Let $A = \begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}$, $B = \begin{bmatrix}4 & 0\\ 1 & 5\end{bmatrix}$, $C = \begin{bmatrix}2 & 0\\ 1 & -2\end{bmatrix}$ and $a = 4$, $b = -2$. Show that: (a) $A + (B + C) = (A + B) + C$ (b) $A(BC) = (AB)C$ (c) $(a + b)B = aB + bB$ (d) $a(C - A) = aC - aA$ (e) $(A^T)^T = A$ (f) $(bA)^T = b\,A^T$ (g) $(AB)^T = B^T A^T$ (h) $(A - B)C = AC - BC$ (i) $(A - B)^T = A^T - B^T$.

Let A = 1 2\-1 3 , B = 4 0\1 5 , C = 2 0\1 -2 and a = 4, b = -2. Show that: (a) A + (B + C) = (A + B) + C (b)

NCERT Exemplar · Q32

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

(a)

A + (B + C) = (A + B) + C

(b)

A(BC) = (AB)C

(c)

(a + b)B = aB + bB

(d)

a(C - A) = aC - aA

(e)

(A^T)^T = A

(f)

(bA)^T = b\,A^T

(g)

(AB)^T = B^T A^T

(h)

(A - B)C = AC - BC

(i)

(A - B)^T = A^T - B^T

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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 …

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Idea: matrix addition/subtraction is entry-wise, multiplication is row-by-column, and transpose swaps rows and columns. Compute both sides of each identity and check they match. Here $A=\begin{bmatrix}1&2\\-1&3\end{bmatrix}$ , $B=\begin{bmatrix}4&0\\1&5\end{bmatrix}$ , $C=\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ , $a=4$ , $b=-2$ .

(a) $B+C=\begin{bmatrix}6&0\\2&3\end{bmatrix}\Rightarrow A+(B+C)=\begin{bmatrix}7&2\\1&6\end{bmatrix}$ ; and $A+B=\begin{bmatrix}5&2\\0&8\end{bmatrix}\Rightarrow (A+B)+C=\begin{bmatrix}7&2\\1&6\end{bmatrix}$ . ✓

(b) $BC=\begin{bmatrix}8&0\\7&-10\end{bmatrix}\Rightarrow A(BC)=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}$ ; and $AB=\begin{bmatrix}6&10\\-1&15\end{bmatrix}\Rightarrow (AB)C=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}$ . ✓

(c) $a+b=2$ , so $(a+b)B=\begin{bmatrix}8&0\\2&10\end{bmatrix}$ ; and $aB+bB=\begin{bmatrix}16&0\\4&20\end{bmatrix}+\begin{bmatrix}-8&0\\-2&-10\end{bmatrix}=\begin{bmatrix}8&0\\2&10\end{bmatrix}$ . ✓

(d) $C-A=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$ , so $a(C-A)=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}=aC-aA$ . ✓

(e) $A^{T}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}\Rightarrow (A^{T})^{T}=\begin{bmatrix}1&2\\-1&3\end{bmatrix}=A$ . ✓

(f) $bA=\begin{bmatrix}-2&-4\\2&-6\end{bmatrix}\Rightarrow (bA)^{T}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}=b\,A^{T}$ . ✓ …

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All nine identities (a)–(i) are true. We verify each by direct computation with $A=\begin{bmatrix}1&2\\-1&3\end{bmatrix}$ , $B=\begin{bmatrix}4&0\\1&5\end{bmatrix}$ , $C=\begin{bmatrix}2&0\\1&-2\end{bmatrix}$ , $a=4$ , $b=-2$ .

The plan is simple: matrices obey the same associative, distributive and transpose laws as ordinary numbers (the one thing you cannot do is swap the order in a product). To verify each law here, we compute the left side and the right side separately and check they are the identical matrix.

(a) $A+(B+C)=(A+B)+C$ — addition is associative

B+C=\begin{bmatrix}4+2&0+0\\1+1&5-2\end{bmatrix}=\begin{bmatrix}6&0\\2&3\end{bmatrix},\quad A+(B+C)=\begin{bmatrix}7&2\\1&6\end{bmatrix}.

A+B=\begin{bmatrix}5&2\\0&8\end{bmatrix},\quad (A+B)+C=\begin{bmatrix}7&2\\1&6\end{bmatrix}.

Both equal $\begin{bmatrix}7&2\\1&6\end{bmatrix}$ .

(b) $A(BC)=(AB)C$ — multiplication is associative

BC=\begin{bmatrix}8&0\\7&-10\end{bmatrix},\quad A(BC)=\begin{bmatrix}1\cdot8+2\cdot7&1\cdot0+2(-10)\\-1\cdot8+3\cdot7&0+3(-10)\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}.

AB=\begin{bmatrix}6&10\\-1&15\end{bmatrix},\quad (AB)C=\begin{bmatrix}12+10&-20\\-2+15&-30\end{bmatrix}=\begin{bmatrix}22&-20\\13&-30\end{bmatrix}.

Both equal $\begin{bmatrix}22&-20\\13&-30\end{bmatrix}$ .

(c) $(a+b)B=aB+bB$

$a+b=2$ , so $(a+b)B=\begin{bmatrix}8&0\\2&10\end{bmatrix}$ , and $aB+bB=\begin{bmatrix}16&0\\4&20\end{bmatrix}+\begin{bmatrix}-8&0\\-2&-10\end{bmatrix}=\begin{bmatrix}8&0\\2&10\end{bmatrix}$ .

(d) $a(C-A)=aC-aA$

$C-A=\begin{bmatrix}1&-2\\2&-5\end{bmatrix}$ , so $a(C-A)=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$ , and $aC-aA=\begin{bmatrix}8&0\\4&-8\end{bmatrix}-\begin{bmatrix}4&8\\-4&12\end{bmatrix}=\begin{bmatrix}4&-8\\8&-20\end{bmatrix}$ .

(e) $(A^{T})^{T}=A$

$A^{T}=\begin{bmatrix}1&-1\\2&3\end{bmatrix}$ ; transposing again gives $\begin{bmatrix}1&2\\-1&3\end{bmatrix}=A$ .

(f) $(bA)^{T}=b\,A^{T}$

$bA=\begin{bmatrix}-2&-4\\2&-6\end{bmatrix}$ , so $(bA)^{T}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$ , and $b\,A^{T}=-2\begin{bmatrix}1&-1\\2&3\end{bmatrix}=\begin{bmatrix}-2&2\\-4&-6\end{bmatrix}$ .

(g) $(AB)^{T}=B^{T}A^{T}$ — the order reverses …

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Q.Let $A = \begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}$ , $B = \begin{bmatrix}4 & 0\\ 1 & 5\end{bmatrix}$ , $C = \begin{bmatrix}2 & 0\\ 1 & -2\end{bmatrix}$ and $a = 4$ , $b = -2$ . Show that:

Concept understanding — Matrix Addition

Matrix Addition: The Intuition

The Precise Statement

The One Rule You Cannot Break

Properties That Feel Natural

A Quick Check …

(a) $A+(B+C)=(A+B)+C$ — addition is associative

(b) $A(BC)=(AB)C$ — multiplication is associative

(c) $(a+b)B=aB+bB$

(d) $a(C-A)=aC-aA$

(e) $(A^{T})^{T}=A$

(f) $(bA)^{T}=b\,A^{T}$

(g) $(AB)^{T}=B^{T}A^{T}$ — the order reverses …

Method: Verifying matrix algebra laws by computing both sides

Steps

Common Mistakes

Q.Let A=[12−13]A = \begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}A=[1−1​23​], B=[4015]B = \begin{bmatrix}4 & 0\\ 1 & 5\end{bmatrix}B=[41​05​], C=[201−2]C = \begin{bmatrix}2 & 0\\ 1 & -2\end{bmatrix}C=[21​0−2​] and a=4a = 4a=4, b=−2b = -2b=−2. Show that:

(a) A+(B+C)=(A+B)+CA+(B+C)=(A+B)+CA+(B+C)=(A+B)+C — addition is associative

(b) A(BC)=(AB)CA(BC)=(AB)CA(BC)=(AB)C — multiplication is associative

(c) (a+b)B=aB+bB(a+b)B=aB+bB(a+b)B=aB+bB

(d) a(C−A)=aC−aAa(C-A)=aC-aAa(C−A)=aC−aA

(e) (AT)T=A(A^{T})^{T}=A(AT)T=A

(f) (bA)T=b AT(bA)^{T}=b\,A^{T}(bA)T=bAT

(g) (AB)T=BTAT(AB)^{T}=B^{T}A^{T}(AB)T=BTAT — the order reverses …

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Q.Let $A = \begin{bmatrix}1 & 2\\ -1 & 3\end{bmatrix}$ , $B = \begin{bmatrix}4 & 0\\ 1 & 5\end{bmatrix}$ , $C = \begin{bmatrix}2 & 0\\ 1 & -2\end{bmatrix}$ and $a = 4$ , $b = -2$ . Show that:

(a) $A+(B+C)=(A+B)+C$ — addition is associative

(b) $A(BC)=(AB)C$ — multiplication is associative

(c) $(a+b)B=aB+bB$

(d) $a(C-A)=aC-aA$

(e) $(A^{T})^{T}=A$

(f) $(bA)^{T}=b\,A^{T}$

(g) $(AB)^{T}=B^{T}A^{T}$ — the order reverses …

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