Determinant and Inverse of Matrices

Finding the inverse of a matrix is very important in many areas of science. For example, decrypting a coded message uses the inverse of a matrix. Determinant may be used to answer this problem. Indeed, let A be a square matrix. We know that A is invertible if and only if $\det(A) \neq 0$. Also if A has order n, then the cofactor Ai,j is defined as the determinant of the square matrix of order (n-1) obtained from A by removing the row number i and the column number j multiplied by (-1)i+j. Recall

\begin{displaymath}\det(A) = \sum_{j=1}^{j=n} a_{ij} A_{ij}\end{displaymath}

for any fixed i, and

\begin{displaymath}\det(A) = \sum_{i=1}^{i=n} a_{ij} A_{ij}\end{displaymath}

for any fixed j. Define the adjoint of A, denoted adj(A), to be the transpose of the matrix whose ijth entry is Aij.

Example. Let

\begin{displaymath}A = \left(\begin{array}{rrr}
1&3&2\\
-1&0&2\\
3&1&-1\\
\end{array}\right).\end{displaymath}

We have

\begin{displaymath}adj(A) = \left(\begin{array}{rrr}
-2&5&-1\\
5&-7&8\\
6&-4&3...
...{array}{rrr}
-2&5&6\\
5&-7&-4\\
-1&8&3\\
\end{array}\right).\end{displaymath}

Let us evaluate $A \cdot adj(A)$. We have

\begin{displaymath}A \cdot adj(A) = \left(\begin{array}{rrr}
1&3&2\\
-1&0&2\\
...
...n{array}{rrr}
11&0&0\\
0&11&0\\
0&0&11\\
\end{array}\right).\end{displaymath}

Note that $\det(A) = 11$. Therefore, we have

\begin{displaymath}A \cdot adj(A) = \det(A) I_3.\end{displaymath}

Is this formula only true for this matrix, or does a similar formula exist for any square matrix? In fact, we do have a similar formula.

Theorem. For any square matrix A of order n, we have

\begin{displaymath}A \cdot adj(A) = \det(A) I_n.\end{displaymath}

In particular, if $\det(A) \neq 0$, then

\begin{displaymath}A^{-1} = \frac{1}{\det(A)} adj(A).\end{displaymath}

For a square matrix of order 2, we have

\begin{displaymath}adj \left(\begin{array}{rr}
a&b\\
c&d\\
\end{array}\right) ...
...^T = \left(\begin{array}{rr}
d&-b\\
-c&a\\
\end{array}\right)\end{displaymath}

which gives

\begin{displaymath}\left(\begin{array}{rr}
a&b\\
c&d\\
\end{array}\right)^{-1}...
...bc} \left(\begin{array}{rr}
d&-b\\
-c&a\\
\end{array}\right).\end{displaymath}

This is a formula which we used on a previous page.

On the next page, we will discuss the application of the above formulas to linear systems.

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Author: M.A. Khamsi

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