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Example.   
+
----
 +
 
 +
<b>Example.</b>  
 
<br>
 
<br>
 
<br>
 
<br>
<math>A = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right)</math>  
+
<math>A = \left(\begin{array}{cc}4&3\\3&2\end{array}\right)</math>  
 
<br>
 
<br>
<math>A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)</math>   
+
<math>A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right)</math>   
  
 
<br>
 
<br>
  
<math>AA^{-1} = \left(\begin{array}{cccc}4&3\\3&2\end{array}\right)</math><math>\left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right) = </math> <math>\left(\begin{array}{cccc}1&0\\0&1\end{array}\right)</math>  
+
<math>AA^{-1} = \left(\begin{array}{cc}4&3\\3&2\end{array}\right)</math><math>\left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) = </math> <math>\left(\begin{array}{cc}1&0\\0&1\end{array}\right)</math>  
  
 
<br>
 
<br>
 
and
 
and
 
<br>
 
<br>
<math>A^{-1} = \left(\begin{array}{cccc}-2&3\\3&-4\end{array}\right)</math><math>\left(\begin{array}{cccc}4&3\\3&2\end{array}\right) = </math><math>\left(\begin{array}{cccc}1&0\\0&1\end{array}\right)</math>   
+
<math>A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right)</math><math>\left(\begin{array}{cc}4&3\\3&2\end{array}\right) = </math><math>\left(\begin{array}{cc}1&0\\0&1\end{array}\right)</math>   
  
 
----
 
----
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<p> If A is a nonsingular matrix, then <i>A</i><sup>T</sup> is nonsingular and (<i>A</i><sup>-1</sup>)<sup>T</sup> = (<i>A</i><sup>T</sup>)<sup>-1</sup>. </p>
 
<p> If A is a nonsingular matrix, then <i>A</i><sup>T</sup> is nonsingular and (<i>A</i><sup>-1</sup>)<sup>T</sup> = (<i>A</i><sup>T</sup>)<sup>-1</sup>. </p>
  
 +
 +
----
 +
 +
<h2> List of Equivalences </h2>
 +
 +
<p> The following are equivalent for an <i>n</i> x <i>n</i> matrix <i>A</i>:
 +
 +
<br>
 +
<b>1.</b> <i>A</i> is nonsingular.<br>
 +
<b>2.</b> <i>A</i><b>x</b> = <b>0</b> has only the trivial solution.<br>
 +
<b>3.</b> <i>A</i> is row(column equivalent to <i>I</i>. (The reduced row echelon form of <i>A</i> is <i>I</i>).<br>
 +
<b>4.</b> The linear system <i>A</i><b>x</b> = <b>b</b> has a unique solution for every <i>n</i> x 1 matrix <b>b</b>.<br>
 +
<b>5.</b> <i>A</i> is a product of elementary matrices.<br>
 +
</p>
  
 
----
 
----
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<h2> Methods for determining the inverse of a matrix </h2>
 
<h2> Methods for determining the inverse of a matrix </h2>
  
<h4> Shortcut for determining the inverse of a 2 x 2 matrix </h4>
+
<h4> 1. Shortcut for determining the inverse of a 2 x 2 matrix </h4>
  
<p> If      <math>A = \left(\begin{array}{cccc}a&b\\c&d\end{array}\right)</math>  then the inverse of matrix A can be found using:
+
<p> If      <math>A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right)</math>  then the inverse of matrix A can be found using:
 
<br>
 
<br>
<math>A^{-1} = \frac{1}{detA}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cccc}d&-b\\-c&a\end{array}\right)</math>
+
<math>A^{-1} = \frac{1}{detA}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right)</math>
  
 +
</p>
  
 +
<br>
 +
<br>
  
 +
----
  
 +
<p> <b>Example</b>
  
 +
<br>
 +
<br>
 +
 +
<math>A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right)</math>
 +
 +
<br>
 +
 +
<math> detA = ad - bc = 1 \times 4 - 2 \times 3 = -2 </math>
 +
 +
<br>
 +
 +
<math>A^{-1} = \frac{1}{-2}\left(\begin{array}{cc}4&-2\\-3&1\end{array}\right)</math>
 +
 +
<br>
 +
 +
<math>A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right)</math>
 +
 +
</p>
 +
 +
----
 +
 +
----
 +
 +
<h4> 2. Augmented Matrix Method (Can be used for any n x n matrix) </h4>
 +
 +
<p> Use Gauss-Jordan Elimination to transform [ A | I ] into [ I | A<sup>-1</sup> ]. </p>
 +
 +
<math>\left(\begin{array}{ccc|ccc}a&b&c&1&0&0\\d&e&f&0&1&0\\g&h&i&0&0&1\end{array}\right)</math>
 +
 +
<p> Perform elementary row operations to transform the A matrix into the identity matrix.  Whatever row operations are carried out on the A matrix, should also be carried out on the augmented Identity matrix.  The final state of the matrix on the left is the inverse of the A matrix. </p>
 +
 +
 +
<h2> Theorem  1</h2>
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 +
<p> An <i>n</i> x <i>n</i> matrix <i>A</i> is singular if and only if <i>A</i> is row equivalent to a matrix <i>B</i> that has a row of zeros. (That is, the reduced row echelon form of <i>A</i> has a row of zeros.) </p>
 +
 +
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<h2> Theorem 2 </h2>
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<p> If <i>A</i> and <i>B</i> are <i>n</i> x <i>n</i> matrices such that <i>A</i><i>B</i> = <i>I</i>, then <i>B</i><i>A</i> = <i>I</i>. Thus <i>B</i> = <i>A</i><sup>-1</sup>. </p>
 +
 +
----
 +
 +
 +
<p> <b>Example.</b>
 +
 +
 +
Find A<sup>-1</sup> of
 +
 +
<br>
 +
 +
<math>A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right)</math>
 +
 +
<br>
 +
<br>
 +
<math>\left(\begin{array}{cc|cc}1&2&1&0\\3&4&0&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&-2&-3&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right)\longrightarrow
 +
\left(\begin{array}{cc|cc}1&0&-2&1\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right)</math>
 +
 +
<br>
 +
<br>
 +
 +
so
 +
 +
<br>
 +
 +
<math>A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right)</math>
 +
 +
</p>
 +
 +
 +
 +
 +
----
 +
 +
----
 +
 +
<h4> 3. Adjoint Matrix Method (Can be used for any n x n matrix) </h4>
 +
 +
 +
<p>
 +
<math>A^{-1} = \frac{1}{detA}(adjoint(A)) = \frac{1}{detA}(cofactor(A^{T}))</math>
 +
</p>
 +
 +
----
 +
 +
<p> <b>Example.</b>
 +
 +
 +
Find A<sup>-1</sup> of
 +
 +
<br>
 +
 +
<math>A = \left(\begin{array}{ccc}1&2&3\\0&4&5\\1&0&6\end{array}\right)</math>
 +
 +
<br>
 +
 +
<math> detA = 4 \times (6 - 3) - 5 \times (0 - 2) = 12 + 10 = 22 </math>
 +
 +
<br>
 +
 +
Adjoint Matrix:
 +
 +
<br>
 +
 +
<math>adjoint(A) = \left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right)</math>
 +
<br>
  
 +
<math>A^{-1} = \frac{1}{22}\left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) = \left(\begin{array}{ccc}\frac{12}{11}&\frac{-6}{11}&\frac{-1}{11}\\\frac{5}{22}&\frac{3}{22}&\frac{-5}{11}\\\frac{-2}{11}&\frac{1}{11}&\frac{2}{11}\end{array}\right)</math>
  
  
 
[[Category:MA265Fall2012Alvarado]]
 
[[Category:MA265Fall2012Alvarado]]

Latest revision as of 11:32, 11 December 2012

Inverse of a Matrix

An n x n matrix A is said to have an inverse provided there exists an n x n matrix B such that AB = BA = In. We call B the inverse of A and denote it as A-1. Thus, AA-1 = A-1A = In. In this case, A is also called nonsingular.



Example.

$ A = \left(\begin{array}{cc}4&3\\3&2\end{array}\right) $
$ A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) $


$ AA^{-1} = \left(\begin{array}{cc}4&3\\3&2\end{array}\right) $$ \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) = $ $ \left(\begin{array}{cc}1&0\\0&1\end{array}\right) $


and
$ A^{-1} = \left(\begin{array}{cc}-2&3\\3&-4\end{array}\right) $$ \left(\begin{array}{cc}4&3\\3&2\end{array}\right) = $$ \left(\begin{array}{cc}1&0\\0&1\end{array}\right) $


Theorem 1

The inverse of a matrix, if it exists, is unique

Theorem 2

If A and B are both nonsingular n x n matrices (i.e. invertible), then AB is nonsingular and (AB)-1 = B-1A-1.

Corollary 1

If A1, A2, ..., Ar are n x n nonsingular matrices, then A1A2...Ar is nonsingular an (A1A2...Ar)-1 = Ar-1Ar-1-1...A1-1.

Theorem 3

If A is a nonsingular matrix, then A-1 is nonsingular and (A-1)-1 = A.

Theorem 4

If A is a nonsingular matrix, then AT is nonsingular and (A-1)T = (AT)-1.



List of Equivalences

The following are equivalent for an n x n matrix A:
1. A is nonsingular.
2. Ax = 0 has only the trivial solution.
3. A is row(column equivalent to I. (The reduced row echelon form of A is I).
4. The linear system Ax = b has a unique solution for every n x 1 matrix b.
5. A is a product of elementary matrices.


Methods for determining the inverse of a matrix

1. Shortcut for determining the inverse of a 2 x 2 matrix

If $ A = \left(\begin{array}{cc}a&b\\c&d\end{array}\right) $ then the inverse of matrix A can be found using:
$ A^{-1} = \frac{1}{detA}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right) = \frac{1}{ad - bc}\left(\begin{array}{cc}d&-b\\-c&a\end{array}\right) $




Example

$ A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right) $
$ detA = ad - bc = 1 \times 4 - 2 \times 3 = -2 $
$ A^{-1} = \frac{1}{-2}\left(\begin{array}{cc}4&-2\\-3&1\end{array}\right) $
$ A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right) $



2. Augmented Matrix Method (Can be used for any n x n matrix)

Use Gauss-Jordan Elimination to transform [ A | I ] into [ I | A-1 ].

$ \left(\begin{array}{ccc|ccc}a&b&c&1&0&0\\d&e&f&0&1&0\\g&h&i&0&0&1\end{array}\right) $

Perform elementary row operations to transform the A matrix into the identity matrix. Whatever row operations are carried out on the A matrix, should also be carried out on the augmented Identity matrix. The final state of the matrix on the left is the inverse of the A matrix.


Theorem 1

An n x n matrix A is singular if and only if A is row equivalent to a matrix B that has a row of zeros. (That is, the reduced row echelon form of A has a row of zeros.)


Theorem 2

If A and B are n x n matrices such that AB = I, then BA = I. Thus B = A-1.



Example. Find A-1 of
$ A = \left(\begin{array}{cc}1&2\\3&4\end{array}\right) $

$ \left(\begin{array}{cc|cc}1&2&1&0\\3&4&0&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&-2&-3&1\end{array}\right)\longrightarrow\left(\begin{array}{cc|cc}1&2&1&0\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right)\longrightarrow \left(\begin{array}{cc|cc}1&0&-2&1\\0&1&\frac{3}{2}&\frac{-1}{2}\end{array}\right) $

so
$ A^{-1} = \left(\begin{array}{cc}-2&1\\\frac{3}{2}&\frac{-1}{2}\end{array}\right) $





3. Adjoint Matrix Method (Can be used for any n x n matrix)


$ A^{-1} = \frac{1}{detA}(adjoint(A)) = \frac{1}{detA}(cofactor(A^{T})) $


Example. Find A-1 of
$ A = \left(\begin{array}{ccc}1&2&3\\0&4&5\\1&0&6\end{array}\right) $
$ detA = 4 \times (6 - 3) - 5 \times (0 - 2) = 12 + 10 = 22 $
Adjoint Matrix:
$ adjoint(A) = \left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) $
$ A^{-1} = \frac{1}{22}\left(\begin{array}{ccc}24&-12&-2\\5&3&-5\\-4&2&4\end{array}\right) = \left(\begin{array}{ccc}\frac{12}{11}&\frac{-6}{11}&\frac{-1}{11}\\\frac{5}{22}&\frac{3}{22}&\frac{-5}{11}\\\frac{-2}{11}&\frac{1}{11}&\frac{2}{11}\end{array}\right) $

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