Through the “means test,” a bankruptcy court can evaluate an…

Through the “means test,” a bankruptcy court can evaluate an individual debtor’s monthly income in recent months is compared to median income in the geographic area where the person lives.  If debtor’s income is below the median income, the debtor is allowed to file Chapter 7, as there is no presumption of bankruptcy abuse.  Otherwise, they file for Chapter 13.

Dr. Patel just finished an appointment with Horace, who shar…

Dr. Patel just finished an appointment with Horace, who shared he has been suffering from mysterious symptoms that lab reports showed to be a sexually transmitted disease.  Dr. Patel retells Horace’s story during her lunch break in the hospital cafeteria “Can you believe this charmer?  This is Horace Smith’s third STD in the past year!  Avoid him on Tinder.” Which of the following is most accurate based on the facts above?

Consider $$A = \begin{bmatrix}&8 & 2 & -2 & 0 &5 \\&12 & 3 &…

Consider $$A = \begin{bmatrix}&8 & 2 & -2 & 0 &5 \\&12 & 3 & -3 & 6 &0 \\&4 & 1 & -1 & 3 &5 \\&0 & 0 & 0 & 1 &5\\&6 & \frac{3}{2} & -\frac{3}{2} & 3 & 0 \end{bmatrix}$$   a) Find the nullspace of A (Nul(A) = span\{…\}).   b) Find a basis for the column space of A.   c) Is A invertible? Justify your answer using 3 different reasons using the Invertible Matrix Theorem.  

The two parts of this problem are independent.   a) Show tha…

The two parts of this problem are independent.   a) Show that if $$||\vec{u}-\vec{v}||^2 = ||\vec{u}+\vec{v}||^2$$ then $$\vec{u}$$ and $$\vec{v}$$ are orthogonal.   b) Let $$\{\vec{u}_1, \vec{u}_2, \vec{u}_3, \vec{u}_4\}$$ be an orthogonal basis for $$R^4$$. Let W be Span $$\{\vec{u}_1, \vec{u}_2, \vec{u}_3\}$$. Write $$\vec{x}$$ as the sum of two vectors, one in W and the other perpendicular to W. $$\vec{u}_1 = \begin{bmatrix}&1 \\&1 \\&0 \\&-1\end{bmatrix}$$, $$\vec{u}_2 = \begin{bmatrix}&1 \\&0 \\&1 \\&1\end{bmatrix}$$, $$\vec{u}_3 = \begin{bmatrix}&0 \\&-1 \\&1 \\&-1\end{bmatrix}$$, and $$\vec{x} = \begin{bmatrix}&-2 \\&3 \\&6 \\&-4\end{bmatrix}$$