Mathematics, 26.06.2020 16:01, jstyopin

Compute the determinant using a cofactor expansion across the first row. Also compute the determinant by a cofactor expansion down the second column. StartAbsoluteValue Start 3 By 3 Matrix 1st Row 1st Column 3 2nd Column 0 3rd Column 3 2nd Row 1st Column 2 2nd Column 3 3rd Column 3 3rd Row 1st Column 0 2nd Column 4 3rd Column negative 2 EndMatrix EndAbsoluteValue

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answered: kobz

Step-by-step explanation:

It is given that

$\Delta=\begin{vmatrix}3&0&3\\2 &3&3\\0 &4&-2\end{vmatrix}$

By cofactor expansion across the first row, we get

$\Delta=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$

$\Delta=3\left[(-1)^{1+1}\begin{vmatrix}3&3\\4&-2\end{vmatrix}\right]+0\left[(-1)^{1+2}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{1+3}\begin{vmatrix}2&3\\0&4\end{vmatrix}\right]$

$\Delta=3\left[-18\right]+0\left[(-1)(-4)\right]+3\left[8\right]$

$\Delta=-54+0+24$

$\Delta=-30$

Therefore, the value of determinant is -30.

By cofactor expansion across the second column, we get

$\Delta=a_{12}C_{12}+a_{22}C_{22}+a_{32}C_{32}$

$\Delta=0\left[(-1)^{2+1}\begin{vmatrix}2&3\\0&-2\end{vmatrix}\right]+3\left[(-1)^{2+2}\begin{vmatrix}3&3\\0&-2\end{vmatrix}\right]+4\left[(-1)^{3+2}\begin{vmatrix}3&3\\2&3\end{vmatrix}\right]$

$\Delta=0\left[(-1)(-4)\right]+3\left[(-6)\right]+4\left[(-1)3\right]$

$\Delta=-18-12$

$\Delta=-30$

Therefore, the value of determinant is -30.

answered: Guest

answer: i think the answer is etheir c or d but i don't know which one but have a good day!

step-by-step explanation:

answered: Guest

1. f(x)=x²+10x+16

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=1(as, a> 0 the parabola is open upward), b=10. by putting the values.

-b/2a = -10/2(1) = -5

f(-b/2a)= f(-5)= (-5)²+10(-5)+16= -9

so, vertex = (-5, -9)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+16, we get point (0,16).

now find x-intercept put y=0 in the above equation. 0= x²+10x+16

x²+10x+16=0 ⇒x²+8x+2x+16=0 ⇒x(x+8)+2(x+8)=0 ⇒(x+8)(x+2)=0 ⇒x=-8 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

2. f(x)=−(x−3)(x+1)

by multiplying the factors, the general form is f(x)= -x²+2x+3.

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=2. by putting the values.

-b/2a = -2/2(-1) = 1

f(-b/2a)= f(1)=-(1)²+2(1)+3= 4

so, vertex = (1, 4)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+3, we get point (0, 3).

now find x-intercept put y=0 in the above equation. 0= -x²+2x+3.

-x²+2x+3=0 the factor form is already given in the question so, ⇒-(x-3)(x+1)=0 ⇒x=3 , x=-1

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

3. f(x)= −x²+4

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=-1(as, a< 0 the parabola is open downward), b=0. by putting the values.

-b/2a = -0/2(-1) = 0

f(-b/2a)= f(0)= −(0)²+4 =4

so, vertex = (0, 4)

now, find y- intercept put x=0 in the above equation. f(0)= −(0)²+4, we get point (0, 4).

now find x-intercept put y=0 in the above equation. 0= −x²+4

−x²+4=0 ⇒-(x²-4)=0 ⇒ -(x-2)(x+2)=0 ⇒x=2 , x=-2

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

4. f(x)=2x²+16x+30

use the formula to find the vertex = (-b/2a, f(-b/2a)) , here in the above equation a=2(as, a> 0 the parabola is open upward), b=16. by putting the values.

-b/2a = -16/2(2) = -4

f(-b/2a)= f(-4)= 2(-4)²+16(-4)+30 = -2

so, vertex = (-4, -2)

now, find y- intercept put x=0 in the above equation. f(0)= 0+0+30, we get point (0, 30).

now find x-intercept put y=0 in the above equation. 0=2x²+16x+30

2x²+16x+30=0 ⇒2(x²+8x+15)=0 ⇒x²+8x+15=0 ⇒x²+5x+3x+15=0 ⇒x(x+5)+3(x+5)=0 ⇒(x+5)(x+3)=0 ⇒x=-5 , x= -3

from vertex, y-intercept and x-intercept you can easily plot the graph of given parabolic equation. the graph is attached below.

5. y=(x+2)²+4

the general form of parabola is y=a(x-h)²+k , where vertex = (h,k)

if a> 0 parabola is opened upward.

if a< 0 parabola is opened downward.

compare the given equation with general form of parabola.