๐ Understanding Matrix Equations: AX = B and XA = B
Matrix equations are fundamental in linear algebra. They provide a concise way to represent systems of linear equations and offer powerful tools for solving them. The equations $AX = B$ and $XA = B$ represent two different scenarios where we are trying to find an unknown matrix $X$. Understanding the conditions under which these equations have solutions and how to find those solutions is crucial.
๐ Historical Context
The study of matrices and linear equations dates back to ancient times, with early forms appearing in Babylonian and Chinese mathematics. However, the modern notation and systematic treatment of matrix equations were developed in the 19th century by mathematicians like Arthur Cayley and James Sylvester. These developments were driven by the need to solve complex systems of linear equations arising in various fields such as physics and engineering.
๐ Key Principles
- ๐ Matrix Multiplication: Understanding how to multiply matrices is essential. The dimensions must be compatible. For $AX = B$, the number of columns in $A$ must equal the number of rows in $X$. For $XA = B$, the number of columns in $X$ must equal the number of rows in $A$.
- ๐ข Invertibility: A square matrix $A$ is invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$, where $I$ is the identity matrix. Invertibility plays a crucial role in solving matrix equations.
- โ๏ธ Determinants: The determinant of a square matrix determines its invertibility. If the determinant is non-zero, the matrix is invertible. $det(A) \neq 0 \implies A \text{ is invertible}$.
- ๐ Rank: The rank of a matrix is the number of linearly independent rows or columns. The rank is useful in determining the existence and uniqueness of solutions to matrix equations.
- ๐ก Transpose: The transpose of a matrix $A$, denoted as $A^T$, is obtained by interchanging its rows and columns. This operation is useful in manipulating matrix equations.
๐งฎ Solving AX = B
If $A$ is a square and invertible matrix, we can solve $AX = B$ by multiplying both sides by $A^{-1}$ on the left:
$A^{-1}AX = A^{-1}B \implies IX = A^{-1}B \implies X = A^{-1}B$
Example 1:
Solve $AX = B$ where $A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 5 \\ 3 \end{bmatrix}$.
First, find the inverse of $A$. $det(A) = (2)(1) - (1)(1) = 1$.
$A^{-1} = \frac{1}{1} \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$.
Then, $X = A^{-1}B = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 5 \\ 3 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$.
๐ Solving XA = B
If $A$ is a square and invertible matrix, we can solve $XA = B$ by multiplying both sides by $A^{-1}$ on the right:
$XAA^{-1} = BA^{-1} \implies XI = BA^{-1} \implies X = BA^{-1}$
Example 2:
Solve $XA = B$ where $A = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 7 & 4 \\ 2 & 1 \end{bmatrix}$.
We already know $A^{-1} = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}$.
Then, $X = BA^{-1} = \begin{bmatrix} 7 & 4 \\ 2 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 3 & 1 \\ 1 & 0 \end{bmatrix}$.
โ๏ธ Solved Problems
Problem 1: Solve $AX = B$ where $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 2 \end{bmatrix}$.
Solution: $A^{-1} = \frac{1}{(1)(4)-(2)(3)} \begin{bmatrix} 4 & -2 \\ -3 & 1 \end{bmatrix} = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix}$.
$X = A^{-1}B = \begin{bmatrix} -2 & 1 \\ 3/2 & -1/2 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 0 \\ 1/2 \end{bmatrix}$.
Problem 2: Solve $XA = B$ where $A = \begin{bmatrix} 1 & 0 \\ 2 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}$.
Solution: $A^{-1} = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix}$.
$X = BA^{-1} = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix} = \begin{bmatrix} -5 & 4 \\ -7 & 6 \end{bmatrix}$.
๐ Practice Quiz
- โ Question 1: Solve for X: $\begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}X = \begin{bmatrix} 1 \\ 0 \end{bmatrix}$
- โ Question 2: Solve for X: $X\begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix} = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$
- โ Question 3: If $A = \begin{bmatrix} 4 & 5 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$, find $X$ in $AX = B$.
- โ Question 4: If $A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 1 \\ 2 & 0 \end{bmatrix}$, find $X$ in $XA = B$.
๐ Conclusion
Mastering matrix equations $AX = B$ and $XA = B$ involves understanding matrix multiplication, invertibility, and applying the appropriate techniques to solve for the unknown matrix $X$. By practicing and applying these concepts, you can confidently tackle a wide range of linear algebra problems. Keep practicing and exploring!