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Amiraneli [1.4K]
2 years ago
9

Find the -2 -4 -3 -1 -9-8 determinant OF -15 - 5 6

Mathematics
1 answer:
sesenic [268]2 years ago
5 0

Determinants are defined only for square matrices, so I assume you are given the 3×3 matrix

A = \begin{bmatrix}-2 & -4 & -3 \\ -1 & -9 & -8 \\ -15 & -5 & 6 \end{bmatrix}

Let's compute the determinant by taking the cofactor expansion along the first column:

\det A = -2 \det\begin{bmatrix}-9 & -8 \\ -5 & 6\end{bmatrix} + \det\begin{bmatrix}-4 & -3 \\ -5 & 6\end{bmatrix} - 15 \det \begin{bmatrix}-4 & -3 \\ -9 & -8\end{bmatrix}

\det A = -2 ((-9)\times6-(-8)\times(-5)) + ((-4)\times6 - (-3)\times(-5)) \\ ~~~~~~~~~~~~~~~- 15 ((-4)\times(-8)-(-3)\times(-9))

\det A = -2(-54-40) + (-24 - 15) - 15 (32 - 27)

\det A = \boxed{74}

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Answer:

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Step-by-step explanation:

(a) Since k(x) is a function of two functions f(x) and g(x) [ k(x)=f(x)g(x) ], so for differentiating k(x) we need to use <u>product rule</u>,i.e., \frac{\mathrm{d} [f(x)\times g(x)]}{\mathrm{d} x}=\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}

this will give <em>k'(x)=f'(x)g(x) + f(x)g'(x)</em>

on substituting the value x=0, we will get the value of k'(0)

{for expressing the value in terms of numbers first we need to know the value of f(0), g(0), f'(0) and g'(0) in terms of numbers}{If f(0)=0 and g(0)=0, and f'(0) and g'(0) exists then k'(0)=0}

(b) m(x) is a function of two functions f(x) and g(x) [ m(x)=\frac{1}{2}\times\frac{f(x)}{g(x)} ]. Since m(x) has a function g(x) in the denominator so we need to use <u>division rule</u> to differentiate m(x). Division rule is as follows : \frac{\mathrm{d} \frac{f(x)}{g(x)}}{\mathrm{d} x}=\frac{\frac{\mathrm{d} f(x)}{\mathrm{d} x}\times g(x) + f(x)\times\frac{\mathrm{d} g(x)}{\mathrm{d} x}}{g^{2}(x)}

this will give <em>m'(x) = \frac{1}{2}\times\frac{f'(x)g(x) - f(x)g'(x)}{g^{2}(x) }</em>

on substituting the value x=5, we will get the value of m'(5).

{for expressing the value in terms of numbers first we need to know the value of f(5), g(5), f'(5) and g'(5) in terms of numbers}

{NOTE : in m(x), g(x) ≠ 0 for all x in domain to make m(x) defined and even m'(x) }

{ NOTE : \frac{\mathrm{d} f(x)}{\mathrm{d} x}=f'(x) }

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Step-by-step explanation:

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