Respuesta :
[tex]\mathbf A=\begin{bmatrix}3&-1\\-10&9\end{bmatrix}[/tex]
The inverse is given by
[tex]\mathbf A^{-1}=\dfrac1{\det\mathbf A}\mathrm{adj}\,\mathbf A[/tex]
where [tex]\det\mathbf A[/tex] is the determinant of the matrix, and [tex]\mathrm adj\,\mathbf A[/tex] is the adjugate matrix, or transpose of the cofactor matrix.
[tex]\det\mathbf A=\begin{vmatrix}3&-1\\-10&9\end{vmatrix}=3(9)-(-1)(-10)=17[/tex]
[tex]\mathrm{adj}\,\mathbf A=\begin{bmatrix}9&10\\1&3\end{bmatrix}^\top=\begin{bmatrix}9&1\\10&3\end{bmatrix}[/tex]
So the inverse is
[tex]\mathbf A^{-1}=\dfrac1{17}\begin{bmatrix}9&1\\10&3\end{bmatrix}[/tex]
The inverse is given by
[tex]\mathbf A^{-1}=\dfrac1{\det\mathbf A}\mathrm{adj}\,\mathbf A[/tex]
where [tex]\det\mathbf A[/tex] is the determinant of the matrix, and [tex]\mathrm adj\,\mathbf A[/tex] is the adjugate matrix, or transpose of the cofactor matrix.
[tex]\det\mathbf A=\begin{vmatrix}3&-1\\-10&9\end{vmatrix}=3(9)-(-1)(-10)=17[/tex]
[tex]\mathrm{adj}\,\mathbf A=\begin{bmatrix}9&10\\1&3\end{bmatrix}^\top=\begin{bmatrix}9&1\\10&3\end{bmatrix}[/tex]
So the inverse is
[tex]\mathbf A^{-1}=\dfrac1{17}\begin{bmatrix}9&1\\10&3\end{bmatrix}[/tex]