Skip to content
Exercise 3.2 · Q1

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

(i) A+BA + B
(ii) A−BA - B
(iii) 3A−C3A - C
(iv) ABAB
(v) BABA
Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★
8% · 14/182 Questions
✓ Free question

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=[3717]A+B = \begin{bmatrix}3 & 7 \\ 1 & 7\end{bmatrix}, A−B=[115−3]A-B = \begin{bmatrix}1 & 1 \\ 5 & -3\end{bmatrix}, 3A−C=[8762]3A-C = \begin{bmatrix}8 & 7 \\ 6 & 2\end{bmatrix}, AB=[−626−119]AB = \begin{bmatrix}-6 & 26 \\ -1 & 19\end{bmatrix}, BA=[1110112]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 AA, BB, and CC are all 2×22 \times 2, so addition and subtraction are straightforward.

Scalar multiplication (like 3A3A) means multiplying every entry of AA by 33. Then you subtract CC entry‑wise.

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


Step‑by‑step

1. A+BA + B

Add corresponding entries:

A+B=[2+14+33+(−2)2+5]=[3717]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 2×22 \times 2, so it’s valid.

2. A−BA - B

Subtract entry‑wise:

A−B=[2−14−33−(−2)2−5]=[115−3]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=53 - (-2) = 3 + 2 = 5. A common slip is forgetting to change the sign when subtracting a negative number.

Watch out

A−BA - B is not the same as B−AB - A. Subtraction is not commutative. If you swapped, you’d get [−1−1−53]\begin{bmatrix} -1 & -1 \\ -5 & 3 \end{bmatrix}.

3. 3A−C3A - C

First multiply every entry of AA by 33:

3A=[3⋅23⋅43⋅33⋅2]=[61296]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 CC entry‑wise:

3A−C=[6−(−2)12−59−36−4]=[8762]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)=86 - (-2) = 8.

4. ABAB

Multiply AA (first matrix) by BB (second matrix). AA has rows, BB has columns.

  • Entry (1,1): row 1 of AA [24]\begin{bmatrix}2 & 4\end{bmatrix} dot column 1 of BB [1−2]\begin{bmatrix}1 \\ -2\end{bmatrix}

    =2⋅1+4⋅(−2)=2−8=−6= 2 \cdot 1 + 4 \cdot (-2) = 2 - 8 = -6

  • Entry (1,2): row 1 of AA dot column 2 of BB [35]\begin{bmatrix}3 \\ 5\end{bmatrix}

    =2⋅3+4⋅5=6+20=26= 2 \cdot 3 + 4 \cdot 5 = 6 + 20 = 26

  • Entry (2,1): row 2 of AA [32]\begin{bmatrix}3 & 2\end{bmatrix} dot column 1 of BB

    =3⋅1+2⋅(−2)=3−4=−1= 3 \cdot 1 + 2 \cdot (-2) = 3 - 4 = -1

  • Entry (2,2): row 2 of AA dot column 2 of BB

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

So

AB=[−626−119]AB = \begin{bmatrix} -6 & 26 \\ -1 & 19 \end{bmatrix}

For 2×22 \times 2 matrices,

[abcd][efgh]=[ae+bgaf+bhce+dgcf+dh]\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} e & f \\ g & h \end{bmatrix} = \begin{bmatrix} ae+bg & af+bh \\ ce+dg & cf+dh \end{bmatrix}.

5. BABA

Now multiply BB by AA. The order is reversed, so the result will likely be different.

  • Entry (1,1): row 1 of BB [13]\begin{bmatrix}1 & 3\end{bmatrix} dot column 1 of AA [23]\begin{bmatrix}2 \\ 3\end{bmatrix}

    =1⋅2+3⋅3=2+9=11= 1 \cdot 2 + 3 \cdot 3 = 2 + 9 = 11

  • Entry (1,2): row 1 of BB dot column 2 of AA [42]\begin{bmatrix}4 \\ 2\end{bmatrix}

    =1⋅4+3⋅2=4+6=10= 1 \cdot 4 + 3 \cdot 2 = 4 + 6 = 10

  • Entry (2,1): row 2 of BB [−25]\begin{bmatrix}-2 & 5\end{bmatrix} dot column 1 of AA

    =(−2)⋅2+5⋅3=−4+15=11= (-2) \cdot 2 + 5 \cdot 3 = -4 + 15 = 11

  • Entry (2,2): row 2 of BB dot column 2 of AA

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

Thus

BA=[1110112]BA = \begin{bmatrix} 11 & 10 \\ 11 & 2 \end{bmatrix}

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

Important

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


✓Final answer

The results are A+B=[3717]A+B = \begin{bmatrix}3 & 7 \\ 1 & 7\end{bmatrix}, A−B=[115−3]A-B = \begin{bmatrix}1 & 1 \\ 5 & -3\end{bmatrix}, 3A−C=[8762]3A-C = \begin{bmatrix}8 & 7 \\ 6 & 2\end{bmatrix}, AB=[−626−119]AB = \begin{bmatrix}-6 & 26 \\ -1 & 19\end{bmatrix}, and BA=[1110112]BA = \begin{bmatrix}11 & 10 \\ 11 & 2\end{bmatrix}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.