If \( A \) is a matrix and \( |A| \neq 0 \), then the solution of the equation \( XA = B \) is:
\( X = B A^{-1} \)
Solving Matrix Equations: Finding X in XA=B
Let's analyze the given matrix equation and find the solution for the matrix \(X\).
We are given the matrix equation:
\( XA = B \)
We are also given that \(A\) is a matrix and its determinant \(|A| \neq 0\). This condition is very important because it tells us that the matrix \(A\) is invertible. If \(|A| \neq 0\), then the inverse matrix, denoted as \(A^{-1}\), exists.
To find the matrix \(X\), we need to isolate \(X\) in the equation \(XA = B\). Since matrix multiplication is not commutative (in general, \(AB \neq BA\)), the side on which we multiply by the inverse matrix matters.
We have \(XA = B\). We want to eliminate \(A\) from the left side of \(X\). To do this, we must multiply by \(A^{-1}\) on the right side of both sides of the equation.
Starting with the equation:
\( XA = B \)
Multiply both sides by \(A^{-1}\) on the right:
\( (XA)A^{-1} = BA^{-1} \)
Using the associative property of matrix multiplication, we can group the terms on the left side:
\( X(AA^{-1}) = BA^{-1} \)
We know that the product of a matrix and its inverse is the identity matrix, \(I\). So, \(AA^{-1} = I\).
\( XI = BA^{-1} \)
Multiplying any matrix \(X\) by the identity matrix \(I\) results in the matrix \(X\) itself (i.e., \(XI = X\)).
\( X = BA^{-1} \)
So, the solution for the matrix \(X\) in the equation \(XA = B\) is \(X = BA^{-1}\).
Let's compare this result with the given options:
- Option 1: \( X = A^{-1} B \)
- Option 2: \( X = \frac{1}{|A|} B \)
- Option 3: \( X = B A^{-1} \)
- Option 4: \( X = AB^{-1} \)
Our derived solution \(X = BA^{-1}\) matches Option 3.
It is crucial to understand why multiplying by \(A^{-1}\) on the right is necessary. If we had multiplied by \(A^{-1}\) on the left, we would get \(A^{-1}(XA) = A^{-1}B\), which simplifies to \((A^{-1}X)A = A^{-1}B\). This does not isolate \(X\).
Therefore, the correct solution for the equation \(XA = B\) when \(|A| \neq 0\) is \(X = BA^{-1}\).
Revision Table: Key Matrix Concepts
| Concept | Description | Notation/Property |
|---|---|---|
| Matrix Determinant | A scalar value calculated from the elements of a square matrix. Indicates properties of the matrix. | \(|A|\) or \(\det(A)\) |
| Invertible Matrix | A square matrix \(A\) for which an inverse matrix \(A^{-1}\) exists such that \(AA^{-1} = A^{-1}A = I\). | \(|A| \neq 0\) is the condition for invertibility. |
| Inverse Matrix | For an invertible matrix \(A\), the inverse \(A^{-1}\) is unique. Used to solve matrix equations similar to how division is used with scalars. | \(AA^{-1} = I\) and \(A^{-1}A = I\) |
| Identity Matrix | A square matrix with 1s on the main diagonal and 0s elsewhere. Acts like the number 1 in scalar multiplication (\(AI = IA = A\)). | \(I\) or \(I_n\) (for size n) |
| Matrix Multiplication Associativity | Matrix multiplication is associative, meaning the grouping of matrices doesn't affect the result: \((AB)C = A(BC)\). | \((AB)C = A(BC)\) |
| Matrix Multiplication Non-commutativity | In general, the order of matrix multiplication matters: \(AB \neq BA\). | \(AB \neq BA\) (generally) |
Additional Information on Solving Matrix Equations
Solving matrix equations like \(AX = B\) or \(XA = B\) is a fundamental skill in linear algebra. The existence of the matrix inverse is key to solving these types of equations.
Consider the equation \(AX = B\), where \(|A| \neq 0\). To solve for \(X\), we multiply by \(A^{-1}\) on the left side:
\( A^{-1}(AX) = A^{-1}B \)
Using associativity:
\( (A^{-1}A)X = A^{-1}B \)
\( IX = A^{-1}B \)
\( X = A^{-1}B \)
Compare this to the problem we solved, \(XA = B\), which resulted in \(X = BA^{-1}\). This highlights the importance of multiplying by the inverse on the correct side due to the non-commutative nature of matrix multiplication.
The condition \(|A| \neq 0\) is essential. If \(|A| = 0\), the matrix \(A\) is singular or non-invertible, and \(A^{-1}\) does not exist. In such cases, the equation might have no solution or infinitely many solutions, but it cannot be solved using the inverse method described here.
Similar Questions
If \( A = \begin{bmatrix} x & -y \\ y & x \end{bmatrix} \) and \( A + A' = 2I \), then \( x \) is equal to:
If \( A \) is a square matrix such that \( A^2 = 2I \), then the value of \( (A - I)^2 + (A + I)^2 - 7A \) is:
If \( A = \begin{bmatrix} 0 & -3 & 2 \\ a & b & 8 \\ -2 & c & 0 \end{bmatrix} \) is a skew-symmetric matrix, then:
Important Questions from Matrices
Three defective bulbs are mixed with 8 good ones. If three bulbs are drawn one by one with replacement, the probabilities of getting exactly 1 defective, more than 2 defective, no defective and more than 1 defective respectively are :
If A, B, and C are three singular matrices given by:
\[ A = \begin{bmatrix} 1 & 4 \\ 3 & 2a \end{bmatrix} \]
\[ B = \begin{bmatrix} 3b & 5 \\ a & 2 \end{bmatrix} \]
and
\[ C = \begin{bmatrix} a + b + c & c + 1 \\ a + c & c \end{bmatrix} \]
then the value of abc is:
If \( P = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix} \) and \( Q = \begin{bmatrix} 2 & -4 & 1 \end{bmatrix} \) are two matrices, then \( (PQ)' \) will be:
The matrix
\[ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} \]
is a:
- \( \text{Scalar matrix} \)
- \( \text{Diagonal matrix} \)
- \( \text{Skew-symmetric matrix} \)
- \( \text{Symmetric matrix} \)
Choose the correct answer from the options given below:
If A is a square matrix and I is an identity matrix such that \(A^2 = A\), then \(A(I - 2A)^3 + 2A^3\) is equal to :
