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Whitepunk [10]
3 years ago
6

Determine if the matrix below is invertible. Use as few calculations as possible. Justify your answer. [Start 4 By 4 Matrix 1st

Row 1st Column 4 2nd Column 5 3rd Column 7 4st Column 5 2nd Row 1st Column 0 2nd Column 1 3rd Column 4 4st Column 6 3rd Row 1st Column 0 2nd Column 0 3rd Column 3 4st Column 8 4st Row 1st Column 0 2nd Column 0 3rd Column 0 4st Column 1 EndMatrix ]
Mathematics
1 answer:
Len [333]3 years ago
4 0

Answer:

Yes, it is invertible

Step-by-step explanation:

We need to find in the matrix determinant is different from zero, since iif it is, that the matrix is invertible.

Let's use co-factor expansion to find the determinant of this 4x4 matrix, using the column that has more zeroes in it as the co-factor, so we reduce the number of determinant calculations for the obtained sub-matrices.We pick the first column for that since it has three zeros!

Then the determinant of this matrix becomes:

4\,*Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] +0+0+0

And the determinant of these 3x3 matrix is very simple because most of the cross multiplications render zero:

Det\left[\begin{array}{ccc}1&4&6\\0&3&8\\0&0&1\end{array}\right] =1 \,(3\,*\,1-0)+4\,(0-0)+6\,(0-0)=3

Therefore, the Det of the initial matrix is : 4 * 3 = 12

and then the matrix is invertible

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The hyperbolic cos (cosh) is given by
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Answer:

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Step-by-step explanation:

I've circles the numbers that you actually need to solve this problem.

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