Chapter 2: Q30Q (page 93)
Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)
Short Answer
Theorem 2(d) is proved.
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Chapter 2: Q30Q (page 93)
Prove Theorem 2(d). (Hint: The \(\left( {i,j} \right)\)- entry in \(\left( {rA} \right)B\) is \(\left( {r{a_{i1}}} \right){b_{1j}} + ... + \left( {r{a_{in}}} \right){b_{nj}}\).)
Theorem 2(d) is proved.
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Show that \({I_n}A = A\) when \(A\) is \(m \times n\) matrix. (Hint: Use the (column) definition of \({I_n}A\).)
Suppose \(\left( {B - C} \right)D = 0\), where Band Care \(m \times n\) matrices and \(D\) is invertible. Show that B = C.
In exercise 5 and 6, compute the product \(AB\) in two ways: (a) by the definition, where \(A{b_{\bf{1}}}\) and \(A{b_{\bf{2}}}\) are computed separately, and (b) by the row-column rule for computing \(AB\).
\(A = \left( {\begin{aligned}{*{20}{c}}{ - {\bf{1}}}&{\bf{2}}\\{\bf{5}}&{\bf{4}}\\{\bf{2}}&{ - {\bf{3}}}\end{aligned}} \right)\), \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{3}}&{ - {\bf{2}}}\\{ - {\bf{2}}}&{\bf{1}}\end{aligned}} \right)\)
Suppose A, B, and Care \(n \times n\) matrices with A, X, and \(A - AX\) invertible, and suppose
\({\left( {A - AX} \right)^{ - 1}} = {X^{ - 1}}B\) …(3)
Suppose A, B,and Care invertible \(n \times n\) matrices. Show that ABCis also invertible by producing a matrix Dsuch that \(\left( {ABC} \right)D = I\) and \(D\left( {ABC} \right) = I\).
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