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Suppose A is a 2 × 2 matrix with the sum of the entries in row i equal to zero for 1 ≤ i ≤ 2. Prove that A is not invertible.

Dont even know how to go about this. Any help would be appreciated. Thanks.

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closed as off-topic by user223391, Dietrich Burde, Namaste, hardmath, Math Lover Oct 17 '17 at 17:25

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  • $\begingroup$ what's the determinant of such matrix? $\endgroup$ – Yanko Oct 17 '17 at 13:28
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    $\begingroup$ another way to do this is that the vector (1,1) must be in the kernel. $\endgroup$ – Yanko Oct 17 '17 at 13:29
  • $\begingroup$ @yanko the determinant would be 0 to make it not invertible. $\endgroup$ – Aisling Crafferty Oct 17 '17 at 13:30
  • $\begingroup$ exactly, just calculate the determinant and you're done. Look at M. Winter's hint. $\endgroup$ – Yanko Oct 17 '17 at 14:14
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Let

$$M=\begin{pmatrix}a&b\\c&d\end{pmatrix}.$$

How can the first row consisting of the numbers $a$ and $b$ have zero sum? Only if $b=-a$. Same for the second row: $d=-c$. So your matrix looks actually like this:

$$M=\begin{pmatrix}a&-a\\c&-c\end{pmatrix}.$$

Any clue why this could be non-invertible?

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Hint: A square matrix is non-invertible (singular) iff its columns are linearly dependent.

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The easiest way to see it is by its determinant.

Let

$$A=\begin{pmatrix}a&-a\\b&-b\\ \end{pmatrix}$$

Then $$\operatorname{det}A=-ab-(-ab)=0$$

We used the fact that $a_{ij}+a_{i(j+1)}=0\Rightarrow a_{ij}=-a_{i(j+1)} $

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$$A=\begin{bmatrix} a_1&a_2 \\ a_3 & a_4\end{bmatrix}$$

Since the first line is zero :

$$a_1+a_2=0 \iff a_1=-a_2$$ $$a_3+a_4=0 \iff a_3=-a_4$$

$$\begin{bmatrix} a_1&a_2 \\ a_3 & a_4\end{bmatrix}=\begin{bmatrix} a_1&-a_1 \\ a_3 & -a_3\end{bmatrix}$$

$$det(A)=-a_3a_1+a_3a_1=0$$

A is not invertible...

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