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Find \(A^{T} ;\) make note if \(A\) is upper/lower triangular, diagonal, symmetric and/or skew symmetric. \(\left[\begin{array}{ccc}0 & 3 & -2 \\ 3 & -4 & 1 \\ -2 & 1 & 0\end{array}\right]\)

Short Answer

Expert verified
The matrix is symmetric.

Step by step solution

01

Identify the Matrix

The given matrix is \( A = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \). It is a 3x3 matrix.
02

Calculate the Transpose

The transpose of a matrix, denoted as \( A^T \), is obtained by swapping the rows and columns. Thus, for the given matrix, the transpose is \( A^T = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \). Notably, in this case, \( A^T = A \).
03

Determine if Matrix is Symmetric or Skew-Symmetric

A matrix is symmetric if \( A = A^T \). Comparing the original matrix with its transpose, we see that \( A = A^T \), therefore, the matrix is symmetric. For skew-symmetric, \( A = -A^T \), which doesn't apply here as the conditions do not hold.
04

Analyze Matrix Triangularity

A matrix is upper triangular if all elements below the main diagonal are 0 and lower triangular if all elements above the main diagonal are 0. In this matrix, there are non-zero elements both above and below the main diagonal, so it is neither upper nor lower triangular.
05

Check for Diagonal Matrix

A diagonal matrix has non-zero elements only on the main diagonal and zero elsewhere. This matrix has non-zero elements off the main diagonal, so it is not diagonal.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

symmetric matrix
A symmetric matrix is a square matrix that is equal to its transpose. In other words, for a matrix \( A \) to be symmetric, it must satisfy the condition \( A = A^T \).
When you look at the elements of a symmetric matrix, they mirror themselves across the main diagonal. This diagonal runs from the top left to the bottom right of the matrix.
For example, in our matrix \[ A = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \] When we calculate the transpose, \[ A^T = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \] we see that the matrix is unchanged. This confirms that the matrix is symmetric. Remember that not every square matrix is symmetric, only those that equal their transpose.
triangular matrix
A triangular matrix is a special type of square matrix where certain elements are always zero. There are two main types of triangular matrices:

  • Upper Triangular Matrix: Only the elements on or above the main diagonal are non-zero. All elements below the main diagonal must be zero.
  • Lower Triangular Matrix: Only the elements on or below the main diagonal are non-zero. All elements above the main diagonal are zero.
In our given matrix \[ A = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \] both above and below the diagonal have non-zero elements, so this matrix is neither upper nor lower triangular.
Recognizing triangular matrices is important because they often allow for more efficient mathematical calculations, especially when solving systems of linear equations.
diagonal matrix
A diagonal matrix is another special type of square matrix where most of the elements are zero, except for those on the main diagonal.
Specifically, a diagonal matrix \( D \) will have non-zero elements only from the top-left to the bottom-right positions, and all other elements off this diagonal must be zero. Mathematically, it looks like this:\[ D = \begin{bmatrix} d_1 & 0 & \ 0 & d_2 & 0 \ 0 & 0 & d_3 \end{bmatrix} \] In our exercise, the matrix \[ A = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \] clearly has non-zero values outside of the diagonal, making it not a diagonal matrix. Diagonal matrices are significant due to their simplicity in computational routines, like matrix multiplication and finding determinants.
transpose calculation
Understanding the transpose of a matrix is essential in linear algebra. The transpose of a matrix \( A \), denoted as \( A^T \), involves swapping its rows with columns.
This means that the element located in row \( i \) and column \( j \) in the original matrix will appear in row \( j \) and column \( i \) in the transposed matrix. Here’s the formal representation:\[ A = \begin{bmatrix} a & b \ c & d \end{bmatrix} \] time will transpose to:\[ A^T = \begin{bmatrix} a & c \ b & d \end{bmatrix} \] In our exercise, computing the transpose of \[ A = \begin{bmatrix} 0 & 3 & -2 \ 3 & -4 & 1 \ -2 & 1 & 0 \end{bmatrix} \] yielded a matrix that was identical to the original, reinforcing its symmetric nature. Mastering transpose calculations is crucial as it forms the basis for higher-level operations and properties in matrix algebra.

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Most popular questions from this chapter

Find the determinant of the \(2 \times 2\) matrix. \(\left[\begin{array}{cc}-1 & -7 \\ -5 & 9\end{array}\right]\)

Find the determinant of the given matrix using cofactor expansion along any row or column you choose. \(\left[\begin{array}{ccc}1 & 2 & 3 \\ -5 & 0 & 3 \\ 4 & 0 & 6\end{array}\right]\)

Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{ccc}4 & -4 & 0 \\ 5 & 1 & -1 \\ 3 & -1 & 2\end{array}\right], \quad \vec{b}=\left[\begin{array}{c}16 \\ 22 \\\ 8\end{array}\right]\)

Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{ccc}7 & -4 & 25 \\ -2 & 1 & -7 \\ 9 & -7 & 34\end{array}\right]\) \(\vec{b}=\left[\begin{array}{c}-1 \\ -3 \\ 5\end{array}\right]\)

Matrices \(A\) and \(\vec{b}\) are given. (a) Give \(\operatorname{det}(A)\) and \(\operatorname{det}\left(A_{i}\right)\) for all \(i\). (b) Use Cramer's Rule to solve \(A \vec{x}=\vec{b}\). If Cramer's Rule cannot be used to find the solution, then state whether or not a solution exists. \(A=\left[\begin{array}{cc}-8 & 16 \\ 10 & -20\end{array}\right], \quad \vec{b}=\left[\begin{array}{c}-48 \\ 60\end{array}\right]\)

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