Question

Find

a)-A raised to -1+3B raised to T

b)B raised to -1+(A raised to T+A raised to -1)

Answer 1

Find:

a)-A raised to -1+3B raised to T

$let\\A=\begin{bmatrix} a & b \\ c & d \end{bmatrix}\\B=\begin{bmatrix} i & j\\ k & l \end{bmatrix}$

then

$-A=\begin{bmatrix} -a & -b \\ -c & -d \end{bmatrix} \implies -A^{-1}=\frac{1}{ad-cb}\begin{bmatrix} -d& b \\ c &- a \end{bmatrix}$

$B^{T}=\begin{bmatrix} i & k\\ j& l \end{bmatrix} \implies 3B^{T}=\begin{bmatrix} 3 i & 3k\\ 3 j& 3l \end{bmatrix}$

$-A^{-1}+3B^T=\frac{1}{ad-cb}\begin{bmatrix} -d& b \\ c &- a \end{bmatrix}+\begin{bmatrix} 3 i & 3k\\ 3 j& 3l \end{bmatrix}$

$-A^{-1}+3B^T=\begin{bmatrix} \frac{3i(ad-cd)-d}{ad-cd} & \frac{3k(ad-cd)+b}{ad-cd}\\ \frac{3j(ad-cd)+c}{ad-cd} & \frac{3l(ad-cd)-a}{ad-cd} \end{bmatrix}$

b)B raised to -1+(A raised to T+A raised to -1)

$B^{-1}=\frac{1}{il-kj}\begin{bmatrix} l&- j\\ - k &i \end{bmatrix};\phantom{i}A^{T}=\begin{bmatrix} a& c\\ b &d \end{bmatrix};A^{-1}=\frac{1}{ad-cb}\begin{bmatrix} d& -b \\ -c &a \end{bmatrix}$

$B^{-1}+A^T+A^{-1}=\begin{bmatrix} \frac{l}{il-kj}+a+\frac{d}{ad-cb}& \frac{j}{kj-il}+c+\frac{b}{cb-ad}\\ \frac{k}{kj-il}+b+\frac{c}{cb-ad} & \frac{i}{il-kj}+d+\frac{a}{ad-cb} \end{bmatrix}$