Given $A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}$. Is $(AB)' = B'A'$?

> [!TLDR] > Computing directly, $(AB)'$ and $B'A'$ both equal $\begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}$, so the reversal law $(AB)'=B'A'$ is verified. ## The property being checked The transpose of a product reverses the order of the factors: $(AB)'=B'A'$ (**not** $A'B'$). We verify it for the given matrices. ## Step 1 — compute $AB$ $A$ is $2\times 3$, $B$ is $3\times 2$, so $AB$ is $2\times 2$. - $(1,1)$: $2(1)+4(2)+0(1)=2+8+0=10$ - $(1,2)$: $2(4)+4(8)+0(3)=8+32+0=40$ - $(2,1)$: $3(1)+9(2)+6(1)=3+18+6=27$ - $(2,2)$: $3(4)+9(8)+6(3)=12+72+18=102$ $$AB=\begin{bmatrix} 10 & 40 \\ 27 & 102 \end{bmatrix}.$$ ## Step 2 — transpose it Swap rows and columns: $$(AB)'=\begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}.$$ ## Step 3 — form $B'$ and $A'$ and multiply $$B'=\begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3 \end{bmatrix}\ (2\times 3),\qquad A'=\begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix}\ (3\times 2).$$ Then $B'A'$ is $2\times 2$: - $(1,1)$: $1(2)+2(4)+1(0)=10$ - $(1,2)$: $1(3)+2(9)+1(6)=27$ …

Given A = 2 4 0 \3 9 6 and B = 1 4 \2 8 \1 3 . Is (AB)' = B'A'?

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A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}

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Computing directly, $(AB)'$ and $B'A'$ both equal $\begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}$ , so the reversal law $(AB)'=B'A'$ is verified.

The property being checked

The transpose of a product reverses the order of the factors: $(AB)'=B'A'$ (not $A'B'$ ). We verify it for the given matrices.

Step 1 — compute $AB$

$A$ is $2\times 3$ , $B$ is $3\times 2$ , so $AB$ is $2\times 2$ .

$(1,1)$ : $2(1)+4(2)+0(1)=2+8+0=10$
$(1,2)$ : $2(4)+4(8)+0(3)=8+32+0=40$
$(2,1)$ : $3(1)+9(2)+6(1)=3+18+6=27$
$(2,2)$ : $3(4)+9(8)+6(3)=12+72+18=102$

AB=\begin{bmatrix} 10 & 40 \\ 27 & 102 \end{bmatrix}.

Step 2 — transpose it

Swap rows and columns:

(AB)'=\begin{bmatrix} 10 & 27 \\ 40 & 102 \end{bmatrix}.

Step 3 — form $B'$ and $A'$ and multiply

B'=\begin{bmatrix} 1 & 2 & 1 \\ 4 & 8 & 3 \end{bmatrix}\ (2\times 3),\qquad A'=\begin{bmatrix} 2 & 3 \\ 4 & 9 \\ 0 & 6 \end{bmatrix}\ (3\times 2).

Then $B'A'$ is $2\times 2$ :

$(1,1)$ : $1(2)+2(4)+1(0)=10$
$(1,2)$ : $1(3)+2(9)+1(6)=27$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
$\begin{bmatrix} 0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0 \end{bmatrix}$
›Reveal solutionSolution
A skew-symmetric matrix has zeros on the diagonal and satisfies $A^T = -A$ . Using this property we find $a = -2$ , $b = 3$ , $c = -4$ , then multiply the two given $2 \times 2$ matrices to get $\begin{bmatrix} 2 & -8 \\ -8 & 2 \end{bmatrix}$ .

The key idea: a skew-symmetric matrix is one where the transpose equals the negative of the original. That single condition forces every diagonal entry to be zero and every off-diagonal pair $(i,j)$ and $(j,i)$ to be negatives of each other. Once we extract $a$ , $b$ , $c$ from that, the rest is just matrix multiplication.

Let’s work through it.

Apply the skew-symmetric condition.

For a matrix $A$ to be skew-symmetric, we need $A^T = -A$ . That means:

All diagonal entries must be $0$ (already satisfied here: $0, 0, 0$ ).

For any $i \neq j$ , the entry at $(i,j)$ must be the negative of the entry at $(j,i)$ .

Our matrix is

$A = \begin{bmatrix} 0 & 2 & a \\ b & 0 & 4 \\ -3 & c & 0 \end{bmatrix}.$

Compare the $(1,2)$ and $(2,1)$ positions. $A_{12} = 2$ and $A_{21} = b$ . Skew-symmetry says $A_{21} = -A_{12}$ , so

$b = -2.$

Compare the $(1,3)$ and $(3,1)$ positions. $A_{13} = a$ and $A_{31} = -3$ . Then $A_{31} = -A_{13}$ gives

$-3 = -a \quad\Rightarrow\quad a = 3.$

Compare the $(2,3)$ and $(3,2)$ positions. $A_{23} = 4$ and $A_{32} = c$ . So $A_{32} = -A_{23}$ means

$c = -4.$
So we have $a = 3$ , $b = -2$ , $c = -4$ .

Watch out

A common mistake is to forget the minus sign and set $A_{ij} = A_{ji}$ instead of $A_{ij} = -A_{ji}$ . That would give the wrong values. Always write $A^T = -A$ explicitly.

Now form the two $2 \times 2$ matrices. The first matrix is $\begin{bmatrix} a & b \\ b & a \end{bmatrix} = \begin{bmatrix} 3 & -2 \\ -2 & 3 \end{bmatrix}.$ …
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TG EAPCET 2024Set eng-2024-05-11-FN1 markMCQ
$A = \begin{bmatrix} 1 & 2 \\ 2 & 1 \end{bmatrix}$
›Reveal solutionSolution
The key idea is that $(A+B)(A-B) = A^2 - B^2$ holds only when $A$ and $B$ commute ( $AB = BA$ ). Using this condition, we solve for $x$ and $y$ , then compute the trace of $C$ .

We start with the matrix identity:
$(A+B)(A-B) = A^2 - AB + BA - B^2.$
For this to equal $A^2 - B^2$ , we need $-AB + BA = 0$ , i.e., $AB = BA$ . So the given equation is not automatically true for matrices; it’s true if and only if $A$ and $B$ commute. That’s the hidden condition.

Write down the matrices

$A = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix}, \quad B = \begin{pmatrix} x & y \\ 1 & 2 \end{pmatrix}.$

Compute $AB$ and $BA$

$AB = \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} \begin{pmatrix} x & y \\ 1 & 2 \end{pmatrix} = \begin{pmatrix} 1\cdot x + 2\cdot 1 & 1\cdot y + 2\cdot 2 \\ 2\cdot x + 1\cdot 1 & 2\cdot y + 1\cdot 2 \end{pmatrix} = \begin{pmatrix} x+2 & y+4 \\ 2x+1 & 2y+2 \end{pmatrix}.$ $BA = \begin{pmatrix} x & y \\ 1 & 2 \end{pmatrix} \begin{pmatrix} 1 & 2 \\ 2 & 1 \end{pmatrix} = \begin{pmatrix} x\cdot 1 + y\cdot 2 & x\cdot 2 + y\cdot 1 \\ 1\cdot 1 + 2\cdot 2 & 1\cdot 2 + 2\cdot 1 \end{pmatrix} = \begin{pmatrix} x+2y & 2x+y \\ 5 & 4 \end{pmatrix}.$

Set $AB = BA$ Equating entry by entry:

Top-left: $x+2 = x+2y$ → $2 = 2y$ → $y = 1$ .

Top-right: $y+4 = 2x+y$ → $4 = 2x$ → $x = 2$ .

Bottom-left: $2x+1 = 5$ → with $x=2$ , $2(2)+1=5$ checks out. …
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TG EAPCET 2023Set eng-2023-05-12-FN1 markMCQ
$3 \times 3$
›Reveal solutionSolution
The key idea is to simplify the given equation using properties of the trace: $\operatorname{Tr}(P^T) = \operatorname{Tr}(P)$ , $\operatorname{Tr}(P - P^T) = 0$ , and $\operatorname{Tr}(P + P^T) = 2\operatorname{Tr}(P)$ . Substituting these reduces the equation to a simple quadratic in $t = \operatorname{Tr}(P)$ , giving $t = -1$ as the only non‑zero solution.

We are told $P$ is a $3 \times 3$ square matrix and $\operatorname{Tr}(P) \neq 0$ . The equation is:
$\operatorname{Tr}(P - P^T) + \operatorname{Tr}(P + P^T) + \frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P^T)} + \operatorname{Tr}(P) \times \operatorname{Tr}(P^T) = 0$
The trace is a linear map, and the trace of a transpose equals the trace of the original matrix. That’s the central fact that will collapse most of this expression.

Simplify the first two terms using linearity and transpose properties.

For any square matrix $A$ , $\operatorname{Tr}(A^T) = \operatorname{Tr}(A)$ .

So $\operatorname{Tr}(P - P^T) = \operatorname{Tr}(P) - \operatorname{Tr}(P^T) = \operatorname{Tr}(P) - \operatorname{Tr}(P) = 0$ .

Similarly, $\operatorname{Tr}(P + P^T) = \operatorname{Tr}(P) + \operatorname{Tr}(P^T) = \operatorname{Tr}(P) + \operatorname{Tr}(P) = 2\operatorname{Tr}(P)$ .

Thus the first two terms become $0 + 2\operatorname{Tr}(P)$ .

Handle the fraction term.

Since $\operatorname{Tr}(P^T) = \operatorname{Tr}(P)$ , we have

$\frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P^T)} = \frac{\operatorname{Tr}(P)}{\operatorname{Tr}(P)} = 1,$
provided $\operatorname{Tr}(P) \neq 0$ (which is given). So this term is simply $1$ .

Handle the product term.

$\operatorname{Tr}(P) \times \operatorname{Tr}(P^T) = \operatorname{Tr}(P) \times \operatorname{Tr}(P) = [\operatorname{Tr}(P)]^2$ .

Substitute everything back into the equation. …
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Q.Given $A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}$ . Is $(AB)' = B'A'$ ?

Concept understanding — Matrix Transpose

Matrix Transpose

The intuition

The precise definition

A worked look

Properties you must know

The property being checked

Step 1 — compute $AB$

Step 2 — transpose it

Step 3 — form $B'$ and $A'$ and multiply

Method: Verifying the transpose-of-a-product rule

Steps

Common Mistakes

Q.Given A=[240396]A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}A=[23​49​06​] and B=[142813]B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}B=​121​483​​. Is (AB)′=B′A′(AB)' = B'A'(AB)′=B′A′?

The property being checked

Step 1 — compute ABABAB

Step 2 — transpose it

Step 3 — form B′B'B′ and A′A'A′ and multiply

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Q.Given $A = \begin{bmatrix} 2 & 4 & 0 \\ 3 & 9 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 4 \\ 2 & 8 \\ 1 & 3 \end{bmatrix}$ . Is $(AB)' = B'A'$ ?

Step 1 — compute $AB$

Step 3 — form $B'$ and $A'$ and multiply

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