Find the value(s) of parameter a that will make det(A)=0?

One way to find the determinant of a 3 by 3 matrix is to recopy the first two columns (attaching them on the right), add the products along the down-and-right diagonals, and subtract the products along the up-and-right diagonals. We have

0 = 15 + 16 + a^2 - 6a - 10 - 4a

0 = a^2 - 10a + 21

0 = (a - 3)(a - 7)

a = 3 or a = 7.

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det A = 5(3-2) - 4(a-4) + a(a-6) = 0

..............15 - 10 - 4a + 16 + a^2 - 6a = 0

.............. a^2 - 10a + 21 = 0

......................10 ± √((-10)² - 4(1)(21))

.............. a = ---------------------------------

....................................2(1)

......................10 ± √16

.............. a = -------------

...........................2

......................10 ± 4

.............. a = ----------

...........................2

.............. a = 7 or 3..............ANS