+0  
 
0
652
1
avatar+76 

Let \(\mathbf{A}\) and \(\mathbf{B}\) be invertible matrices such that \(\mathbf{A} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \text{ and } \mathbf{B} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}.\)
Calculate \((\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix}.\)

 Apr 17, 2020
 #1
avatar+26367 
+2

Let \(A\) and \(B\) be invertible matrices such that \(\mathbf{A} \begin{pmatrix} 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 1 \\ 2 \end{pmatrix} \text{ and } \mathbf{B} \begin{pmatrix} 1 \\ 3 \end{pmatrix} = \begin{pmatrix} 2 \\ -1 \end{pmatrix}\).
Calculate \((\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix}\).

 

\(\begin{array}{lcll} \text{Let }v_1 &=& \begin{pmatrix} 2 \\ -1 \end{pmatrix} \\\\ \text{Let }v_2 &=& \begin{pmatrix} 1 \\ 2 \end{pmatrix} \\\\ \text{Let }v_3 &=& \begin{pmatrix} 1 \\ 3 \end{pmatrix} \\\\ \text{Let }Av_1 &=& v_2 \\\\ \text{Let }Bv_3 &=& v_1 \\ \end{array}\)

 

Formula: \(\boxed{(AB)^{-1}=B^{-1}A^{-1}}\)

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ (\mathbf{A}\mathbf{B})^{-1} \begin{pmatrix} 1 \\ 2 \end{pmatrix} } \\ &=& (\mathbf{A}\mathbf{B})^{-1} v_2 \quad &| \quad \boxed{(AB)^{-1}=B^{-1}A^{-1}} \\ &=& B^{-1}A^{-1}v_2 \quad &| \quad v_2=Av_1 \\ &=& B^{-1}A^{-1} Av_1 \quad &| \quad A^{-1} A=I\quad \text{identity matrix} \\ &=& B^{-1}Iv_1 \quad &| \quad Iv_1=v_1 \\ &=& B^{-1}v_1 \quad &| \quad v_1=Bv_3 \\ &=& B^{-1}Bv_3 \quad &| \quad B^{-1} B=I\quad \text{identity matrix} \\ &=& Iv_3 \quad &| \quad Iv_3=v_3 \\ &=&\mathbf{ v_3 } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline \mathbf{ (\mathbf{A}\mathbf{B})^{-1} \dbinom12 = \dbinom13 } \\ \hline \end{array}\)

 

laugh

 Apr 17, 2020

4 Online Users

avatar
avatar