Determinants

Dear Student, after a discussion on Matrices , I would like to post the notes for Determinants here.

Definition: The determinant of a $1 × 1$ matrix $A$ = $(a)$ is the number $a$ itself.

The determinant of a $2 \times 2$ matrix  $A = \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{pmatrix}$ is 

$\text{det}(A)=|A|=$   \begin{vmatrix}a_{11}&a_{12}\\a_{21}&a_{22} \end{vmatrix} $=a_{11}a_{22}-a_{12}a_{21}$

The determinant of $3 \times 3$ matrix 
$A = \begin{pmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{pmatrix}$ is the quantity

$\text{det}(A)= $ \begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix} 

$=a_{11}a_{22}a_{33}+ a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}$

Although a determinant, strictly speaking, is a single number, the word "determinant" is occasionally used to refer to the matrix that gave rise to it. Thus, we might refer to the element 7 in the determinant  $\begin{vmatrix}1& 7\\8&0\end{vmatrix}$ meaning, of course, the element 7 in the matrix $\begin{pmatrix}1& 7\\8&0\end{pmatrix}$. We shall also speak of $3 \times 3$ determinants, or a third order determinant, meaning the determinants arising from $3 \times 3$ matrices.
Evaluating $3 \times 3$ Determinant: Consider

$\text{det}(A)=$ $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix} =a_{11}a_{22}a_{33}+ a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}$

We get 
$a_{11}a_{22}a_{33}+ a_{12}a_{23}a_{31}+a_{13}a_{21}a_{32}-a_{11}a_{23}a_{32}-a_{12}a_{21}a_{33}-a_{13}a_{22}a_{31}$
$=(a_{11}a_{22}a_{33}-a_{11}a_{23}a_{32})-(a_{12}a_{21}a_{33}-a_{12}a_{23}a_{31})+(a_{13}a_{21}a_{32}-a_{13}a_{22}a_{31})$
$=a_{11}(a_{22}a_{33}-a_{23}a_{32})-a_{12}(a_{21}a_{33}-a_{23}a_{31})+a_{13}(a_{21}a_{32}-a_{22}a_{31})$

$=a_{11}\begin{vmatrix} a_{22}&a_{23}\\a_{32}&a_{33}\end{vmatrix} -a_{12}\begin{vmatrix} a_{21}&a_{23}\\a_{31}&a_{33}\end{vmatrix} +a_{13}\begin{vmatrix} a_{21}&a_{22}\\a_{31}&a_{32}\end{vmatrix}$

The above expression is called expansion by the elements of first row of matrix $A$. The sum of the products of the elements by the $2 \times 2$ determinants thus formed, and each element provided with a proper sign, yields the $3 \times 3$ determinant. This is a general rule, and it works for any row or column.

The proper sign is given by this checkerboard of signs: $\begin{matrix} +&-&+\\-&+&-\\+&-&+\end{matrix}$

For example, the sign in the $(1, 1)$ position is $+$. To show a further example of how this works, we shall expand a $3 \times 3$ determinant by the elements of its second row:

$\text{det}(A)=$ $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix} = -a_{21}\begin{vmatrix} a_{12}&a_{13}\\a_{32}&a_{33}\end{vmatrix} +a_{22}\begin{vmatrix} a_{11}&a_{13}\\a_{31}&a_{33}\end{vmatrix} -a_{23}\begin{vmatrix} a_{11}&a_{12}\\a_{31}&a_{32}\end{vmatrix}$

we shall expand a $3 \times 3$ determinant by the elements of  third row:

$\text{det}(A)=$ $\begin{vmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33} \end{vmatrix} = a_{31}\begin{vmatrix} a_{12}&a_{13}\\a_{22}&a_{23}\end{vmatrix} -a_{32}\begin{vmatrix} a_{11}&a_{13}\\a_{21}&a_{23}\end{vmatrix} +a_{33}\begin{vmatrix} a_{11}&a_{12}\\a_{21}&a_{22}\end{vmatrix}$

Exercise:

1. Find 
1. $\begin{vmatrix} 6&3\\9&4 \end{vmatrix}$
2. $\begin{vmatrix} 2&-1\\0&3 \end{vmatrix}$
3. $\begin{vmatrix} x&3\\y&z \end{vmatrix}$
4. $\begin{vmatrix}1&2\\4&8  \end{vmatrix}$
5. $\begin{vmatrix}x&z+1\\x+2&z+3 \end{vmatrix}$
2. Expand $\begin{vmatrix} 1&2&1\\4&1&-4\\0&1&2 \end{vmatrix}$ by the elements of first column.
3. Show that $\begin{vmatrix} 1&2\\3&4\end{vmatrix} \times \begin{vmatrix} 5&6\\7&8\end{vmatrix}=4$
4. Show that $\begin{vmatrix} 1&2&4\\0&1&1\\1&1&1\end{vmatrix}=-2$, $\begin{vmatrix} 1&1&0\\0&1&1\\0&0&1\end{vmatrix}=1$ and $\begin{vmatrix} 1&0&2\\1&2&5\\6&8&0\end{vmatrix}=-48$
5. Solve for $x$: $\begin{vmatrix}x&2\\1&3\end{vmatrix}=2$, $\begin{vmatrix}x&1\\0&x\end{vmatrix}=4$,$\begin{vmatrix}x&1&1\\0&x&1\\0&0&x\end{vmatrix}=8$

Properties of $n \times n$ determinants:

(1) $\text{det}(A) = \text{det} (A^{\prime})$.

(2) If $B$ results from $A$ by interchanging a pair of rows or columns, $\text{det}(B) = - \text{det}(A)$.

(3) If $B$ results from $A$ by multiplying the elements of a row or column by a factor $k$ then $\text{det}(B) = k \text{det}(A)$.

(4) If two rows (or columns) of $A$ are identical or proportional, then $\text{det}(A) = 0$.

(5) If a row (or a column) of $A$ is identically 0, (that is, all the elements are zcro), then $\text{det}(A) = 0$

(6) If the $i^{th}$ row (or column) of $B$ equals the $i^{th}$ row (or column) of $A$ plus a multiple of the $j^{th}$ row (or column) of $A$ ­ and otherwise $B$ and $A$ are identical, then $\text{det}(B) = \text{det}(A)$. (Here $i \ne­ j$).

(7) If $I$ is the identity matrix, then $\text{det} I = 1$.

(8) If matrix $D$ is diagonal matrix $D=\begin{pmatrix}a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\end{pmatrix}$ then $|D| =a_{11}\times a_{22} \times a_{33}$

Occasionally it is convenient to evaluate a determinant by making use of Property (6) to reduce a row or column to as many zeros as possible and then to expand by the elements of that row or column.

Here is how it works:

Given a matrix  $D=\begin{pmatrix}1&1&6\\2&1&3\\4&2&2\end{pmatrix}$ to find $|D|=\begin{vmatrix}1&1&6\\2&1&3\\4&2&2\end{vmatrix}$

Subtract first row from second row, by property (6) we get  $|D|=\begin{vmatrix}1&1&6\\1&0&-3\\4&2&2\end{vmatrix}$

Multiply first row by -2 and add to third row, by property (6) we get 

$|D|=\begin{vmatrix}1&1&6\\1&0&-3\\2&0&-10\end{vmatrix}$

Now expand by second column we get

$|D|=-1\times \begin{vmatrix}1&-3\\2&-10\end{vmatrix}+0\times \begin{vmatrix}1&6\\2&-10\end{vmatrix}-0\times \begin{vmatrix}1&6\\1&-3\end{vmatrix}$

$|D|=-1\times (1\times -10-2\times -3)+0-0$

$|D| = -1\times -4 = 4$

Exercise: Evaluate the following determinants by making use of property (6)

1. $\begin{vmatrix}1&2&6\\3&0&1\\1&1&2\end{vmatrix}$
2. $\begin{vmatrix}2&1&2\\4&1&3\\6&3&1\end{vmatrix}$
3. $\begin{vmatrix}2&2&1\\1&2&2\\2&1&2\end{vmatrix}$

DETERMINANTS, MATRICES AND MULTIPLICATION

If $A$ and $B$ are two $n \times n$ matrices then $\text{det}(AB)=\text{det}(A) \times \text{det}(B)$ or using alternative symbol $|AB| = |A| \times |B|$

Exercise: Check the law of multiplication of determinants in following cases

1. $A= \begin{vmatrix}3&1\\2&1\end{vmatrix}$, $B=\begin{vmatrix}1&-4\\2&6\end{vmatrix}$
2. $A= \begin{vmatrix}2&9&1\\6&0&3\\1&1&0\end{vmatrix}$, $B=\begin{vmatrix}1&0&0\\0&1&0\\0&0&2\end{vmatrix}$

Definition: MINOR

Given an $n \times n$ matrix $A = (a_{ij})$, $i=1,2,\ldots, n$, $j=1,2,\ldots, n$ the minor of the element $a_{ij}$ is the determinant of the $(n-1) \times(n-1)$ matrix that results when the $i^{th}$ row and $j^{th}$ column of $A$ are deleted. If $M_{ij}$ is the resultant matrix then, minor of  $a_{ij}$ is $|M_{ij}|$

Example: If $A= \begin{pmatrix}3&0&6\\9&1&1\\2&0&-5\end{pmatrix}$, then 

$M_{22} =\begin{pmatrix}3&6\\2&-5\end{pmatrix}$ $\therefore$ $|M_{22}|=3\times-5 -2\times 6 = -27$

$M_{12} =\begin{pmatrix}9&1\\2&-5\end{pmatrix}$ $\therefore$ $|M_{12}|=9\times-5 -1\times 2 = -47$

The minor of element $a_{22}=1$ is $-27$ and minor of element $a_{12} =0$ is $-47$

Example: if  $A= \begin{pmatrix}2&3\\1&-4\end{pmatrix}$, then

$M_{12}=(1)_{1\times 1}$, $\therefore$ $|M_{12}| = 1$

$M_{22}=(2)_{1\times 1}$, $\therefore$ $|M_{22}| = 2$

The minor of element $a_{12}=3$ is $1$ and minor of element $a_{22} =-4$ is $2$

Definition: COFACTOR

 Given an $n \times n$ matrix $A=(a_{ij})$,$i=1,2,\ldots, n$, $j=1,2,\ldots, n$  the cofactor $c_{ij}$  of the element $a_{ij}$ is defined by $c_{ij} = (-1)^{i+j}\times \text{Minor of }a_{ij} = (-1)^{i+j}\times |M_{ij}|$

Where $M_{ij}$ is the $(n-1)\times(n-1)$ matrix, the results when the $i^{th}$ row and $j^{th}$ column of $A$ are deleted

Example: If $A= \begin{pmatrix}3&0&6\\9&1&1\\2&0&-5\end{pmatrix}$, then 

$M_{22} =\begin{pmatrix}3&6\\2&-5\end{pmatrix}$ $\therefore$ $|M_{22}|=3\times-5 -2\times 6 = -27$ Then $c_{22} = (-1)^{2+2} \times -27 = -27$

$M_{12} =\begin{pmatrix}9&1\\2&-5\end{pmatrix}$ $\therefore$ $|M_{12}|=9\times-5 -1\times 2 = -47$ Then $c_{12} = (-1)^{1+2} \times -47 = 47$

Example: if  $A= \begin{pmatrix}2&3\\1&-4\end{pmatrix}$, then

$M_{12}=(1)_{1\times 1}$, $\therefore$ $|M_{12}| = 1$ Then $c_{12} = (-1)^{1+2} \times 1 = -1$

$M_{22}=(2)_{1\times 1}$, $\therefore$ $|M_{22}| = 2$ Then $c_{22} = (-1)^{2+2} \times 2 =  2$

COFACTOR EXPANSION

The expansion of a $3 \times 3$ determinant according to the elements of a row or column can be formulated in this way.

If $\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}$

then   $|A| = \begin{vmatrix} a_{11}& a_{12}& a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix}$

Determinant $|A|$                                                     Expansion by the element of 

$|A| = a_{11}c_{11}+a_{12}c_{12}+a_{13}c_{13}$                           First Row

$|A| = a_{21}c_{21}+a_{22}c_{22}+a_{23}c_{23}$                           Second Row

$|A| = a_{31}c_{31}+a_{32}c_{32}+a_{33}c_{33}$                           Third Row

Determinant $|A|$                                                     Expansion by the element of 

$|A| = a_{11}c_{11}+a_{21}c_{21}+a_{31}c_{31}$                           First Column

$|A| = a_{12}c_{12}+a_{22}c_{22}+a_{32}c_{32}$                           Second Column

$|A| = a_{13}c_{13}+a_{23}c_{23}+a_{33}c_{33}$                           Third Column

Definition: INVERSE OF MATRIX

The matrix $A=\begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{pmatrix}$ has an inverse if and only if $|A| = \begin{vmatrix} a_{11}& a_{12}& a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33}\end{vmatrix} \ne 0$

The inverse if given by 

$A^{-1} = \text{Inv}(A) = \dfrac{1}{|A|} \begin{pmatrix} c_{11}&c_{12}&c_{13}\\ c_{21}&c_{22}&c_{23}\\ c_{31}&c_{32}&c_{33}\end{pmatrix}$ the quantities $c_{ij}$ are cofactor of $a_{ij}$

The matrix $\begin{pmatrix} c_{11}&c_{12}&c_{13}\\ c_{21}&c_{22}&c_{23}\\ c_{31}&c_{32}&c_{33}\end{pmatrix}$ is called an Adjoint matrix of matrix $A$. Therefore $A^{-1} = \text{Inv}(A) = \dfrac{\text{Adj}(A)}{|A|}$

The matrix $\begin{pmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\end{pmatrix}$ has inverse if and only if $|A| = a_{11}a_{22}-a_{21}a_{12}\ne 0$ and the inverse is given by the formula $A^{-1} = \text{Inv}(A) =\dfrac{1}{a_{11}a_{22}-a_{21}a_{12}}\begin{pmatrix} a_{22}&-a_{12}\\ -a_{21}&a_{11}\end{pmatrix}$

Given a square matrix $A$, a second square matrix $B$ (of the same order)    is called an inverse of $A$ if the following equations both hold:

$AB = I$ and $BA = I$

Note that,

(a)    If a matrix $A$ has an inverse $B$, then the matrix $B$ has an inverse $A$.

(b)  The unit matrix $I$ is its own inverse. $I × I = I$

Exercise: 

1. Verify that $ \begin{pmatrix}-6&7\\7&-8\end{pmatrix}$ is the inverse of $ \begin{pmatrix}8&7\\7&6\end{pmatrix}$
2. Verify that $ \begin{pmatrix}-1/3&0\\0&5\end{pmatrix}$ is the inverse of $ \begin{pmatrix}3&0\\0&1/5\end{pmatrix}$
3. Find the inverse of $ \begin{pmatrix}5&3\\3&2\end{pmatrix}$
4. Find the inverse of $A= \begin{pmatrix}0&1&0\\1&0&0\\0&0&1\end{pmatrix}$
5. Find the inverse of matrix D, where $D =\begin{pmatrix}1&1&6\\2&1&3\\4&2&2\end{pmatrix} $

 INVERSE MATRIX OF SPECIAL MATRICES

      If  $D$ is diagonal matrix $D=\begin{pmatrix}a_{11}&0&0\\0&a_{22}&0\\0&0&a_{33}\end{pmatrix}$ and if $a_{ii} \ne 0$, for $i=1,2,3$  then $D^{-1} = \begin{pmatrix}\dfrac{1}{a_{11}}&0&0\\0&\dfrac{1}{a_{22}}&0\\0&0&\dfrac{1}{a_{33}}\end{pmatrix}$

If a matrix is upper triangular or lower triangular, then it is not too difficult to compute its inverse.  The inverse will also be upper triangular or lower triangular.

        

      Definition: SINGULAR MATRIX

     An n×n matrix $A = (a_{ij})$,$i=1,2,\ldots, n$, $j=1,2,\ldots, n$  is called singular matrix if     $|A|=0$   otherwise it is called nonsingular.

      Exercise:

Which of the following matrices are singular and which are nonsingular?

$A= \begin{pmatrix}1&1\\2&3\end{pmatrix}$,$B= \begin{pmatrix}1&2\\5&8\end{pmatrix}$,$C= \begin{pmatrix}-1&6\\3&9\end{pmatrix}$,$D= \begin{pmatrix}0&0\\0&1\end{pmatrix}$,$E= \begin{pmatrix}2&3\\4&5\end{pmatrix}$

ALGEBRA OF INVERSE

1. If matrix $A$ has an inverse, then $A^{-1}$  has an inverse and $(A^{-1})^{-1} = A$
2. If matrix $A$ has an inverse and $k$ is a constant that is different from zero, then $kA$ has an inverse and $(kA)^{-1} = k^{-1} A^{-1} = \dfrac{1}{k}A^{-1}$
3. If matrices $A$ and $B$ has inverse, so does $AB$ and $(AB)^{-1}=B^{-1} A^{-1}$ This says that the inverse of the product is the product of the inverses, taken in reverse order. The reversal of order is important to keep in mind.
4. If the matrices $A_{1},A_{2},\ldots, A_{m}$ all have inverses, so does the product $A_{1}\times A_{2}\times,\ldots, \times A_{m}$ and $(A_{1}\times A_{2}\times,\ldots, \times A_{m})^{-1}=A_{m}^{-1}\times A_{m-1}^{-1}\times,\ldots, \times A_{1}^{-1}$
5. If matrix $A$ has an inverse, so does $A^{\prime}$  and $(A^{\prime})^{-1} = (A^{-1})^{\prime}$
6. if matrix $A$ is non- singular then $|A^{-1}| = \dfrac{1}{|A|}$

Exercise:

If $A= \begin{pmatrix}1&6\\2&1\end{pmatrix}$, $B= \begin{pmatrix}1&1\\3&2\end{pmatrix}$ then prove by direct computation,

1. $(AB)^{-1}=B^{-1} A^{-1}$
2. $(A^{\prime})^{-1} = (A^{-1})^{\prime}$
3. $(A^{-1})^{-1} = A$

Back to Basic Mathematics

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