Require help with this question
1 answer:
A) There are a number of ways to compute the determinant of a 3x3 matrix. Since k is on the bottom row, it is convenient to compute the cofactors of the numbers on the bottom row. Then the determinant is ...
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
1×(2×-1 -3×1) -k×(3×-1 -2×1) +2×(3×3 -2×2) = 5 -5k
bi) Π₁ can be written using r = (x, y, z).
Π₁ ⇒ 3x +2y +z = 4
bii) The cross product of the coefficients of λ and μ will give the normal to the plane. The dot-product of that with the constant vector will give the desired constant.
Π₂ ⇒ ((1, 0, 2)×(1, -1, -1))•(x, y, z) = ((1, 0, 2)×(1, -1, -1))•(1, 2, 3)
Π₂ ⇒ 2x +3y -z = 5
c) If the three planes form a sheath, the ranks of their coefficient matrix and that of the augmented matrix must be 2. That is, the determinant must be zero. The value of k that makes the determinant zero is found in part (a) to be -1.
A common approach to determining the rank of a matrix is to reduce it to row echelon form. Then the number of independent rows becomes obvious. (It is the number of non-zero rows.) This form for k=-1 is shown in the picture.
You might be interested in
First, lets create a equation for our situation. Let 
be the months. We know four our problem that <span>Eliza started her savings account with $100, and each month she deposits $25 into her account. We can use that information to create a model as follows:
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
. So lets replace the values in our formula to find the average of our function:
![\frac{25(10)+100-[25(2)+100]}{10-2}](https://tex.z-dn.net/?f=%20%5Cfrac%7B25%2810%29%2B100-%5B25%282%29%2B100%5D%7D%7B10-2%7D%20)
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>
</span>
<span>
We want to find the average value of that function </span>from the 2nd month to the 10th month, so its average value in the interval [2,10]. Remember that the formula for finding the average of a function over an interval is:
We can conclude that <span>the average rate of change in Eliza's account from the 2nd month to the 10th month is $25.</span>