If A is an orthogonal matrix of order 3 and \({\rm{B}} = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ { - 3}&0&2\\ 2&5&0 \end{array}} \right]\) , then which of the following is/are correct? 1. |AB| = ± 47 2. AB = BA
1 only
This question asks us to evaluate two statements about matrix multiplication and determinants involving an orthogonal matrix A and a specific matrix B.
First, let's recall what an orthogonal matrix is and its key properties.
An orthogonal matrix is a square matrix whose transpose is equal to its inverse. If A is an orthogonal matrix, then:
A crucial property related to the determinant of an orthogonal matrix can be derived from \(A^T A = I\). Taking the determinant of both sides:
\(|A^T A| = |I|\)
We know that the determinant of a product of matrices is the product of their determinants: \(|AB| = |A| |B|\). Also, the determinant of a transpose is equal to the determinant of the original matrix: \(|A^T| = |A|\). The determinant of the identity matrix is 1.
So, \(|A^T| |A| = 1\), which means \(|A| |A| = 1\), or \(|A|^2 = 1\).
Therefore, the determinant of an orthogonal matrix is always either 1 or -1: \(|A| = \pm 1\).
Statement 1 claims that \(|AB| = \pm 47\). We can use the property \(|AB| = |A| |B|\).
We know \(|A| = \pm 1\) for an orthogonal matrix A.
Now, let's calculate the determinant of the given matrix B:
\({\rm{B}} = \left[ {\begin{array}{*{20}{c}} 1&2&3\\ { - 3}&0&2\\ 2&5&0 \end{array}} \right]\)
We can compute the determinant by expanding along the first row:
\(|B| = 1 \times \det \left[ {\begin{array}{*{20}{c}} 0&2\\ 5&0 \end{array}} \right] - 2 \times \det \left[ {\begin{array}{*{20}{c}} { - 3}&2\\ 2&0 \end{array}} \right] + 3 \times \det \left[ {\begin{array}{*{20}{c}} { - 3}&0\\ 2&5 \end{array}} \right]\)
\(|B| = 1 \times (0 \times 0 - 2 \times 5) - 2 \times ((-3) \times 0 - 2 \times 2) + 3 \times ((-3) \times 5 - 0 \times 2)\)
\(|B| = 1 \times (0 - 10) - 2 \times (0 - 4) + 3 \times (-15 - 0)\)
\(|B| = 1 \times (-10) - 2 \times (-4) + 3 \times (-15)\)
\(|B| = -10 + 8 - 45\)
\(|B| = -2 - 45\)
\(|B| = -47\)
Now, we can find \(|AB|\):
\(|AB| = |A| |B|\)
\(|AB| = (\pm 1) \times (-47)\)
\(|AB| = \mp 47\), which is equivalent to \(\pm 47\).
So, Statement 1, \(|AB| = \pm 47\), is correct.
Statement 2 claims that AB = BA.
Matrix multiplication is generally not commutative. That is, for two matrices A and B, AB is typically not equal to BA. For AB = BA to hold, the matrices A and B must commute. There is no general property stating that an orthogonal matrix A will commute with an arbitrary matrix B.
The statement AB = BA must hold for *any* orthogonal matrix A of order 3 and the given matrix B for Statement 2 to be correct. Since matrix multiplication is not commutative in general, and there is no special relationship given between all 3x3 orthogonal matrices and the specific matrix B that would force them to commute, this statement is incorrect.
For example, let A be a simple orthogonal matrix like the identity matrix \(I_3\). \(I_3\) is orthogonal because \(I_3^T = I_3\) and \(I_3 I_3 = I_3\). In this case, \(AB = I_3 B = B\) and \(BA = B I_3 = B\), so AB = BA holds. However, the statement must hold for *any* orthogonal matrix. Consider a rotation matrix in 3D (which is an orthogonal matrix). Multiplying a general matrix B by a rotation matrix A will generally result in a different matrix than multiplying A by B in the reverse order.
Therefore, Statement 2, AB = BA, is incorrect.
Based on our analysis:
Thus, only Statement 1 is correct.
| Statement | Evaluation | Correctness |
|---|---|---|
| 1. |AB| = ± 47 | \(|A| = \pm 1\), \(|B| = -47\). So \(|AB| = |A||B| = (\pm 1)(-47) = \mp 47 = \pm 47\). | Correct |
| 2. AB = BA | Matrix multiplication is not generally commutative. AB = BA is not guaranteed for an arbitrary orthogonal matrix A and the given matrix B. | Incorrect |
| Property | Description | Mathematical Notation |
|---|---|---|
| Orthogonal Matrix | A square matrix whose transpose is its inverse. | \(A^T = A^{-1}\) or \(A^T A = I\) |
| Determinant of Orthogonal Matrix | The determinant of an orthogonal matrix is either 1 or -1. | \(|A| = \pm 1\) |
| Determinant of a Product | The determinant of the product of two matrices is the product of their determinants. | \(|AB| = |A| |B|\) |
| Matrix Commutation | Two matrices A and B commute if their product is independent of the order of multiplication. | AB = BA (This is not generally true) |
Orthogonal matrices have many important applications in various fields, especially where transformations that preserve length and angles are involved. Some key areas include:
The property \(|A| = \pm 1\) for an orthogonal matrix A implies that such transformations preserve volume. If \(|A|=1\), the transformation corresponds to a rotation (preserving orientation). If \(|A|=-1\), it corresponds to a reflection possibly combined with rotation (reversing orientation).
If \(A=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\), then what is the value of det(I + AA'), where I is the 3 × 3 identity matrix?
If \(A=\left[\begin{array}{lll} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 4 \end{array}\right]\), then which of the following statements are correct?
1. An will always be singular for any positive integer n.
2. An will always be a diagonal matrix for any positive integer n.
3. An will always be a symmetric matrix for any positive integer n.
Select the correct answer using the code given below:
If \(A=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\), then what is 23A3 - 19A2 - 4A equal to ?
If \(A_k=\left[\begin{array}{cc} k-1 & k \\ k-2 & k+1 \end{array}\right] \), then what is det(A1) + det(A2) + det(A3) + ... + det(A100) equal to ?
Consider the following in respect of the matrix \({\rm{A}} = \left( {\begin{array}{*{20}{c}} { - 1}&1\\ 1&{ - 1} \end{array}} \right):\)
1. A 2= -A
2. A 3= 4A
Which of the above is/are correct?