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tino4ka555 [31]
3 years ago
13

By remainder theorem, find the remainder of p(x) when it is divided by g(x), where p(x)= x^3-2x^2-4x-1, g(x)= x+1

Mathematics
1 answer:
Anuta_ua [19.1K]3 years ago
7 0

By the remainder theorem the remainder when dividing p(x) by x-a is p(a).

Dividing by x+1 means we want p(-1).

p(-1) = (-1)³ - 2(-1)² -4(-1) - 1 = -1 - 2 + 4 - 1 = 0

Answer: 0

They go in evenly so x+1 is a factor of p(x).

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Since the definition of the inverse matrix states that the inverse of matrix A is a matrix B such that:

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we have to assume the form of such matrix. In our case we have the matrix cA, c\neq 0 and so, the constant c must be somehow eliminated from the equation. The most logical way to do so is to include \frac{1}{c} in the inverse. If we choose matrix B to be B=\frac{1}{c}\cdot A^{-1}, we will have this:

cA\cdot \frac{1}{c}\cdot A^{-1}=c\cdot \frac{1}{c}\cdot A\cdot A^{-1}=1\cdot I=I and

\frac{1}{c}\cdot A^{-1}\cdot cA=\frac{1}{c}\cdot c\cdot A^{-1}\cdot A=1\cdot I=I.

We can form the matrix B like this because we know from the text of the problem that the inverse matrix of A exists and that c is a nonzero number.

<u><em>Here is another way to solve this using the formula of the inverse matrix</em></u>

Since we know that the matrix A is invertible, it follows that its determinant is different from zero. Using the formula for the inverse matrix:

A^{-1}=\frac{1}{\det (A)}\cdot \text{Adj} (A)

we will assume the form of an inverse matrix of cA. We need to obtain the formula for the inverse of cA, so we first need to find \det (cA)\ \text{and}\ \text{Adj} (cA). Since the matrix cA is obtained from matrix A by multiplying every term with c, while calculating determinant we have a constant c that can be extracted from every column (or row) in front. Therefore, we have that

\det (cA)=c^n\cdot \det (A).

On the other hand, \text{Adj} (cA) consists of minors of the matrix cA. Therefore, when we extract the constant in front of such (n-1 \times n-1) determinants, we have c^{n-1} in each column (row). Including all this into the formula we have that:

(cA)^{-1}=\frac{1}{c^n\cdot \det (A)}\cdot c^{n-1} \text{Adj } (A)=\frac{1}{c\cdot \det (A)} \cdot \text{Adj} A=\frac{1}{c}\cdot A^{-1}.

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4 years ago
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