
Which of the following statements is true?
A. Non-singular square matrix does not have a unique inverse
B. Determinant of a non-singular matrix is zero
C. If\[{A^T} = A\], then A
is a square matrix
D. if\[\left| A \right| \ne 0\],
then
\[A = {\left| {{a_{ij}}} \right|_{n \times n}}\]
Answer
163.5k+ views
Hint: Since we know that the determinant of a singular matrix is always zero. Also square matrix contains 2 elements in rows as well as columns so by this information we can get our answer.
Formula Used:A transpose contains 4 elements if it’s a square matrix and a
matrix A also contains 4 elements if it’s a square matrix.
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.
Complete step by step solution:Now first of all we will see all the solutions and will see that
which is true
A singular matrix has a determinant value equal to zero, and a
non- singular matrix has a determinant whose value is a non- zero value. The
singular matrix does not have an inverse, and only a non- singular matrix
has an inverse matrix.
Hence we can clearly see that option A is not correct.
Now we will see other option which is
A matrix can be singular, only if it has a determinant of
zero. A matrix with a non-zero determinant certainly means a non-singular
matrix. In case the matrix has an inverse, then the matrix multiplied by its
inverse will give you the identity matrix.
Hence option B is also wrong
Now, coming to the third option.
We already know that if \[{A^T}
= A\]then the matrix will be a square matrix.
As we had seen in above formula, Let 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, Option ‘C’ is correct
Note:We have to remember many definitions as in this question.
Non-singular square matrices have a unique inverse. And Determinant of a
singular matrix is zero. Without such information we can’t do such questions
And always remember that If\[{A^T}
= A\], then A is a square matrix.
Formula Used:A transpose contains 4 elements if it’s a square matrix and a
matrix A also contains 4 elements if it’s a square matrix.
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.
Complete step by step solution:Now first of all we will see all the solutions and will see that
which is true
A singular matrix has a determinant value equal to zero, and a
non- singular matrix has a determinant whose value is a non- zero value. The
singular matrix does not have an inverse, and only a non- singular matrix
has an inverse matrix.
Hence we can clearly see that option A is not correct.
Now we will see other option which is
A matrix can be singular, only if it has a determinant of
zero. A matrix with a non-zero determinant certainly means a non-singular
matrix. In case the matrix has an inverse, then the matrix multiplied by its
inverse will give you the identity matrix.
Hence option B is also wrong
Now, coming to the third option.
We already know that if \[{A^T}
= A\]then the matrix will be a square matrix.
As we had seen in above formula, Let 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, Option ‘C’ is correct
Note:We have to remember many definitions as in this question.
Non-singular square matrices have a unique inverse. And Determinant of a
singular matrix is zero. Without such information we can’t do such questions
And always remember that If\[{A^T}
= A\], then A is a square matrix.
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