Chapter 2: Q13Q (page 93)
An \(m \times n\) upper triangular matrix is one whose enteries below the main diagonal are 0’s (as in Exercise 8). When is a square upper triangular matrix invertible? Justify your answer.
An upper triangular matrix is invertible as it can be reduced in the echelon form.
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In Exercise 10 mark each statement True or False. Justify each answer.
10. a. A product of invertible \(n \times n\) matrices is invertible, and the inverse of the product of their inverses in the same order.
b. If A is invertible, then the inverse of \({A^{ - {\bf{1}}}}\) is A itself.
c. If \(A = \left( {\begin{aligned}{*{20}{c}}a&b\\c&d\end{aligned}} \right)\) and \(ad = bc\), then A is not invertible.
d. If A can be row reduced to the identity matrix, then A must be invertible.
e. If A is invertible, then elementary row operations that reduce A to the identity \({I_n}\) also reduce \({A^{ - {\bf{1}}}}\) to \({I_n}\).
Generalize the idea of Exercise 21(a) [not 21(b)] by constructing a \(5 \times 5\) matrix \(M = \left[ {\begin{array}{*{20}{c}}A&0\\C&D\end{array}} \right]\) such that \({M^2} = I\). Make C a nonzero \(2 \times 3\) matrix. Show that your construction works.
Suppose Tand Ssatisfy the invertibility equations (1) and (2), where T is a linear transformation. Show directly that Sis a linear transformation. [Hint: Given u, v in \({\mathbb{R}^n}\), let \[{\mathop{\rm x}\nolimits} = S\left( {\mathop{\rm u}\nolimits} \right),{\mathop{\rm y}\nolimits} = S\left( {\mathop{\rm v}\nolimits} \right)\]. Then \(T\left( {\mathop{\rm x}\nolimits} \right) = {\mathop{\rm u}\nolimits} \), \[T\left( {\mathop{\rm y}\nolimits} \right) = {\mathop{\rm v}\nolimits} \]. Why? Apply Sto both sides of the equation \(T\left( {\mathop{\rm x}\nolimits} \right) + T\left( {\mathop{\rm y}\nolimits} \right) = T\left( {{\mathop{\rm x}\nolimits} + y} \right)\). Also, consider \(T\left( {cx} \right) = cT\left( x \right)\).]
If \(A = \left( {\begin{aligned}{*{20}{c}}1&{ - 2}\\{ - 2}&5\end{aligned}} \right)\) and \(AB = \left( {\begin{aligned}{*{20}{c}}{ - 1}&2&{ - 1}\\6&{ - 9}&3\end{aligned}} \right)\), determine the first and second column of B.
Let \(A = \left( {\begin{aligned}{*{20}{c}}{\bf{2}}&{\bf{5}}\\{ - {\bf{3}}}&{\bf{1}}\end{aligned}} \right)\) and \(B = \left( {\begin{aligned}{*{20}{c}}{\bf{4}}&{ - {\bf{5}}}\\{\bf{3}}&k\end{aligned}} \right)\). What value(s) of \(k\), if any will make \(AB = BA\)?
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