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victus00 [196]
4 years ago
9

Find -3A+6B Matrices

Mathematics
1 answer:
natima [27]4 years ago
6 0
\bf A=
\begin{bmatrix}
-3&5&-6\\9&-5&3
\end{bmatrix}\qquad \qquad 
B=
\begin{bmatrix}
-2&6&7\\2&-1&-6
\end{bmatrix}\\\\
-------------------------------\\\\
-3A=
\begin{bmatrix}
9&-15&18\\-27&15&-9
\end{bmatrix}\qquad \qquad 6B=
\begin{bmatrix}
-12&36&42\\12&-6&-36
\end{bmatrix}\\\\
-------------------------------\\\\
-3A+6B\implies 
\begin{bmatrix}
9-12&\qquad -15+36&\qquad 18+42\\
-27+12&\qquad 15-6&\qquad -9-36
\end{bmatrix}
\\\\\\
-3A+6B\implies 
\begin{bmatrix}
-3&21&60\\
-15&9&-45
\end{bmatrix}
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Answer: k=18\dfrac{8}{11}.

Step-by-step explanation:

If a line passing through two points, then

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The endpoints of segment AB have coordinates \left(1\dfrac{1}{22} , 2\dfrac{1}{4}\right) and  \left(-2\dfrac{1}{4} , k\right).

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B=\left(-2\dfrac{1}{4} , k\right)=\left(-\dfrac{9}{4} , k\right).

Slope of AB =\dfrac{k-\frac{9}{4}}{-\frac{9}{4}-\dfrac{23}{22}}

=\dfrac{\frac{4k-9}{4}}{\frac{-99-46}{44}}

=\dfrac{4k-9}{4}\times \dfrac{44}{-145}

=4k-9\times \dfrac{11}{-145}

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Product of slopes of two perpendicular segments is -1.

Slope of MN × Slope of AB = -1

\dfrac{1}{5}\times \dfrac{44k-99}{-145}=-1

\dfrac{44k-99}{-725}=-1

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44k=725+99

k=\dfrac{824}{44}

k=\dfrac{206}{11}

k=18\dfrac{8}{11}

Therefore, the value of k is k=18\dfrac{8}{11}.

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