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If $A=\begin{bmatrix} 10^{30}+5& 10^{20}+4 &10^{20}+6 \\ 10^{4}+2 & 10^{8}+7 &10^{10}+2n \\ 10^{4}+8&10^{6}+4 &10^{15}+9 \end{bmatrix}$ for all $n\in \mathbb{N},$ Then

$(a)\;\;A$ is invertible for all $n\in \mathbb{N}$

$(b)\;\;A$ is not invertible for all $n\in \mathbb{N}$

$(c)\;\;A$ may or may not be invertible depending on the values of $n\in \mathbb{N}$

$(d)\;$ Data Insufficient

What I try

If $A$ is invertiable, then $\det(A)\neq 0$

$$A=\begin{vmatrix} 10^{30}+5& 10^{20}+4 &10^{20}+6 \\ 10^{4}+2 & 10^{8}+7 &10^{10}+2n \\ 10^{4}+8&10^{6}+4 &10^{15}+9 \end{vmatrix}$$

Expanding along $1^\mathrm{st}$ row

$\displaystyle A=\bigg(10^{30}+5\bigg)\bigg[\bigg(10^8+7\bigg)\bigg(10^{15}+9\bigg)-\bigg(10^6+4\bigg)\bigg(10^{10}+2n\bigg)\bigg]-\bigg(10^{20}+4\bigg)\bigg[\bigg(10)^4+2\bigg)\bigg(10^{15}+9\bigg)-\bigg(10^{10}+2n\bigg)\bigg(10^8+4\bigg)\bigg]+\bigg(10^{20}+6\bigg)\bigg[\bigg(10^4+2\bigg)\bigg(10^6+4\bigg)-\bigg(10^8+7\bigg)\bigg(10^4+8\bigg)\bigg]$

How do I simplify such a huge calculation?

Please help me.

amWhy
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jacky
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  • I would start by opening the square brackets and rearranging the terms ... Perhaps there will be some cancellations? – Matti P. Feb 11 '20 at 10:53
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    a) is true. Hint: Work modulo 2. – Dirk Feb 11 '20 at 11:20
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    If you don't know what modulo 2 means: decide whether the determinant will be even or odd. If it is is, it is certainly not 0. – lisyarus Feb 11 '20 at 11:36
  • Instead of computing determinant of $A$, can we just compute the term, in which $n$ appears, i.e. $2n\left(\left(10^{20}+4\right)\left(10^{8}+4\right)−\left(10^{30}+5\right)\left(10^{6}+4\right)\right)$, in order to investigate the effect of the values of $n$ on the determinant? If the answer to my question is "yes", can we consider only the two predominant parts of this particular term, i.e. $-\left(10^{36}-10^{28}\right)$, which is nowhere near zero, to make ourselves a decision? – YNK Feb 11 '20 at 15:22

3 Answers3

4

Using the modulo 2 hint, the principal diagonal is odd and since all the other elements are even, every other component of the determinant, including the 2n term, is even, so the determinant is odd and therefore not zero.

(Added later)

Note that the applies to any matrix with all the diagonal elements odd and the off diagonal elements even.

marty cohen
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1

Following @Dirk's hint. we can actually take each component modulo $2$, since the determinant is a polynomial function of $A$'s entries with integer coefficients. So the determinant has the same parity as $\det I_3=1$. For each $n$, $\det A$ is odd so isn't $0$, making the answer (a).

J.G.
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Write for $A$ $$ A=\left[ \begin{array}{cc} o_1\textrm{ }e_1\textrm{ }e_2\\ e_3\textrm{ }o_2\textrm{ }e_4\\ e_5\textrm{ }e_6\textrm{ }o_3 \end{array}\right], $$ where $o_i=$odd, $i=1,2,3$ and $e_i=$even, $i=1,\ldots,6$. Then $$ \textrm{det}(A)=\left|A\right|=e_1e_4e_5+e_2e_3e_6-e_4e_6o_1-e_2e_5o_2-e_1e_3o_3+o_1o_2o_3=\textrm{odd.} $$ Hence $|A|\neq 0$ and thus $A$ is invertable

Nikos Bagis
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