(i) Use the formula for the determinant of a 2×2 matrix.
[tex]\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc[/tex]
[tex]\implies \det(A) = \begin{vmatrix}4 & -3 \\ 2 & 5\end{vmatrix} = 4\times5 - (-3)\times2 = \boxed{26}[/tex]
(ii) The adjugate matrix is the transpose of the cofactor matrix of A. (These days, the "adjoint" of a matrix X is more commonly used to refer to the conjugate transpose of X, which is not the same.)
The cofactor of the (i, j)-th entry of A is the determinant of the matrix you get after deleting the i-th row and j-th column of A, multiplied by [tex](-1)^{i+j}[/tex]. If C is the cofactor matrix of A, then
[tex]C = \begin{pmatrix}5&-2\\3&4\end{pmatrix}[/tex]
Then the adjugate of A is the transpose of C,
[tex]\mathrm{adj}(A) = C^\top = \boxed{\begin{pmatrix}5&3\\-2&4\end{pmatrix}}[/tex]
(iii) The inverse of A is equal to 1/det(A) times the adjugate:
[tex]A^{-1} = \dfrac1{\det(A)} \mathrm{adj}(A) = \boxed{\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix}}[/tex]
(iv) The system of equations translates to the matrix equation
[tex]A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}[/tex]
Multiplying both sides on the left by the inverse of A gives
[tex]A^{-1}\left(A\begin{pmatrix}x\\y\end{pmatrix}\right)=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}[/tex]
[tex]\left(A^{-1}A\right)\begin{pmatrix}x\\y\end{pmatrix}=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}[/tex]
[tex]\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix} \begin{pmatrix}6\\16\end{pmatrix}[/tex]
[tex]\begin{pmatrix}x\\y\end{pmatrix}=\boxed{\begin{pmatrix}3\\2\end{pmatrix}}[/tex]
