Gunakan matriks persegi B dengan det(B)tidak sama dengan nol untuk menunjukkan bahwaA.(B invers)^-1= BB. ( B transpos)^-1 =(B invers)^t Tolong buatkan pakai ang

Misalkan matriks dengan determinan tak nol dengan dimensi 2x2:
[tex]$\begin{align}B&=\left[\begin{array}{cc}1&2\\1&3\end{array}\right]\end{align}[/tex]

Menggunakan pendefinisian invers matriks:
[tex]\boxed{\left[\begin{array}{cc}a&b\\c&d\end{array}\right]^{-1}=\frac{1}{ad-bc}\left[\begin{array}{cc}-d&b\\c&-a\end{array}\right]}[/tex]

Dengan memulai pembuktian pertama:
[tex]$\begin{align}(B^{-1})^{-1}&=\left(\left[\begin{array}{cc}1&2\\1&3\end{array}\right]^{-1}\right)^{-1} \\ &=\left(\frac{1}{1.3-1.2}\left[\begin{array}{cc}3&-2\\-1&1\end{array}\right]\right)^{-1} \\ &=\left[\begin{array}{cc}3&-2\\-1&1\end{array}\right]^{-1} \\ &=\frac{1}{3.1-(-1)(-2)}\left[\begin{array}{cc}1&2\\1&3\end{array}\right] \\ &=\left[\begin{array}{cc}1&2\\1&3\end{array}\right] \\ &=B\end{align}[/tex]

Serta untuk pembuktian kedua:
[tex]$\begin{align}(B^T)^{-1}&=\left(\left[\begin{array}{cc}1&2\\1&3\end{array}\right]^T\right)^{-1} \\ &=\left[\begin{array}{cc}1&1\\2&3\end{array}\right]^{-1} \\ &=\frac{1}{1.3-2.1}\left[\begin{array}{cc}3&-1\\-2&1\end{array}\right] \\ &=\left[\begin{array}{cc}3&-1\\-2&1\end{array}\right]\end{align}[/tex]

Untuk ruas kanan:
[tex]$\begin{align}(B^{-1})^T&=\left(\frac{1}{1.3-1.2}\left[\begin{array}{cc}3&-2\\-1&1\end{array}\right]\right)^T \\ &=\left[\begin{array}{cc}3&-2\\-1&1\end{array}\right]^T \\ &=\left[\begin{array}{cc}3&-2\\-1&1\end{array}\right]\end{align}[/tex]

Yang mana diperoleh bukti bahwa:
[tex](B^T)^{-1}=(B^{-1})^T[/tex]