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How do you solve for the values of a so that det(A) = 0? I’m trying to do this using the cofactor expansion method but I’m getting a very convoluted algebraic equation and I’m not sure it’s the correct way to go about this. Thanks

8 Answers

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One way would be to expand it like you normally would and set that to 0. Then solve the quadratic eqn for all of the 'a's.
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By expanding along the first row, the determinant is

det(A)= a(-4a+27) -7(a²-72) +9(-3a+32)

Tidying it up:

det(A) = -11a² + 792

Therefore det(A)=0 if and only if a²= 72, i.e. when a=±6√2
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Suppose the columns are linearly dependent, and without loss of generality, we assume of the form:

a = 7x + 9y

a = -4x + 9y

8 = -3x + ay

Subtracting the first two rows from each other immediately tells us that x=0.  So we now have the two equations:

a = 9y

8 = ay

Substituting y = a/9 into the second equations yields:

a^2 = 72,

and hence that

a = -\sqrt{72}, \sqrt{72}.
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As the other comments say, one way would be to just calculate the determinant directly.

Alternatively, assuming a != 0 (this can be treated as a special case), you can use ERO's to transform A into an (upper) triangular matrix, and then easily calculate the determinant.
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Use Sarrus' rule.
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Subtract the first row from the second row, leaving other entries unaltered, so only the second row becomes [0 -11 0].

This doesn’t change the determinant since we may split the determinant after the alteration into determinants of two matrices, one of which is the original matrix, then the other one would have its first and second row differing only by a factor of -1 and hence its determinant is 0.

Use cofactor expansion on the second row, noting that 0 in the second row contributes nothing, we immediately get a=6sqrt(2) or -6sqrt(2)
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Use gaussian elimination. The last row will have at the last column an equation.

For the matrix to be non-inversible, that equation should be 0.
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There are a lot of good ideas posted here, but there is a shortcut.

Note that the first column and the third column are very similar, in that they have two of the same value on the first and second rows. One way that you can tell if a matrix has a determinant of zero is if any of its rows or columns are linearly dependent on one another. So all you have to do is make the first row some multiple of the third.

This results in a much simpler system of equations. If the third column is some value c times the first then we have:

a = 9c  
8 = ac

Which has solutions c = +/- 2 sqrt(2)/3, and a = +/- 6 sqrt(2)
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