Question

1. Solve the following simultaneous equations using the matrix method.

Answer 1

(i) Use the formula for the determinant of a 2×2 matrix.

$\begin{vmatrix}a&b\\c&d\end{vmatrix} = ad-bc$

$\implies \det(A) = \begin{vmatrix}4 & -3 \\ 2 & 5\end{vmatrix} = 4\times5 - (-3)\times2 = \boxed{26}$

(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 $(-1)^{i+j}$. If C is the cofactor matrix of A, then

$C = \begin{pmatrix}5&-2\\3&4\end{pmatrix}$

Then the adjugate of A is the transpose of C,

$\mathrm{adj}(A) = C^\top = \boxed{\begin{pmatrix}5&3\\-2&4\end{pmatrix}}$

(iii) The inverse of A is equal to 1/det(A) times the adjugate:

$A^{-1} = \dfrac1{\det(A)} \mathrm{adj}(A) = \boxed{\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix}}$

(iv) The system of equations translates to the matrix equation

$A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}6\\16\end{pmatrix}$

Multiplying both sides on the left by the inverse of A gives

$A^{-1}\left(A\begin{pmatrix}x\\y\end{pmatrix}\right)=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\left(A^{-1}A\right)\begin{pmatrix}x\\y\end{pmatrix}=A^{-1} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac5{26}&\frac3{26}\\\\-\frac1{13}&\frac2{13}\end{pmatrix} \begin{pmatrix}6\\16\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=\boxed{\begin{pmatrix}3\\2\end{pmatrix}}$