Q.$AA'$ is always a symmetric matrix for any matrix $A$.

Concept and Intuition

Symmetry of a matrix $M$ means $M^T=M$. Checking this for $M=AA^T$ using the transpose-reversal rule $(XY)^T=Y^TX^T$ resolves the question immediately and generally — it doesn't matter what specific entries $A$ has.

Step-by-Step Solution

1. Let $M = AA^T$.
2. $M^T = (AA^T)^T = (A^T)^T \cdot A^T = A\cdot A^T = M$. …

Concept and Intuition

There's a universal matrix identity worth memorizing: for any matrix $A$ (of any shape, as long as $AA^T$ is defined), $AA^T$ is always symmetric, and so is $A^TA$. This follows immediately from the transpose-of-a-product rule, so you don't even need to know $A$'s specific entries to answer this correctly — though verifying by direct computation is a good check.

Step-by-Step Solution

1. General proof: $(AA^T)^T = (A^T)^T \, A^T = A\,A^T$. Since $(AA^T)^T = AA^T$, the matrix $AA^T$ is symmetric by definition — true for every matrix $A$.
2. Direct check with the given $A = \begin{bmatrix}3&-3&4\\2&-3&4\\0&-1&1\end{bmatrix}$: computing row-dot-row products gives $AA^T = \begin{bmatrix}34&31&7\\31&29&7\\7&7&2\end{bmatrix}$, which is indeed symmetric (entries mirror across the diagonal).
3. Check it's not singular (ruling out option C): $\det(AA^T) = 34(29\cdot2-7\cdot7) - 31(31\cdot2-7\cdot7)+7(31\cdot7-29\cdot7) = 34(9)-31(13)+7(14) = 306-403+98=1 \neq 0$. Not singular.
4. Check it's not the inverse of $A$ (ruling out option D): $AA^T$'s entries (up to 34) are nothing like $A$'s original small entries, and there's no reason $AA^T$ should equal $A^{-1}$ in general — this is easily dismissed by inspection. …

Concept and Intuition

For any two symmetric matrices A and B, their commutator-related combinations always split into a symmetric part ($AB+BA$) and a skew-symmetric part ($AB-BA$) — this is analogous to how any square matrix splits into symmetric and skew-symmetric pieces. Recognising $X$ as symmetric ($X^T=X$) and $Y$ as skew-symmetric ($Y^T=-Y$) lets us compute $(XY)^T$ purely from these properties, without needing to know A and B explicitly.

Step-by-Step Solution

1. Show $X=AB+BA$ is symmetric: $X^T=(AB+BA)^T=(AB)^T+(BA)^T=B^TA^T+A^TB^T$. Since $A^T=A$ and $B^T=B$, this is $BA+AB=X$. So $X^T=X$.
2. Show $Y=AB-BA$ is skew-symmetric: $Y^T=(AB-BA)^T=(AB)^T-(BA)^T=B^TA^T-A^TB^T=BA-AB=-(AB-BA)=-Y$. So $Y^T=-Y$. …

Concept and Intuition

When two matrix expressions that share a common "core" product ($A^TA$) are both set equal to the same matrix ($I$), subtracting them cancels the shared quadratic term and isolates the linear terms — exactly like solving simultaneous equations. This lets us extract a clean condition on $A$ itself.

Step-by-Step Solution

1. Expand $(A^T-\tfrac12I)(A-\tfrac12I) = A^TA - \tfrac12A^T - \tfrac12A + \tfrac14I$. This equals $I$: call this Equation 1.
2. Expand $(A^T+\tfrac12I)(A+\tfrac12I) = A^TA + \tfrac12A^T + \tfrac12A + \tfrac14I$. This also equals $I$: call this Equation 2.
3. Subtract Equation 1 from Equation 2: $\left(A^TA+\tfrac12A^T+\tfrac12A+\tfrac14I\right) - \left(A^TA-\tfrac12A^T-\tfrac12A+\tfrac14I\right) = I - I = 0$.
4. The $A^TA$ and $\tfrac14I$ terms cancel, leaving $A^T+\tfrac12A^T+\tfrac12A + \tfrac12A = 0$... more carefully: $(\tfrac12A^T+\tfrac12A) - (-\tfrac12A^T-\tfrac12A) = A^T+A = 0$.
5. So $A^T = -A$, which is exactly the definition of a skew-symmetric matrix. …

Concept and Intuition

Any square matrix $A$ can be written uniquely as $A=P+Q$ where $P=\dfrac{A+A^T}{2}$ is symmetric and $Q=\dfrac{A-A^T}{2}$ is skew-symmetric. The defining properties are $P^T=P$ and $Q^T=-Q$. The question asks for $P^T-Q^T$, not $P-Q$, so we should substitute these properties directly rather than compute $P$ and $Q$ separately.

Step-by-Step Solution

1. Since $P$ is symmetric, $P^T=P$.
2. Since $Q$ is skew-symmetric, $Q^T=-Q$.
3. Therefore $P^T-Q^T = P-(-Q) = P+Q$.
4. But $P+Q=A$ by definition of the decomposition.
5. So $P^T-Q^T=A=\begin{bmatrix}2 & 0 & -3\\ 4 & 3 & 1\\ -5 & 7 & 2\end{bmatrix}$. …

We start with the key idea: any square matrix can be uniquely written as the sum of a symmetric and a skew-symmetric matrix. For a given matrix $M$, the symmetric part is $\frac{M + M^T}{2}$ and the skew-symmetric part is $\frac{M - M^T}{2}$. Here, $S$ is obtained from $A$, and $T$ from $B$, but with additional conditions that adjust them slightly.

1. Find the symmetric matrix $S$ from $A$ The natural symmetric part of $A$ is $\frac{A + A^T}{2}$. Compute:

The trace of this matrix is $1 + (-2) + (-1) = -2$. But we are told the trace of $S$ is $-4$. So $S$ is not exactly this; it must be a scalar multiple? Actually, the phrase "obtained from $A$" means $S$ is the symmetric matrix derived from $A$ by the usual decomposition, but the trace condition tells us we need to adjust. Wait — let's re-read: "Let $S$ be a symmetric matrix obtained from $A$". This typically means $S = \frac{A + A^T}{2}$. But then trace is $-2$, not $-4$. So perhaps $S$ is the symmetric part of some matrix related to $A$? Or maybe $S$ is simply the symmetric part of $A$, and the trace condition is a red herring? No, it's given explicitly.

Let’s check: If $S = \frac{A + A^T}{2}$, trace = $-2$. But problem says trace of $S = -4$. So $S$ must be something else. Possibly $S$ is the symmetric matrix obtained from $A$ after some operation? Or maybe $S$ is the symmetric part of $A$ multiplied by 2? That would give trace $-4$. Indeed, $2 \times \frac{A + A^T}{2} = A + A^T$ is symmetric, and its trace is $2 + (-4) + (-2) = -4$. So $S = A + A^T$ works. Let’s verify:

Trace = $2 -4 -2 = -4$. Perfect. So $S$ is the sum, not the average.

2. Find the skew-symmetric matrix $T$ from $B$ The skew-symmetric part of $B$ is $\frac{B - B^T}{2}$. Compute:

$\frac{B - B^T}{2} = \frac{1}{2} \begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 1 \\ 0 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0.5 & 0 \\ -0.5 & 0 & 0.5 \\ 0 & -0.5 & 0 \end{bmatrix}$ …

Concept and Intuition

For any square matrix $A$, the matrix $S=A-A^T$ satisfies $S^T = A^T-A = -(A-A^T)=-S$, i.e. $S$ is skew-symmetric. A key structural fact from matrix theory: for a skew-symmetric matrix $S$ of odd order $n$, $\det(S)=\det(S^T)=\det(-S)=(-1)^n\det(S)=-\det(S)$ (since $n$ is odd), which forces $\det(S)=-\det(S)$, i.e. $2\det(S)=0$, so $\det(S)=0$. This holds for any odd-order skew-symmetric matrix, so the specific numbers in $A$ are a red herring.

Step-by-Step Solution

1. $A$ is a $3\times 3$ matrix (order 3, odd).
2. Let $S=A-A^T$. Then $S^T = A^T - (A^T)^T = A^T - A = -S$, so $S$ is skew-symmetric.
3. For any skew-symmetric $S$: $\det(S^T)=\det(S)$ (determinant is invariant under transpose) but also $S^T=-S$ so $\det(S^T)=\det(-S)=(-1)^3\det(S)=-\det(S)$ (order 3 is odd).
4. Combining: $\det(S) = -\det(S) \Rightarrow 2\det(S)=0 \Rightarrow \det(S)=0$. …

Concept and Intuition

A matrix is symmetric iff every off-diagonal pair of entries mirrored across the main diagonal is equal ($a_{ij}=a_{ji}$). Since $x$ and $y$ appear in specific off-diagonal slots here, symmetry alone completely determines their values — there is no freedom left. We then just have to check whether that single resulting matrix happens to be singular.

Step-by-Step Solution

1. $A=\begin{pmatrix}1&2&1\\2&2&x\\y&1&2\end{pmatrix}$. Compare $A$ with $A^T$: entry $(1,3)=1$ must equal $(3,1)=y \Rightarrow y=1$. Entry $(2,3)=x$ must equal $(3,2)=1 \Rightarrow x=1$. Entry $(1,2)=2$ already equals $(2,1)=2$, no new condition.
2. So symmetry forces the unique pair $x=1,\,y=1$.
3. Substitute: $A=\begin{pmatrix}1&2&1\\2&2&1\\1&1&2\end{pmatrix}$. …

Concept and Intuition

For a symmetric matrix, $A=A^T$ means mirror-image entries across the main diagonal must be equal. For a skew-symmetric matrix, $B=-B^T$ means mirror-image entries are negatives of each other, and every diagonal entry must be zero (since $b_{ii}=-b_{ii}\Rightarrow b_{ii}=0$). A key structural fact then finishes the problem: any skew-symmetric matrix of odd order has determinant exactly zero (because $|B|=|B^T|=|-B|=(-1)^n|B|$, and for odd $n$ this forces $|B|=-|B|\Rightarrow|B|=0$). Since determinants are multiplicative, $|AB|=|A|\,|B|=0$ too, collapsing the whole expression to just $|A|$.

Step-by-Step Solution

1. Use symmetry of $A$ ($a_{ij}=a_{ji}$): comparing $(1,2)$ vs $(2,1)$ gives $x=2$; $(1,3)$ vs $(3,1)$ gives $-3=y\Rightarrow y=-3$; $(2,3)$ vs $(3,2)$ gives $z=5$.
2. Use skew-symmetry of $B$ ($b_{ij}=-b_{ji}$, diagonal $=0$): $(3,3)$ entry $s=0$ (forced by skew-symmetry); $(1,2)=2,(2,1)=p\Rightarrow p=-2$; $(1,3)=q,(3,1)=-3\Rightarrow q=3$; $(2,3)=-4,(3,2)=r\Rightarrow r=4$.
3. Since $B$ is $3\times3$ (odd order) and skew-symmetric, $|B|=0$ identically.
4. So $|AB|=|A||B| = |A|\cdot 0 = 0$, and the required expression is $|A|+|B|-|AB| = |A|+0-0 = |A|$.
5. Compute $|A|$ with $A=\begin{bmatrix}-1&2&-3\\2&4&5\\-3&5&-6\end{bmatrix}$: expanding along row 1, $|A| = -1(4\cdot(-6)-5\cdot5) - 2(2\cdot(-6)-5\cdot(-3)) + (-3)(2\cdot5-4\cdot(-3))$ …

Concept and Intuition

A skew-symmetric matrix always has zero entries on its main diagonal ($A_{ii}=-A_{ii}\Rightarrow A_{ii}=0$). When you multiply any diagonal matrix $D$ by $A$, the $i$-th diagonal entry of the product $DA$ is $D_{ii}A_{ii}=D_{ii}\cdot0=0$. So $\mathrm{Tr}(DA)=0$ for ANY diagonal $D$ (or product of diagonals) and skew-symmetric $A$ — this collapses three of the four terms in the given trace expression to zero immediately, without needing to know $A$ explicitly.

Step-by-Step Solution

1. $D_1D_2A$: since $D_1D_2$ is diagonal (product of two diagonal matrices is diagonal) and $A$ is skew-symmetric (zero diagonal), $\mathrm{Tr}(D_1D_2A)=0$.
2. $D_1A$: diagonal times skew-symmetric → $\mathrm{Tr}(D_1A)=0$.
3. $D_2A$: same reasoning → $\mathrm{Tr}(D_2A)=0$.
4. $D_1D_2=\mathrm{diag}(3a,3b,3c)$ (since $D_2=3I$), so $\mathrm{Tr}(D_1D_2)=3a+3b+3c$. …

Concept and Intuition

A $3\times3$ skew-symmetric matrix $A$ satisfies $A^T=-A$, which forces every diagonal entry to be its own negative (hence $0$), and forces the three entries below the diagonal to be the negatives of the three entries above it. So a skew-symmetric $3\times3$ matrix is fully determined by just 3 free numbers — the upper-triangular entries $p=A_{12},\,q=A_{13},\,r=A_{23}$ — and the matrix looks like

Step-by-Step Solution

1. The set of values we must use is $\{0,a,-a,b,-b,c,-c\}$ — 7 values total.
2. The diagonal is forced to be $0,0,0$, which accounts for the value $0$ (used three times).
3. The 6 off-diagonal positions carry exactly the values $\{p,q,r,-p,-q,-r\}$. For all six of $a,-a,b,-b,c,-c$ to appear (each matrix has exactly 6 off-diagonal slots for exactly 6 required nonzero values), this set must equal $\{a,-a,b,-b,c,-c\}$ exactly — a bijection, no repeats and no omissions.
4. This means $\{p,q,r\}$ must contain exactly one representative from each pair $\{a,-a\}$, $\{b,-b\}$, $\{c,-c\}$ (if two of $p,q,r$ were, say, both $\pm a$, some value would repeat and another would be missing). …