Question

Tolong dibantu matriks​

Answer 1

Given

`A = \begin{bmatrix}-1&2\\1&4\end{bmatrix} \text{ and } B = \begin{bmatrix}1&3\\3&2\end{bmatrix} \text{ and } C = \begin{bmatrix}2&-1\\2&0\end{bmatrix}`

we have

(1)

`AB = \begin{bmatrix}-1&2\\1&4\end{bmatrix}\begin{bmatrix}1&3\\3&2\end{bmatrix} \\\\ AB = \begin{bmatrix}-1*1+2*3 & -1*3+2*2 \\ 1*1+4*3 & 1*3 + 4*2\end{bmatrix} \\\\ AB = \begin{bmatrix}5&1\\13&11\end{bmatrix}`

`BA = \begin{bmatrix}1&3\\3&2\end{bmatrix}\begin{bmatrix}-1&2\\1&4\end{bmatrix} \\\\ BA = \begin{bmatrix}1*(-1)+3*1 & 1*2+3*4 \\ 3*(-1)+2*1 & 3*2+2*4\end{bmatrix} \\\\ BA = \begin{bmatrix}2&14\\-1&14\end{bmatrix}`

So AB ≠ BA. This is generally true for arbitrary matrices A and B, since matrix multplication is not commutative.

(2)

`2B \cdot 2C = 2^2 \cdot BC = 4 BC`

so we have 2BC = 2B•2C = 4BC if and only if BC is the zero matrix.

`BC = \begin{bmatrix}1&3\\3&2\end{bmatrix}\begin{bmatrix}2&-1\\2&0\end{bmatrix} \\\\ BC = \begin{bmatrix}1*3+3*2 & 1*(-1)+3*0\\3*3+2*2&3*(-1)+2*0\end{bmatrix} \\\\ BC = \begin{bmatrix}9&-1\\13&-3\end{bmatrix}`

So 2BC ≠ 2B•2C.

(3) Yes, (AB)C = A(BC) because matrix multplication is associative. You don't need to compute this, but just to confirm this result:

`(AB)C = \begin{bmatrix}5&1\\13&11\end{bmatrix}\begin{bmatrix}2&-1\\2&0\end{bmatrix} \\\\ (AB)C = \begin{bmatrix}5*3+1*2&5*(-1)+1*0\\13*3+11*2&13*(-1)+11*0\end{bmatrix} \\\\ (AB)C = \begin{bmatrix}17&-5\\61&-13\end{bmatrix}`

`A(BC) = \begin{bmatrix}-1&2\\1&4\end{bmatrix}\begin{bmatrix}9&-1\\13&-3\end{bmatrix} \\\\ A(BC) = \begin{bmatrix}(-1)*9+2*13 & (-1)*(-1)+2*(-3)\\1*9+4*13&1*(-1) + 4*(-3)\end{bmatrix} \\\\ A(BC) = \begin{bmatrix}17&-5\\61&-13\end{bmatrix}`