Answer:
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Answer:
The matrix is not invertible.
Step-by-step explanation:
We are given the following matrix in the question:
![A =\left[\begin{array}{ccc}-5&0&1\\-1&3&2\\0&10&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-5%260%261%5C%5C-1%263%262%5C%5C0%2610%266%5Cend%7Barray%7D%5Cright%5D)
Condition for invertible matrix:
A matrix is invertible if and only if the the determinant is non-zero.
We can find the determinant of the matrix as:
![|A| = -5[(3)(6)-(2)(10)]-0[(-1)(6)-(2)(0)] + 1[(-1)(10)-(3)(0)]\\|A| = -5(18-20)+(-10)\\|A| = 10-10\\|A| = 0](https://tex.z-dn.net/?f=%7CA%7C%20%3D%20-5%5B%283%29%286%29-%282%29%2810%29%5D-0%5B%28-1%29%286%29-%282%29%280%29%5D%20%2B%201%5B%28-1%29%2810%29-%283%29%280%29%5D%5C%5C%7CA%7C%20%3D%20-5%2818-20%29%2B%28-10%29%5C%5C%7CA%7C%20%3D%2010-10%5C%5C%7CA%7C%20%3D%200)
Since the determinant of the given matrix is zero, the given matrix is not invertible.
Y=0 would be a horizontal line. An asymptote is a line that a function approaches, but never reaches. Exponential functions such as these are a smooth curve. If both numbers are positive numbers greater than or equal to 1, the curve increases. If at least one of the numbers is a positive number between 0 and 1, the curve decreases. If <em>a</em> is a negative number, the curve decreases as well. If either <em>a</em> or <em>b</em> is zero, then the graph would stay constant at 0. However, as long as neither <em>a</em> nor <em>b</em> is zero, then this graph will never touch that point. The only way to get an answer of y=0 is to multiply by 0. If neither <em>a</em> nor <em>b</em> is zero, this won't happen.
