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?
Best Answer
(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.