Chapter 2: Q17Q (page 93)
If \(A\) is invertible, then the columns of \({A^{ - {\bf{1}}}}\) are linearly independent. Explain why?
The columns of the inverse matrix are linearly independent.
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[M] For block operations, it may be necessary to access or enter submatrices of a large matrix. Describe the functions or commands of your matrix program that accomplish the following tasks. Suppose A is a \(20 \times 30\) matrix.
[Note: It may not be necessary to specify the zero blocks in B.]
Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{ - {\bf{3}}}\\{ - {\bf{4}}}&{\bf{6}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{8}}&{\bf{4}}\\{\bf{5}}&{\bf{5}}\end{aligned}} \right)\) and \(C = \left( {\begin{aligned}{*{20}{c}}{\bf{5}}&{ - {\bf{2}}}\\{\bf{3}}&{\bf{1}}\end{aligned}} \right)\). Verfiy that \(AB = AC\) and yet \(B \ne C\).
A useful way to test new ideas in matrix algebra, or to make conjectures, is to make calculations with matrices selected at random. Checking a property for a few matrices does not prove that the property holds in general, but it makes the property more believable. Also, if the property is actually false, you may discover this when you make a few calculations.
37. Construct a random \({\bf{4}} \times {\bf{4}}\) matrix Aand test whether \(\left( {A + I} \right)\left( {A - I} \right) = {A^2} - I\). The best way to do this is to compute \(\left( {A + I} \right)\left( {A - I} \right) - \left( {{A^2} - I} \right)\) and verify that this difference is the zero matrix. Do this for three random matrices. Then test \(\left( {A + B} \right)\left( {A - B} \right) = {A^2} - {B^{\bf{2}}}\) the same way for three pairs of random \({\bf{4}} \times {\bf{4}}\) matrices. Report your conclusions.
Suppose \(AD = {I_m}\) (the \(m \times m\) identity matrix). Show that for any b in \({\mathbb{R}^m}\), the equation \(A{\mathop{\rm x}\nolimits} = {\mathop{\rm b}\nolimits} \) has a solution. (Hint: Think about the equation \(AD{\mathop{\rm b}\nolimits} = {\mathop{\rm b}\nolimits} \).) Explain why Acannot have more rows than columns.
Let Abe an invertible \(n \times n\) matrix, and let B be an \(n \times p\) matrix. Show that the equation \(AX = B\) has a unique solution \({A^{ - 1}}B\).
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