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Marta_Voda [28]
2 years ago
9

If A is a 2 × 2 matrix, then A × I = and I × A =

Mathematics
1 answer:
krok68 [10]2 years ago
3 0

Since the multiplication between two matrices is not <em>commutative</em>, then \vec A\, \times\,\vec I \ne \vec I \,\times \,\vec A, regardless of the dimensions of \vec A.

<h3>Is the product of two matrices commutative?</h3>

In linear algebra, we define the product of two matrices as follows:

\vec C = \vec A \,\times \vec B, where \vec A \in \mathbb{R}_{m\times p}, \vec B \in \mathbb{R}_{p\times n} and \vec C \in \mathbb{R}_{m \times n}     (1)

Where each element of the matrix is equal to the following dot product:

c_{ij} = \left[\begin{array}{cccc}a_{i1}&a_{i2}&\ldots&a_{ip}\end{array}\right]\,\bullet\,\left[\begin{array}{ccc}b_{1j}\\b_{2j}\\\vdots\\b_{pj}\end{array}\right], where 1 ≤ i ≤ m and 1 ≤ j ≤ n.     (2)

Because of (2), we can infer that the product of two matrices, no matter what dimensions each matrix may have, is not <em>commutative</em> because of the nature and characteristics of the definition itself, which implies operating on a row of the <em>former</em> matrix and a column of the <em>latter</em> matrix.

Such <em>"arbitrariness"</em> means that <em>resulting</em> value for c_{ij} will be different if the order between \vec A and \vec B is changed and even the dimensions of \vec C may be different. Therefore, the proposition is false.

To learn more on matrices: brainly.com/question/9967572

#SPJ1

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