Problem 11
Let \(A=\left[\begin{array}{rr}1 & 1 \\ -2 & -1 \\ 1 & 2\end{array}\right], B=\left[\begin{array}{rrr}1 & -1 & -2 \\ 2 & 1 & -2\end{array}\right],\) and \(C=\left[\begin{array}{rrr}1 & 1 & -3 \\ -1 & 2 & 0 \\ -3 & -1 & 0\end{array}\right] .\) Find the following if possible. (a) \(A B\) (b) \(B A\) (c) \(A C\) (d) \(C A\) (e) \(C B\) (f) \(B C\)
Problem 26
Suppose \(A\) and \(B\) are square matrices of the same size. Which of the following are necessarily true? (a) \((A-B)^{2}=A^{2}-2 A B+B^{2}\) (b) \((A B)^{2}=A^{2} B^{2}\) (c) \((A+B)^{2}=A^{2}+2 A B+B^{2}\) (d) \((A+B)^{2}=A^{2}+A B+B A+B^{2}\) (e) \(A^{2} B^{2}=A(A B) B\) (f) \((A+B)^{3}=A^{3}+3 A^{2} B+3 A B^{2}+B^{3}\) \((g) \quad(A+B)(A-B)=A^{2}-B^{2}\)
Problem 29
Show that the main diagonal of every skew symmetric matrix consists of only zeros. Recall that the main diagonal consists of every entry of the matrix which is of the form a \(_{i i} .\)
Problem 33
Suppose \(A B=A C\) and \(A\) is a non invertible \(n \times n\) matrix. Does it follow that \(B=C ?\) Explain why or why not.
Problem 44
Using the inverse of the matrix, find the solution to the systems: \((a)\) $$ \left[\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 1 \\ 2 \end{array}\right] $$ \((b)\) $$ \left[\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \end{array}\right] $$ Now give the solution in terms of a and b to $$ \left[\begin{array}{ll} 2 & 4 \\ 1 & 1 \end{array}\right]\left[\begin{array}{l} x \\ y \end{array}\right]=\left[\begin{array}{l} a \\ b \end{array}\right] $$
Problem 57
Let \(A=\left[\begin{array}{rrr}1 & 2 & 1 \\ 0 & 5 & 1 \\ 2 & -1 & 4\end{array}\right] .\) Suppose a row operation is applied to \(A\) and the result is $$ B=\left[\begin{array}{rrr} 1 & 2 & 1 \\ 2 & -1 & 4 \\ 0 & 5 & 1 \end{array}\right] $$ (a) Find the elementary matrix \(E\) such that \(E A=B\). (b) Find the inverse of \(E, E^{-1},\) such that \(E^{-1} B=A\).