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

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 $$Y$$.

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 $$X=0+Y=Y$$.

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

• Regarding a), $2^n\ne2$ unless $n=1$ Commented Apr 30, 2023 at 23:41

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.

• $-I$ shouldn’t be a reflection, as any reflection can be written in a suitable basis as the diagonal matrix with all $1$’s and one $-1$. Commented May 1, 2023 at 3:10
• @Chris If I'm not mistaken, reflection through the origin is simply the map $v\mapsto -v$, which can be represented by $-I$. It's not a reflection about a line or hyperplane. Commented May 1, 2023 at 3:14
• It says in the problem statement reflection in a given plane, perhaps we interpreted that differently. Commented May 1, 2023 at 16:32
• @Chris I see. It does make more sense to interpret the statement as reflection about a plane that passes through the origin. Commented May 2, 2023 at 1:29