# Linear Algebra Review Questions

So I have a test on Monday and my professor posted a couple of non-graded review questions that she said we should look over. Anyhow, I have a couple of questions that I'd like answered if that's okay. I'm not looking for complex proofs or anything, just a simple explanation. :)

*If A is invertible, is A-Transpose also invertible? If so, what is (A^T)^-1?

My guess is that A-Transpose would also be invertible because of the determinant. Though I don't think my teacher will like this answer because we haven't covered determinants.

*Why are elementary matrices important for the Inverse Algorithm?

I'm assuming that the answer here is simple, in that elementary row operations are used for row-reduction so that's why they are important.

*Is it possible that all solutions of a homogeneous system of ten linear equations in twelve variables are multiples of one fixed nonzero explanation?

I'm assuming this is possible, but I'm not sure why....

*Suppose that A is a 5x11 matrix. What is the smallest dimension for Nul(A)?

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For your first question, I assume that you have discussed the property that $\left(AB\right)^{t} = B^t A^t$ for all square matrices $A$, $B$. If you know that $A$ is invertible, then you know that there exists a matrix $A^{-1}$ such that $AA^{-1} = I$. Now consider applying the property discussed.

For your second question, it is tough to know exactly what someone would expect, but your answer certainly seems reasonable. (I would probably be looking for a bit more of an explanation if I assigned that type of question, however. Namely, I would probably want you to actually discuss/describe the inverse algorithm.)

For your third question, the answer is no. This relates to the explanation provided below for your fourth question. The dimension of the solution space of the homogeneous system is the nullity of coefficient matrix $A$, and in this case, the matrix of coefficients of the system has to have a nullity that is at least two-dimensional.

For your fourth question, my explanation would depend a bit on what you know and how you are viewing matrices at this point. If you are viewing $A$ as a linear transformation from an 11-dimensional vector space to a 5-dimensional vector space then it is a simple observation that the image of $A$ (i.e. the rank of A or the number of linearly independent rows of $A$) is at most a 5-dimensional vector space. If the rank of $A$ (i.e. the dimension of the image of $A$) drops and $A$ is not onto, then the dimension of the nullity $A$ must necessarily increase. With that observation, then the nullity must be at least 6-dimensional.

I expect that the argument that you are supposed to make in this instance is based on a general rule that the rank of $A$ and the nullity of $A$ must add up to the number of columns in $A$.

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$$I_n=\left(AA^{-1}\right)^T=\left(A^{-1}\right)^TA^T$$ so $A^T$ is invertible and its inverse is $\left(A^{-1}\right)^T$.

For your last question: let $$f\colon \Bbb R^{11}\rightarrow \Bbb R^5,\quad x\mapsto A x$$ the linear transformation represented by the matrix $A$ in the standard basis, then by the rank-nullity theorem the smallest dimension for $\ker(A)=\ker(f)$ corresponds with the biggest dimension of the image of $f$ which's $5$ so the answer is $11-5=6$.

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