To solve the problem, we need to identify which type of matrix can be both symmetric and skew-symmetric.
1. Understanding Symmetric and Skew-Symmetric Matrices:
A matrix \( A \) is symmetric if:
\( A^T = A \)
A matrix \( A \) is skew-symmetric if:
\( A^T = -A \)
Now, if a matrix is both symmetric and skew-symmetric, then:
\( A = A^T = -A \Rightarrow A = -A \)
This implies:
\( 2A = 0 \Rightarrow A = 0 \)
So, the only matrix that satisfies both conditions is the null matrix (all elements are zero).
2. Evaluating Each Option:
(A) Unit Matrix → Not possible. It's symmetric but not skew-symmetric.
(B) Diagonal Matrix → Could be symmetric, but not necessarily skew-symmetric.
(C) Null Matrix → Satisfies both \( A = A^T \) and \( A = -A \). → Correct
(D) Row Matrix → Not necessarily square, and thus not even eligible for symmetric/skew-symmetric.
3. Conclusion:
The null matrix is the only matrix that is both symmetric and skew-symmetric.
Final Answer:
The correct answer is Null Matrix.
Determine whether each of the following relations are reflexive, symmetric, and transitive.
Show that the relation R in the set R of real numbers, defined as
R = {(a, b): a ≤ b2 } is neither reflexive nor symmetric nor transitive.
Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive.
In the matrix A= \(\begin{bmatrix} 2 & 5 & 19&-7 \\ 35 & -2 & \frac{5}{2}&12 \\ \sqrt3 & 1 & -5&17 \end{bmatrix}\),write:
I. The order of the matrix
II. The number of elements
III. Write the elements a13, a21, a33, a24, a23
If a matrix has 24 elements, what are the possible order it can have? What, if it has 13 elements?
If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?
Construct a 3×4 matrix, whose elements are given by
I. \(a_{ij}=\frac{1}{2}\mid -3i+j\mid\)
II. \(a_{ij}=2i-j\)
Find the value of x, y, and z from the following equation:
I.\(\begin{bmatrix} 4&3&\\x&5\end{bmatrix}=\begin{bmatrix}y&z\\1&5\end{bmatrix}\)
II. \(\begin{bmatrix}x+y&2\\5+z&xy\end{bmatrix}=\begin{bmatrix}6&2\\5&8\end{bmatrix}\)
III. \(\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}=\begin{bmatrix}9\\5\\7\end{bmatrix}\)