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True or false (with a counterexample if false and a reason if true):
(a) A $4 \times 4$ matrix with a row of zeros is not invertible.
(b) Every matrix with $1's$ down the main diagonal is invertible.
(c) If A is invertible then $A^{-1}$ and $A^2$ are invertible.

Are my answers correct, and how do I prove my answer for (c)?

For (a) I think it's true because if you have a row of zeroes equal to a number not zero, then that does not make sense, making the matrix false, if the row of zeroes equals zero then it would simply become a $3\times4$ matrix.

For (b) I think it's false because you can have a square matrix full of $1's$ and we know that it isn't invertible because if a matrix has the same column or row twice then it is not an inverse.

For (c) I think the answer is true because if $A$ is invertible then isn't there a rule that states that $A^{-1}$ and $A^2$ are also invertile? How would I prive that?

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    $\begingroup$ $(a)$ No, it would not simply become a $3\times 4$-matrix. $(b)$ is good, and $(c)$ is a duplicate. $\endgroup$ Feb 19 '16 at 19:54
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    $\begingroup$ Just check to see if the determinant is zero or not. $\endgroup$
    – akech
    Feb 19 '16 at 19:54
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    $\begingroup$ I think you have to state what you are or aren't allowed to use. Most of these can be answered by computing determinants. $\endgroup$
    – BCLC
    Feb 20 '16 at 10:47
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    $\begingroup$ What does 'down' in 'down the main diagonal' mean? $a_{ii} = 1?$ $\endgroup$
    – BCLC
    Feb 20 '16 at 10:48
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    $\begingroup$ @idknuttin so $a_{ii} = 1$? $\endgroup$
    – BCLC
    Feb 20 '16 at 16:01
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(a) Yes this is true, but you cannot think of a $4 \times 4$ matrix with a row of all zeroes as a $3 \times 4$ matrix. But the determinant of any matrix with an all zero row is zero, hence the matrix is not invertible.

(b) Correct

(c) follows straight from the definition. A matrix $A$ is said to be invertible if there exists a matrix $A^{-1}$ such that $AA^{-1} = A^{-1}A = I$. Note by this definition, $A^{-1}$ is also invertible with inverse $A$. To show that $A^2$ is invertible, just note that $$A^2(A^{-1})^2 = AAA^{-1}A^{-1} = AIA^{-1} = AA^{-1} = I.$$ You can check that $(A^{-1})^2A^2 = I$ as well.

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The answer for (b) is good. (If you want to be very precise you'll have to state that you exclude the case of $1 \times 1$ matrices, but often this goes without saying.)

The answer for (a) might be good idea, but not expressed too well. A way to argue this is that $A$ invertible, means that $Ax = b$ has a solution whatever the $b$. But this is false in this case, as the result of $Ax_0$ will always yield a vector with $0$ in the row that is the $0$-row in the matrix.

For (c). Just show that there is a matrix $C$ such that $A^{-1}C=CA^{-1}=I$ and likewise there is a matrix $D$ such that $A^2D = D A^2 = I$. (In the first case take $C = A$ and in the second $D = (A^{-1})^2$.)

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For (a) : your answer is correct but the motivation is wrong. A $4 \times 4$matrix with a row of zeros is a truly $4 \times 4$ matrix but its determinat is null, and we know that

a matrix is invertible if and only if its determinat is not null.

You can use this fact also:

For (b) : Your answer is correct because the determinant is an alternating multilinear function of the rows ( or columns) of the matrix, so it is null if two rows (or columns) are equal.

for (c) : Use the fact that, for two matrices $A,B$ we have: $\det(AB)=\det(A) \det(B)$. This means that we have: $$ \det(A^{-1})\det (A)=\det(A^{-1}A)=\det (I)=1 \quad \Rightarrow \quad \det(A^{-1})=\det(A)^{-1} $$ and $$ \det (A^2)=\left(\det(A)\right)^2 $$

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