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3 years ago

Does anyone know the answer bc I've done the math problem multiple times and can't figure it out?

Mathematics

2 answers:

Olin [163]3 years ago

7 0

ANSWER: C 4/20

EXPLANATION: since you are trying to find the total your going to add both fractions together. You also have to follow the rule of having the same denominator when adding or subtracting only… Since the denominators are the same you just add straight accross the numerator and the denominator stays the same… hope this was helpful;)

lapo4ka [179]3 years ago

6 0

Answer:

Step-by-step explanation:

it is c because $1/20 + 3/20$

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3. for each item, decide whether or not the given expression is defined. for each item that is defined, compute the result. (a)

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The results of given matrices can be obtained using matrix multiplication.

<h3>Find the results of the given matrices:</h3>

Here in the question it is given that,

A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ ,

E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

We have to find AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE.

AB = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$

a₁₁ = 1×2 + (-1)×5 + 2×4 = 5, a₁₂ = 1×(-1) + (-1)×1 + 2×6 = 10, a₁₃ = 1×3 + (-1)×2 + 2×(-2) = -3, a₂₁ = 3×2 + 1×5 + 4×4 = 27, a₂₂ = 3×(-1) + 1×1 + 4×6 = 22, a₂₃ = 3×3 + 1×2 + 4×(-2) = 3

AB = $\left[\begin{array}{ccc}5&10&-3\\27&22&3\end{array}\right]$

BC = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$

a₁₁ = 2×1 + (-1)×(-1) + 3×2 = 9, a₂₁ = 5×1 + 1×(-1) + 2×2 = 8, a₃₁ = 4×1 + 6×(-1) + (-2)×2 = -6

BC = $\left[\begin{array}{ccc}9\\8\\-6\end{array}\right]$

CA, CA is not defined since dimension of the matrices are 3×1 and 2×3

$A^{T}E = \left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = 1×(2-i) + 3×(-i) = 2-4i, a₁₂ = 1x(1+i) + 3×(2+4i) = 7+13i, a₂₁ = -1×(2-i) + 1×(-i) = -2, a₂₂ = -1×(1+i) + 1×(2+4i) = 1+3i, a₃₁ = 2×(2-i) + 4×(-i) = 4-6i, a₃₂ = 2×(1+i) + 4×(2+4i) = 10+18i

$A^{T}E = \left[\begin{array}{ccc}2-4i&7+13i\\-2&1+3i\\4-6i&10+18i\end{array}\right]$

CD = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$

a₁₁ = 1×2 = 2, a₁₂ = 1×(-2) = -2, a₁₃ = 1×3 = 3, a₂₁ = -1×2 = -2, a₂₂ = -1×(-2) = 2, a₂₃ = -1×3 = -3,a₃₁= 2×2 = 4, a₃₂ = 2×(-2) = -4, a₃₃ = 2×3 = 6

CD = $\left[\begin{array}{ccc}2&-2&3\\-2&2&-3\\4&-4&6\end{array}\right]$

$C^{T} A^{T} =\left[\begin{array}{ccc}1&-1&2\end{array}\right]\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right]$

a₁₁ = 1×1 + (-1)×(-1) + 2×2 = 6, a₁₂ = 1×3 + (-1)×1 + 2×4 = 10

$C^{T}A^{T}=\left[\begin{array}{ccc}6&10\end{array}\right]$

F² = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

a₁₁ = i×i + (1-3i)×0 = -1,a₁₂ = i×(1-3i) + (1-3i)×(4+i) = 10-10i, a₂₁= 0×i + (4+i)×0 = 0, a₂₂ = 0×(1-3i) + (4+i)×(4+i) = 15+8i

F² = $\left[\begin{array}{ccc}-1&10-10i\\0&15+8i\end{array}\right]$

$BD^{T}=\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]\left[\begin{array}{ccc}2\\-2\\3\end{array}\right]$

a₁₁ = 2×2 + (-1)×(-2) + 3×3 = 15, a₂₁ = 5×2 + 1×(-2) + 2×3 = 14, a₃₁ = 4×2 + 6×(-2) + (-2)×3 = -10

$BD^{T}= \left[\begin{array}{ccc}15\\14\\-10\end{array}\right]$

$A^{T} A=\left[\begin{array}{ccc}1&3\\-1&1\\2&4\end{array}\right] \left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$

a₁₁ = 1×1 + 3×3 = 10, a₁₂ = 1×(-1) + 3×1 = 2, a₁₃ = 1×2 + 3×4 = 14, a₂₁ = -1×1 + 1×3 = 2, a₂₂ = -1×(-1) + 1×1 = 2, a₂₃ = -1×2 + 1×4 = 2, a₃₁ = 2×1 + 4×3 = 14, a₃₂ = 2×(-1) + 4×1 = 2, a₃₃ = 2×2 + 4×4 = 20

$A^{T} A=\left[\begin{array}{ccc}10&2&14\\2&2&2\\14&2&20\end{array}\right]$

FE = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$ $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$

a₁₁ = i×(2-i) + (1-3i)×(-i) = -2+i, a₁₂ = i×(1+i) + (1-3i)×(2+4i) = 13-i, a₂₁ = 0×(2-i) + (4+i)×(-i) = 1-4i, a₂₂ = 0×(1+i) + (4+i)×(2+4i) = 4+18i

FE = $\left[\begin{array}{ccc}-2+i&13-i\\1-4i&4+18i\end{array}\right]$

Hence we can obtain the results of the required matrices using matrix multiplication.

Disclaimer: The question was given incomplete on the portal. Here is the complete question.

Question: Let A = $\left[\begin{array}{ccc}1&-1&2\\3&1&4\end{array}\right]$ , B = $\left[\begin{array}{ccc}2&-1&3\\5&1&2\\4&6&-2\end{array}\right]$ , C = $\left[\begin{array}{ccc}1\\-1\\2\end{array}\right]$ , D = $\left[\begin{array}{ccc}2&-2&3\end{array}\right]$ , E = $\left[\begin{array}{ccc}2-i&1+i\\-i&2+4i\end{array}\right]$ , F = $\left[\begin{array}{ccc}i&1-3i\\0&4+i\end{array}\right]$

For each item, decide whether or not the given expression is defined. for each item that is defined, compute the result.

AB, BC, CA, CD, $C^{T} A^{T}$ , F², $BD^{T}$ , $A^{T} A$ and FE

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