what makes this matrix true?

$\displaystyle

(\begin{array}{|ccc|}x&2&-1\\0&y&w\end{array}

+ \begin{array}{|ccc|}0&1&2\\0&0&2\end{array})

* \begin{array}{|ccc|}1&1\\3&z\\4&2\end{array}

= \begin{array}{|ccc|}12&4\\2&2\end{array}

$

please help///

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- Apr 25th 2010, 03:14 PMAnemoriWhat makes this matrix true?
what makes this matrix true?

$\displaystyle

(\begin{array}{|ccc|}x&2&-1\\0&y&w\end{array}

+ \begin{array}{|ccc|}0&1&2\\0&0&2\end{array})

* \begin{array}{|ccc|}1&1\\3&z\\4&2\end{array}

= \begin{array}{|ccc|}12&4\\2&2\end{array}

$

please help///

- Apr 25th 2010, 03:27 PMpickslides
Perform the operations as the problem suggests.

1. add the matrices inside the brackets.

2. multiply the result with the remaining LHS matrix.

Show your work to there, i will then help further. - Apr 25th 2010, 07:00 PMAnemori
- Apr 26th 2010, 03:00 AMpickslides
I get

$\displaystyle \left(\begin{array}{ccc}x&3&1\\0&y&w+2\end{array}\ right)

\times \left(\begin{array}{ccc}1&1\\3&z\\4&2\end{array}\r ight)

= \left(\begin{array}{cc}x+9+4&x+3z+2\\3y+4(w+2)&yz+ 2(w+2)\end{array}\right)$

Now solve

$\displaystyle

\left(\begin{array}{cc}x+9+4&x+3z+2\\3y+4(w+2)&yz+ 2(w+2)\end{array}\right) = \left(\begin{array}{cc}12&4\\2&2\end{array}\right)

$

Might need to do some expanding on the LHS first! - Apr 26th 2010, 06:10 AMAnemori
- Apr 26th 2010, 01:58 PMpickslides
- Apr 26th 2010, 04:44 PMAnemori
Thanks (Clapping).. i got it...