Which statement is correct about matrix multiplication for square matrices? A) It satisfies the associative and commutative properties, but not the distributive property. B) It satisfies the associative and distributive properties, but not the commutative property. C) It satisfies the commutative property, but not the associative and distributive properties. D) It satisfies the distributive property, but not the associative and commutative properties.

Respuesta :


Checking the existence conditions for the multiplication of matrices, the following properties are valid:

1- associative: 
[tex](A*B)* C = A*(B*C) [/tex]

2- distributive in relation to addition: 
[tex]A*(B + C) = A*B + A*C\:\:or\:\:(A + B)*C = A*C + B*C [/tex]

3- neutral element: 
[tex]A*I_{n} = I_{n}*A = A[/tex], where [tex]I_{n}[/tex] is the identity matrix of order [tex]n[/tex]   
p.s:. For the multiplication of matrices is not worth the commutative property.

Therefore: 
Answer 
B) It satisfies the associative and distributive properties, but not the commutative property.

Answer:

The correct answer is option B.

Step-by-step explanation:

In order to obtain the correct answer we only need to recall the basic properties of the multiplication of square matrices. Let us analyze each option.

A. The multiplication of square matrices is associative, but not commutative. About this last statement just check

[tex]\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix}\begin{pmatrix}0 & 0\\ 1 & 0 \end{pmatrix} = \begin{pmatrix}0 & 0\\ 0 & 0 \end{pmatrix}[/tex]

but

[tex]\begin{pmatrix}0 & 0\\ 1 & 0 \end{pmatrix}\begin{pmatrix}1 & 0\\ 0 & 0 \end{pmatrix} = \begin{pmatrix}0 & 0\\ 1 & 0 \end{pmatrix}[/tex].

So, the statement is False.

B. The multiplication of square matrices is associative and distributive. Also, is not commutative, as we have seen previously. So, the statement is True.

C. We have seen that the multiplication of square matrices is not commutative. Then, the statement is False.

D. As the product of square matrices is associative, the statement is False.

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