Question:
Which ONE of the following options CORRECTLY matches the matrices to their properties?
Matrix Property \( P = \begin{bmatrix} -30 & 17 & 5 \\ 17 & 0 & -12 \\ 5 & -12 & 4 \end{bmatrix} \) 1: Singular \( Q = \begin{bmatrix} 0 & 1 & 9 \\ 7 & 0 & 2 \\ 12 & 3 & 0 \end{bmatrix} \) 2: Triangular \( R = \begin{bmatrix} 2/3 & 1/3 & 2/3 \\ 0 & 2/3 & -1/3 \\ 0 & 4/3 & -2/3 \end{bmatrix} \) 3: Symmetric \( S = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} \) 4: Trace free
Which ONE of the following options CORRECTLY matches the matrices to their properties?
| Matrix | Property |
|---|---|
| \( P = \begin{bmatrix} -30 & 17 & 5 \\ 17 & 0 & -12 \\ 5 & -12 & 4 \end{bmatrix} \) | 1: Singular |
| \( Q = \begin{bmatrix} 0 & 1 & 9 \\ 7 & 0 & 2 \\ 12 & 3 & 0 \end{bmatrix} \) | 2: Triangular |
| \( R = \begin{bmatrix} 2/3 & 1/3 & 2/3 \\ 0 & 2/3 & -1/3 \\ 0 & 4/3 & -2/3 \end{bmatrix} \) | 3: Symmetric |
| \( S = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} \) | 4: Trace free |
Show Hint
Check the easy properties first: look for zero entries below the diagonal (triangular), mirrored entries (symmetric), and a diagonal sum of zero (trace free). What is left over must be singular.
Updated On: Aug 5, 2026
- P-3; Q-1; R-4; S-2
- P-4; Q-1; R-3; S-2
- P-3; Q-4; R-1; S-2
- P-4; Q-1; R-2; S-3
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Question:
We are given four 3x3 matrices, P, Q, R and S, and four properties: singular, triangular, symmetric, and trace free.
We need to identify which property belongs to which matrix, using the definitions of each property.
Step 2: Recalling the definitions:
A matrix is symmetric if it equals its own transpose, meaning the entry in row i column j equals the entry in row j column i for every pair.
A matrix is triangular if all the entries either above or below the main diagonal are zero.
A matrix is trace free if the sum of its main diagonal entries is zero.
A matrix is singular if its determinant is zero, which also means it has no inverse.
Step 3: Checking matrix P:
\[ P = \begin{bmatrix} -30 & 17 & 5 \\ 17 & 0 & -12 \\ 5 & -12 & 4 \end{bmatrix} \]
Compare entries across the diagonal: \( P_{12}=17=P_{21} \), \( P_{13}=5=P_{31} \), \( P_{23}=-12=P_{32} \).
Since every mirrored pair matches, P is symmetric.
Step 4: Checking matrix Q:
\[ Q = \begin{bmatrix} 0 & 1 & 9 \\ 7 & 0 & 2 \\ 12 & 3 & 0 \end{bmatrix} \]
The diagonal entries are 0, 0 and 0, so the trace is \( 0+0+0=0 \), which makes Q trace free.
Q is not symmetric since \( Q_{12}=1 \ne 7=Q_{21} \), and it is not triangular since it has nonzero entries on both sides of the diagonal.
Step 5: Checking matrix S:
\[ S = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} \]
Every entry below the main diagonal is zero, so S is an upper triangular matrix.
Step 6: Checking matrix R by elimination and by its determinant:
By elimination, since P, Q and S are already used up by symmetric, trace free and triangular, R must be the singular one, let us verify.
\[ R = \begin{bmatrix} 2/3 & 1/3 & 2/3 \\ 0 & 2/3 & -1/3 \\ 0 & 4/3 & -2/3 \end{bmatrix} \]
Expanding the determinant along the first column, only the top left entry contributes since the other two entries in that column are zero.
\[ |R| = \frac{2}{3}\left[ \left(\frac{2}{3}\right)\left(-\frac{2}{3}\right) - \left(-\frac{1}{3}\right)\left(\frac{4}{3}\right) \right] = \frac{2}{3}\left[ -\frac{4}{9} + \frac{4}{9} \right] = \frac{2}{3}(0) = 0 \]
Since the determinant is zero, R is indeed singular, confirming our elimination.
Step 7: Matching to the given options:
We found P-3 (symmetric), Q-4 (trace free), R-1 (singular), S-2 (triangular), which is exactly option (C).
Option (A) swaps Q and S and puts R at trace free, which is wrong since Q is trace free, not R.
Option (B) assigns P to trace free and swaps other pairs incorrectly.
Option (D) assigns S to symmetric, but S has zeros only below the diagonal, so it is triangular, not symmetric.
Final Answer:
The correct pairing is P-3, Q-4, R-1, S-2. \[ \boxed{\text{Option (C): } P{-}3,\ Q{-}4,\ R{-}1,\ S{-}2} \]
We are given four 3x3 matrices, P, Q, R and S, and four properties: singular, triangular, symmetric, and trace free.
We need to identify which property belongs to which matrix, using the definitions of each property.
Step 2: Recalling the definitions:
A matrix is symmetric if it equals its own transpose, meaning the entry in row i column j equals the entry in row j column i for every pair.
A matrix is triangular if all the entries either above or below the main diagonal are zero.
A matrix is trace free if the sum of its main diagonal entries is zero.
A matrix is singular if its determinant is zero, which also means it has no inverse.
Step 3: Checking matrix P:
\[ P = \begin{bmatrix} -30 & 17 & 5 \\ 17 & 0 & -12 \\ 5 & -12 & 4 \end{bmatrix} \]
Compare entries across the diagonal: \( P_{12}=17=P_{21} \), \( P_{13}=5=P_{31} \), \( P_{23}=-12=P_{32} \).
Since every mirrored pair matches, P is symmetric.
Step 4: Checking matrix Q:
\[ Q = \begin{bmatrix} 0 & 1 & 9 \\ 7 & 0 & 2 \\ 12 & 3 & 0 \end{bmatrix} \]
The diagonal entries are 0, 0 and 0, so the trace is \( 0+0+0=0 \), which makes Q trace free.
Q is not symmetric since \( Q_{12}=1 \ne 7=Q_{21} \), and it is not triangular since it has nonzero entries on both sides of the diagonal.
Step 5: Checking matrix S:
\[ S = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} \]
Every entry below the main diagonal is zero, so S is an upper triangular matrix.
Step 6: Checking matrix R by elimination and by its determinant:
By elimination, since P, Q and S are already used up by symmetric, trace free and triangular, R must be the singular one, let us verify.
\[ R = \begin{bmatrix} 2/3 & 1/3 & 2/3 \\ 0 & 2/3 & -1/3 \\ 0 & 4/3 & -2/3 \end{bmatrix} \]
Expanding the determinant along the first column, only the top left entry contributes since the other two entries in that column are zero.
\[ |R| = \frac{2}{3}\left[ \left(\frac{2}{3}\right)\left(-\frac{2}{3}\right) - \left(-\frac{1}{3}\right)\left(\frac{4}{3}\right) \right] = \frac{2}{3}\left[ -\frac{4}{9} + \frac{4}{9} \right] = \frac{2}{3}(0) = 0 \]
Since the determinant is zero, R is indeed singular, confirming our elimination.
Step 7: Matching to the given options:
We found P-3 (symmetric), Q-4 (trace free), R-1 (singular), S-2 (triangular), which is exactly option (C).
Option (A) swaps Q and S and puts R at trace free, which is wrong since Q is trace free, not R.
Option (B) assigns P to trace free and swaps other pairs incorrectly.
Option (D) assigns S to symmetric, but S has zeros only below the diagonal, so it is triangular, not symmetric.
Final Answer:
The correct pairing is P-3, Q-4, R-1, S-2. \[ \boxed{\text{Option (C): } P{-}3,\ Q{-}4,\ R{-}1,\ S{-}2} \]
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