Let A and B be two non zero square matrics and AB and BA both are defined. It means
Both matrices (A) and (B) have same order
This question asks about the implications when two non-zero square matrices, A and B, can be multiplied in both orders, meaning both AB and BA are defined. Let's break down the conditions required for matrix multiplication.
For two matrices, say P and Q, to be multiplied to form the product PQ, the number of columns in matrix P must be equal to the number of rows in matrix Q.
Let's assume the order of matrix A is $m \times n$ and the order of matrix B is $p \times q$.
The problem statement specifies that A and B are square matrices. A square matrix is a matrix where the number of rows equals the number of columns.
Now let's use the information that A is $m \times m$ and B is $p \times p$ along with the conditions for AB and BA being defined:
Both conditions lead to the same conclusion: the number of rows (and thus columns, since they are square) of matrix A must be equal to the number of rows (and columns) of matrix B. This means matrix A and matrix B must have the same order.
Let's examine the given options based on our conclusion:
| Option | Statement | Analysis |
|---|---|---|
| 1 | No. of columns of A $\ne$ No. of rows of B | If this were true, AB would not be defined. This contradicts the problem statement. |
| 2 | No. of rows of A $\ne$ No. of columns of B | If this were true, BA would not be defined. This contradicts the problem statement. |
| 3 | Both matrices (A) and (B) have same order | Our analysis shows that if A is $m \times m$ and B is $p \times p$, and both AB and BA are defined, then $m=p$. This means they have the same order. This aligns with our findings. |
| 4 | Both matrices (A) and (B) does not have same order | This contradicts our finding that they must have the same order. |
Based on the analysis, the only statement that must be true is that both matrices A and B have the same order.
When two square matrices A and B are such that both the products AB and BA are defined, it necessarily implies that the matrices A and B must be of the same order. If matrix A is of order $m \times m$ and matrix B is of order $p \times p$, then the condition for both AB and BA to be defined forces $m=p$.
| Operation | Condition for definition | Order of resulting matrix (if A is $m \times n$, B is $p \times q$) |
|---|---|---|
| A $\times$ B (AB) | Number of columns of A = Number of rows of B ($n=p$) | $m \times q$ |
| B $\times$ A (BA) | Number of columns of B = Number of rows of A ($q=m$) | $p \times n$ |
Square matrices are important in linear algebra because they have special properties, particularly when their order is the same. For example, if two square matrices A and B of the same order are multiplied, the resulting matrix (AB or BA) is also a square matrix of the same order. If A is $m \times m$ and B is $m \times m$, then:
This confirms that if A and B are square matrices of the same order, both AB and BA are always defined.
The eigenvalues of the 3 × 3 matrix M = \(\left(\begin{array}{lll}\rm a^2 & \rm a b & \rm a c \\ \rm a b & \rm b^2 & \rm b c \\ \rm a c & \rm b c &\rm c^2\end{array}\right)\) are
A generic 3 × 3 real matrix A has eigenvalues 0, 1 and 6, and I is the 3 × 3 identity matrix. The quantity/quantities that cannot be determined from this information is/are the
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Let A be a non-singular diagonalisable matrix of order 3 with eignvalues λ1, λ2, λ3. A -1 is diagonalisable if:
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Then det A = 0, since all elements in column II are zero
Reason (R): Laplace expansion permits evaluation of a determinant along any row or column