1. Matrix multiplication is generally commutative.
2. The product of two complex conjugates is always a real number.
3. When multiplying polynomials, the degree of the product is the sum of the degrees of the factors.
4. The distributive property applies to scalar multiplication of a matrix, meaning \(k(A + B) = kA + kB\).
5. Multiplying a non-zero polynomial by a constant (non-zero scalar) changes its degree.
✏️ 2. Fill in the Blanks
1. To multiply two matrices \(A\) and \(B\), the number of columns in matrix \(A\) must equal the number of in matrix \(B\).
2. The product of a complex number \(a + bi\) and its conjugate \(a - bi\) is \(a^2 + \).
3. When multiplying polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial, a process often referred to as the property.
4. For any scalar \(k\) and matrices \(A\) and \(B\), the property \(k(AB) = (kA)B = A(kB)\) is known as the property of scalar multiplication.
5. The identity matrix, when multiplied by any square matrix \(A\), results in .
🔗 3. Matching
« For real numbers, the order of factors does not change the product (e.g., \(ab = ba\)).
« Multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products (e.g., \(a(b + c) = ab + ac\)).
« For a complex number \(a + bi\), its conjugate is \(a - bi\).
« The process of multiplying every element of a matrix by a single real number.
« The sum of the degrees of the individual non-zero polynomials being multiplied.
✍️ 4. Short Answer Questions
1. Explain why matrix multiplication is generally not commutative.
💡 Suggested Answer: Matrix multiplication is generally not commutative because the order of multiplication often affects the result. For two matrices \(A\) and \(B\), \(AB\) is typically not equal to \(BA\). In many cases, if \(AB\) is defined, \(BA\) might not even be defined due to incompatible dimensions, or if both are defined, their elements will be different.
2. What is the result of multiplying a complex number \(z = a + bi\) by \(i^2\)?
💡 Suggested Answer: Since \(i^2 = -1\), multiplying a complex number \(z = a + bi\) by \(i^2\) results in \(-1 \cdot (a + bi) = -a - bi\). This is the additive inverse of the complex number.
2. Given matrices \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}\), find the element in the first row, second column of the product \(AB\).
3. What is the product of the complex numbers \((3 + 2i)(1 - 4i)\)?
📝 6. Open-Ended Questions
1. Multiply the polynomials \((x^2 - 2x + 1)(3x^2 + x - 2)\) and simplify the result.
💡 Solution Steps:
To multiply the polynomials, we distribute each term of the first polynomial by each term of the second polynomial:\
\[ (x^2 - 2x + 1)(3x^2 + x - 2) \]
First, multiply \(x^2\) by each term in the second polynomial:\
\(x^2(3x^2 + x - 2) = 3x^4 + x^3 - 2x^2\)\
Next, multiply \(-2x\) by each term in the second polynomial:\
\(-2x(3x^2 + x - 2) = -6x^3 - 2x^2 + 4x\)\
Finally, multiply \(1\) by each term in the second polynomial:\
\(1(3x^2 + x - 2) = 3x^2 + x - 2\)\
Now, combine all the results and group like terms:\
\( (3x^4 + x^3 - 2x^2) + (-6x^3 - 2x^2 + 4x) + (3x^2 + x - 2) \)\
Group terms by degree:\
\(3x^4 \quad\text{(only one } x^4 \text{ term)}\
\(x^3 - 6x^3 = -5x^3 \quad\text{(for } x^3 \text{ terms)}\
\(-2x^2 - 2x^2 + 3x^2 = -x^2 \quad\text{(for } x^2 \text{ terms)}\
\(4x + x = 5x \quad\text{(for } x \text{ terms)}\
\(-2 \quad\text{(constant term)}\
Combining these, the simplified product is:\
\[ 3x^4 - 5x^3 - x^2 + 5x - 2 \]
2. Given matrices \(A = \begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix}\) and \(B = \begin{pmatrix} -1 & 4 \\ 2 & 0 \end{pmatrix}\), calculate the product \(AB\).
💡 Solution Steps:
To calculate the product \(AB\), we multiply the rows of matrix \(A\) by the columns of matrix \(B\). The resulting matrix will also be a \(2 \times 2\) matrix.\
Let \(AB = C = \begin{pmatrix} c_{11} & c_{12} \\ c_{21} & c_{22} \end{pmatrix}\)\
Calculate \(c_{11}\) (first row of \(A\) times first column of \(B\)):\
\(c_{11} = (2)(-1) + (1)(2) = -2 + 2 = 0\)\
Calculate \(c_{12}\) (first row of \(A\) times second column of \(B\)):\
\(c_{12} = (2)(4) + (1)(0) = 8 + 0 = 8\)\
Calculate \(c_{21}\) (second row of \(A\) times first column of \(B\)):\
\(c_{21} = (0)(-1) + (3)(2) = 0 + 6 = 6\)\
Calculate \(c_{22}\) (second row of \(A\) times second column of \(B\)):\
\(c_{22} = (0)(4) + (3)(0) = 0 + 0 = 0\)\
Therefore, the product matrix \(AB\) is:\
\[ AB = \begin{pmatrix} 0 & 8 \\ 6 & 0 \end{pmatrix} \]
3. Find the product of the complex numbers \((5 - 3i)\) and \((2 + 6i)\) and express the answer in the form \(a + bi\).
💡 Solution Steps:
To find the product of two complex numbers, we use the distributive property (similar to FOIL for binomials):\
\[ (5 - 3i)(2 + 6i) \]
Multiply the first terms: \(5 \cdot 2 = 10\)\
Multiply the outer terms: \(5 \cdot 6i = 30i\)\
Multiply the inner terms: \(-3i \cdot 2 = -6i\)\
Multiply the last terms: \(-3i \cdot 6i = -18i^2\)\
Now, combine these results:\
\(10 + 30i - 6i - 18i^2\)\
Recall that \(i^2 = -1\). Substitute this into the expression:\
\(10 + 30i - 6i - 18(-1)\
\(10 + 30i - 6i + 18\)\
Group the real parts and the imaginary parts:\
\((10 + 18) + (30i - 6i)\
\(28 + 24i\)\
The product of the complex numbers \((5 - 3i)\) and \((2 + 6i)\) is \(28 + 24i\).
Name Surname: .................................. Date: .... / .... / 202...
Multiplication Worksheet
SCORE
A. True (T) / False (F)
( .... )
Matrix multiplication is generally commutative.
( .... )
The product of two complex conjugates is always a real number.
( .... )
When multiplying polynomials, the degree of the product is the sum of the degrees of the factors.
( .... )
The distributive property applies to scalar multiplication of a matrix, meaning \(k(A + B) = kA + kB\).
( .... )
Multiplying a non-zero polynomial by a constant (non-zero scalar) changes its degree.
B. Fill in the Blanks
1)
To multiply two matrices \(A\) and \(B\), the number of columns in matrix \(A\) must equal the number of .................... in matrix \(B\).
2)
The product of a complex number \(a + bi\) and its conjugate \(a - bi\) is \(a^2 + ....................\).
3)
When multiplying polynomials, each term of the first polynomial must be multiplied by each term of the second polynomial, a process often referred to as the .................... property.
4)
For any scalar \(k\) and matrices \(A\) and \(B\), the property \(k(AB) = (kA)B = A(kB)\) is known as the .................... property of scalar multiplication.
5)
The identity matrix, when multiplied by any square matrix \(A\), results in .....................
C. Matching Concepts
( .... )
For real numbers, the order of factors does not change the product (e.g., \(ab = ba\)).
- Degree of a Polynomial Product
( .... )
Multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products (e.g., \(a(b + c) = ab + ac\)).
- Complex Conjugate
( .... )
For a complex number \(a + bi\), its conjugate is \(a - bi\).
- Distributive Property
( .... )
The process of multiplying every element of a matrix by a single real number.
- Scalar Multiplication (Matrices)
( .... )
The sum of the degrees of the individual non-zero polynomials being multiplied.
- Commutative Property of Multiplication
D. Short Answer Questions
1)
Explain why matrix multiplication is generally not commutative.
2)
What is the result of multiplying a complex number \(z = a + bi\) by \(i^2\)?
E. Multiple Choice Questions
1)
Multiply the polynomials: \((2x - 3)(x^2 + 4x - 1)\)
Given matrices \(A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}\) and \(B = \begin{pmatrix} 5 & 6 \\ 7 & 8 \end{pmatrix}\), find the element in the first row, second column of the product \(AB\).
A) 19B) 22C) 26D) 30
3)
What is the product of the complex numbers \((3 + 2i)(1 - 4i)\)?