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Kazeer [188]
4 years ago
11

Find the product, AB, if possible.

Mathematics
1 answer:
Eddi Din [679]4 years ago
4 0
A=  \left[\begin{array}{cc}3&-2\\6&0\end{array}\right] 
\\
\\B=  \left[\begin{array}{cc}0&-2\\2&6\end{array}\right] 
\\
\\AB= \left[\begin{array}{cc}3&-2\\6&0\end{array}\right]  \left[\begin{array}{cc}0&-2\\2&6\end{array}\right] 


\\
\\AB= \left[\begin{array}{cc}3\times0-2\times2&3(-2)-2\times6\\6\times0+0\times2&6(-2)+0\times6\end{array}\right] 
\\
\\AB=  \left[\begin{array}{cc}-4&-18\\0&-12\end{array}\right]
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Case 1:   b^{2}-4ac>0, if the discriminant is greater than 0, it means the quadratic equation has two real distinct roots.

Case 2: b^{2}-4ac, if the discriminant is less than 0, it means the quadratic equation has no real roots.

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