\). Compute \({S^k}\) for \(k = {\bf{2}},...,{\bf{6}}\).
Let \(A = \left( {\begin{aligned}{*{20}{c}}1&1&1\\1&2&3\\1&4&5\end{aligned}} \right)\), and \(D = \left( {\begin{aligned}{*{20}{c}}2&0&0\\0&3&0\\0&0&5\end{aligned}} \right)\). Compute \(AD\) and \(DA\). Explain how the columns or rows of A change when A is multiplied by D on the right or on the left. Find a \(3 \times 3\) matrix B, not the identity matrix or the zero matrix, such that \(AB = BA\).
In Exercises 1–9, assume that the matrices are partitioned conformably for block multiplication. Compute the products shown in Exercises 1–4.
1. \(\left[ {\begin{array}{*{20}{c}}I&{\bf{0}}\\E&I\end{array}} \right]\left[ {\begin{array}{*{20}{c}}A&B\\C&D\end{array}} \right]\)
What do you think about this solution?
We value your feedback to improve our textbook solutions.
