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If det [a 10 e b d 0] = [2 10 3 6 -1 0],find a,d and e

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Answer:

The given equation is:

det [a 10 e b d 0] = [2 10 3 6 -1 0]

The determinant of a 3x3 matrix is calculated as follows:

|a b c|

|d e f| = a(ei - fh) - b(di - fg) + c(dh - eg)

Applying this formula to the given matrix, we get:

det [a 10 e b d 0] = a(d0 - b-1) - 10(ed - b0) + e(10*-1 - d*3)

= ab + 10b - 10ed - 3e

Substituting the given values, we get:

ab + 10b - 10ed - 3e = 2(6-0) - 10(3-(-1)) + 3(10-(-1))

ab + 10b - 10ed - 3e = 12 + 40 + 33

ab + 10b - 10ed - 3e = 85

We can simplify this equation by factoring out b and e:

b(a + 10) - e(10d + 3) = 85 - 10b

We can solve for a, d, and e by setting up a system of equations using the given values of the determinant:

ab + 10b - 10ed - 3e = 85

From this equation, we can solve for b:

b(a + 10) - e(10d + 3) = 85 - 10b

Substituting the value of b from the determinant equation, we get:

(a+10)(-1) - e(10d+3) = -8.5

Simplifying, we get:

-a - 10e - 3.3d = -8.5 ...(1)

Also, from the determinant equation, we have:

6a + 10d + 3e = 29

Solving for e in terms of a and d, we get:

e = (29 - 6a - 10d)/3 ...(2)

Substituting equation (2) into equation (1), we get:

-a - 10[(29 - 6a - 10d)/3] - 3.3d = -8.5

Simplifying and multiplying both sides by -3, we get:

3a + 20d - 29e = 25.5

Substituting equation (2) into this equation, we get:

3a + 20d - 29[(29 - 6a - 10d)/3] = 25.5

Simplifying, we get:

a + 6d - 29 = 1.5

a + 6d = 30.5

We have two equations in two variables:

-a - 10e - 3.3d = -8.5

a + 6d = 30.5

Solving for a and d using any method (substitution, elimination, etc.), we get:

a = 7

d = 4

Substituting these values into equation (2), we get:

e = (29 - 6a - 10d)/3 = (29 - 6(7) - 10(4))/3 = -3

Therefore, a = 7, d = 4, and e = -3.

Explanation:

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