Q.If $A$ and $B$ are symmetric matrices, prove that $AB - BA$ is a skew symmetric matrix.

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 $Y^T=-Y$.
3. Now compute $(XY)^T$ using the reversal rule $(XY)^T=Y^TX^T$.
4. Substitute: $Y^TX^T=(-Y)(X)=-YX$.
5. So $(XY)^T=-YX$.

Common Mistakes

• Forgetting the matrix product transpose reverses order: $(XY)^T=Y^TX^T$, not $X^TY^T$.
• Mixing up which combination (sum or difference) is symmetric vs. skew-symmetric — it's the sum $AB+BA$ that's symmetric and the difference $AB-BA$ that's skew-symmetric, for symmetric A, B.

ANSWER: C

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. $M^T = (AA^T)^T = (A^T)^T \cdot A^T = A\cdot A^T = M$.
3. Since $M^T=M$, $M=AA^T$ is symmetric — always, for any square matrix $A$.

Common Mistakes

• Thinking the answer depends on whether $A$ itself is symmetric — it doesn't; $AA^T$ is symmetric unconditionally.

ANSWER: B

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, $|A| = -1(4\cdot(-6)-5\cdot5) - 2(2\cdot(-6)-5\cdot(-3)) + (-3)(2\cdot5-4\cdot(-3))$ $= -1(-24-25) - 2(-12+15) - 3(10+12) = 49 - 6 - 66 = -23$.
6. Check against the options numerically: $xyz+q+r = (2)(-3)(5) + 3 + 4 = -30+7=-23$ — matches exactly. (The other options: $xyz+pqr=-30-24=-54$; $xyz/(pq) = -30/(-6)=5$; $xyz+pq+rs=-30-6+0=-36$ — none match.)

Common Mistakes

• Trying to compute $|AB|$ directly by multiplying the matrices — completely unnecessary once you recognise $|B|=0$ from the odd-order skew-symmetric property.
• Sign slips when reading off $x,y,z,p,q,r,s$ from the symmetry/skew-symmetry conditions — it helps to write out both matrices fully with the solved values before computing.

ANSWER: B

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 $P^T-Q^T=A=\begin{bmatrix}2 & 0 & -3\\ 4 & 3 & 1\\ -5 & 7 & 2\end{bmatrix}$.

Common Mistakes

• Confusing $P^T-Q^T$ with $P-Q$ (which instead equals $A^T$, not $A$).
• Wasting time computing $P$ and $Q$ explicitly from $A$ when the symmetry properties give the answer immediately.

ANSWER: B

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:

The non-zero elements here are $0.5$ and $-0.5$, but the problem says the non-zero elements of $T$ are $-1$ and $1$. So again, we need a scaling. Multiply by 2: $T = B - B^T$ gives:

Non-zero elements are $1$ and $-1$, matching the condition. So $T = B - B^T$.

3. Add $S$ and $T$

This matches option (A)? Wait, check option (A): $\begin{bmatrix} 2 & 5 & 0 \\ 3 & -4 & 3 \\ 0 & 1 & -2 \end{bmatrix}$. Yes, that’s exactly it.

But let’s double-check the other options: (C) has $-3$ in the (2,2) position, not $-4$. So (A) is correct.

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.
6. (As a consistency check, substituting $A^T=-A$ back into Equation 1 gives $-A^2 = \tfrac34I$, i.e. $A^2=-\tfrac34I$ — this rules out $A$ being any real scalar multiple of $I$, since $(cI)^2=c^2I\ge0$ can never equal a negative multiple of $I$. So options claiming $A=\pm\tfrac34I$ are impossible, reinforcing that "skew-symmetric" is the correct general characterization.)

Common Mistakes

• Trying to solve for a specific numeric matrix $A$ rather than recognizing the question only asks for a general structural property (skew-symmetry) satisfied by all valid $A$.
• Sign errors when subtracting the two expanded equations.

ANSWER: C

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.
5. Also rule out skew-symmetric (option B): a skew-symmetric matrix must have all zero diagonal entries, but $AA^T$ has diagonal entries $34, 29, 2$ — clearly not skew-symmetric.

Common Mistakes

• Not recognizing the general $AA^T$-is-always-symmetric identity and instead trying to guess based on properties specific to this particular $A$ (like the fact that $A^{-1}=A$ for this matrix, which is a red herring here since the question asks about $AA^T$, not $A^2$).
• Confusing symmetric with skew-symmetric — a skew-symmetric matrix requires zero diagonal, which is clearly not the case here.

ANSWER: A

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).
5. Count the choices: assign which of $p,q,r$ (i.e., which matrix position: $(1,2),(1,3),(2,3)$) gets the $a$-pair, which gets the $b$-pair, which gets the $c$-pair — that's $3!=6$ ways. Independently, each of the three chosen values can be $+$ or $-$ its letter — $2^3=8$ ways.
6. Total matrices $=6\times8=48$.

Common Mistakes

• Forgetting that choosing the sign of an upper-triangle entry automatically fixes the sign of its mirrored lower entry (they are not independent choices).
• Allowing repeated pairs (e.g., both $p$ and $q$ from $\{a,-a\}$), which would fail to use all seven required values.

ANSWER: D