Question

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

(a) LetA=(4 −3)
(2 5)

(i) Determine the determinant of A

(ii) Determine the adjoint of A

(iii) Determine the inverse of A

(iv) Hence, using the matrix method solve the following simultaneous
equations
4 − 3 = 6
2 + 5 = 16

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*5 - (-3)*2 = \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) = \frac1{\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}}`