If [tex]\mathbf A[/tex] is invertible, then we can simply choose [tex]\mathbf B=\mathbf A^{-1}[/tex] because for any invertible matrix [tex]\mathbf X[/tex] by definition we must have [tex]\mathbf{XX}^{-1}=\mathbf X^{-1}\mathbf X=\mathbf I[/tex].
We have [tex]\det\mathbf A\neq0[/tex], as
[tex]\det\mathbf A=\begin{vmatrix}1&7\\0&1\end{vmatrix}=1\times1-0\times7=1[/tex]
which means [tex]\mathbf A[/tex] is non-singular and has an inverse. The inverse itself would be
[tex]\mathbf A^{-1}=\dfrac1{\det\mathbf A}\begin{bmatrix}1&0\\-7&1\end{bmatrix}^\top=\begin{bmatrix}1&-7\\0&1\end{bmatrix}[/tex]
