Determinant, invertible matrices, and rank - Help with true/false questions in linear algebra

Question

I'm having trouble with some true/false questions in my linear algebra class and was hoping someone could help me out.

Here are the questions:

a) For all square matrices $A$, $\det(2A)=2\det(A)$.

b) For all square matrices $A$, $\det(A^T)=\det(A)$.

c) If a $3\times3$ matrix $A$ is invertible, then $\mathrm{rank}(A)=3$.

d) Assume $A$ is the transformation matrix for reflecting vectors in a given plane through the origin. Then $\det(A)=1$.

e) Let $A$ be a square matrix. If the homogeneous equation system $AX=0$ has a unique solution, then the non-homogeneous equation system $AX=Y$ has a unique solution for all $Y$.

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I tried to solve them but I'm not sure if I did it right. Here are my attempted solutions:

• a) For all square matrices $A$, $\det(2A)=2\det(A)$.

Solution: True. This can be proven using the properties of determinants, specifically that $\det(kA)=k^n*\det(A)$, where $k$ is a scalar and $n$ is the dimension of $A$. Since $A$ is square, it has the same number of rows and columns, which means $n$ is the same for $A$ and $2A$. Therefore, $\det(2A)=2^n*\det(A)=2\det(A)$.

• b) For all square matrices $A$, $\det(A^T)=\det(A)$.

Solution: True. This is a well-known property of determinants, that the determinant of a matrix is the same as the determinant of its transpose.

• c) If a $3\times3$ matrix $A$ is invertible, then $\mathrm{rank}(A)=3$.

Solution: True. An invertible matrix is one that has a unique solution to the equation $Ax=b$ for any non-zero $b$. Since $A$ is $3\times3$, its rank is at most $3$. If $A$ is invertible, then its rank must be exactly $3$, since otherwise the equation $Ax=0$ would have non-zero solutions, which contradicts the invertibility of $A$.

• d) Assume $A$ is the transformation matrix for reflecting vectors in a given plane through the origin. Then $\det(A)=1$.

Solution: False. The transformation matrix for reflecting vectors in a given plane through the origin is actually $-I$ or $I$, where $I$ is the identity matrix. Both of these matrices have determinant $-1$ or $1$, respectively, so $\det(A)$ cannot be $1$.

• e) Let $A$ be a square matrix. If the homogeneous equation system $AX=0$ has a unique solution, then the non-homogeneous equation system $AX=Y$ has a unique solution for all $Y$.

Solution: True. If the homogeneous equation system $AX=0$ has a unique solution, then the only solution is the trivial one, $X=0$. Since the non-homogeneous equation system $AX=Y$ is equivalent to the equation $AX-Y=0$, this means that $AX=Y$ has a unique solution for any $Y$, since the only solution to $AX-Y=0$ is $X=0$, which implies that $AX=Y$ has a unique solution given by $X=0+Y=Y$.

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Can someone please tell me if my answers are correct and if not, where I went wrong?

Thanks in advance!

Answer 1

a) False, because $2^n\neq 2$ unless $n=1$.

b) True

c) True

d) False, but your reasoning is incorrect. First, $I$ does not reflect vectors about a plane through the origin, it is the identity and it does nothing. Also, $-I$ does not reflect vectors about a plane through the origin if $n>1$, although it reflects vectors about the origin. To disprove the statement, the matrix $$ \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} $$ reflects vectors about the $yz$-plane, but has determinant $-1$.

e) True, but your reasoning is incorrect. You seem to think that $X=0$ is a solution to $AX-Y=0$ for any $Y$. That is simply not true if $Y\neq 0$. Just plug in $X=0$ and see what happens. The correct way to argue is to note that $AX=0$ having a unique solution implies that the null space of $A$ is trivial, so $A$ is invertible. Then $X=A^{-1}Y$ is a solution to $AX=Y$. To show uniqueness, note that if $Z$ is another solution, i.e. $AZ=Y$, then we have $Z=A^{-1}Y$ by left multiplying $A^{-1}$ on both sides. This shows that any solution is equal to $A^{-1}Y$, proving the uniqueness.