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vagabundo [1.1K]
3 years ago
14

If the size of the matrix B is 4 × 5, the size of the matrix D is 4 × 2 and the order of the matrix E is 5 × 4, then the order o

f the matrix D^T × (BE) is:
Mathematics
1 answer:
kvasek [131]3 years ago
7 0

Answer:

The order of the matrix D^T × (BE) is 2 × 4

Step-by-step explanation:

In the multiplication of matrix, the number of columns in the first matrix must be equal to the number of rows of the second matrix. The resulting matrix from the product will have the same number of rows as the first matrix and the same number of column as the second matrix. For example, given Matrix A of order (m × n) and Matrix Z of order ( n × k), the product of the two matrices will give a matrix ( say Y) of order (m × k).

AZ = Y

(m x n)(n x k) = (m x k)

Now, From the question,

The order of Matrix B is 4 × 5,

the order of the matrix D is 4 × 2

and the order of the matrix E is 5 × 4.

To determine the order of the matrix D^T × (BE)

First, will find the order of (BE)

Order of BE = (4 × 5) (5 × 4)

Order of BE = (4 × 4)

Hence, the order of matrix (BE) is 4 × 4.

Now, we will determine the order of matrix D^T

D^T means the transpose of matrix D.

The transpose of a matrix is defined as a new matrix whose rows are the columns of the original matrix and the columns of the new matrix are the rows of the original matrix. This means given Matrix A of order (m × n), the transpose of Matrix A (A^T) will have an order of (n × m).

The order of the matrix D is 4 × 2

Hence, the order of the transpose of matrix D (D^T) will be 2 × 4

Now, the order of the matrix D^T × (BE) will be

Order of the matrix D^T × (BE) = (2 × 4) × (4 × 4)

Order of the matrix D^T × (BE) = 2 × 4

Hence, the order of the matrix D^T × (BE) is 2 × 4.

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\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Explanation:</h2>

Here we have the following expression:

3\sqrt{22}\sqrt{58}\sqrt{18}

So we need to simplify that radical expression. By property of radicals we know that:

\sqrt[n]{a}\sqrt[n]{b}=\sqrt[n]{ab}

So:

3\sqrt{22}\sqrt{58}\sqrt{18}=3\sqrt{22\times 58 \times 18}=3\sqrt{22968}

The prime factorization of 22968 is:

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Hence:

3\sqrt{22968}=3\sqrt{2^3\cdot 3^2\cdot11\cdot 29}=3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29}

By property:

\sqrt[n]{a^n}=a

So:

3\sqrt{2^2\cdot 3^2\cdot 2\cdot 11\cdot 29} \\ \\ 3(2)(3)\sqrt{2\cdot 11\cdot 29}=18\sqrt{638}

Finally:

\boxed{3\sqrt{22}\sqrt{58}\sqrt{18}=18\sqrt{638}}

<h2>Learn more:</h2>

Radical expressions: brainly.com/question/13452541

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