Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ Find each of the following: (i) $A + B$ (ii) $A - B$ (iii) $3A - C$ (iv) $AB$ (v) $BA$

Let A = 2 4 \3 2 , B = 1 3 \-2 5 , C = -2 5 \3 4 F – Class 12 Mathematics

Exercise 3.2 · Q1

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

(i)

A + B

(ii)

A - B

(iii)

3A - C

(iv)

AB

(v)

BA

Sikkim CbseNCERTSubjective· 3mImportance★★★★★

8% · 14/182 Questions

✓ Free question

Concept understanding — Matrix Addition

Matrix Addition: The Intuition

You run a fruit stall and record apples and bananas sold each morning and afternoon in a table:

Time	Apples	Bananas
Morning	10	5
Afternoon	8	12

That's a matrix — a rectangular array of numbers. Your friend's stall has its own table for the same day (morning: 6 apples, 7 bananas; afternoon: 4 apples, 9 bananas). To get the combined sales, you add the numbers in the same position: morning apples with morning apples, afternoon bananas with afternoon bananas, and so on.

That's matrix addition: you add corresponding entries — numbers in the same row and column.

The Precise Statement

(A + B)_{ij} = A_{ij} + B_{ij}

If $A$ and $B$ have the same size (same number of rows and columns), their sum $A + B$ is a matrix of that size where each entry is the sum of the corresponding entries.

Example:

A = \begin{bmatrix} 2 & 5 \\ 1 & -3 \end{bmatrix}, \quad B = \begin{bmatrix} 0 & 4 \\ 7 & 2 \end{bmatrix}

A + B = \begin{bmatrix} 2+0 & 5+4 \\ 1+7 & -3+2 \end{bmatrix} = \begin{bmatrix} 2 & 9 \\ 8 & -1 \end{bmatrix}

The One Rule You Cannot Break

Watch out

You can only add matrices with the exact same dimensions. A $2 \times 3$ matrix cannot be added to a $3 \times 2$ matrix — the positions don't match.

Properties That Feel Natural

Matrix addition behaves like ordinary number addition, inheriting these from the addition of individual entries:

Commutative: $A + B = B + A$
Associative: $(A + B) + C = A + (B + C)$
Zero matrix: there's a matrix $O$ (all zeros) with $A + O = A$

A Quick Check

P = \begin{bmatrix} 1 & 0 & 2 \\ -1 & 3 & 4 \end{bmatrix}, \quad Q = \begin{bmatrix} 0 & 2 & 1 \\ 5 & -2 & 0 \end{bmatrix}

Both are $2 \times 3$ , so:

P + Q = \begin{bmatrix} 1+0 & 0+2 & 2+1 \\ -1+5 & 3+(-2) & 4+0 \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 1 & 4 \end{bmatrix}

That's all there is to it — matrix addition is just entry-wise addition, the simplest operation in linear algebra and the foundation for everything that follows.

Matrix Addition, requiring matrices of the same order and defined entry-by-entry, is one of the first operations taught in the CBSE Class 12 Matrices chapter, and "matrix addition rules and properties class 12" is a widely searched revision topic. Its commutative and associative properties, inherited directly from ordinary number addition, are frequently tested as short-answer board exam questions.

Matrix addition and subtraction are element-wise operations; scalar multiplication scales every entry; matrix multiplication follows the row‑by‑column rule. For the given matrices, $A+B = \begin{bmatrix}3 & 7 \\ 1 & 7\end{bmatrix}$ , $A-B = \begin{bmatrix}1 & 1 \\ 5 & -3\end{bmatrix}$ , $3A-C = \begin{bmatrix}8 & 7 \\ 6 & 2\end{bmatrix}$ , $AB = \begin{bmatrix}-6 & 26 \\ -1 & 19\end{bmatrix}$ , $BA = \begin{bmatrix}11 & 10 \\ 11 & 2\end{bmatrix}$ .

Why these operations work the way they do

Matrix addition and subtraction are the simplest: you just combine corresponding entries. Think of two matrices as two tables of numbers sitting side by side — adding them means adding the number in the top‑left of the first to the top‑left of the second, and so on. This only makes sense when both matrices have the same shape (same number of rows and columns). Here $A$ , $B$ , and $C$ are all $2 \times 2$ , so addition and subtraction are straightforward.

Scalar multiplication (like $3A$ ) means multiplying every entry of $A$ by $3$ . Then you subtract $C$ entry‑wise.

Matrix multiplication is different. When you multiply $AB$ , you take each row of $A$ and “dot” it with each column of $B$ . The entry in row $i$ , column $j$ of $AB$ is the sum of products: $( \text{row } i \text{ of } A ) \cdot ( \text{column } j \text{ of } B )$ . This is not commutative — $AB$ and $BA$ usually give different results, as you’ll see.

Step‑by‑step

1. $A + B$

Add corresponding entries:

A + B = \begin{bmatrix} 2+1 & 4+3 \\ 3+(-2) & 2+5 \end{bmatrix} = \begin{bmatrix} 3 & 7 \\ 1 & 7 \end{bmatrix}

Tip

Always check that the matrices have the same dimensions before adding. Here both are $2 \times 2$ , so it’s valid.

2. $A - B$

Subtract entry‑wise:

A - B = \begin{bmatrix} 2-1 & 4-3 \\ 3-(-2) & 2-5 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 5 & -3 \end{bmatrix}

Notice that $3 - (-2) = 3 + 2 = 5$ . A common slip is forgetting to change the sign when subtracting a negative number.

Watch out

$A - B$ is not the same as $B - A$ . Subtraction is not commutative. If you swapped, you’d get $\begin{bmatrix} -1 & -1 \\ -5 & 3 \end{bmatrix}$ .

3. $3A - C$

First multiply every entry of $A$ by $3$ :

3A = \begin{bmatrix} 3 \cdot 2 & 3 \cdot 4 \\ 3 \cdot 3 & 3 \cdot 2 \end{bmatrix} = \begin{bmatrix} 6 & 12 \\ 9 & 6 \end{bmatrix}

Now subtract $C$ entry‑wise:

3A - C = \begin{bmatrix} 6 - (-2) & 12 - 5 \\ 9 - 3 & 6 - 4 \end{bmatrix} = \begin{bmatrix} 8 & 7 \\ 6 & 2 \end{bmatrix}

Again, watch the double negative: $6 - (-2) = 8$ .

4. $AB$

Multiply $A$ (first matrix) by $B$ (second matrix). $A$ has rows, $B$ has columns.

Entry (1,1): row 1 of $A$ $\begin{bmatrix}2 & 4\end{bmatrix}$ dot column 1 of $B$ $\begin{bmatrix}1 \\ -2\end{bmatrix}$

$= 2 \cdot 1 + 4 \cdot (-2) = 2 - 8 = -6$
Entry (1,2): row 1 of $A$ dot column 2 of $B$ $\begin{bmatrix}3 \\ 5\end{bmatrix}$

$= 2 \cdot 3 + 4 \cdot 5 = 6 + 20 = 26$
Entry (2,1): row 2 of $A$ $\begin{bmatrix}3 & 2\end{bmatrix}$ dot column 1 of $B$

$= 3 \cdot 1 + 2 \cdot (-2) = 3 - 4 = -1$
Entry (2,2): row 2 of $A$ dot column 2 of $B$

$= 3 \cdot 3 + 2 \cdot 5 = 9 + 10 = 19$

AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}

2 \times 2

5. $BA$

Now multiply $B$ by $A$ . The order is reversed, so the result will likely be different.

Entry (1,1): row 1 of $B$ $\begin{bmatrix}1 & 3\end{bmatrix}$ dot column 1 of $A$ $\begin{bmatrix}2 \\ 3\end{bmatrix}$

$= 1 \cdot 2 + 3 \cdot 3 = 2 + 9 = 11$
Entry (1,2): row 1 of $B$ dot column 2 of $A$ $\begin{bmatrix}4 \\ 2\end{bmatrix}$

$= 1 \cdot 4 + 3 \cdot 2 = 4 + 6 = 10$
Entry (2,1): row 2 of $B$ $\begin{bmatrix}-2 & 5\end{bmatrix}$ dot column 1 of $A$

$= (-2) \cdot 2 + 5 \cdot 3 = -4 + 15 = 11$
Entry (2,2): row 2 of $B$ dot column 2 of $A$

$= (-2) \cdot 4 + 5 \cdot 2 = -8 + 10 = 2$

Thus

BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}

Compare with $AB$ — they are clearly not the same. This is a key property: matrix multiplication is not commutative.

Important

$AB \neq BA$ in general. Always pay attention to the order when multiplying matrices.

✓Final answer

A+B = \begin{bmatrix}3 & 7 \\ 1 & 7\end{bmatrix}

←Previous question · 3.4.6Example 19 — In a legislative assembly election, a political group hired a public r…→Next question · 3.4.6Q2 — Compute the following: (i) a b \-b a + a b b a (ii) a²+b² b²+c² a²+c² …

Q.Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ Find each of the following:

Concept understanding — Matrix Addition

Matrix Addition: The Intuition

The Precise Statement

The One Rule You Cannot Break

Properties That Feel Natural

A Quick Check

Why these operations work the way they do

Step‑by‑step

1. $A + B$

2. $A - B$

3. $3A - C$

4. $AB$

5. $BA$

Method: Combining Matrices Under Addition, Scalar Multiplication, and Multiplication

Steps

Common Mistakes

Q.Let A=[2432]A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}A=[23​42​], B=[13−25]B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}B=[1−2​35​], C=[−2534]C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}C=[−23​54​] Find each of the following:

Why these operations work the way they do

Step‑by‑step

1. A+BA + BA+B

2. A−BA - BA−B

3. 3A−C3A - C3A−C

4. ABABAB

5. BABABA

🎓Unlock everything free for 14 days

Q.Let $A = \begin{bmatrix} 2 & 4 \\ 3 & 2 \end{bmatrix}$ , $B = \begin{bmatrix} 1 & 3 \\ -2 & 5 \end{bmatrix}$ , $C = \begin{bmatrix} -2 & 5 \\ 3 & 4 \end{bmatrix}$ Find each of the following:

1. $A + B$

2. $A - B$

3. $3A - C$

4. $AB$

5. $BA$

Unlock everything free for 14 days