Q.For the matrices A and B, verify that $(AB)' = B'A'$, where

The key idea is the transpose of a product: $(AB)' = B'A'$. We verify this by computing both sides for each pair.

(i) $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

First, $AB$ is a $3 \times 3$ matrix:

So $(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$.

Now compute $B'A'$: $B' = \begin{bmatrix} -1 \\ 2 \\ 1 \end{bmatrix}$, $A' = \begin{bmatrix} 1 & -4 & 3 \end{bmatrix}$. Then

Both sides match.

(ii) $A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$ …

Why This Works

The property $(AB)' = B'A'$ is a fundamental rule of matrix transposition. It says: if you multiply two matrices and then transpose the result, it's the same as transposing each matrix individually and multiplying them in reverse order. This reverse order is crucial — matrix multiplication is not commutative, so the order matters. For the given matrices, $A$ is a column vector (3×1) and $B$ is a row vector (1×3), so $AB$ is a 3×3 matrix. Its transpose will be a 3×3 matrix, and $B'A'$ will also be 3×3. Let's verify step by step for each case.

Case (i): $A = \begin{bmatrix} 1 \\ -4 \\ 3 \end{bmatrix}$, $B = \begin{bmatrix} -1 & 2 & 1 \end{bmatrix}$

1. Compute $AB$

$A$ is 3×1, $B$ is 1×3, so $AB$ is 3×3. Multiply each row of $A$ (only one element per row) by each column of $B$ (only one column, but we treat it as a row of entries):

2. Compute $(AB)'$

Transpose means swap rows and columns. The first row becomes the first column, etc.:

3. Compute $B'$ and $A'$

$B$ is 1×3, so $B'$ is 3×1:

$A$ is 3×1, so $A'$ is 1×3:

4. Compute $B'A'$

Now $B'$ is 3×1 and $A'$ is 1×3, so their product is 3×3. Multiply:

5. Compare

$(AB)' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$ and $B'A' = \begin{bmatrix} -1 & 4 & -3 \\ 2 & -8 & 6 \\ 1 & -4 & 3 \end{bmatrix}$. They are identical. So $(AB)' = B'A'$ holds.

Case (ii): $A = \begin{bmatrix} 0 \\ 1 \\ 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 5 & 7 \end{bmatrix}$

1. Compute $AB$

Again, $A$ is 3×1, $B$ is 1×3, product is 3×3: …

Method: Verifying the reversal law $(AB)'=B'A'$

To verify the transpose-of-a-product rule, compute $AB$ then transpose it, and separately compute $B'A'$; respecting the reversal of order is the key.

Steps

Step 1: Compute $AB$ and transpose it

Multiply $A$ by $B$ using the row-by-column rule, then transpose the result to get $(AB)'$. …

Common Mistakes

Mistake 1: Writing $(AB)'=A'B'$ instead of $B'A'$

Why it's wrong: the transpose reverses the order of the factors; $A'B'$ may not even be defined (here $A'$ is $1\times3$ and $B'$ is $3\times1$, giving a scalar, not the $3\times3$ result). Correct approach: use $B'A'$ — the first factor, transposed, goes last.

Mistake 2: Multiplying rows by rows instead of rows by columns …