Question

(a) Prove that.${\left(\mathrm{AB}\right)}^{\mathbf{t}}\mathbf{=}{\mathbf{B}}^{\mathbf{t}}{\mathbf{A}}^{\mathbf{t}}$ Hint: See$\left(9.10\right)$.

(b) Verify (9.11), that is, show that (9.10) applies to a product of any number of matrices. Hint: Use (9.10)and (9.8).

Short answer

a) It is proven that ${\left(\mathrm{AB}\right)}^{\mathrm{t}}={\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}}.$.

b) It is proven that ${\left(\mathrm{ABCD}\right)}^{\mathrm{T}}={\left(\mathrm{A}\left(\mathrm{BCD}\right)\right)}^{\mathrm{T}}$.

Step-by-step solution

Property used.

Use this property to solve the problem.

$$\begin{gathered} {\mathbf{A}}^{\mathbf{t}}\mathbf{=}{\left(\stackrel{\rightharpoonup }{A}\right)}^{\mathbf{T}} \\ \mathbf{ABC}\mathbf{=}\mathbf{A}\left(\mathrm{BC}\right)\mathbf{=}\left(\mathrm{AB}\right)\mathbf{C} \end{gathered}$$

Prove the given function (AB)t =BtAt ..

a)

Taking Left hand side:

$$\begin{gathered} {\left(AB\right)}_{ik=}^t=\left(\overline{AB}\right)ki\left(\because A^t={\left(\overline{A}\right)}^T\right) \\ =\underset{}{\sum _j{\overline{A}}_{kj}{\overline{B}}_{ji}} \\ =\sum _j{B}_{jk}^t{A}_{ij}^t \\ =\sum _j{B}_{ij}^t{A}_{jk}^t \end{gathered}$$

Solve further

$$\begin{gathered} ={\left({\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}}\right)}_{\mathrm{ik}} \\ {\left(\mathrm{AB}\right)}^{\mathrm{t}}={\mathrm{B}}^{\mathrm{t}}{\mathrm{A}}^{\mathrm{t}} \end{gathered}$$

Hence proved.

Prove the given function(ABCD)T =(ABCD)T.

b)

Taking Left hand side:

$$\begin{gathered} {\left(\mathrm{ABCD}\right)}^{\mathrm{T}}={\left(\mathrm{A}\left(\mathrm{BCD}\right)\right)}^{\mathrm{T}}\left\{\because \mathrm{ABC}=\mathrm{A}\left(\mathrm{BC}\right)=\left(\mathrm{AB}\right)\mathrm{C}\right\} \\ ={\left(\mathrm{BCD}\right)}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\left\{\mathrm{use}{\left(\mathrm{AB}\right)}^{\mathrm{T}}={\mathrm{B}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}}\right\} \\ ={\left(\mathrm{B}\left(\mathrm{CD}\right)\right)}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}} \\ ={\left(\mathrm{CD}\right)}^{\mathrm{T}}{\mathrm{B}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}} \\ \mathrm{Solve}\mathrm{the}\mathrm{equation}\mathrm{further} \\ ={\mathrm{D}}^{\mathrm{T}}{\mathrm{C}}^{\mathrm{T}}{\mathrm{B}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}} \\ {\left(\mathrm{ABCD}\right)}^{\mathrm{T}}={\mathrm{D}}^{\mathrm{T}}{\mathrm{C}}^{\mathrm{T}}{\mathrm{B}}^{\mathrm{T}}{\mathrm{A}}^{\mathrm{T}} \end{gathered}$$

Hence proved.