Answer:
1. The matrix A isn't the inverse of matrix B.
2. |B|=12, |A|=12
Step-by-step explanation:
1. We want to know if matrix A is the inverse of matrix B, this means that if you do the product between B and A you have to obtain the identity matrix.
We have:
![A=\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D)
and
![B=\left[\begin{array}{cc}3&2\\1&4\end{array}\right]](https://tex.z-dn.net/?f=B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D)
A and B are 2×2 matrices (2 rows and 2 columns), if you multiply them you have to obtain a 2×2 matrix.
Then if A is the inverse of B:
Where,
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
Observation:
If you have two matrices:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]\\and\\B=\left[\begin{array}{cc}e&f\\g&h\end{array}\right]\\\\\\A.B=\left[\begin{array}{cc}(a.e+b.g)&(a.f+b.h)\\(c.e+d.g)&(c.f+d.h)\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%5C%5Cand%5C%5CB%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CA.B%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%28a.e%2Bb.g%29%26%28a.f%2Bb.h%29%5C%5C%28c.e%2Bd.g%29%26%28c.f%2Bd.h%29%5Cend%7Barray%7D%5Cright%5D)
Now:
![B.A=\left[\begin{array}{cc}3&2\\1&4\end{array}\right].\left[\begin{array}{cc}4&-2\\-1&3\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}4.3+(-2).1&4.2+(-2).4\\(-1).3+3.1&(-1).2+3.4\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}12-2&8-8\\-3+3&-2+12\end{array}\right]\\\\\\B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%262%5C%5C1%264%5Cend%7Barray%7D%5Cright%5D.%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4%26-2%5C%5C-1%263%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D4.3%2B%28-2%29.1%264.2%2B%28-2%29.4%5C%5C%28-1%29.3%2B3.1%26%28-1%29.2%2B3.4%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D12-2%268-8%5C%5C-3%2B3%26-2%2B12%5Cend%7Barray%7D%5Cright%5D%5C%5C%5C%5C%5C%5CB.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D)
![B.A=\left[\begin{array}{cc}10&0\\0&10\end{array}\right]\neq \left[\begin{array}{cc}1&0\\0&1\end{array}\right]=I\\\\\\B.A\neq I](https://tex.z-dn.net/?f=B.A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D10%260%5C%5C0%2610%5Cend%7Barray%7D%5Cright%5D%5Cneq%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%3DI%5C%5C%5C%5C%5C%5CB.A%5Cneq%20I)
Then, the matrix A isn't the inverse of matrix B.
2. If you have a matrix A:
![A=\left[\begin{array}{cc}a&b\\c&d\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D)
The determinant of the matrix is:
Then the determinant of B is:
The determinant of A is:
Y = mx+b that’ll help you answer it
Answer:
268 mg
Step-by-step explanation:
Let A₀ = the original amount of caffeine
The amount remaining after one half-life is ½A₀.
After two half-lives, the amount remaining is ½ ×½A₀ = (½)²A₀.
After three half-lives, the amount remaining is ½ ×(½)²A₀ = (½)³A₀.
We can write a general formula for the amount remaining:
A =A₀(½)ⁿ
where n is the number of half-lives
.
n = t/t_½
Data:
A₀ = 800 mg
t₁ = 10 a.m.
t₂ = 7 p.m.
t_½ = 5.7 h
Calculations:
(a) Calculate t
t = t₂ - t₁ = 7 p.m. - 10 a.m. = 19:00 - 10:00 = 9:00 = 9.00 h
(b) Calculate n
n = 9.00/5.7 = 1.58
(b) Calculate A
A = 800 × (½)^1.58 = 800 × 0.334 = 268 mg
You will still have 268 mg of caffeine in your body at 10 p.m.
