Answer:
2.3 : 0.83
Step-by-step explanation:
Simplify
Answer:
This shows 3 pivot position matrixes.
Step-by-step explanation:
The given matrix is:
![\left[\begin{array}{ccc}1&-2&-5\\0&4&3\\-3&3&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-5%5C%5C0%264%263%5C%5C-3%263%260%5Cend%7Barray%7D%5Cright%5D)
The option D is correct for this matrix.
The matrix is invertible and the given matrix has 3 pivot positions.
The matrix is invertible if its determinant is nonzero.
Multiply the 3rd row by 1/3.we get:
![\left[\begin{array}{ccc}1&-2&-5\\0&4&3\\-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-2%26-5%5C%5C0%264%263%5C%5C-1%261%260%5Cend%7Barray%7D%5Cright%5D)
Now, add the first row with third row:
![\left[\begin{array}{ccc}0&-1&-5\\0&4&3\\-1&1&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%26-1%26-5%5C%5C0%264%263%5C%5C-1%261%260%5Cend%7Barray%7D%5Cright%5D)
Replace third row by first row:
![\left[\begin{array}{ccc}-1&1&0\\0&4&3\\0&-1&-5\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%261%260%5C%5C0%264%263%5C%5C0%26-1%26-5%5Cend%7Barray%7D%5Cright%5D)
This shows 3 pivot position matrixes.
Hence, a matrix is invertible and has 3 pivot positions.
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Answer:
The answer is B (2).
Step-by-step explanation:
AB = BC
3x-4=5x-10
-4=2x-10
6=2x
3=x
AB = 3x-4
AB = 3 (3)-4
AB = 9-4
AB =5
Therefore, AB is 5.
BC = 5x-10
BC = 5(3)-10
BC = 15-10
BC= 5