Determinants

We learnt about basic matrix operations . In this chapter we will learn about another very important and widely used property of matrices, called determinant. 
 
I. Introduction
The determinant of a square matrix is a scalar value that is computed from its entries. For non-square matrices, determinants are not defined.

For a matrix with a single element, that is, a $1 \times 1$ matrix, the determinant is the element itself.
$\det(\begin{bmatrix} a \end{bmatrix}) = a$

Determinant of a square matrix of order $2 \times 2$ is as follows:
If $A = \begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{bmatrix}$, then

$|A| = a_{11} \cdot a_{22} - a_{12} \cdot a_{22}$

Conceptually, the magnitude of the determinant of a $2 \times 2$ matrix, is the area enclosed by the parallelogram formed by the two column (or row) vectors of the matrix.
For example, the $2 \times 2$ matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$. can be treated as a combination of two column vectors
$\begin{bmatrix} \begin{bmatrix} a \\ c \end{bmatrix} & \begin{bmatrix} b \\ d \end{bmatrix} \end{bmatrix}$ or as a combination of two row vectors like $\begin{bmatrix} \begin{bmatrix} a & b \end{bmatrix} \\ \begin{bmatrix} c & d \end{bmatrix} \end{bmatrix}$. Any $n \times n$ square matrix can be treated as a combination of $n$, $n$-dimensional column vectors or $n$, $n$-dimensional row vectors.

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The two column vectors can be represented as $v_1 =$
$a \hat{i} + c \hat{j}$
 and $v_2 = $
$b \hat{i} + d \hat{j}$
, the tips of which, in the Cartesian plane will be at $(a, c)$ and $(b, d)$ respectively.
It is easy to show that the parallelogram, drawn as shown in the following widget, will have the fourth point $D$ at $(a + c, b + d)$ and will have an area of $ad - bc$

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We will see a brief proof of this here. The above widget shows the concept using column vectors, therefore, for this proof we will consider the matrix as a combination of row vectors just to show that it works for both cases.

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Consider the matrix:
$\begin{bmatrix} p & q \\ r & s \end{bmatrix}$

Representing this as a combination of row vectors, we get:
$\begin{bmatrix} \begin{bmatrix} p & q\end{bmatrix} \\ \begin{bmatrix} r & s\end{bmatrix} \end{bmatrix}$

Let's call these two vectors as $\overrightarrow{OM} = \begin{bmatrix} p & q\end{bmatrix}$ and $\overrightarrow{ON} = \begin{bmatrix} r & s\end{bmatrix}$.
Now, we can plot these two vectors as follows:


The coordinates of the points $M$ and $N$ are $(p, q)$ and $(r, s)$ respectively. 
We draw a line passing through the point $M$,  parallel to $\overrightarrow{ON}$ and a line passing through $N$ parallel to $\overrightarrow{OM}$. Let these two lines meet at $P$. The coordinates of the point $P$ will be $(p + r, q + s)$ To find the area of the parallelogram $OMPN$, we draw perpendiculars to both axes from points $M$, $P$ and $N$ respectively as shown above.

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We can find the area of $OMPN$ using:
$[OMPN] = [OBPD] - [\triangle OAN] - [ANPB] - [\triangle OCM] - [CMPD]$
$[OBPD] = OB \times OD = (p + r)(q + s) = pq + ps + rq + rs$
$[\triangle OAN] = \dfrac{1}{2} OA \times AN = \dfrac{1}{2} r  s$
$[ANPB] = \dfrac{1}{2} AB (AN + BP) = \dfrac{1}{2} p (s + q + s)$
$[\triangle OCM] = \dfrac{1}{2} OC \times MC = \dfrac{1}{2} q p$
$[CMPD] = \dfrac{1}{2} CD \times (MC + PD) = \dfrac{1}{2} s \times (p + p + r)$

$\therefore [OMPN]$
$= [OBPD] - [\triangle OAN] - [ANPB] - [\triangle OCM] - [CMPD]$
$=  (pq + ps + rq + rs) - \dfrac{1}{2} r s -  \dfrac{1}{2} p (2s + q) - \dfrac{1}{2} q p - \dfrac{1}{2} s \times (2p + r)$
$= (pq + ps + rq + rs) - \dfrac{1}{2}rs - ps - \dfrac{1}{2}pq - \dfrac{1}{2}pq - ps - \dfrac{1}{2}sr$
$= pq + ps + rq + rs - 2ps - pq - rs = qr - ps$

The determinant of the given matrix, as we saw before, is given by $ps - qr$. 
Since, $|ps - qr| = |qr - ps|$. therefore, the magnitude of the two quantities are the same.

Similarly the magnitude of the determinant of a $3 \times 3$ matrix, is the volume of the parallelepiped formed by the three row or column vectors of that matrix.

The symbols commonly used for denoting the determinant of any matrix $A = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\  a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\  \vdots & \vdots & \ddots & \vdots \\  a_{n,1} & a_{n,2} & \cdots & a_{n, n}   \end{bmatrix}$   are as follows:

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$i)$ The matrix elements enclosed inside two plain vertical lines:
    $\begin{vmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\  a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\  \vdots & \vdots & \ddots & \vdots \\  a_{n,1} & a_{n,2} & \cdots & a_{n, n}   \end{vmatrix}$

$ii)$ The matrix with a $\det$ prefix:
      $\det \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\  a_{2,1} & a_{2,2} & \cdots & a_{2, n} \\  \vdots & \vdots & \ddots & \vdots \\  a_{n,1} & a_{n,2} & \cdots & a_{n, n}   \end{bmatrix}$

$iii)$ The name of the matrix surrounded by two vertical lines
        $|A|$

$iv)$ The $\Delta$ symbol with the name of the matrix as suffix:
       $\Delta_{A}$

II. Submatrix, Minors And Cofactors
Before we get onto finding determinants, we need to understand about minors and cofactors of a matrix.
A submatrix of a given matrix, $A$, is a matrix that is obtained by eliminating any collection or rows and columns from $A$. A submatrix can be obtained from not only square matrices, but any matrix of any random order. Since we are learning submatrices for the purpose of finding determinants, we will focus mostly on square matrices.

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Although there is no standard notation for submatrices, we will use the notation $S_{i,j}$ to denote a submatrix which is obtained by eliminating the $i \xasuper{th}$ row and the $j \xasuper{th}$ column from a given input matrix $A$.

For example, if we have:
$A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ as our original matrix, then submatrix, $S_{2, 3}$, of $A$ will be obtained by eliminating the $2\xasuper{nd}$ row and the $3 \xasuper{rd}$ column of $A$.

$\therefore S_{2, 3} = \begin{bmatrix} 1 & 2 & {\color{red}\xcancel{3}} \\ {\color{red}\xcancel{4}} & {\color{red}\xcancel{5}} & {\color{red}\xcancel{6}} \\ 7 & 8 & {\color{red}\xcancel{9}} \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix}$

The minor $m_{ij}$ of a matrix $A$ is given by the determinant of the submatrix $S_{ij}$ of $A$.
In our example $S_{2,3}$ of $A$ was the matrix $\begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix}$, therefore,
$m_{2,3} = \det \begin{bmatrix} 1 & 2 \\ 7 & 8 \end{bmatrix} = 1 \times 8 - 2 \times 7 = -6$

Now that we know what a minor is, understanding cofactor should just be a small step away.
The cofactor, $c_{i, j}$, of any element $a_{ij}$ of $A$ is the its minor $m_{i,j}$, multiplied by $(-1)^{i + j}$.
That is:
$c_{ij} = (-1)^{i + j} \cdot m_{ij}$

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A cofactor matrix is a matrix formed by replacing each element, $a_{ij}$, of a matrix with the corresponding cofactor, $c_{ij}$, of $a_{i,j}$.

For our example matrix of $\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}$ the cofactor values can be computed in the following way:

$c_{1,1} = (-1)^{1 + 1} \cdot \det(S_{1,1}) = 1 \cdot \begin{vmatrix} 5 & 6 \\ 8 & 9\end{vmatrix} = -3$
$c_{1,2} = (-1)^{1 + 2} \cdot \det(S_{1,2}) = (-1) \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9\end{vmatrix} = 6$
$c_{1,3} = (-1)^{1 + 3} \cdot \det(S_{1,3}) = 1 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8\end{vmatrix} = -3$
$c_{2,1} = (-1)^{2 + 1} \cdot \det(S_{2,1}) = (-1) \cdot \begin{vmatrix} 2 & 3 \\ 8 & 9\end{vmatrix} = 6$
$c_{2,2} = (-1)^{2 + 2} \cdot \det(S_{2,2}) = 1 \cdot \begin{vmatrix} 1 & 3 \\ 7 & 9\end{vmatrix} = -12$
$c_{2,3} = (-1)^{2 + 3} \cdot \det(S_{2,3}) = (-1) \cdot \begin{vmatrix} 1 & 2 \\ 7 & 8\end{vmatrix} = 6$
$c_{3,1} = (-1)^{3 + 1} \cdot \det(S_{2,1}) = 1 \cdot \begin{vmatrix} 2 & 3 \\ 5 & 6\end{vmatrix} = -3$
$c_{3,2} = (-1)^{3 + 2} \cdot \det(S_{2,1}) = (-1) \cdot \begin{vmatrix} 1 & 3 \\ 4 & 6\end{vmatrix} = 6$
$c_{3,3} = (-1)^{3 + 3} \cdot \det(S_{2,1}) = 1 \cdot \begin{vmatrix} 1 & 2 \\ 4 & 5\end{vmatrix} = -3$

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Therefore, we get the cofactor matrix, $C$ of $A$, as:

$\begin{bmatrix} -3 & 6 & -3 \\ 6 & -12 & 6 \\ -3 & 6 & -3 \end{bmatrix}$

III. Finding Determinants
Now that we have familiarized ourselves with the concepts of minors and cofactors of a matrix, we can go ahead a learn about finding the determinant.

To find the determinant of an $n \times n$ matrix $A$ we need to perform the following steps:
- Select any single row or column of the matrix. This is called the reference row or column. Let us say we selected the $k\xasuper{th}$ row of the matrix as the reference row.

- For each of the $n$ element $a_{kj}$ of the reference row/column, compute the cofactor, $c_{kj}$ of $a_{kj}$. If instead of a row a reference column is selected, then the cofactors $c_{ik}$ of each element $a_{ik}$ of the $k \xasuper{th}$ column needs to be computed.

- For any selected row $k$,  The determinant of $A$ is given by:
$|A| = \sum\limits_{j = 1}^{n}  a_{kj} \times c_{kj}$. The result will be same for any selected column as well.

This method of finding the determinant is called the Laplace Expansion of matrices.

Note that although the order of each submatrix $S_{ij}$ is one less than the order of $A$, it may still be higher than $1$. In that case the determinant of each $S_{ij}$ can be calculated by repeating the Laplace Expansion for each $S_{i,j}$.

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The following interactive example shows how the determinant of a $3 \times 3$ matrix is computed. You can select either row-wise or column-wise calculation and then click on any row or column, to view the calculation steps.

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IV. Elementary Row & Column Transformation
Now we will see some properties of matrices involving row and column manipulations, that does not change the value of the determinant, but can reduce the calculation time significantly.
Any transformation of a row involving adding or subtracting multiples of one or more other rows, does not alter the value of a matrix. For example, if we add $n$ times the second row of a matrix to the first row of a matrix, and replace the first row values with the new values then the determinant of the matrix does not change. This is written as $R_1 = R_1 \pm n \times R_2$. This is true for column transformations as well.


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We can take a simple example to understand it better.
Let us consider the following $3 \times 3$ matrix:
$A = \begin{bmatrix} 16 & 7 & 13 \\ 6 & 10 & 7 \\ 10 & 7 & 9\end{bmatrix}$
Using the first row to calculate the determinant, we get:
$|A|$
$= 16 \times (10 \times 9 - 7 \times 7) - 7 \times (6 \times 9 - 7 \times 10) + 13 \times (6 \times 7 - 10 \times 10)$
$= 14$ 

We can observe that $a_{2, 1} = 6$, which can be written as $2 \times (a_{2, 2} - a_{2, 3})$. 
Therefore, if we perform the transformation $C_1 = C_1 - 2(C2 - C3)$, then $a_{2, 1}$ will become $0$. Performing this transformation on $C_1$, we get:
$C_1 = \begin{bmatrix} 16 \\ 6 \\ 10 \end{bmatrix} - 2 \times \left(\begin{bmatrix} 7 \\ 10 \\ 7 \end{bmatrix} - \begin{bmatrix} 13 \\ 7 \\ 9 \end{bmatrix} \right) = \begin{bmatrix} 16 - 2(7 - 13) \\ 6 - 2 (10 - 7) \\ 10 - 2 (7 - 9) \end{bmatrix} = \begin{bmatrix} 28 \\ 0 \\ 14 \end{bmatrix}$.
Therefore, we get the new matrix $B$ as:
$B = \begin{bmatrix} 28 & 7 & 13 \\ 0 & 10 & 7 \\ 14 & 7 & 9 \end{bmatrix}$

Calculating the determinant of this matrix using the first column, we get:
$|B| = 28(10 \times 9 - 7 \times 7) - 0(7 \times 9 - 13 \times 7) + 14(7 \times 7 - 13 \times 10)$
$= 14$
Observe, that although $A \ne B$, $|A| = |B|$, and since one of the elements in the first column was $0$, our calculation was much shorter.

Now, we can see that in matrix $B$, $b_{1, 1} = 2 \times b_{3, 1}$
Therefore, the row transformation $R_1 = R_1 - 2 \times R_3$, will give us $b_{1, 1} = 0$

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Performing this transformation we get the first row as:
$R_1 = R_1 - 2 \times R_3$
$= \begin{bmatrix} 28 & 7 & 13 \end{bmatrix} - 2 \times \begin{bmatrix} 14 & 7 & 9 \end{bmatrix} = \begin{bmatrix} 0 & -7 & -5 \end{bmatrix}$

Therefore, we get the new matrix $C$ as:
$C = \begin{bmatrix} 0 & -7 & -5 \\ 0 & 10 & 7 \\ 14 & 7 & 9 \end{bmatrix}$

Now that we have two elements of the first column as $0$s, we can find the determinant of $C$ using the first column as:
$|C| = 14((-7) \times 7 - 10 \times (-5)) = 14$

Now we can see that although $A \ne B \ne C$, we get $|A| = |B| = |C|$

Matrices that have a determinant value of $0$ are called degenerate or singular matrices. Observe that for a non-degenerate matrix, it will never be possible to reduce a complete column or a row to all zeros using elementary transformations, whereas for a degenerate matrix such a reduction should be possible.
A few points to remember regarding elementary transforms:
The row or column that is being transformed must be included in the transform with a coefficient of $1$. 
$R_2 = R_1 + 4R_3$ is an invalid transformation.

$R_2 = 3R_2 + 2R_1$ is an invalid transformation, since this will change the determinant $\Delta$ to $3\Delta$. If such a transformation is applied, the final determinant value needs to be multiplied by the coefficients inverse, like in this case $\dfrac{1}{3}$

Multiplication of two rows or columns is not allowed.
$C_3 = C_3 - C_1 \cdot C_2$ is not a valid transformation.


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Here is a thinking assignment.
Why doesn't a valid row/column transformation change the value of the determinant, although it changes the matrix?
Hint: Think about a $2 \times 2$ matrix and the fact that the determinant represents the area of the parallelogram as shown in the first section.

Now it is time for some practice. Use the following widget to practice calculating determinants of matrices of different orders. Remember, for matrices of order $\ge 3$, performing row or column transformation will make your calculations much faster.

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V. Properties Of Determinants
Scalar Multiplication Of Row Or Column
Multiplying a single row or a single column of a matrix by a scalar quantity, $k$, will multiply the determinant by $k$. This is easy to see. Earlier we saw that the determinant of a matrix can be computed by selecting any row or column of the matrix. Now, if we multiply any row or column of the matrix by a scalar quantity $k$ and find the determinant using that row or column, then each term of the determinant expression will get multiplied by $k$. Therefore, the determinant of the matrix will also get multiplied by $k$.

Scalar Multiplication Of Matrix
When we multiply a matrix $A_{(n, n)}$ by a scalar quantity $k$, the resultant matrix will have a determinant of $k^n \Delta$. This is also fairly simple to see, because multiplying each row or column by $k$ contributes a multiplication factor of $k$ to the determinant. Therefore, multiplying all rows (or columns) by $k$ will contribute $n$ multiplying factors, each equal to $k$, since the matrix has $n$ rows and columns..
Therefore, $|kA_{(n,n)}| = k^n |A_{(n, n)}|$.

Determinant Of Transpose
Since determinant can be computed using any row or column, leading to the same result, interchanging the rows or columns of a matrix will not change the determinant. Therefore,
$|A| = |A^T|$ 

Determinant Of Matrix Products
If we have two square matrices, $A$ and $B$, of the same order, then:
$\Delta (AB) = \Delta (BA) = \Delta(A) \cdot \Delta(B)$

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Note that this rule does not hold for matrix addition subtraction, that is, in general:
$|A \pm B| \ne |A| \pm |B|$

Determinant Of Lower Or Upper Triangular Matrix
The determinant of a lower or upper triangular matrix is equal to the product of the diagonal elements.

Let us consider the following lower triangular matrix:
$A = \begin{bmatrix} a_{1,1} & a_{1,2} & \cdots & a_{1, n} \\  0 & a_{2,2} & \cdots & a_{2, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n}   \end{bmatrix}$

If we use the first column of the matrix to find the determinant, the only non-zero term in the first column is $a_{1,1}$, therefore,
$|A| = a_{1,1} \times \begin{vmatrix} a_{2,2} & a_{2,3} & \cdots & a_{2, n} \\  0 & a_{3,3} & \cdots & a_{3, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n} \end{vmatrix}$     $...eqn\ (i)$

Now we can observe that $\begin{bmatrix} a_{2,2} & a_{2,3} & \cdots & a_{2, n} \\  0 & a_{3,3} & \cdots & a_{3, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n} \end{bmatrix}$ is also a lower triangular matrix.

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Again, using the first column to calculate the determinant:
$\begin{vmatrix} a_{2,2} & a_{2,3} & \cdots & a_{2, n} \\  0 & a_{3,3} & \cdots & a_{3, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n} \end{vmatrix} = a_{2,2} \times \begin{vmatrix} a_{3,3} & a_{3,4} & \cdots & a_{3, n} \\  0 & a_{4,4} & \cdots & a_{4, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n} \end{vmatrix}$

Substituting the above in $eqn\ (i)$ we get:

$|A| = a_{1,1} \times a_{2,2} \times \begin{vmatrix} a_{3,3} & a_{3,4} & \cdots & a_{3, n} \\  0 & a_{4,4} & \cdots & a_{4, n} \\  \vdots & \vdots & \ddots & \vdots \\  0 & 0 & \cdots & a_{n, n} \end{vmatrix}$

Likewise, if we continue till we are left with just the bottom right element $a_{n,n}$, we will get:
$|A| = a_{1,1} \times a_{2,2} \times ... \times a_{n-1,n-1}\times \begin{vmatrix} a_{n, n} \end{vmatrix}$

Since $|a_{n, n}| = a_{n, n}$, we get:
$|A| = a_{1,1} \times a_{2,2} \times ... \times a_{n-1,n-1}\times a_{n, n} = \prod\limits_{i = 1}^{n} a_{i,i}$

Determinant Of Inverses
We will learn more about this when we learn about inverse of a matrix, but for now it should suffice to know that the inverse of a matrix $A$, denoted by $A^{-1}$, is such that $AA^{-1} = I$, where $I$ is the unit matrix.
The unit matrix has a determinant of $1$, since it can be considered as either a lower or upper triangular matrix, hence the determinant is equal to the product of the diagonal elements which are all ones.

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$AA^{-1} = I$
$\Rightarrow |AA^{-1}| = |I|$
$\Rightarrow |A||A^{-1}| = 1$
$\Rightarrow |A^{-1}| = \dfrac{1}{|A|}$

Now, let us try the following problem:
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