Consider the following statements in respect of square matrices A and B of same order : 1. If AB is a null matrix, then at least one of A and B is a null matrix. 2. If AB is an identity matrix, then BA = AB. Which of the above statements is/are correct?
2 only
Analyzing Matrix Properties: Null and Identity Matrices
We are asked to evaluate two statements about square matrices A and B of the same order.
Analyzing Statement 1: AB is a null matrix implies A or B is a null matrix
Statement 1 says: If AB is a null matrix, then at least one of A and B is a null matrix.
Let's consider an example of two non-null square matrices A and B whose product AB is a null matrix. This would be a counterexample to the statement.
Consider the following 2x2 matrices:
\[ A = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \] \ \[ B = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \]
Neither matrix A nor matrix B is a null matrix (a matrix where all entries are zero). Now let's calculate their product AB:
\[ AB = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} (0)(0) + (1)(0) & (0)(1) + (1)(0) \\ (0)(0) + (0)(0) & (0)(1) + (0)(0) \end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]
The product AB is the null matrix. In this example, AB is a null matrix, but neither A nor B is a null matrix.
This counterexample shows that Statement 1 is false.
Analyzing Statement 2: AB is an identity matrix implies BA = AB
Statement 2 says: If AB is an identity matrix, then BA = AB.
If A and B are square matrices of the same order and AB is the identity matrix (denoted by I), this means that B is the inverse of A, and A is the inverse of B. This is written as \(B = A^{-1}\) or \(A = B^{-1}\).
A fundamental property of matrix inverses for square matrices is that if AB = I, then it is also true that BA = I. This property does not necessarily hold for non-square matrices, but it is true for square matrices of the same order.
Since AB = I and BA = I, it follows directly that AB = BA.
Therefore, Statement 2 is correct.
Summary of Analysis
- Statement 1: If AB is a null matrix, then at least one of A and B is a null matrix. (False)
- Statement 2: If AB is an identity matrix, then BA = AB. (True)
Based on the analysis, only Statement 2 is correct.
| Statement | Assessment | Reasoning |
|---|---|---|
| 1. If AB is a null matrix, then at least one of A and B is a null matrix. | Incorrect | Counterexample exists where AB is null, but A and B are non-null. |
| 2. If AB is an identity matrix, then BA = AB. | Correct | For square matrices, AB=I implies BA=I. Thus, BA = AB. |
Comparing our findings with the given options:
- 1. 1 only (Incorrect, Statement 1 is false)
- 2. 2 only (Correct, Statement 2 is true)
- 3. Both 1 and 2 (Incorrect, Statement 1 is false)
- 4. Neither 1 nor 2 (Incorrect, Statement 2 is true)
Revision Table: Matrix Properties Review
| Concept | Description | Important Note for Square Matrices |
|---|---|---|
| Null Matrix (O) | A matrix where all elements are zero. | If A and B are matrices, AB=O does NOT necessarily mean A=O or B=O (unlike scalar multiplication). |
| Identity Matrix (I) | A square matrix with ones on the main diagonal and zeros elsewhere. Acts like '1' in multiplication. | AI = IA = A for any matrix A of compatible size. |
| Matrix Inverse (\(A^{-1}\)) | For a square matrix A, its inverse \(A^{-1}\) satisfies \(AA^{-1} = A^{-1}A = I\). | If AB=I for square matrices A, B, then BA=I is automatically true. B is the inverse of A, and A is the inverse of B. |
| Commutativity | AB = BA. Matrix multiplication is generally NOT commutative (AB ≠ BA). | However, if AB=I for square matrices, then AB=BA=I, so they commute in this specific case. |
Additional Information: Matrix Multiplication
Matrix multiplication is a fundamental operation in linear algebra. Here are some key points relevant to the question:
- Definition: The product AB of two matrices A and B is defined only if the number of columns in A is equal to the number of rows in B.
- Order of Product: If A is an \(m \times n\) matrix and B is an \(n \times p\) matrix, the product AB is an \(m \times p\) matrix. The element in the \(i\)-th row and \(j\)-th column of AB is the sum of the products of corresponding elements from the \(i\)-th row of A and the \(j\)-th column of B.
- Non-Commutativity: In general, matrix multiplication is not commutative, meaning AB ≠ BA, even if both products are defined and of the same order. For example, if A and B are square matrices of the same order, AB and BA are both defined and are of the same order, but they are usually not equal.
- Properties: Matrix multiplication is associative \((A(BC) = (AB)C)\) and distributive over matrix addition \((A(B+C) = AB+AC)\) and \(( (A+B)C = AC+BC)\).
- Zero Product: As seen in the analysis of Statement 1, it is possible for the product of two non-zero matrices to be the zero matrix (null matrix). This is different from multiplication of real numbers, where if \(xy=0\), then either \(x=0\) or \(y=0\).
Understanding these properties is crucial for working with matrices, especially when dealing with concepts like null matrices, identity matrices, and inverses.
Similar Questions
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
Let \(A = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}\) and (mI + nA) 2= A where m, n are positive real numbers and I is the identify matrix. What is (m + n) equal to?
Let \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} {\rm x + y}& \rm y\\ {\rm 2x}&{\rm x - y} \end{array}} \right],\;\rm B = \left[ {\begin{array}{*{20}{c}} 2\\ { - 1} \end{array}} \right]\) and \(\rm C = \left[ {\begin{array}{*{20}{c}} 3\\ 2 \end{array}} \right]\) . If AB = C, then what is the value of the determinant of the matrix A?
If \(A = \left[ {\begin{array}{c} 1&{ - 1}\\ { - 1}&1 \end{array}} \right],\) then the expression A 3- 2A 2is
If \(A = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&3\\ 3&4 \end{array}} \right)\) and \(B = \left( {\begin{array}{*{20}{c}} 1&2\\ 2&1 \end{array}} \right),\) then which one of the following is correct?
A square matrix A is called orthogonal if_______ where A’ is the transpose of A.
Consider the following in respect of matrices A and B of same order:
1) A 2– B 2= (A + B) (A – B)
2) (A – I) (I + A) = O ⇔ A 2= I
Where I is the identity matrix and O is the null matrix.
Which of the above is/are correct?If \({\rm{f}}\left( {\rm{x}} \right) = \left[ {\begin{array}{c} {\cos {\rm{x}}}&{ - \sin {\rm{x}}}&0\\ {\sin {\rm{x}}}&{\cos {\rm{x}}}&0\\ 0&0&1 \end{array}} \right]\) , then which of the following are correct?
1. f(θ) × f(ϕ) = f(θ + ϕ)
2. The value of the determinant of the matrix f(θ) × f(ϕ) is 1
3. The determinant of f(x) is an even function.
Select the correct answer using the code given below:
Consider the following in respect of the matrices:
A = [m n], B = [-n -m] & \( C = \begin{bmatrix} m \\ -m\end{bmatrix} \)
1. CA = CB
2. AC = BC
3. C(A + B) = CA + CB
Which of the above statements is/are correct?
Important Questions from Operations on Matrices
The adjoint of matrix \(\left[ {\begin{array}{*{20}{c}} a&b\\ c&d \end{array}} \right]\)is
Consider the following in respect of matrices A, B and C of same order:
1) (A + B + C)' = A' + B’ + C’
2) (AB)’ = A’B’
3) (ABC)’ = C’B’A’
Where A’ is the transpose of the matrix A.
Which of the above are correct?If \({\rm{A}} = \left[ {\begin{array}{*{20}{c}} 1&0&{ - 2}\\ 2&{ - 3}&4 \end{array}} \right]\) , then the matrix X for which 2X + 3A = 0 holds true is
Find a matrix X such that 2A + B + X = 0 , where
\(A=\begin{bmatrix} -1 & 2 \\\ 3 & 4 \end{bmatrix} \ \text{and} \;\rm B =\ \begin{bmatrix} 3 & -2 \\\ 1 & 5 \end{bmatrix} \ ?\)
The solution of the matrix equation \(\left[ {\begin{array}{*{20}{c}} 2&{ - 1}&3\\ 1&1&1\\ 1&{ - 1}&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} x\\ y\\ z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 9\\ 6\\ 2 \end{array}} \right]\) is:
