Matrix multiplication worksheet with exercises and explanations for students.

Here are the step-by-step calculations and solutions for the problems on the worksheet.

Section I: Matrix Multiplication

1. Fill in the blanks
Based on the rules of matrix multiplication:
1. Two matrices, $A$ and $B$, can be multiplied together iff the number of columns in $A$ is equal to the number of rows in $B$.
2. The multiplication of two matrices, $A$ and $B$, will give rise to a matrix $AB$ with the number of rows of $A$ and the number of columns of $B$, i.e., $A_{m \times n} \times B_{n \times p} = AB_{m \times p}$.

Practice 1: Compute the products $AB$ and $BA$

(i)
$A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}$, $B = \begin{pmatrix} -1 & 4 \\ 2 & 0.5 \end{pmatrix}$

* Calculate $AB$:
* Row 1, Col 1: $(1)(-1) + (2)(2) = -1 + 4 = 3$
* Row 1, Col 2: $(1)(4) + (2)(0.5) = 4 + 1 = 5$
* Row 2, Col 1: $(3)(-1) + (1)(2) = -3 + 2 = -1$
* Row 2, Col 2: $(3)(4) + (1)(0.5) = 12 + 0.5 = 12.5$
* $AB = \begin{pmatrix} 3 & 5 \\ -1 & 12.5 \end{pmatrix}$

* Calculate $BA$:
* Row 1, Col 1: $(-1)(1) + (4)(3) = -1 + 12 = 11$
* Row 1, Col 2: $(-1)(2) + (4)(1) = -2 + 4 = 2$
* Row 2, Col 1: $(2)(1) + (0.5)(3) = 2 + 1.5 = 3.5$
* Row 2, Col 2: $(2)(2) + (0.5)(1) = 4 + 0.5 = 4.5$
* $BA = \begin{pmatrix} 11 & 2 \\ 3.5 & 4.5 \end{pmatrix}$

(ii)
$A = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$, $B = \begin{pmatrix} 2 & 0.5 & -1 \\ 1 & 3 & 4 \end{pmatrix}$

* Calculate $AB$:
* Row 1, Col 1: $(2)(2) + (3)(1) = 4 + 3 = 7$
* Row 1, Col 2: $(2)(0.5) + (3)(3) = 1 + 9 = 10$
* Row 1, Col 3: $(2)(-1) + (3)(4) = -2 + 12 = 10$
* Row 2, Col 1: $(1)(2) + (2)(1) = 2 + 2 = 4$
* Row 2, Col 2: $(1)(0.5) + (2)(3) = 0.5 + 6 = 6.5$
* Row 2, Col 3: $(1)(-1) + (2)(4) = -1 + 8 = 7$
* $AB = \begin{pmatrix} 7 & 10 & 10 \\ 4 & 6.5 & 7 \end{pmatrix}$

* Calculate $BA$:
* Matrix $B$ has 3 columns, but Matrix $A$ has 2 rows. Since $3 \neq 2$, the multiplication is not possible.

(iii)
$A = \begin{pmatrix} 1 & 2 & 2 & 2 \\ 0 & 4 & 0 & 3 \end{pmatrix}$, $B = \begin{pmatrix} 1 & 1 \\ 2 & 0 \\ 3 & 4 \end{pmatrix}$

* Calculate $AB$:
* Matrix $A$ has 4 columns. Matrix $B$ has 3 rows. Since $4 \neq 3$, the multiplication is not possible.

* Calculate $BA$:
* Row 1, Col 1: $(1)(1) + (1)(0) = 1$
* Row 1, Col 2: $(1)(2) + (1)(4) = 6$
* Row 1, Col 3: $(1)(2) + (1)(0) = 2$
* Row 1, Col 4: $(1)(2) + (1)(3) = 5$
* Row 2, Col 1: $(2)(1) + (0)(0) = 2$
* Row 2, Col 2: $(2)(2) + (0)(4) = 4$
* Row 2, Col 3: $(2)(2) + (0)(0) = 4$
* Row 2, Col 4: $(2)(2) + (0)(3) = 4$
* Row 3, Col 1: $(3)(1) + (4)(0) = 3$
* Row 3, Col 2: $(3)(2) + (4)(4) = 6 + 16 = 22$
* Row 3, Col 3: $(3)(2) + (4)(0) = 6$
* Row 3, Col 4: $(3)(2) + (4)(3) = 6 + 12 = 18$
* $BA = \begin{pmatrix} 1 & 6 & 2 & 5 \\ 2 & 4 & 4 & 4 \\ 3 & 22 & 6 & 18 \end{pmatrix}$

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Section II: Laws of Matrix Multiplication

1. Associative Law
Given $A = \begin{pmatrix} 3 & -1 \\ 4 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 2 & 5 \\ 1 & 3 \end{pmatrix}$, $C = \begin{pmatrix} 1 & 0 \\ 2 & 3 \end{pmatrix}$

* (i) Find $A(BC)$:
* First calculate $BC$:
* Row 1: $(2)(1)+(5)(2)=12$, $(2)(0)+(5)(3)=15$
* Row 2: $(1)(1)+(3)(2)=7$, $(1)(0)+(3)(3)=9$
* $BC = \begin{pmatrix} 12 & 15 \\ 7 & 9 \end{pmatrix}$
* Now calculate $A(BC)$:
* Row 1: $(3)(12)+(-1)(7)=36-7=29$, $(3)(15)+(-1)(9)=45-9=36$
* Row 2: $(4)(12)+(0)(7)=48$, $(4)(15)+(0)(9)=60$
* $A(BC) = \begin{pmatrix} 29 & 36 \\ 48 & 60 \end{pmatrix}$

* (ii) Find $(AB)C$:
* First calculate $AB$:
* Row 1: $(3)(2)+(-1)(1)=5$, $(3)(5)+(-1)(3)=12$
* Row 2: $(4)(2)+(0)(1)=8$, $(4)(5)+(0)(3)=20$
* $AB = \begin{pmatrix} 5 & 12 \\ 8 & 20 \end{pmatrix}$
* Now calculate $(AB)C$:
* Row 1: $(5)(1)+(12)(2)=29$, $(5)(0)+(12)(3)=36$
* Row 2: $(8)(1)+(20)(2)=48$, $(8)(0)+(20)(3)=60$
* $(AB)C = \begin{pmatrix} 29 & 36 \\ 48 & 60 \end{pmatrix}$

2. Distributive Law
Given $A = \begin{pmatrix} 2 & -1 \\ 1 & 0 \end{pmatrix}$, $B = \begin{pmatrix} 2 & 4 \\ 2 & 3 \end{pmatrix}$, $C = \begin{pmatrix} 1 & 0 \\ 0 & 3 \end{pmatrix}$

* (i) Find $A(B+C)$:
* First calculate $B+C$:
* $\begin{pmatrix} 2+1 & 4+0 \\ 2+0 & 3+3 \end{pmatrix} = \begin{pmatrix} 3 & 4 \\ 2 & 6 \end{pmatrix}$
* Now calculate $A(B+C)$:
* Row 1: $(2)(3)+(-1)(2)=4$, $(2)(4)+(-1)(6)=2$
* Row 2: $(1)(3)+(0)(2)=3$, $(1)(4)+(0)(6)=4$
* $A(B+C) = \begin{pmatrix} 4 & 2 \\ 3 & 4 \end{pmatrix}$

* (ii) Find $AB + AC$:
* Calculate $AB$:
* Row 1: $(2)(2)+(-1)(2)=2$, $(2)(4)+(-1)(3)=5$
* Row 2: $(1)(2)+(0)(2)=2$, $(1)(4)+(0)(3)=4$
* $AB = \begin{pmatrix} 2 & 5 \\ 2 & 4 \end{pmatrix}$
* Calculate $AC$:
* Row 1: $(2)(1)+(-1)(0)=2$, $(2)(0)+(-1)(3)=-3$
* Row 2: $(1)(1)+(0)(0)=1$, $(1)(0)+(0)(3)=0$
* $AC = \begin{pmatrix} 2 & -3 \\ 1 & 0 \end{pmatrix}$
* Add them:
* $\begin{pmatrix} 2+2 & 5+(-3) \\ 2+1 & 4+0 \end{pmatrix} = \begin{pmatrix} 4 & 2 \\ 3 & 4 \end{pmatrix}$
* $AB + AC = \begin{pmatrix} 4 & 2 \\ 3 & 4 \end{pmatrix}$

* (iii) Find $(B+C)A$:
* Use the sum from part (i): $B+C = \begin{pmatrix} 3 & 4 \\ 2 & 6 \end{pmatrix}$
* Multiply by $A$:
* Row 1: $(3)(2)+(4)(1)=10$, $(3)(-1)+(4)(0)=-3$
* Row 2: $(2)(2)+(6)(1)=10$, $(2)(-1)+(6)(0)=-2$
* $(B+C)A = \begin{pmatrix} 10 & -3 \\ 10 & -2 \end{pmatrix}$

* (iv) Find $BA + CA$:
* Calculate $BA$:
* Row 1: $(2)(2)+(4)(1)=8$, $(2)(-1)+(4)(0)=-2$
* Row 2: $(2)(2)+(3)(1)=7$, $(2)(-1)+(3)(0)=-2$
* $BA = \begin{pmatrix} 8 & -2 \\ 7 & -2 \end{pmatrix}$
* Calculate $CA$:
* Row 1: $(1)(2)+(0)(1)=2$, $(1)(-1)+(0)(0)=-1$
* Row 2: $(0)(2)+(3)(1)=3$, $(0)(-1)+(3)(0)=0$
* $CA = \begin{pmatrix} 2 & -1 \\ 3 & 0 \end{pmatrix}$
* Add them:
* $\begin{pmatrix} 8+2 & -2+(-1) \\ 7+3 & -2+0 \end{pmatrix} = \begin{pmatrix} 10 & -3 \\ 10 & -2 \end{pmatrix}$
* $BA + CA = \begin{pmatrix} 10 & -3 \\ 10 & -2 \end{pmatrix}$

Final Answer:
Section I
1. 1. columns, rows
2. rows, columns

Practice 1
(i) $AB = \begin{pmatrix} 3 & 5 \\ -1 & 12.5 \end{pmatrix}$, $BA = \begin{pmatrix} 11 & 2 \\ 3.5 & 4.5 \end{pmatrix}$
(ii) $AB = \begin{pmatrix} 7 & 10 & 10 \\ 4 & 6.5 & 7 \end{pmatrix}$, $BA$ is not possible.
(iii) $AB$ is not possible, $BA = \begin{pmatrix} 1 & 6 & 2 & 5 \\ 2 & 4 & 4 & 4 \\ 3 & 22 & 6 & 18 \end{pmatrix}$

Section II
1. (i) $\begin{pmatrix} 29 & 36 \\ 48 & 60 \end{pmatrix}$
(ii) $\begin{pmatrix} 29 & 36 \\ 48 & 60 \end{pmatrix}$

2. (i) $\begin{pmatrix} 4 & 2 \\ 3 & 4 \end{pmatrix}$
(ii) $\begin{pmatrix} 4 & 2 \\ 3 & 4 \end{pmatrix}$
(iii) $\begin{pmatrix} 10 & -3 \\ 10 & -2 \end{pmatrix}$
(iv) $\begin{pmatrix} 10 & -3 \\ 10 & -2 \end{pmatrix}$