Question

Please help!!!

1. Select the best answer for the question.
Multiply matrix D = [5,-2,1 ] by matrix E= [12] A. [4 8] B. 2 C. [5 -2 1 ,10 4 2] D. [5 10, -2 -4, 1 2]

2. Subtract [-6 -4,6 0, 6 4] - [-5 5, -4 -4,6 -4]

3. Add [1 -4,3 5] + [-2 6,-2 4]

4. Find the determinant of A= [ 4 -7, 3 -2]

A. 13
B. −2
C. 2
D. 29


5. What is matrix A + matrix B?
A=[3 0, 2 -1] B= [ 2 8, .6 3]

6. What is 4C?
c= [12 03/2, 1 -6 7]

7. Evaluate the determinant for the following matrix: [-5 1, 5 4]

8. Complete the multiplication: 2EA E= [12] A= [3 0, 2 -1]

9. Multiply [5 0 , 3 -5] x [2 -1, 2 -2]

Answer 1

The set of problems requires understanding matrix operations, such as addition, subtraction, multiplication, and finding determinants. The answers include a mix of matrices that cannot be multiplied, matrix arithmetic resulting in new matrices, and the calculation of determinants.

Step-by-step explanation:

The problem set involves matrix operations and the determination of scalars from matrices and vectors. Here is the breakdown of the solutions:

Question 1:

Matrix D = [5, -2, 1] cannot be multiplied by matrix E = [12] since they do not conform to the rules for matrix multiplication.

Question 2:

The subtraction of the given matrices results in matrix [-1, -9, 0, 4, 0, 8].

Question 3:

The addition of [1, -4, 3, 5] and [-2, 6, -2, 4] results in matrix [-1, 2, 1, 9].

Question 4:

The determinant of matrix A = [4, -7, 3, -2] is found using the formula ad - bc, which yields 29.

Question 5:

Matrix A + matrix B is calculated as [3, 0, 2, -1] + [2, 8, 0.6, 3] to give [5, 8, 2.6, 2].

Question 6:

If C = [12, 9/2, 1, -6, 7], then 4C is [48, 18, 4, -24, 28].

Question 7:

The determinant of matrix [-5, 1, 5, 4] is calculated as (-5)(4) - (1)(5) which equals -25 + 5, giving -20.

Question 8:

Matrix E cannot be multiplied by matrix A directly due to their dimensions. A scalar multiplication by 2 followed by matrix A should be applied, leading to [6, 0, 4, -2].

Question 9:

The product of matrices [5, 0, 3, -5] and [2, -1, 2, -2] is calculated by performing the matrix multiplication, resulting in [10, -5, 16, -4].

Answer 2

Question 1:


`\left[\begin{array}{ccc}5\\-2\\1\end{array}\right] \left[\begin{array}{ccc}12\end{array}\right] = \left[\begin{array}{ccc}(5)(12)+(-2)(12)+(1)(1)\end{array}\right]`
=
`\left[\begin{array}{ccc}60-24+12\end{array}\right] = \left[\begin{array}{ccc}48\end{array}\right]`
Answer: A
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Question 2

`\left[\begin{array}{ccc}-6&-4\\6&0\\6&4\end{array}\right] - \left[\begin{array}{ccc}-5&5\\-4&-4\\6&-4\end{array}\right]`

`\left[\begin{array}{ccc}(-6--5)&(-4-5)\\(6--4)&(0--4)\\(6-6)&(4--4)\end{array}\right] = \left[\begin{array}{ccc}-1&-9\\10&4\\0&8\end{array}\right]`
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Question 3


`\left[\begin{array}{ccc}1&-4\\3&5\end{array}\right] + \left[\begin{array}{ccc}-2&6\\-2&4\end{array}\right]`
=
`\left[\begin{array}{ccc}(1+-2)&(-4+6)\\(3+-2)&(5+4)\end{array}\right]`
=
`\left[\begin{array}{ccc}-1&2\\1&9\end{array}\right]`
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Question 4


`A= \left[\begin{array}{ccc}4&-7\\3&-2\end{array}\right]`

`det A=(4)(-2)-(-7)(3)=-8+21=13`

Answer: A
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Question 5

`\left[\begin{array}{ccc}3&0\\2&-1\end{array}\right] + \left[\begin{array}{ccc}2&8\\6&3\end{array}\right]`
=
`\left[\begin{array}{ccc}3+2&0+8\\2+6&(-1)+3\end{array}\right]`
=
`\left[\begin{array}{ccc}5&8\\8&2\end{array}\right]`
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Question 6

`4 \left[\begin{array}{ccc}12&0& (3)/(2) \\1&-6&7\end{array}\right] = \left[\begin{array}{ccc}48&0&6\\4&-24&28\end{array}\right]`

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Question 7


`C = \left[\begin{array}{ccc}-5&1\\5&4\end{array}\right]`

`det C = (-5)(4) - (1)(5) = -20-5 = -25`
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Question 8

`2 \left[\begin{array}{ccc}12\end{array}\right] \left[\begin{array}{ccc}3&0\\2&-1\end{array}\right]`

`\left[\begin{array}{ccc}24\end{array}\right] \left[\begin{array}{ccc}3&6\\2&-1\end{array}\right]`

`\left[\begin{array}{ccc}72&0\\48&-24\end{array}\right]`
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Question 9

`\left[\begin{array}{ccc}5&0\\3&-5\end{array}\right] \left[\begin{array}{ccc}2&-1\\2&-2\end{array}\right]`

`\left[\begin{array}{ccc}(5)(2)+(0)(2)&(5)(-1)+(0)(-2)\\(3)(2)+(-5)(2)&(3)(-1)+(-5)(-2)\end{array}\right]`

`\left[\begin{array}{ccc}10+0&-5+0\\6-10&-3+10\end{array}\right] = \left[\begin{array}{ccc}10&-5\\-4&7\end{array}\right]`