Answer:
Therefore, the inverse of given matrix is
![=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cbegin%7Bpmatrix%7D%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%5C%5C%20-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B15%7D%5Cend%7Bpmatrix%7D)
Step-by-step explanation:
The inverse of a square matrix
is
such that
where I is the identity matrix.
Consider, ![A = \left[\begin{array}{ccc}4&3\\3&6\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D4%263%5C%5C3%266%5Cend%7Barray%7D%5Cright%5D)
![\mathrm{Matrix\:can\:only\:be\:inverted\:if\:it\:is\:non-singular,\:that\:is:}](https://tex.z-dn.net/?f=%5Cmathrm%7BMatrix%5C%3Acan%5C%3Aonly%5C%3Abe%5C%3Ainverted%5C%3Aif%5C%3Ait%5C%3Ais%5C%3Anon-singular%2C%5C%3Athat%5C%3Ais%3A%7D)
![\det \begin{pmatrix}4&3 \\3&6\end{pmatrix}\ne 0](https://tex.z-dn.net/?f=%5Cdet%20%5Cbegin%7Bpmatrix%7D4%263%20%5C%5C3%266%5Cend%7Bpmatrix%7D%5Cne%200)
![\mathrm{Find\:2x2\:matrix\:inverse\:according\:to\:the\:formula}:\quad \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}^{-1}=\frac{1}{\det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}}\begin{pmatrix}d\:&\:-b\:\\ -c\:&\:a\:\end{pmatrix}](https://tex.z-dn.net/?f=%5Cmathrm%7BFind%5C%3A2x2%5C%3Amatrix%5C%3Ainverse%5C%3Aaccording%5C%3Ato%5C%3Athe%5C%3Aformula%7D%3A%5Cquad%20%5Cbegin%7Bpmatrix%7Da%5C%3A%26%5C%3Ab%5C%3A%5C%5C%20c%5C%3A%26%5C%3Ad%5C%3A%5Cend%7Bpmatrix%7D%5E%7B-1%7D%3D%5Cfrac%7B1%7D%7B%5Cdet%20%5Cbegin%7Bpmatrix%7Da%5C%3A%26%5C%3Ab%5C%3A%5C%5C%20c%5C%3A%26%5C%3Ad%5C%3A%5Cend%7Bpmatrix%7D%7D%5Cbegin%7Bpmatrix%7Dd%5C%3A%26%5C%3A-b%5C%3A%5C%5C%20-c%5C%3A%26%5C%3Aa%5C%3A%5Cend%7Bpmatrix%7D)
![=\frac{1}{\det \begin{pmatrix}4&3\\ 3&6\end{pmatrix}}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B%5Cdet%20%5Cbegin%7Bpmatrix%7D4%263%5C%5C%203%266%5Cend%7Bpmatrix%7D%7D%5Cbegin%7Bpmatrix%7D6%26-3%5C%5C%20-3%264%5Cend%7Bpmatrix%7D)
![\mathrm{Find\:the\:matrix\:determinant\:according\:to\:formula}:\quad \det \begin{pmatrix}a\:&\:b\:\\ c\:&\:d\:\end{pmatrix}\:=\:ad-bc](https://tex.z-dn.net/?f=%5Cmathrm%7BFind%5C%3Athe%5C%3Amatrix%5C%3Adeterminant%5C%3Aaccording%5C%3Ato%5C%3Aformula%7D%3A%5Cquad%20%5Cdet%20%5Cbegin%7Bpmatrix%7Da%5C%3A%26%5C%3Ab%5C%3A%5C%5C%20c%5C%3A%26%5C%3Ad%5C%3A%5Cend%7Bpmatrix%7D%5C%3A%3D%5C%3Aad-bc)
![4\cdot \:6-3\cdot \:3=15](https://tex.z-dn.net/?f=4%5Ccdot%20%5C%3A6-3%5Ccdot%20%5C%3A3%3D15)
![=\frac{1}{15}\begin{pmatrix}6&-3\\ -3&4\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B15%7D%5Cbegin%7Bpmatrix%7D6%26-3%5C%5C%20-3%264%5Cend%7Bpmatrix%7D)
![=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cbegin%7Bpmatrix%7D%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%5C%5C%20-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B15%7D%5Cend%7Bpmatrix%7D)
Therefore, the inverse of given matrix is
![=\begin{pmatrix}\frac{2}{5}&-\frac{1}{5}\\ -\frac{1}{5}&\frac{4}{15}\end{pmatrix}](https://tex.z-dn.net/?f=%3D%5Cbegin%7Bpmatrix%7D%5Cfrac%7B2%7D%7B5%7D%26-%5Cfrac%7B1%7D%7B5%7D%5C%5C%20-%5Cfrac%7B1%7D%7B5%7D%26%5Cfrac%7B4%7D%7B15%7D%5Cend%7Bpmatrix%7D)
Answer:
P(t)=25000+1.12t
Step-by-step explanation:
Now can you help me?
A basketball is thrown upwards. The height f(t), in feet, of the basketball at time t, in seconds, is given by the following function:
f(t) = −16t2 + 16t + 32
Which of the following is a reasonable domain of the graph of the function when the basketball falls from its maximum height to the ground?
Answer:
im in online school
Step-by-step explanation:
(a+5)(a+2) = a^2 + 7a + 10
look at what adds to make 7 and multiplies to make 10
as you know one is 5, the other is 2 and 5 + 2 = 7 so both positive
(a+2)