A matrix which is upper and lower triangular matrix both, must be:
(a) An identity matrix
(b) A scalar matrix
(c) A null matrix
(d) A diagonal matrix
Answer
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Hint: An upper triangular matrix is the one whose elements with i>j are zero while a lower triangular matrix is the one whose elements with i
Complete step-by-step answer:
Upper triangular matrix is the one whose elements ${{a}_{ij}}$ are zero if i>j and may or may not be zero for $i\le j$ . An example of such matrix is:
$\left| \begin{matrix}
2 & 7 & 6 \\
0 & 3 & 5 \\
0 & 0 & 4 \\
\end{matrix} \right|$
Lower triangular matrix on the other hand is the one whose elements having i$\left| \begin{matrix}
2 & 0 & 0 \\
5 & 3 & 0 \\
6 & 7 & 4 \\
\end{matrix} \right|$
Now let us move to the solution to the next question. We know if a matrix is a lower triangular as well as upper triangular matrix, all the elements with i>j or j>I are zero. So, the only elements which are non-zero are those which have i=j.
So, the matrix must be of the type:
$\left| \begin{matrix}
j & 0 & 0 \\
0 & k & 0 \\
0 & 0 & l \\
\end{matrix} \right|$
So, let us check which option suits it the most. First, it is not necessarily an identity matrix, as all the diagonal elements are not necessarily 1. It is not a scalar matrix as well, because it is not necessary that j, k, l are equal.
Also, it is not a constraint that j, k and l are 0, so the matrix is not a null matrix as well. However, all the elements other than the diagonal elements are zero, so it is surely a diagonal matrix.
Hence, the answer to the above question is option (d).
Note: Remember that there is a difference between possibility and surety. We have not concluded that the matrix was identity, scalar or null matrix because we didn’t have any constraint which confirms what the values of j, k and l are. Yes, but it is a diagonal element irrespective of the values of j,k and l, which is a surety, so we answered that it is a diagonal matrix.
Complete step-by-step answer:
Upper triangular matrix is the one whose elements ${{a}_{ij}}$ are zero if i>j and may or may not be zero for $i\le j$ . An example of such matrix is:
$\left| \begin{matrix}
2 & 7 & 6 \\
0 & 3 & 5 \\
0 & 0 & 4 \\
\end{matrix} \right|$
Lower triangular matrix on the other hand is the one whose elements having i
2 & 0 & 0 \\
5 & 3 & 0 \\
6 & 7 & 4 \\
\end{matrix} \right|$
Now let us move to the solution to the next question. We know if a matrix is a lower triangular as well as upper triangular matrix, all the elements with i>j or j>I are zero. So, the only elements which are non-zero are those which have i=j.
So, the matrix must be of the type:
$\left| \begin{matrix}
j & 0 & 0 \\
0 & k & 0 \\
0 & 0 & l \\
\end{matrix} \right|$
So, let us check which option suits it the most. First, it is not necessarily an identity matrix, as all the diagonal elements are not necessarily 1. It is not a scalar matrix as well, because it is not necessary that j, k, l are equal.
Also, it is not a constraint that j, k and l are 0, so the matrix is not a null matrix as well. However, all the elements other than the diagonal elements are zero, so it is surely a diagonal matrix.
Hence, the answer to the above question is option (d).
Note: Remember that there is a difference between possibility and surety. We have not concluded that the matrix was identity, scalar or null matrix because we didn’t have any constraint which confirms what the values of j, k and l are. Yes, but it is a diagonal element irrespective of the values of j,k and l, which is a surety, so we answered that it is a diagonal matrix.
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