1. Tardigrade
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  3. Mathematics
  4. If A = [1&-2 3&0], B = [-1&4 2&3] and C = [0&1 -1&0] then 5A - 3B + 2C is equal to

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Q. If $A = \begin{bmatrix}1&-2\\ 3&0\end{bmatrix}, B = \begin{bmatrix}-1&4\\ 2&3\end{bmatrix} $ and $C = \begin{bmatrix}0&1\\ -1&0\end{bmatrix} $ then 5A - 3B + 2C is equal to

Matrices

Solution:

Given $A = \begin{bmatrix}1&-2\\ 3&0\end{bmatrix} , B = \begin{bmatrix}-1&4\\ 2&3\end{bmatrix} C = \begin{bmatrix}0&1\\ -1&0\end{bmatrix}$
Consider 5A - 3B + 2C
$= 5 \begin{bmatrix}1&-2\\ 3&0\end{bmatrix} - 3 \begin{bmatrix}-1&4\\ 2&3\end{bmatrix} + 2 \begin{bmatrix}0&1\\ -1&0\end{bmatrix} $
$= \begin{bmatrix}5 &-10\\ 15&0\end{bmatrix} - \begin{bmatrix}-3&12\\ 6&9\end{bmatrix} + \begin{bmatrix}0&2\\ -2&0\end{bmatrix} $
$ = \begin{bmatrix}8&-22\\ 9&-9\end{bmatrix} + \begin{bmatrix}0&2\\ -2&0\end{bmatrix} = \begin{bmatrix}8&-20\\ 7&-9\end{bmatrix} $