Question

Consider the matrix

A = 1 4

2 3

(a) Compute A^-1

(b) Find det(A^-1)

(c) Deduce a relation (if it exists) between det(A) and det(A^-1)

Answer 1

Solution.

$$A=\begin{pmatrix} 1 & 4\\ 2 & 3 \end{pmatrix}$$

(a)

$det A=3-8=-5.$

$A^{-1}=-\frac{1}{5}\begin{pmatrix} 3 & -2 \\ -4 & 1 \end{pmatrix}^T=-\frac{1}{5}\begin{pmatrix} 3 & -4 \\ -2& 1 \end{pmatrix}=\begin{pmatrix} -0.6 & 0.8 \\ 0.4 & -0.2 \end{pmatrix}.$

(b)

$det (A^{-1})=0.12-0.32=-0.2=-\frac{1}{5}.$

$det(A^{-1})=\frac{1}{det A}=-\frac{1}{5}.$

(c)

$det(A^{-1})=\frac{1}{det A}.$