Yes, we can obtain a diagonal matrix by multiplying two non diagonal matrix.
Consider the matrix multiplication below
![\left[\begin{array}{cc}a&b\\c&d\end{array}\right] \left[\begin{array}{cc}e&f\\g&h\end{array}\right] = \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=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%26b%5C%5Cc%26d%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7De%26f%5C%5Cg%26h%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7Da%20e%2Bb%20g%26a%20f%2Bb%20h%5C%5Cc%20e%2Bd%20g%26c%20f%2Bd%20h%5Cend%7Barray%7D%5Cright%5D%20)
For the product to be a diagonal matrix,
a f + b h = 0 ⇒ a f = -b h
and c e + d g = 0 ⇒ c e = -d g
Consider the following sets of values
The the matrix product becomes:
![\left[\begin{array}{cc}1&2\\3&4\end{array}\right] \left[\begin{array}{cc}\frac{1}{3}&-1\\-\frac{1}{4}&\frac{1}{2}\end{array}\right] = \left[\begin{array}{cc}\frac{1}{3}-\frac{1}{2}&-1+1\\1-1&-3+2\end{array}\right]= \left[\begin{array}{cc}-\frac{1}{6}&0\\0&-1\end{array}\right]](https://tex.z-dn.net/?f=%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%262%5C%5C3%264%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D%26-1%5C%5C-%5Cfrac%7B1%7D%7B4%7D%26%5Cfrac%7B1%7D%7B2%7D%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Cfrac%7B1%7D%7B3%7D-%5Cfrac%7B1%7D%7B2%7D%26-1%2B1%5C%5C1-1%26-3%2B2%5Cend%7Barray%7D%5Cright%5D%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D-%5Cfrac%7B1%7D%7B6%7D%260%5C%5C0%26-1%5Cend%7Barray%7D%5Cright%5D)
Thus, as can be seen we can obtain a diagonal matrix that is a product of non diagonal matrices.
The diameter would be twice 5, or 10
Answer:
18 eggs would cost $3.06 :)
Step-by-step explanation:
Respuesta:
(2945.411; 3054.589)
Explicación paso a paso:
Dado ;
Tamaño de la muestra, n = 50
Media, xbar = 3000
Desviación estándar, s = 200
Nivel de confianza, Zcrítico al 95% = 1,96
El intervalo de confianza se define como:
Xbar ± margen de error
Margen de error = Zcrítico * s / sqrt (n)
Margen de error = 1,96 * 200 / sqrt (50)
Margen de error = 54.589
Límite inferior = (3000 - 54.589) = 2945.411
Límite superior = (3000 + 54.589) = 3054.589
(2945.411; 3054.589)
Answer:
The length of the rectangle is 15 inches and the width is 33 inches
Step-by-step explanation:
The dimensions for the rectangle are as follows:
L=L and W=2L+3
The equation for perimeter of a rectangle is:
P=2l+2w
To Solve:
1. Plug values into perimeter equation
96=2(L)+2(2L+3)
2. Simplify the equation above.
96=2L+4L+6
3. Combine like terms:
96=6L+6
4. Solve for L by subtracting 6 from both sides:
90=6L
5. Solve for L by dividing both sides by 6:
15=L.
6. Plug 15 into the equation for the width:
2(15)+3=w
30+3=w.
