
For a matrix A, \[AI=A\] and $A{{A}^{T}} =I$ is true for
A. If A is a square matrix
B. If A is a non- singular matrix
C. If A is a symmetric matrix
D. If A is any matrix
Answer
163.2k+ views
Hint:
If we multiply a matrix with its transpose we will get a square identity matrix only but it is not applicable in a symmetric, skew-symmetric matrix or any other type of matrix. That is why it is true with a square matrix only.
Formula Used:
Let a matrix A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]and then its transpose will be equal to
\[{A^T} = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]
Now as we clearly see that both matrix A and its transpose are 2×2 matrices so it will be a square matrix.
So if we multiply A matrix with its transpose we will get an identity matrix and it is applicable only if A is a square matrix.
Complete Step-by-Step Solution:
So we can see other options whether it is satisfying these conditions too
A non-singular matrix is a square matrix whose determinant is not equal to zero. The non-singular matrix is an invertible matrix, and its inverse can be computed as it has a determinant value.
So as we can clearly see that if a matrix is nonsingular then, this condition doesn’t work.
Now, square matrix B which is of size n × n is considered to be symmetric if and only if \[BT = B\]. Consider the given matrix B, that is, a square matrix that is equal to the transposed form of that matrix, called a symmetric matrix.
So we can clearly see that if \[A’=A\] then this given condition will not satisfy the identity matrix.
So this option is also not correct.
So at last we have only one option left which is
For a matrix \[A\], \[AI = A\] and \[A{A^T} = I\] is true only if \[A\] is a square matrix.
Hence option A is correct.
Note :
We have to remember the following definitions of matrix like non-singular matrix, symmetric matrix and like that for solving such questions and always remember that whenever such a condition happens which is AI=A then this will only be true for square matrix only.
If we multiply a matrix with its transpose we will get a square identity matrix only but it is not applicable in a symmetric, skew-symmetric matrix or any other type of matrix. That is why it is true with a square matrix only.
Formula Used:
Let a matrix A= \[\left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]and then its transpose will be equal to
\[{A^T} = \left[ {\begin{array}{*{20}{c}}{{a_{11}}}&{{a_{12}}}\\{{a_{21}}}&{{a_{22}}}\end{array}} \right]\]
Now as we clearly see that both matrix A and its transpose are 2×2 matrices so it will be a square matrix.
So if we multiply A matrix with its transpose we will get an identity matrix and it is applicable only if A is a square matrix.
Complete Step-by-Step Solution:
So we can see other options whether it is satisfying these conditions too
A non-singular matrix is a square matrix whose determinant is not equal to zero. The non-singular matrix is an invertible matrix, and its inverse can be computed as it has a determinant value.
So as we can clearly see that if a matrix is nonsingular then, this condition doesn’t work.
Now, square matrix B which is of size n × n is considered to be symmetric if and only if \[BT = B\]. Consider the given matrix B, that is, a square matrix that is equal to the transposed form of that matrix, called a symmetric matrix.
So we can clearly see that if \[A’=A\] then this given condition will not satisfy the identity matrix.
So this option is also not correct.
So at last we have only one option left which is
For a matrix \[A\], \[AI = A\] and \[A{A^T} = I\] is true only if \[A\] is a square matrix.
Hence option A is correct.
Note :
We have to remember the following definitions of matrix like non-singular matrix, symmetric matrix and like that for solving such questions and always remember that whenever such a condition happens which is AI=A then this will only be true for square matrix only.
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