If $A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$, then verify $(BA)^2 \neq B^2 A^2$.

If A = 3 -4 \1 1 \2 0 and B = 2 1 2 \1 2 4 , then verify (BA)² ≠ B² A².

NCERT Exemplar · Q14

A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}

Telangana TsbieShort· 3mImportance★★★★★

52% · 95/182 Questions

🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

$BA$ is a $2\times2$ matrix with $(BA)^2 = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$ , whereas $B^2$ is undefined because $B$ is $2\times3$ ; so $(BA)^2 \neq B^2A^2$ .

The first thing to fix is the shapes. $A$ is $3\times2$ and $B$ is $2\times3$ . The product $BA$ is $(2\times3)(3\times2) = 2\times2$ — square, so it can be squared. But $B^2 = B\cdot B$ would be $(2\times3)(2\times3)$ , and the inner dimensions ( $3$ and $2$ ) don't match, so $B^2$ does not exist. That single observation already tells us the two sides cannot be equal, but let us compute $(BA)^2$ so the left side is explicit.

Step 1 — compute $BA$

BA = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}\begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}.

$(1,1):\ 2\cdot3+1\cdot1+2\cdot2 = 6+1+4 = 11$
$(1,2):\ 2\cdot(-4)+1\cdot1+2\cdot0 = -8+1+0 = -7$
$(2,1):\ 1\cdot3+2\cdot1+4\cdot2 = 3+2+8 = 13$
$(2,2):\ 1\cdot(-4)+2\cdot1+4\cdot0 = -4+2+0 = -2$

So $BA = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}$ .

Step 2 — square $BA$

(BA)^2 = \begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}\begin{bmatrix} 11 & -7 \\ 13 & -2 \end{bmatrix}.

$(1,1):\ 11\cdot11+(-7)\cdot13 = 121-91 = 30$
$(1,2):\ 11\cdot(-7)+(-7)\cdot(-2) = -77+14 = -63$
$(2,1):\ 13\cdot11+(-2)\cdot13 = 143-26 = 117$
$(2,2):\ 13\cdot(-7)+(-2)\cdot(-2) = -91+4 = -87$

So $(BA)^2 = \begin{bmatrix} 30 & -63 \\ 117 & -87 \end{bmatrix}$ .

Step 3 — why $B^2A^2$ fails …

Start your 14-day free trial to unlock the full solution →

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:

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

$(1)(1) + (2)(2) + (0)(2) = 1 + 4 + 0 = 5$ …
Start your 14-day free trial to see the full solution →

←Previous questionQ13 — Find A, if 4 \1 \3 A = -4 8 4 \-1 2 1 \-3 6 3 .→Next questionQ15 — If possible, find BA and AB, where A = 2 1 2 \1 2 4 , B = 4 1 \2 3 \1 …

Q.If $A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$ , then verify $(BA)^2 \neq B^2 A^2$ .

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

Step 1 — compute $BA$

Step 2 — square $BA$

Step 3 — why $B^2A^2$ fails …

Method: Checking a product identity for existence before value

Steps

Common Mistakes

Q.If A=[3−41120]A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}A=​312​−410​​ and B=[212124]B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}B=[21​12​24​], then verify (BA)2≠B2A2(BA)^2 \neq B^2 A^2(BA)2=B2A2.

Step 1 — compute BABABA

Step 2 — square BABABA

Step 3 — why B2A2B^2A^2B2A2 fails …

🎓Unlock everything free for 14 days

Q.If $A = \begin{bmatrix} 3 & -4 \\ 1 & 1 \\ 2 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 2 & 4 \end{bmatrix}$ , then verify $(BA)^2 \neq B^2 A^2$ .

Step 1 — compute $BA$

Step 2 — square $BA$

Step 3 — why $B^2A^2$ fails …

Unlock everything free for 14 days