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

• A. P-3; Q-1; R-4; S-2
• B. P-4; Q-1; R-3; S-2
• C. P-3; Q-4; R-1; S-2
• D. P-4; Q-1; R-2; S-3

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.

Answer 1

The Correct Option is C

Step 1: Picking the easiest checks first:
Instead of testing every matrix against every property in order, it is faster to spot the two easiest properties first: triangular and symmetric are visible just by looking at zero positions and mirrored entries.

Step 2: Spotting the triangular matrix:
$ S = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 4 \\ 0 & 0 & 1 \end{bmatrix} $
Every position below the main diagonal is already 0, nothing needs to be computed, this is an upper triangular matrix. So S matches property 2.

Step 3: Spotting the symmetric matrix:
$ P = \begin{bmatrix} -30 & 17 & 5 \\ 17 & 0 & -12 \\ 5 & -12 & 4 \end{bmatrix} $
Reading across the diagonal, 17 appears twice, 5 appears twice, and -12 appears twice, in mirrored positions. A matrix with this mirror pattern equals its own transpose, so P is symmetric and matches property 3.

Step 4: Spotting the trace free matrix by adding the diagonal:
$ Q = \begin{bmatrix} 0 & 1 & 9 \\ 7 & 0 & 2 \\ 12 & 3 & 0 \end{bmatrix} $
Adding the diagonal entries of Q gives $0+0+0 = 0$, so Q is trace free and matches property 4.

Step 5: Assigning the last matrix by elimination, then confirming with a shortcut:
Only R and property 1 (singular) are left, so by process of elimination R must be singular.
$ R = \begin{bmatrix} 2/3 & 1/3 & 2/3 \\ 0 & 2/3 & -1/3 \\ 0 & 4/3 & -2/3 \end{bmatrix} $
As a quick check, row 3 is $(0, 4/3, -2/3)$, which is exactly row 2, $(0, 2/3, -1/3)$, multiplied by 2.
Since row 3 is a scalar multiple of row 2, the rows are linearly dependent, and a matrix with dependent rows always has a zero determinant, so R is confirmed singular.

Final Answer:
Working from the easiest pattern to the hardest gives the same match: P-3, Q-4, R-1, S-2. \[ \boxed{\text{Option (C)}} \]