# Operations on Determinants

### Operations on Determinants:

1. Transpose Property:

• For any square matrix $A$, $\text{det}\left(A\right)=\text{det}\left({A}^{T}\right)$.
• The determinant of a matrix and its transpose are equal.
2. Row/Column Operations:

• Swapping rows or columns changes the sign of the determinant: $\text{det}\left(A\right)=-\text{det}\left({A}^{\mathrm{\prime }}\right)$, where ${A}^{\mathrm{\prime }}$ is the matrix obtained after swapping rows or columns in $A$.

• Multiplying a row or column by a scalar $k$ multiplies the determinant by $k$: $\text{det}\left(kA\right)={k}^{n}×\text{det}\left(A\right)$ for an $n×n$ matrix.

3. Adding a Multiple of One Row/Column to Another:

• If ${A}^{\mathrm{\prime }}$ is obtained by adding a multiple of one row or column to another in $A$, then $\text{det}\left(A\right)=\text{det}\left({A}^{\mathrm{\prime }}\right)$.
4. Row/Column Expansion:

• Expansion of a determinant along a row or column yields the sum of determinants of smaller matrices obtained by excluding that row or column.

### Application in Matrix Operations:

• The determinant of a square matrix $A$ is nonzero if and only if $A$ is invertible.

• The adjoint matrix $Adj\left(A\right)$ is calculated using cofactors and transposes, which involves determinants.

2. Solving Systems of Equations:

• Determinants play a crucial role in determining the solvability of systems of linear equations.

• Cramer's rule uses determinants to express the solutions of a system in terms of determinants of smaller matrices.

3. Geometry and Transformations:

• The absolute value of the determinant of a matrix representing a linear transformation determines the scaling factor and orientation change in space.

### Multiplying Determinants

Given two matrices $A$ and $B$ with determinants $\mathrm{\mid }A\mathrm{\mid }$ and $\mathrm{\mid }B\mathrm{\mid }$:

1. For $2×2$Matrices:

For matrices $2×2$:

$A=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ $B=\left[\begin{array}{cc}e& f\\ g& h\end{array}\right]$

The product of their determinants is: $\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=\left(ad-bc\right)×\left(eh-fg\right)$

1. For $3×3$ Matrices:

For matrices $3×3$:

$A=\left[\begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\right]$ $B=\left[\begin{array}{ccc}p& q& r\\ s& t& u\\ v& w& x\end{array}\right]$

The product of their determinants is not straightforward. It involves a more complex expansion of the matrices' elements according to the rule:

$\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=\mid \begin{array}{ccc}a& b& c\\ d& e& f\\ g& h& i\end{array}\mid ×\mid \begin{array}{ccc}p& q& r\\ s& t& u\\ v& w& x\end{array}\mid$

1. For Higher-Dimensional Matrices:

Beyond $3×3$ matrices, calculating the product of determinants becomes increasingly intricate due to expansion methods and may involve computational complexity.

### Example:

Let's take two $2×2$ matrices and multiply their determinants:

$A=\left[\begin{array}{cc}2& 3\\ -1& 4\end{array}\right]$ $B=\left[\begin{array}{cc}5& 1\\ 2& 0\end{array}\right]$

Calculating the determinants:

$\mathrm{\mid }A\mathrm{\mid }=\left(2×4\right)-\left(3×\left(-1\right)\right)=8+3=11$

$\mathrm{\mid }B\mathrm{\mid }=\left(5×0\right)-\left(1×2\right)=0-2=-2$

Now, let's find the product of the determinants:

$\mathrm{\mid }A\mathrm{\mid }×\mathrm{\mid }B\mathrm{\mid }=11×\left(-2\right)=-22$

Therefore, the product of the determinants of matrices $A$ and $B$ is $-22$.

### Differentiation of Determinants

Given a square matrix $A$ of size $n×n$ with elements depending on a variable $x$:

$A\left(x\right)=\left[\begin{array}{cccc}{a}_{11}\left(x\right)& {a}_{12}\left(x\right)& \cdots & {a}_{1n}\left(x\right)\\ {a}_{21}\left(x\right)& {a}_{22}\left(x\right)& \cdots & {a}_{2n}\left(x\right)\\ & & \ddots & \\ {a}_{n1}\left(x\right)& {a}_{n2}\left(x\right)& \cdots & {a}_{nn}\left(x\right)\end{array}\right]$

To differentiate the determinant of matrix $A$ with respect to $x$ ($\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }$), we use the following formula:

$\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }\cdot \text{Tr}\left({A}^{-1}\cdot \frac{dA}{dx}\right)$

Where:

• $\mathrm{\mid }A\mathrm{\mid }$denotes the determinant of matrix $A$.
• $\text{Tr}$ represents the trace of a matrix.
• ${A}^{-1}$ is the inverse of matrix $A$.
• $\frac{dA}{dx}$ represents the derivative of matrix $A$ with respect to $x$.

### Example:

Let's consider a $2×2$ matrix $A\left(x\right)$ dependent on $x$:

$A\left(x\right)=\left[\begin{array}{cc}x& {x}^{2}\\ 3x& 2\end{array}\right]$

First, calculate the determinant of matrix $A\left(x\right)$:

$\mathrm{\mid }A\mathrm{\mid }=x×2-{x}^{2}×3x=2x-3{x}^{3}$

Next, find the inverse of matrix $A\left(x\right)$:

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

Now, calculate $\frac{dA}{dx}$, the derivative of matrix $A\left(x\right)$ with respect to $x$:

$\frac{dA}{dx}=\left[\begin{array}{cc}1& 2x\\ 3& 0\end{array}\right]$

Next, find the trace of ${A}^{-1}\cdot \frac{dA}{dx}$:

${A}^{-1}\cdot \frac{dA}{dx}=\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2& -{x}^{2}\\ -3x& x\end{array}\right]\cdot \left[\begin{array}{cc}1& 2x\\ 3& 0\end{array}\right]$

${A}^{-1}\cdot \frac{dA}{dx}=\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2-3{x}^{2}& 2x\\ -3-3{x}^{3}& -3x\end{array}\right]$

Finally, find $\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }$ using the formula:

$\frac{d}{dx}\mathrm{\mid }A\mathrm{\mid }=\left(2x-3{x}^{3}\right)\cdot \text{Tr}\left(\frac{1}{2x-3{x}^{3}}\left[\begin{array}{cc}2-3{x}^{2}& 2x\\ -3-3{x}^{3}& -3x\end{array}\right]\right)$

This formula will yield the derivative of the determinant of matrix $A\left(x\right)$ with respect to $x$.

### Summation of Determinants

Given two square matrices $A$ and $B$ of the same size $n×n$, the sum or difference of their determinants can be found using the following rules:

1. Addition Rule: If $A$ and $B$ are $n×n$ matrices: $\mathrm{\mid }A+B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }$ The determinant of the sum of two matrices is the sum of their determinants.

2. Subtraction Rule: Similarly, if $A$ and $B$ are $n×n$ matrices: $\mathrm{\mid }A-B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }B\mathrm{\mid }$The determinant of the difference of two matrices is the difference of their determinants.

These rules apply when the matrices involved are of the same size.

### Example:

Let's consider two $3×3$ matrices $A$ and $B$:

$A=\left[\begin{array}{ccc}2& 1& 3\\ 4& 0& -1\\ 2& 3& 5\end{array}\right]$ $B=\left[\begin{array}{ccc}1& -2& 0\\ 3& 2& 1\\ -1& 4& 2\end{array}\right]$

First, calculate the determinants of matrices $A$ and $B$:

$\mathrm{\mid }A\mathrm{\mid }=\mid \begin{array}{ccc}2& 1& 3\\ 4& 0& -1\\ 2& 3& 5\end{array}\mid =\left(2×0×5\right)+\left(1×\left(-1\right)×2\right)+\left(3×4×3\right)-\left(3×0×2\right)-\left(2×\left(-1\right)×2\right)-\left(5×4×1\right)=18$

$\mathrm{\mid }B\mathrm{\mid }=\mid \begin{array}{ccc}1& -2& 0\\ 3& 2& 1\\ -1& 4& 2\end{array}\mid =\left(1×2×2\right)+\left(-2×1×\left(-1\right)\right)+\left(0×3×4\right)-\left(0×2×\left(-1\right)\right)-\left(2×3×\left(-1\right)\right)-\left(1×1×4\right)=8$

Now, find the sum and difference of the determinants:

1. Sum of Determinants: $\mathrm{\mid }A+B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }+\mathrm{\mid }B\mathrm{\mid }=18+8=26$

2. Difference of Determinants: $\mathrm{\mid }A-B\mathrm{\mid }=\mathrm{\mid }A\mathrm{\mid }-\mathrm{\mid }B\mathrm{\mid }=18-8=10$

Thus, the sum of the determinants of matrices $A$ and $B$ is $26$, and the difference of their determinants is $10$.