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I need to find a, b, c, d, e, f, g, h (all of which are not zero)

such that for all k is in Real number, show A is invertible or this can't happen

$$A = \left(\begin{array}{ccc} a&b&c\\d&k&e\\f&g&h \end{array}\;\begin{array}{c}\end{array}\right)$$

My answer was this can't happen, but I don't know what the answer is. And I couldn't provide a complete proof to support my answer.

I assume that a, ~, h are all 1. Then use det and element row operation I got 0 determinant for all k.

And the general determinant for this with Sarrus's rule is

det(A) = k(ah-cf) + g(cd - ae) + b(ef - dh)

What should I do next? This is invertible or non-invertible?

Thanks,

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It is possible to choose $a,b,c,d,e,f,g,h$ such that the determinant is non-zero for all $k$, i.e., the matrix is invertible for all $k$.

Note that as you have $$\det(A) = k(ah-cf) + g( cd-ae) + b(ef-dh)$$ First lets cut the dependence on $k$ in the determinant, i.e., set $ah-cf = 0$, i.e., $ah=cf$. Hence, now the determinant is (eliminating $a$) $$\det(A) = g\left(cd-\dfrac{cfe}h\right) + b(ef-dh) = - \dfrac{(dh-ef)(bh-cg)}h$$ Now choose $b,c,d,e,f,g,h$ such that the determinant is non-zero.

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