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jonny [76]
3 years ago
5

Use the following matrices, A, B, C and D to perform each operation.

Mathematics
1 answer:
wolverine [178]3 years ago
4 0

Step-by-step explanation:

\bold{40.}\\\\A+B=\left[\begin{array}{ccc}3&1\\5&7\end{array}\right] +\left[\begin{array}{ccc}4&1\\6&0\end{array}\right] =\left[\begin{array}{ccc}3+4&1+1\\5+6&7+0\end{array}\right]=\left[\begin{array}{ccc}7&1\\11&7\end{array}\right]

\bold{41.}\\\\B-A=\left[\begin{array}{ccc}4&1\\6&0\end{array}\right]-\left[\begin{array}{ccc}3&1\\5&7\end{array}\right]=\left[\begin{array}{ccc}4-3&1-1\\6-5&0-7\end{array}\right]=\left[\begin{array}{ccc}1&0\\1&-7\end{array}\right]

\bold{42.}\\\\3C=3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]=\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]=\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\bold{43.}\\\\CD=\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]\\\\=\left[\begin{array}{ccc}(-2)(-2)+(3)(0)+(1)(3)&(-2)(3)+(3)(-2)+(1)(4)&(-2)(4)+(3)(1)+(1)(-1)\\(-1)(-2)+(0)(0)+(4)(3)&(-1)(3)+(0)(-2)+(4)(4)&(-1)(4)+(0)(1)+(4)(-1)\end{array}\right]\\\\=\left[\begin{array}{ccc}4+0+1&-6-6+4&-8+3-1\\2+0+12&-3+0+16&-4+0-4\end{array}\right]\\\\=\left[\begin{array}{ccc}5&-8&-6\\14&13&-8\end{array}\right]

\bold{44.}\\\\2D+3C=2\left[\begin{array}{ccc}-2&3&4\\0&-2&1\\3&4&-1\end{array}\right]+3\left[\begin{array}{ccc}-2&3&1\\-1&0&4\end{array}\right]\\\\=\left[\begin{array}{ccc}(2)(-2)&(2)(3)&(2)(4)\\(2)(0)&(2)(-2)&(2)(1)\\(2)(3)&(2)(4)&(2)(-1)\end{array}\right]+\left[\begin{array}{ccc}(3)(-2)&(3)(3)&(3)(1)\\(3)(-1)&(3)(0)&(3)(4)\end{array}\right]\\\\=\left[\begin{array}{ccc}-4&6&8\\0&-4&2\\6&8&-2\end{array}\right]+\left[\begin{array}{ccc}-6&9&3\\-3&0&12\end{array}\right]

\large\bold{You\ can\ not\ add\ matrices\ of\ different\ dimensions!!!}

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