Question:
Consider the matrix
Consider the matrix
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A \(3\times 3\) matrix with entries from \(\{-1,0,1\}\) is orthogonal only when it is a signed permutation matrix.
Updated On: Jun 4, 2026
- \(\dfrac{2^4}{3^8}\)
- \(\dfrac{1}{3^5}\)
- \(\dfrac{2^5}{3^9}\)
- \(\dfrac{2^3}{3^7}\)
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The Correct Option is A
Solution and Explanation
Step 1: Understand the condition for orthogonal matrix.
A matrix \(A\) is orthogonal if
\[ A^TA=I \] This means that the columns of \(A\) form an orthonormal set.
Equivalently, each column must have length \(1\), and different columns must be mutually perpendicular.
Step 2: Use the possible values of entries.
Each entry of \(A\) can take values
\[ -1,0,1 \] Since the entries are integers, for a column to have length \(1\), the sum of squares of entries in that column must be \(1\).
That means each column must contain exactly one non-zero entry, which is either \(1\) or \(-1\), and the remaining two entries must be \(0\).
So every column must be of the form
\[ \begin{bmatrix} \pm 1\\ 0\\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0\\ \pm 1\\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0\\ 0\\ \pm 1 \end{bmatrix} \]
Step 3: Count the number of orthogonal matrices.
For \(A\) to be orthogonal, its columns must be distinct standard basis vectors with arbitrary signs.
First, choose the positions of the non-zero entries in the three columns.
This can be done in
\[ 3! \] ways.
For each non-zero entry, we can choose either \(1\) or \(-1\).
So the signs can be chosen in
\[ 2^3 \] ways.
Therefore, the total number of orthogonal matrices is
\[ 3!\cdot 2^3 \] \[ =6\cdot 8=48 \]
Step 4: Count the total number of possible matrices.
There are \(9\) entries in the \(3\times 3\) matrix.
Each entry has \(3\) possible values:
\[ -1,0,1 \] Therefore, the total number of possible matrices is
\[ 3^9 \]
Step 5: Find the required probability.
Since all matrices are equally likely,
\[ P(A\text{ is orthogonal}) = \frac{3!\cdot 2^3}{3^9} \] \[ = \frac{48}{3^9} \] Since
\[ 48=16\cdot 3=2^4\cdot 3 \] we get
\[ \frac{48}{3^9} = \frac{2^4\cdot 3}{3^9} \] \[ = \frac{2^4}{3^8} \]
Step 6: Final conclusion.
Hence, the required probability is
\[ \boxed{\frac{2^4}{3^8}} \]
A matrix \(A\) is orthogonal if
\[ A^TA=I \] This means that the columns of \(A\) form an orthonormal set.
Equivalently, each column must have length \(1\), and different columns must be mutually perpendicular.
Step 2: Use the possible values of entries.
Each entry of \(A\) can take values
\[ -1,0,1 \] Since the entries are integers, for a column to have length \(1\), the sum of squares of entries in that column must be \(1\).
That means each column must contain exactly one non-zero entry, which is either \(1\) or \(-1\), and the remaining two entries must be \(0\).
So every column must be of the form
\[ \begin{bmatrix} \pm 1\\ 0\\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0\\ \pm 1\\ 0 \end{bmatrix}, \quad \begin{bmatrix} 0\\ 0\\ \pm 1 \end{bmatrix} \]
Step 3: Count the number of orthogonal matrices.
For \(A\) to be orthogonal, its columns must be distinct standard basis vectors with arbitrary signs.
First, choose the positions of the non-zero entries in the three columns.
This can be done in
\[ 3! \] ways.
For each non-zero entry, we can choose either \(1\) or \(-1\).
So the signs can be chosen in
\[ 2^3 \] ways.
Therefore, the total number of orthogonal matrices is
\[ 3!\cdot 2^3 \] \[ =6\cdot 8=48 \]
Step 4: Count the total number of possible matrices.
There are \(9\) entries in the \(3\times 3\) matrix.
Each entry has \(3\) possible values:
\[ -1,0,1 \] Therefore, the total number of possible matrices is
\[ 3^9 \]
Step 5: Find the required probability.
Since all matrices are equally likely,
\[ P(A\text{ is orthogonal}) = \frac{3!\cdot 2^3}{3^9} \] \[ = \frac{48}{3^9} \] Since
\[ 48=16\cdot 3=2^4\cdot 3 \] we get
\[ \frac{48}{3^9} = \frac{2^4\cdot 3}{3^9} \] \[ = \frac{2^4}{3^8} \]
Step 6: Final conclusion.
Hence, the required probability is
\[ \boxed{\frac{2^4}{3^8}} \]
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