If the value of third order determinant is 11. Then what is the value of the determinant formed by its cofactors.
(Please show Solution)
CBSE Grade 12

146.41

Step-by-step explanation:

third order determinant = determinant of 3×3 matrix A

given ∣A∣=11

det (cofactor matrix of A) =set (transpare of cofactor amtrix of A) (transpare does not change the det)

=det(adjacent of A)

{det (cofactor matrix of A)}  ^2  = {det (adjacent of A)} ^2

(Using for an n×n det (cofactor matrix of A)=det (A)^n−1 )

we get

det (cofactor matrix of A)^2  = {det(A)  ^3−1 }^2

=(11)^2×2  = 11^4

 =146.41

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

let's first notice, the center of the hyperbola is at the origin, since the focus points are equidistantly 3 units away from the origin, that also means that the distance "c" from the center to either focus is 3.

it should also be notable that, the traverse axis for this hyperbola is vertical, or namely over the y-axis and therefore, the positive fraction will be the one with y-variable.

the traverse "a" component or distance from the center to either vertex is not 2, is more than 2, so we can safely put it at √5 ≈ 2.24.