tentukan invers matriks dri soal di atas

Rumus:

[tex]A^{-1} = \frac{1}{ad-bc} \left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right] [/tex]

Tinggal masukkan:

[tex]ad-bc = \frac{1}{2(a-b)} \frac{1}{2(a+b)} - \frac{1}{2(a+b)} \frac{-1}{2(a+b)} \\ \\ = \frac{1}{4(a-b)(a+b)} + \frac{1}{4(a+b)^2} \\ \\ = \frac{(a+b) + (a - b)}{4(a-b)(a+b)^2} \\ \\ = \frac{2a}{4(a-b)(a+b)^2}[/tex]

[tex]A^{-1} = \frac{4(a-b)(a+b)^2}{2a} \left[\begin{array}{cc}\frac{1}{2(a+b)} & \frac{-1}{2(a+b)} \\ \frac{1}{2(a+b)} & \frac{1}{2(a-b)}\end{array}\right] \\ \\ A^{-1} = \left[\begin{array}{cc}\frac{(a-b)(a+b)}{a} & \frac{-(a-b)(a+b)}{a} \\ \frac{(a-b)(a+b)}{a} & \frac{(a+b)^2}{a}\end{array}\right] \\ \\ [/tex]