User:Jw116104

Linear Algebra Extra Credit

For MATH 511, Submitted 11/9/2010

With examples, explain...

#1

...how to find matrix inverses using row operations.

We augment A with I, then do row operations until we have I|A^-1.

$A=\left[{\begin{array}{ccc}0&2&4\\2&4&2\\3&3&1\end{array}}\right]$

$(A|I)=\left[{\begin{array}{ccc|ccc}0&2&4&1&0&0\\2&4&2&0&1&0\\3&3&1&0&0&1\end{array}}\right]$

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \rightarrow \left[\begin{array}{ccc|ccc} 2 & 4 & 2 & 0 & 1 & 0 \\ 0 & 2 & 4 & 1 & 0 & 0 \\ 3 & 3 & 1 & 0 & 0 & 1 \end{array}\right] }

$\rightarrow \left[{\begin{array}{ccc|ccc}1&2&1&0&1/2&0\\0&2&4&1&0&0\\3&3&1&0&0&1\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&2&1&0&1/2&0\\0&2&4&1&0&0\\0&-3&-2&0&-3/2&1\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&0&-3&-1&1/2&0\\0&1&2&1/2&0&0\\0&0&4&3/2&-3/2&1\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&0&-3&-1&1/2&0\\0&1&2&1/2&0&0\\0&0&1&3/8&-3/8&1/4\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&0&0&1/8&-5/8&3/4\\0&1&0&-1/4&3/4&-1/2\\0&0&1&3/8&-3/8&1/4\end{array}}\right]$

$A^{-1}=\left[{\begin{array}{ccc}1/8&-5/8&3/4\\-1/4&3/4&-1/2\\3/8&-3/8&1/4\end{array}}\right]$

#2

...how to recognize when a square matrix is not invertible using the method from #1.

$A=\left[{\begin{array}{ccc}1&2&1\\2&1&-1\\1&5&4\end{array}}\right]$

$(A|I)=\left[{\begin{array}{ccc|ccc}1&2&1&1&0&0\\2&1&-1&0&1&0\\1&5&4&0&0&1\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&2&1&1&0&0\\0&-3&-3&-2&1&0\\1&3&3&-1&0&1\end{array}}\right]$

$\rightarrow \left[{\begin{array}{ccc|ccc}1&2&1&0&1/2&0\\0&-3&-3&1&0&0\\0&0&0&-3&1&1\end{array}}\right]$

Since the A side of the augmentation now has a zero row, A is not invertible.

#3

...how to express an invertible matrix as a product of elementary matrices using the method from #1.

1. List the row ops used
2. Replace each with its “undo” row operation
3. Convert these to elementary matrices (apply to I) and list left to right

$A=\left[{\begin{array}{ccc}0&2&4\\2&4&2\\3&3&1\end{array}}\right]$

Step 1 $R1\leftrightarrow R2$

$1/2R1\rightarrow R1$

$R3-3R1\rightarrow R3$

$1/2R2\rightarrow R2$

$R1-2R2\rightarrow R1$

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle R3 + 3R2 \rightarrow R3}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle 1/4 R3 \rightarrow R3}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle R2 - 2R3 \rightarrow R2}

$R1+3R3\rightarrow R1$

Step 2 $R1\leftrightarrow R2$

$2R1\rightarrow R1$

$R3+3R1\rightarrow R3$

$2R2\rightarrow R2$

$R1+2R2\rightarrow R1$

$R3-3R2\rightarrow R3$

$4R3\rightarrow R3$

$R2+2R3\rightarrow R2$

$R1-3R3\rightarrow R1$

Step 3 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \left[\begin{array}{ccc} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{array}\right]\ast\left[\begin{array}{ccc} 2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\ast\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 3 & 0 & 1 \end{array}\right]\ast\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{array}\right]<br>\ast\left[\begin{array}{ccc} 1 & 2 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\ast\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & -3 & 1 \end{array}\right]\ast\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 4 \end{array}\right]\ast\left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 1 \end{array}\right]<br>\ast\left[\begin{array}{ccc} 1 & 0 & -3 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right] = \left[\begin{array}{ccc} 1/8 & -5/8 & 3/4 \\ -1/4 & 3/4 & -1/2 \\ 3/8 & -3/8 & 1/4 \end{array}\right] = A^{-1} }

#4

...how to calculate the determinant of a square matrix A.

First we must know that if we have a 2x2 matrix

$A=\left[{\begin{array}{cc}a&b\\c&d\end{array}}\right],$

then det(A) is the scalar ad - bc.

We must also know that Ã_ij is the matrix obtained by deleting row i and column j from A.

Now, the cofactor of the entry of A in row i, column j is the scalar (-1)^i+jdet(Ã_ij).

The determinant of A equals the sum of the products of each entry in row 1 of A multiplied by its cofactor.

So if we have $A=\left[{\begin{array}{cccc}2&0&0&1\\0&1&3&-3\\-2&-3&-5&2\\4&-4&4&-6\end{array}}\right],$

$|A|=(-1)^{2}(2)|{\tilde {A}}_{11}|+(-1)^{3}(0)|{\tilde {A}}_{12}|+(-1)^{4}(0)|{\tilde {A}}_{13}|+(-1)^{5}(1)|{\tilde {A}}_{14}|.$

Therefore,

$|A|=(-1)^{2}(2)\left|{\begin{array}{cccc}1&3&-3\\-3&-5&2\\-4&4&-6\end{array}}\right|+0+0+(-1)^{5}(1)\left|{\begin{array}{cccc}0&1&3\\-2&-3&-5\\4&-4&4\end{array}}\right|$

We would then continue using cofactor expansion on the 3x3 matrices and our problem will only involve taking determinants of 2x2 matrices.

$|A|=2(40)+0+0-1(48)=32.$

#5

...that for every square matrix A there exists a square matrix B such that AB = (|A|)*(I) is a square matrix where the diagonal entries are all |""A""| and non-diagonal entries are zero.

Let B be the square matrix with the same dimension as A whose i,j entry is (-1)^(i+j) det(A_{ji}) (called the transpose of the cofactor matrix, or the classical adjoint of A).

Here is an example with a 2x2 matrix:

$A=\left[{\begin{array}{cc}1&4\\3&2\end{array}}\right],$

|A| = 2 - 12 = -10

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle A^t = \left[\begin{array}{cc} 1 & 3 \\ 4 & 2 \end{array}\right], }

and the cofactor matrix of A^t is

$B=\left[{\begin{array}{cc}2&-4\\-3&1\end{array}}\right].$

$AB=\left[{\begin{array}{cc}1&3\\4&2\end{array}}\right]\left[{\begin{array}{cc}2&-4\\-3&1\end{array}}\right]=\left[{\begin{array}{cc}-10&0\\0&-10\end{array}}\right].$