Answer to Question #211537 in Linear Algebra for Moe

Question

Let $A = \begin{pmatrix} 2 & 1 \\ 5 & 3 \end{pmatrix}$, the A^-1 = 1/p $\begin{pmatrix} m & n \\ s & t \end{pmatrix}$

What is the value of p, n, s, t.

Answer 1

Firstly we should find the determinant of the given matrix A. If the determinant is equal to zero, the finding of the inverse matrix stops, since the matrix is ​​degenerate and its inverse does not exist. Because the rank of the matrix is 2, we should find the product of the elements of the side diagonal and subtract it from the product of the elements of the main diagonal.  

$$det(A)=a_{11}*a_{22}-a_{12}*a_{21}=2*3-1*5=6-5=1$$

p=det(A)=1, so the matrix is not singular, it has the inverse matrix. We should find the algebraic complements of the each element of the matrix and put each of them to the element's place in the resultant inverse matrix. The algebraic complement is the determinant of the matrix from which the row and the column of the sought element is deleted. Because the rank of the matrix is 2, we should simply use the element located diagonally to the given as complement not calculating the determinant because it represents the 1-ranked matrix whose determinant is the single element itself. After that transformations we have the matrix transposed regarding the given.

$$A = \begin{pmatrix} 2 & 1\\ 5 & 3 \end{pmatrix} =>\begin{pmatrix} 3 & 5\\ 1 & 2 \end{pmatrix}$$

So we have m=3, n=5, s=1, t=2, p=1 and the inverse matrix:

$$A^{-1}=1/1*\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}=\begin{pmatrix} 3 & 5 \\ 1 & 2 \end{pmatrix}$$