If \(A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right]\), then which one of the following statements is correct?
A2 is symmetric matrix with det(A2) = 0.
Understanding the Matrix Problem
We are given a matrix \(A\) and asked to determine the properties of \(A^2\). Specifically, we need to find out if \(A^2\) is a symmetric or skew-symmetric matrix and what its determinant is equal to.
The given matrix is:
\[ A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right] \]First, let's identify the type of matrix \(A\).
Identifying the Type of Matrix A
A matrix \(M\) is called symmetric if \(M^T = M\). A matrix \(M\) is called skew-symmetric if \(M^T = -M\).
Let's find the transpose of matrix \(A\), denoted by \(A^T\).
\[ A^T=\left[\begin{array}{rrr} 0 & -3 & -4 \\ 3 & 0 & -5 \\ 4 & 5 & 0 \end{array}\right] \]Now, let's compare \(A\) and \(A^T\).
If we multiply \(A\) by -1, we get:
\[ -A = -1 \times \left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right] = \left[\begin{array}{rrr} 0 & -3 & -4 \\ 3 & 0 & -5 \\ 4 & 5 & 0 \end{array}\right] \]We can see that \(A^T = -A\). Therefore, matrix \(A\) is a skew-symmetric matrix.
Calculating A Squared (\(A^2\))
Next, we need to calculate \(A^2\), which is \(A \times A\).
\[ A^2 = \left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right] \left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right] \]Let's perform the matrix multiplication:
- Element (1,1): \( (0)(0) + (3)(-3) + (4)(-4) = 0 - 9 - 16 = -25 \)
- Element (1,2): \( (0)(3) + (3)(0) + (4)(-5) = 0 + 0 - 20 = -20 \)
- Element (1,3): \( (0)(4) + (3)(5) + (4)(0) = 0 + 15 + 0 = 15 \)
- Element (2,1): \( (-3)(0) + (0)(-3) + (5)(-4) = 0 + 0 - 20 = -20 \)
- Element (2,2): \( (-3)(3) + (0)(0) + (5)(-5) = -9 + 0 - 25 = -34 \)
- Element (2,3): \( (-3)(4) + (0)(5) + (5)(0) = -12 + 0 + 0 = -12 \)
- Element (3,1): \( (-4)(0) + (-5)(-3) + (0)(-4) = 0 + 15 + 0 = 15 \)
- Element (3,2): \( (-4)(3) + (-5)(0) + (0)(-5) = -12 + 0 + 0 = -12 \)
- Element (3,3): \( (-4)(4) + (-5)(5) + (0)(0) = -16 - 25 + 0 = -41 \)
So, the resulting matrix \(A^2\) is:
| Col 1 | Col 2 | Col 3 | |
|---|---|---|---|
| Row 1 | -25 | -20 | 15 |
| Row 2 | -20 | -34 | -12 |
| Row 3 | 15 | -12 | -41 |
\[ A^2 = \left[\begin{array}{rrr} -25 & -20 & 15 \\ -20 & -34 & -12 \\ 15 & -12 & -41 \end{array}\right] \]
Checking Symmetry of A Squared (\(A^2\))
To check if \(A^2\) is symmetric or skew-symmetric, we find its transpose, \((A^2)^T\).
A property of matrix transpose is \((AB)^T = B^T A^T\). Also, \((A^T)^T = A\).
Since \(A\) is skew-symmetric, we know \(A^T = -A\).
Let's find \((A^2)^T\):
\[ (A^2)^T = (A \times A)^T = A^T \times A^T \]Substitute \(A^T = -A\):
\[ (A^2)^T = (-A) \times (-A) = (-1)(-1)(A \times A) = 1 \times A^2 = A^2 \]Since \((A^2)^T = A^2\), the matrix \(A^2\) is a symmetric matrix.
We can also verify this from the calculated matrix \(A^2\):
\[ A^2 = \left[\begin{array}{rrr} -25 & -20 & 15 \\ -20 & -34 & -12 \\ 15 & -12 & -41 \end{array}\right] \]Its transpose is:
\[ (A^2)^T = \left[\begin{array}{rrr} -25 & -20 & 15 \\ -20 & -34 & -12 \\ 15 & -12 & -41 \end{array}\right] \]Indeed, \((A^2)^T = A^2\). So, \(A^2\) is symmetric.
Calculating the Determinant of A (\(\det(A)\))
Now, let's find the determinant of \(A\). We can use the cofactor expansion along the first row.
\[ A=\left[\begin{array}{rrr} 0 & 3 & 4 \\ -3 & 0 & 5 \\ -4 & -5 & 0 \end{array}\right] \] \[ \det(A) = 0 \times \det\left[\begin{array}{rr} 0 & 5 \\ -5 & 0 \end{array}\right] - 3 \times \det\left[\begin{array}{rr} -3 & 5 \\ -4 & 0 \end{array}\right] + 4 \times \det\left[\begin{array}{rr} -3 & 0 \\ -4 & -5 \end{array}\right] \] \[ \det(A) = 0 \times ((0)(0) - (5)(-5)) - 3 \times ((-3)(0) - (5)(-4)) + 4 \times ((-3)(-5) - (0)(-4)) \] \[ \det(A) = 0 \times (0 + 25) - 3 \times (0 + 20) + 4 \times (15 - 0) \] \[ \det(A) = 0 \times 25 - 3 \times 20 + 4 \times 15 \] \[ \det(A) = 0 - 60 + 60 \] \[ \det(A) = 0 \]The determinant of \(A\) is 0.
Calculating the Determinant of A Squared (\(\det(A^2)\))
We know that for any two square matrices \(A\) and \(B\) of the same order, \(\det(AB) = \det(A)\det(B)\). Therefore,
\[ \det(A^2) = \det(A \times A) = \det(A) \times \det(A) = (\det(A))^2 \]Since we found that \(\det(A) = 0\), we can calculate \(\det(A^2)\):
\[ \det(A^2) = (0)^2 = 0 \]So, the determinant of \(A^2\) is 0.
Summary of Findings
Based on our calculations and properties, we found that:
- Matrix \(A\) is skew-symmetric.
- Matrix \(A^2\) is symmetric.
- The determinant of \(A\) is 0.
- The determinant of \(A^2\) is 0.
Evaluating the Options
Let's look at the given options:
- \(A^2\) is symmetric matrix with det(\(A^2\)) = 0.
- \(A^2\) is symmetric matrix with det(\(A^2\)) \(\ne\) 0.
- \(A^2\) is skew-symmetric matrix with det(\(A^2\)) = 0.
- \(A^2\) is skew-symmetric matrix with det(\(A^2\)) \(\ne\) 0.
Our findings show that \(A^2\) is symmetric and det(\(A^2\)) = 0. This matches option 1.
Revision Table: Key Matrix Concepts
| Concept | Definition | Property Used Here |
|---|---|---|
| Symmetric Matrix | \(M^T = M\) | \(A^2\) is symmetric because \((A^2)^T = A^2\). |
| Skew-Symmetric Matrix | \(M^T = -M\) | The given matrix \(A\) is skew-symmetric because \(A^T = -A\). |
| Transpose of a Product | \((AB)^T = B^T A^T\) | Used to show \((A^2)^T = (A \times A)^T = A^T A^T\). |
| Determinant of a Product | \(\det(AB) = \det(A)\det(B)\) | Used to show \(\det(A^2) = \det(A)\det(A)\). |
| Determinant of kA | \(\det(kA) = k^n \det(A)\) for an \(n \times n\) matrix | Used implicitly to understand why \(\det(A)=0\) for skew-symmetric A (3x3). |
Additional Information on Skew-Symmetric Matrices
A skew-symmetric matrix \(A\) has the property \(A^T = -A\). For the determinant, we know \(\det(A^T) = \det(A)\).
Also, for an \(n \times n\) matrix, \(\det(kA) = k^n \det(A)\). In the case of a skew-symmetric matrix, we have \(A^T = -A\), so \(\det(A^T) = \det(-A)\).
Thus, \(\det(A) = \det(-A)\). Using the property \(\det(kA) = k^n \det(A)\) with \(k=-1\) and \(n\) being the order of the matrix:
\[ \det(A) = (-1)^n \det(A) \]If the order \(n\) is even, \(n=2m\), then \( (-1)^n = (-1)^{2m} = ((-1)^2)^m = 1^m = 1 \). The equation becomes \(\det(A) = 1 \times \det(A)\), which is \(\det(A) = \det(A)\). This doesn't give us information about the determinant value itself; a skew-symmetric matrix of even order can have a non-zero determinant (its determinant is the square of its Pfaffian).
If the order \(n\) is odd, \(n=2m+1\), then \( (-1)^n = (-1)^{2m+1} = (-1)^{2m} \times (-1)^1 = 1 \times (-1) = -1 \). The equation becomes \(\det(A) = -1 \times \det(A)\), which is \(\det(A) = -\det(A)\). Rearranging gives \(2 \det(A) = 0\), which means \(\det(A) = 0\).
Since the given matrix \(A\) is a 3x3 matrix (odd order, n=3) and it is skew-symmetric, its determinant must be 0. Our calculation \(\det(A) = 0\) confirms this general property.
This property helps explain why \(\det(A^2) = (\det(A))^2 = (0)^2 = 0\) in this specific problem involving a 3x3 skew-symmetric matrix.
Important Questions from Determinants
If \(A=\left[\begin{array}{rrr} 2 & -1 & 0 \\ -1 & 3 & 0 \\ 1 & 0 & 1 \end{array}\right]\), then what is the value of det[adj(adjA)] ?
If A, B and C are square matrices of order 3 and det(BC) = 2 det(A), then what is the value of det(2A-1BC)?
If \(\left|\begin{array}{ccc} x^2+3 x & x-1 & x+3 \\ x+1 & -2 x & x-4 \\ x-3 & x+4 & 3 x \end{array}\right|\) = ax4 + bx3 + cx2 + dx + e, then what is the value of e?"
If all elements of a third order determinant are equal to 1 or -1, then the value of the determinant is:
If a, b, c are in AP, then what is \(\left|\begin{array}{lll} x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \\ x+a & x+b & x+c \end{array}\right|\) equal to ?
