If A is matrix of order m × n and B is a matrix such that AB' and B'A are both defined, then order of matrix B

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Telangana TsbieMCQ· 1mImportance★★★★★

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For matrix multiplication to be defined, the inner dimensions must match. Given $A$ is $m \times n$ , $AB'$ defined forces $B'$ to have $n$ rows, and $B'A$ defined forces $B'$ to have $m$ columns. So $B'$ is $n \times m$ , meaning $B$ is $m \times n$ . The correct option is (D).

The key idea here is matrix multiplication compatibility: you can multiply two matrices only when the number of columns in the first equals the number of rows in the second. That single rule drives everything.

Let’s unpack what’s given. $A$ is $m \times n$ . We have two products that are both defined: $AB'$ and $B'A$ . Notice that both involve $B'$ , the transpose of $B$ . So instead of guessing $B$ directly, it’s cleaner to figure out the order of $B'$ first, then transpose back to get $B$ .

From $AB'$ being defined

$A$ is $m \times n$ . For $AB'$ to exist, the number of columns in $A$ (which is $n$ ) must equal the number of rows in $B'$ . So $B'$ must have exactly $n$ rows.

Let the order of $B'$ be $n \times p$ for some unknown $p$ .
From $B'A$ being defined

Now $B'$ is $n \times p$ and $A$ is $m \times n$ . For $B'A$ to exist, the number of columns in $B'$ (which is $p$ ) must equal the number of rows in $A$ (which is $m$ ). So $p = m$ .

Therefore $B'$ is $n \times m$ .
Transpose back to find $B$ …

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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-2022-07-18-AN1 markMCQ
$A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix},\ AB = \begin{bmatrix} 1 & 2 & 2 \\ 1 & 1 & 0 \\ 1 & 2 & 1 \end{bmatrix}$
›Reveal solutionSolution
The expression $A^2 + B(A+B)$ simplifies to $(A+B)^2 - AB$ by using the distributive property of matrix multiplication. We calculate $(A+B)^2$ and then subtract $AB$ to find the result $\boxed{\begin{bmatrix} 4 & 6 & 6 \\ 3 & 4 & 2 \\ 1 & 6 & 3 \end{bmatrix}}$ .

The core idea here is to simplify the given expression $A^2 + B(A+B)$ using the properties of matrix algebra, specifically the distributive property of matrix multiplication over addition. We are given $A+B$ and $AB$ , so we should try to express the target expression in terms of these known quantities.

First, let's expand $B(A+B)$ :

$B(A+B) = BA + B^2$

So, the expression we need to evaluate becomes:

$A^2 + B(A+B) = A^2 + BA + B^2$

Now, let's recall the expansion of $(A+B)^2$ for matrices:
$A$
Watch out

It is crucial to remember that matrix multiplication is generally not commutative, meaning $AB \neq BA$ . Therefore, $(A+B)^2$ is not equal to $A^2 + 2AB + B^2$ unless $AB=BA$ .

Comparing the expression we need, $A^2 + BA + B^2$ , with the expansion of $(A+B)^2$ , which is $A^2 + AB + BA + B^2$ , we can see a direct relationship.

If we subtract $AB$ from $(A+B)^2$ , we get:

$(A+B)^2 - AB = (A^2 + AB + BA + B^2) - AB$

$= A^2 + (AB - AB) + BA + B^2$

$= A^2 + 0 + BA + B^2$

$= A^2 + BA + B^2$

This shows that $A^2 + B(A+B) = (A+B)^2 - AB$ . This simplification is key because we are given $A+B$ and $AB$ directly.

Now, we can proceed with the calculations:

Calculate $(A+B)^2$ :

We are given $A+B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$ .

$(A+B)^2 = (A+B)(A+B) = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 0 \\ 0 & 2 & 2 \end{bmatrix}$

Let's perform the matrix multiplication:

First row:

$(2)(2) + (1)(1) + (2)(0) = 4 + 1 + 0 = 5$

$(2)(1) + (1)(2) + (2)(2) = 2 + 2 + 4 = 8$

$(2)(2) + (1)(0) + (2)(2) = 4 + 0 + 4 = 8$

Second row:

$(1)(2) + (2)(1) + (0)(0) = 2 + 2 + 0 = 4$

$(1)(1) + (2)(2) + (0)(2) = 1 + 4 + 0 = 5$ …
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Q.If $A$ is matrix of order $m \times n$ and $B$ is a matrix such that $AB'$ and $B'A$ are both defined, then order of matrix $B$ is
(A) $m \times m$
(B) $n \times n$
(C) $n \times m$
(D) $m \times n$

Concept understanding — Matrix Multiplication Compatibility

Matrix Multiplication Compatibility: The Inner-Dimensions Rule

The Idea: A Row Meets a Column

The Precise Statement

Why the Rule Exists

Quick Check

Method: Deduce an unknown matrix's order from compatibility conditions

Steps

Common Mistakes

Q.If AAA is matrix of order m×nm \times nm×n and BBB is a matrix such that AB′AB'AB′ and B′AB'AB′A are both defined, then order of matrix BBB is (A) m×mm \times mm×m (B) n×nn \times nn×n (C) n×mn \times mn×m (D) m×nm \times nm×n

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Q.If $A$ is matrix of order $m \times n$ and $B$ is a matrix such that $AB'$ and $B'A$ are both defined, then order of matrix $B$ is
(A) $m \times m$
(B) $n \times n$
(C) $n \times m$
(D) $m \times n$

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