Quiz 8 (Solution)

Date: MATH 1090-4 - Fall 2003

1. Let

   $$A=\left[ \begin{array}{ccc} 1 & -1 & 2\\ 4 & 0 & 1 \end{array} \right] \quad B=\left[ \begin{array}{cc} 0&1\\ 1&0 \end{array} \right]$$

   
   
   Perform each of the following computations or state why they can't be done.
   1. $2\cdot B + A\cdot (A^t)$.

      $$\begin{split}2\cdot B + A\cdot (A^t) &= 2\cdot \left[ \begin{... ...t[ \begin{array}{cc} 6&8\\ 8&17 \end{array} \right] \end{split}$$

      
      

   2. $2\cdot B + (A^t)\cdot A$.

      $$\begin{split}2\cdot B + (A^t)\cdot A &=2\cdot \left[ \begin{a... ...ccc} 17&-1&6\\ -1&1&-2\\ 6&-2&5 \end{array} \right] \end{split}$$

      
      
      and we see that the addition cannot be performed because the two matrices have different sizes. Notice that there was no need to actually compute $(A^t)\cdot A$ since we knew that that matrix had size $3\times 3$ and, so, it was different from the size of $2B$.

2. Consider the system of equations

   $$\begin{cases}x+y-z=0\\ x+2y+3z=-5\\ 2x-y-13z=17. \end{cases}$$

   
   
   1. Write the augmented matrix of the system.

      $$\left[ \begin{array}{ccc\vert c} 1&1&-1&0\\ 1&2&3&-5\\ 2&-1&-13&17 \end{array} \right]$$

      
      

   2. Use Gauss-Jordan elimination to ``simplify'' the augmented matrix.

      $$\begin{split}\left[ \begin{array}{ccc\vert c} 1&1&-1&0\\ 1&2&... ... 1&0&0&15\\ 0&1&0&-13\\ 0&0&1&2 \end{array} \right] \end{split}$$

      
      

   3. Use the reduced matrix that you found above to solve the system of equations.

      From the simplified matrix that we found above we have the system

      $$\begin{cases}x=15\\ y=-13\\ z=2 \end{cases}$$

      
      

      Then, the solution of the system of equations is $(x,y,z) = (15,-13,2)$.

   4. According with your computations, how many solutions does the system have?

      In the previous item we found only one solution. Thus, the system has only one solution.