If A and B are symmetric matrices of the same order, then AB − BA is a:
Skew-symmetric matrix
Understanding Matrix Properties: Symmetric and Skew-Symmetric Matrices
This question asks us to determine the nature of the matrix \(AB - BA\) given that matrices \(A\) and \(B\) are symmetric and of the same order. To solve this, we need to understand the definitions of symmetric and skew-symmetric matrices and how transposing matrices works.
What is a Symmetric Matrix?
A square matrix \(M\) is called symmetric if its transpose is equal to the matrix itself. Mathematically, this is written as:
\(M^T = M\)
The question states that \(A\) and \(B\) are symmetric matrices of the same order. This means:
- \(A^T = A\)
- \(B^T = B\)
What is a Skew-Symmetric Matrix?
A square matrix \(M\) is called skew-symmetric if its transpose is equal to the negative of the matrix. Mathematically, this is written as:
\(M^T = -M\)
Analyzing the Expression AB − BA
We need to find out if the matrix \(C = AB - BA\) is symmetric, skew-symmetric, or something else. To do this, we will calculate the transpose of \(C\), which is \(C^T = (AB - BA)^T\).
We will use the following properties of matrix transpose:
- Transpose of a difference: \((X - Y)^T = X^T - Y^T\)
- Transpose of a product: \((XY)^T = Y^T X^T\)
Step-by-Step Transpose Calculation
Let's apply the transpose properties to the expression \(AB - BA\):
Starting with \(C = AB - BA\), we find \(C^T\):
\(C^T = (AB - BA)^T\)
Using the transpose of a difference property:
\(C^T = (AB)^T - (BA)^T\)
Now, using the transpose of a product property for both terms:
\(C^T = (B^T A^T) - (A^T B^T)\)
We know that \(A\) and \(B\) are symmetric, so \(A^T = A\) and \(B^T = B\). Substituting these into the expression:
\(C^T = (B A) - (A B)\)
\(C^T = BA - AB\)
Now, let's compare \(C^T = BA - AB\) with \(C = AB - BA\). We can factor out \(-1\) from \(BA - AB\):
\(BA - AB = -(AB - BA)\)
So, we have:
\(C^T = -(AB - BA)\)
Since \(C = AB - BA\), we can write this as:
\(C^T = -C\)
Conclusion on the Nature of AB − BA
The calculation shows that the transpose of the matrix \(AB - BA\) is equal to the negative of the matrix \(AB - BA\). By definition, this means that the matrix \(AB - BA\) is a skew-symmetric matrix.
Therefore, if A and B are symmetric matrices of the same order, then AB − BA is a skew-symmetric matrix.
Summary of Matrix Types
| Matrix Type | Condition | Example (2x2) |
|---|---|---|
| Symmetric | \(M^T = M\) | \(\begin{pmatrix} a & b \\ b & c \end{pmatrix}\) |
| Skew-Symmetric | \(M^T = -M\) | \(\begin{pmatrix} 0 & a \\ -a & 0 \end{pmatrix}\) |
| Zero Matrix | All elements are 0 | \(\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}\) |
| Identity Matrix | Diagonal elements are 1, others 0 | \(\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\) |
Revision Table: Matrix Transpose and Properties
| Property | Formula |
|---|---|
| Transpose of Transpose | \((A^T)^T = A\) |
| Transpose of Sum/Difference | \((A \pm B)^T = A^T \pm B^T\) |
| Transpose of Scalar Multiple | \((cA)^T = cA^T\) |
| Transpose of Product | \((AB)^T = B^T A^T\) |
Additional Information: Properties of Symmetric and Skew-Symmetric Matrices
Symmetric and skew-symmetric matrices have several interesting properties:
- Any square matrix \(A\) can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix: \(A = \frac{1}{2}(A + A^T) + \frac{1}{2}(A - A^T)\). Here, \(\frac{1}{2}(A + A^T)\) is symmetric and \(\frac{1}{2}(A - A^T)\) is skew-symmetric.
- The diagonal elements of a skew-symmetric matrix are always zero. This is because for a skew-symmetric matrix \(M\), \(M_{ii}^T = -M_{ii}\). Since \(M_{ii}^T = M_{ii}\), we have \(M_{ii} = -M_{ii}\), which implies \(2M_{ii} = 0\), so \(M_{ii} = 0\).
- If \(A\) is any square matrix, then \(A + A^T\) is symmetric, and \(A - A^T\) is skew-symmetric.
- If \(A\) is a symmetric matrix, then \(kA\) is also symmetric for any scalar \(k\).
- If \(A\) is a skew-symmetric matrix, then \(kA\) is also skew-symmetric for any scalar \(k\).
- If \(A\) and \(B\) are symmetric matrices of the same order, then \(AB\) is symmetric if and only if \(AB = BA\) (i.e., \(A\) and \(B\) commute).
Skew-symmetric matrix
Problem Analysis: Nature of AB - BA
We are given two matrices, A and B, which are symmetric and of the same order. We want to determine the nature of the matrix \( P = AB - BA \).
This approach will analyze the individual elements of the resulting matrix P.
Step-by-Step Solution using Index Notation:
- Define Symmetric Matrix using Index Notation:
A matrix M is symmetric if its element at the i-th row and j-th column is equal to the element at the j-th row and i-th column. \[ (M)_{ij} = (M)_{ji} \quad \text{for all } i, j \] Since A and B are symmetric: \[ (A)_{ij} = (A)_{ji} \] \[ (B)_{ij} = (B)_{ji} \] - Define Matrix Multiplication using Index Notation:
The element at the i-th row and j-th column of the product of two matrices X and Y is given by: \[ (XY)_{ij} = \sum_k (X)_{ik} (Y)_{kj} \] where the sum is over all possible values of the column index k of X (which is the same as the row index k of Y). - Define the Matrix P and its Element \( (P)_{ij} \):
Let \( P = AB - BA \). We want to find the element \( (P)_{ij} \). \[ (P)_{ij} = (AB - BA)_{ij} \] Using the property of matrix subtraction and the formula for matrix multiplication: \[ (P)_{ij} = (AB)_{ij} - (BA)_{ij} \] \[ (P)_{ij} = \sum_k (A)_{ik} (B)_{kj} - \sum_k (B)_{ik} (A)_{kj} \] - Calculate the Element \( (P)_{ji} \):
To determine if P is symmetric or skew-symmetric, we need to compare \( (P)_{ij} \) with \( (P)_{ji} \). Let's find \( (P)_{ji} \) by swapping i and j in the formula above: \[ (P)_{ji} = (AB - BA)_{ji} \] \[ (P)_{ji} = (AB)_{ji} - (BA)_{ji} \] \[ (P)_{ji} = \sum_k (A)_{jk} (B)_{ki} - \sum_k (B)_{jk} (A)_{ki} \] - Apply the Symmetric Property of A and B to \( (P)_{ji} \):
Since A and B are symmetric, we can replace \( (A)_{jk} \) with \( (A)_{kj} \), \( (B)_{ki} \) with \( (B)_{ik} \), \( (B)_{jk} \) with \( (B)_{kj} \), and \( (A)_{ki} \) with \( (A)_{ik} \). \[ (P)_{ji} = \sum_k (A)_{kj} (B)_{ik} - \sum_k (B)_{kj} (A)_{ik} \] - Rearrange the Terms in the Expression for \( (P)_{ji} \):
Since the elements like \( (A)_{kj} \) and \( (B)_{ik} \) are just numbers (scalars), their multiplication is commutative (\( a \times b = b \times a \)). We can reorder the terms within each product in the summation: \[ (P)_{ji} = \sum_k (B)_{ik} (A)_{kj} - \sum_k (A)_{ik} (B)_{kj} \] Now, let's compare this to the expression for \( (P)_{ij} \): \[ (P)_{ij} = \sum_k (A)_{ik} (B)_{kj} - \sum_k (B)_{ik} (A)_{kj} \] We can see that: \[ (P)_{ji} = - \left( \sum_k (A)_{ik} (B)_{kj} - \sum_k (B)_{ik} (A)_{kj} \right) \] Therefore: \[ (P)_{ji} = - (P)_{ij} \] - Interpret the Result:
The condition \( (P)_{ji} = - (P)_{ij} \) for all i and j is the definition of a skew-symmetric matrix.- If we had found \( (P)_{ji} = (P)_{ij} \), the matrix would be symmetric.
- If we had found \( (P)_{ij} = 0 \) for all i, j, the matrix would be a zero matrix.
- If P was the identity matrix, \( (P)_{ij} \) would be 1 if \( i = j \) and 0 if \( i \neq j \).
Final Answer (Alternative Approach):
By analyzing the elements of the matrix \( P = AB - BA \) using index notation and the property that A and B are symmetric (\( (A)_{ij} = (A)_{ji}, (B)_{ij} = (B)_{ji} \)), we found that \( (P)_{ij} = - (P)_{ji} \). This is the definition of a Skew-symmetric matrix.
Skew-symmetric matrix
Step 1: Recall Properties of Symmetric Matrices
For symmetric matrices: \[ A^T = A \quad \text{and} \quad B^T = B \] where \( ^T \) denotes the transpose operation.
Step 2: Compute the Transpose of AB − BA
\[ (AB - BA)^T = (AB)^T - (BA)^T = B^T A^T - A^T B^T \] Using the symmetric property: \[ = BA - AB = -(AB - BA) \]
Step 3: Identify the Matrix Type
A matrix \( M \) is skew-symmetric if \( M^T = -M \). Here: \[ (AB - BA)^T = -(AB - BA) \] Therefore, \( AB - BA \) is skew-symmetric.
Step 4: Verify Other Options
- Symmetric: Would require \( (AB-BA)^T = AB-BA \) (false)
- Zero matrix: Only if A and B commute (not generally true)
- Identity matrix: Doesn't satisfy the condition
Final Answer:
\( AB - BA \) is a \[ \boxed{\text{Skew-symmetric matrix}} \].
Key Observations:
- The commutator \( AB-BA \) of symmetric matrices is always skew-symmetric
- This result is fundamental in Lie algebra and quantum mechanics
- The zero matrix case only occurs when A and B commute (AB = BA)
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 :
