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Linear Algebra Fall 2004

Exam I-Solutions

This is a closed-note/closed-book exam. Please show all relevant work for each solution. Solutions given without relevant work will not receive full credit. Please use (if needed) the backs of each page to work the problems and then write your answer on the front of the page under the appropriate problem. Don't forget to check that any discovered solution to a linear system of equations actually satisfies the system.

  1. It can be shown (you do NOT have to verify) that

    $\displaystyle \left(\begin{array}{cccccc} 1&2&-3&-2&4&1\\ 2&5&-8&-1&6&4\\ 1... ...cccccc} 1&0&1&0&24&21\\ 0&1&-2&0&-8&-7\\ 0&0&0&1&2&3\\ \end{array}\right)$

    1. Give the particular solution to the following system of equations when you let all free variables have the value one.


      $\displaystyle x+2y-3z-2s+4t$ $\displaystyle =$ $\displaystyle 1$  
      $\displaystyle 2x+5y-8z-s+6t$ $\displaystyle =$ $\displaystyle 4$  
      $\displaystyle x+4y-7z+5s+2t$ $\displaystyle =$ $\displaystyle 8$  

      Solution:The reduced echelon form tells us that


      $\displaystyle x_1+x_3+24x_5$ $\displaystyle =$ $\displaystyle 21$  
      $\displaystyle x_2-2x_3-8x_5$ $\displaystyle =$ $\displaystyle -7$  
      $\displaystyle x_4+2x_5$ $\displaystyle =$ $\displaystyle 3$  

      So the solution vector can be written with the basic variables written as a function of the free variables as follows.


      $\displaystyle \bar{x}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c} x_1\\ x_2\\ x_3\\ x_4\\ x_5\\ \end{array}... ...ay}\right)+x_5\left(\begin{array}{c} -24\\ 8\\ 0\\ -2\\ 1\\ \end{array}\right)$  

      Now letting $ x_3=x_5=1$ we get the solution vector

      $\displaystyle \bar{x}=\left(\begin{array}{r} -4\\ 3\\ 1\\ 1\\ 1\\ \end{array}\right)$

    2. Write the solution(s) to the corresponding homogeneous system in parametric vector form.

      Solution:We know that once a non-homogeneous system is solved we have also solved the homogeneous system. The vectors with free variable weights span the set of solutions to the homogeneous system. So the set of all solutions to the corresponding homogeneous system is

      $\displaystyle \left\{x_3\left(\begin{array}{r} -1\\ 2\\ 1\\ 0\\ 0\\ \end{... ...,\left(\begin{array}{r} -24\\ 8\\ 0\\ -2\\ 1\\ \end{array}\right)\right\}$

  2. Let $ A=\left(\begin{array}{rrrr} 1&2&1&3\\ 0&1&-3&-10\\ 3&-2&-1&5\\ \end{array}\right)$.

    Perform row operations and transform the matrix $ A$ into

    1. Row Echelon Form.

      Solution:


      $\displaystyle \left(\begin{array}{rrrr} 1&2&1&3\\ 0&1&-3&-10\\ 3&-2&-1&5\\ \end{array}\right)$ $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrr} 1&2&1&3\\ 0&1&-3&-10\\ 0&-8&-4&-4\\ \end{array}\right)$  
             
        $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrr} \mathbf{1}&2&1&3\\ 0&\mathbf{1}&-3&-10\\ 0&0&\mathbf{-28}&-84\\ \end{array}\right)$  

    2. Reduced Row Echelon Form.

      Solution:


      $\displaystyle \left(\begin{array}{rrrr} 1&2&1&3\\ 0&1&-3&-10\\ 0&0&-28&-84\\ \end{array}\right)$ $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrr} 1&2&1&3\\ 0&1&-3&-10\\ 0&0&1&3\\ \end{array}\right)$  
             
        $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrr} 1&2&0&0\\ 0&1&0&-1\\ 0&0&1&3\\ \end{array}\right)$  
             
        $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrr} \mathbf{1}&0&0&2\\ 0&\mathbf{1}&0&-1\\ 0&0&\mathbf{1}&3\\ \end{array}\right)$  

  3. For each of the following augmented matrices decide if the corresponding system is consistent or inconsistent. If the system is consistent, decide how many solutions exist.

    1. $\displaystyle \left(\begin{array}{rrrrrr} 1&-2&6&3&1&9\\ 0&0&2&1&-1&4\\ 0&0&0&0&1&0\\ 0&0&0&0&0&0\\ \end{array}\right)$

      Consistent with an infinite number of solutions.

    2. $\displaystyle \left(\begin{array}{rrrr} 0&1&-2&9\\ 0&0&-1&5\\ 0&0&0&2\\ 0&0&0&0\\ \end{array}\right)$

      Inconsistent

    3. $\displaystyle \left(\begin{array}{rrr} 1&-2&7\\ 0&1&4\\ 0&0&0\\ \end{array}\right)$

      Consistent with a unique solution.

    4. $\displaystyle \left(\begin{array}{rrrrr} 1&0&0&0&2\\ 0&1&0&0&0\\ 0&0&1&0&8\\ 0&0&0&1&-1\\ \end{array}\right)$

      Consistent with a unique solution.

  4. Let $ T$ be the linear transformation defined as $ T(\bar{x})=A\bar{x}$ where

    $\displaystyle A=\left(\begin{array}{rrrrr} 2&2&-1&6&4\\ 4&4&1&10&13\\ 6&6&0... ...{array}{rrrrr} 1&1&0&0&3/2\\ 0&0&1&0&2\\ 0&0&0&1&1/2\\ \end{array}\right)$


    1. What is the domain of $ T$?

      Solution:The domain is $ \mathbb{R}^6$.

    2. Do the columns of $ A$ span $ \mathbb{R}^3$?

      Solution:There is a pivot in every row and so the map is surjective and the columns of $ A$ span $ \mathbb{R}^3$.

    3. Is the transformation $ T$ one-to-one?

      Solution:There are columns without pivots, and so there are free variables and the map is NOT injective.

  5. A hyperplane of $ \mathbb{R}^n$is defined to be a subset of $ \mathbb{R}^n$that is spanned by a set of $ n-1$ linearly independent vectors. Show that the solution set of any single linear homogeneous equation in the variables $ x_1,x_2,\dots,x_n$ (such as $ a_1x_1+a_2x_2+\cdots+a_nx_n=0$) is a hyperplane of $ \mathbb{R}^n$.

    Solution:This is the set of all solutions to the homogeneous system $ A\bar{x}=\bar{0}$ with the $ 1\times n$ matrix $ A=(a_1\;a_2\cdots a_n)$, (assume without loss of generality that $ a_1\not=0$). The matrix is in reduced row echelon form and has one basic variable, $ x_1$, and $ n-1$ free variables. So the solution vector can be written as
    $\displaystyle \bar{x}$ $\displaystyle =$ $\displaystyle \left(\begin{array}{c} x_1\\ x_2\\ \vdots\\ x_n\\ \end{array}\ri... ...} -(a_2x_2+a_3x_3+\cdots a_nx_n)/a_1\\ x_2\\ \vdots\\ x_n\\ \end{array}\right)$  
      $\displaystyle =$ $\displaystyle x_2\left(\begin{array}{c} -a_2/a_1\\ 1\\ 0\\ \vdots\\ 0\\ 0\\ \e... ...n\left(\begin{array}{c} -a_n/a_1\\ 0\\ 0\\ \vdots\\ 0\\ 1\\ \end{array}\right)$  

    and the solution set is spanned by these $ n-1$ linearly independent vectors.

  6. Let $ A=\left(\begin{array}{rrrr} 1&3&0&2\\ 3&-1&-2&1\\ -2&1&4&-5\\ \end{array}\right)$.

    Observe that negative one times the first column, plus the second column, plus negative two times the third column equals the fourth column. Give a non-trivial solution to the equation $ A\bar{x}=\bar{0}$.

    Solution:We know that if we label the column vectors of $ A$ as $ \v _1,v_2,v_3$, and $ \v _4$ we can write

    $\displaystyle -\v _1+\v _2-2\v _3=\v _4$

    or

    $\displaystyle -\v _1+\v _2-2\v _3-\v _4=0$

    and so

    $\displaystyle \bar{x}=\left(\begin{array}{r} -1\\ 1\\ -2\\ -1\\ \end{array}\right)$

    is a solution to the system.

  7. Let $ T:\mathbb{R}^3\mapsto\mathbb{R}^3$ be the linear transformation that reflects all points of $ \mathbb{R}^3$ across the $ (x,z)$-plane. Find the standard matrix of $ T$. Then, without the help of determinants, decide if $ T$ is an invertible transformation.

    Solution:Its easy to see that for any point $ \bar{x}=(x,y,z)$ in $ \mathbb{R}^3$ we get

    $\displaystyle T(x,y,z)=(x,-y,z)$

    So looking at the effect of $ T$ on the vectors $ \bar{e}_1,\bar{e}_2$, and $ \bar{e}_3$ we get the standard matrix

    $\displaystyle A=\left(\begin{array}{rrr} 1&0&0\\ 0&-1&0\\ 0&0&1\\ \end{array}\right)$

    Since the matrix has a pivot in every row and column the transformation is invertible.

  8. Compute the following matrix products only when a given product is defined.

    1. $\displaystyle \left(\begin{array}{rrr} 1&0&-1\\ 0&2&-2\\ 1&1&1\\ \end{arr... ...ight)\left(\begin{array}{rrr} 1&2&0\\ -1&0&1\\ 2&0&0\\ \end{array}\right)$


      Solution:A $ 3\times 3$ matrix times a $ 3\times 3$ matrix results in a $ 3\times 3$ matrix.

      $\displaystyle \left(\begin{array}{rrr} 1&0&-1\\ 0&2&-2\\ 1&1&1\\ \end{arr... ...ht)=\left(\begin{array}{rrr} -1&2&0\\ -6&0&2\\ 2&2&1\\ \end{array}\right)$

    2. $\displaystyle \left(\begin{array}{rrrr} 1&2&3&-2\\ \end{array}\right)\left(\begin{array}{r} -1\\ 1\\ -1\\ 1\\ \end{array}\right)$


      Solution:A $ 1\times 4$ matrix times a $ 4\times 1$ matrix produces a $ 1\times 1$ matrix, which is a scalar.

      $\displaystyle \left(\begin{array}{rrrr} 1&2&3&-2\\ \end{array}\right)\left(\begin{array}{r} -1\\ 1\\ -1\\ 1\\ \end{array}\right)=-4$

    3. $\displaystyle \left(\begin{array}{rr} 1&-1\\ 2&1\\ 0&0\\ \end{array}\righ... ... \end{array}\right)\left(\begin{array}{r} 1\\ 0\\ -1\\ \end{array}\right)$


      Solution:A $ 3\times 2$ matrix times a $ 2\times 3$ matrix times a $ 3\times 1$ matrix produces a $ 3\times 1$ matrix.

      $\displaystyle \left(\begin{array}{rr} 1&-1\\ 2&1\\ 0&0\\ \end{array}\righ... ... \end{array}\right)=\left(\begin{array}{r} -3\\ -9\\ 0\\ \end{array}\right)$

    4. $\displaystyle \left(\begin{array}{rrr} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{array}\right)\left(\begin{array}{rrr} 4&5&6\\ 7&-1&1\\ \end{array}\right)$


      Solution:A $ 3\times 3$ matrix times a $ 2\times 3$ matrix is undefined.

  9. Which of the following sets of vectors are linearly independent? Explain your reasoning!

    1. $\displaystyle \left(\begin{array}{r} 1\\ -1\\ -1\\ 0\\ 0\\ \end{array}\ri... ...ray}\right),\left(\begin{array}{c} 0\\ 0\\ 3\\ 0\\ 0\\ \end{array}\right)$


      Solution: Row reduction of the matrix with these vectors as columns results in

      $\displaystyle \left(\begin{array}{rrr} 1&2&0\\ -1&-1&0\\ -1&1&3\\ 0&0&0\\... ...rray}{rrr} 1&2&0\\ 0&1&0\\ 0&0&3\\ 0&0&0\\ 0&0&0\\ \end{array}\right)$

      and since there is a pivot in every column the vectors are independent.


    2. $\displaystyle \left(\begin{array}{c} 1\\ 5\\ \end{array}\right),\left(\begin... ... 3\\ \end{array}\right),\left(\begin{array}{c} -1\\ 1\\ \end{array}\right)$


      Solution:A set of three vectors in $ \mathbb{R}^2$ is always linearly dependent.

    3. $\displaystyle \left(\begin{array}{c} 1\\ 2\\ 3\\ \end{array}\right),\left(\begin{array}{c} -2\\ -4\\ -6\\ \end{array}\right)$


      Solution:These two vectors are scalar multiples of each other and so are linearly dependent.



  10. Prove that any subset of a linearly independent set of vectors must be linearly independent.

    Solution:Let $ \v _1,\v _2,\dots,\v _n$ be a set of linearly independent vectors. Then

    $\displaystyle c_1\v _1+c_2\v _2+\cdots c_n\v _n=\bar{0}$

    has only the trivial solution. Suppose we take any subset $ \v _i,\v _j,\dots,\v _k$ from the above independent set. Then any linear combination

    $\displaystyle c_i\v _i+c_j\v _j+\cdots c_k\v _k=\bar{0}$

    can be written as a linear combination of the entire set with zero weights one the remaining vectors.

    $\displaystyle c_i\v _i+c_j\v _j+\cdots c_k\v _k+(0\cdot\v _t+\cdots+0\cdot\v _s)=\bar{0}$

    and so the only linear combination of $ \v _i,\v _j,\dots,\v _k$ that equals zero is the trivial linear combination.

  11. Let $ A=\left(\begin{array}{rrrrr} 2&4&-1&5&-2\\ -4&-5&3&-8&1\\ 2&-5&-5&1&8\\ -6&0&7&-3&1\\ \end{array}\right)$.

    Given the following row equivalent matrices, find the $ LU$-decomposition of $ A$. Hint: Remember that, in the forward phase of rwo reduction, row operations are only performed to zero all entries below the pivot positions of $ A$.


    $\displaystyle A$ $\displaystyle =$ $\displaystyle \left(\begin{array}{rrrrr} \mathbf{2}&4&-1&5&-2\\ -4&-5&3&-8&1\\ 2&-5&-5&1&8\\ -6&0&7&-3&1\\ \end{array}\right)$  
           
      $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrrr} \mathbf{2}&4&-1&5&-2\\ 0&\mathbf{3}&1&2&-3\\ 0&-9&-3&-4&10\\ 0&12&4&12&-5\\ \end{array}\right)$  
           
      $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrrr} \mathbf{2}&4&-1&5&-2\\ 0&\mathbf{3}&1&2&-3\\ 0&0&0&\mathbf{2}&1\\ 0&0&0&4&7\\ \end{array}\right)$  
           
      $\displaystyle \sim$ $\displaystyle \left(\begin{array}{rrrrr} \mathbf{2}&4&-1&5&-2\\ 0&\mathbf{3}&1&2&-3\\ 0&0&0&\mathbf{2}&1\\ 0&0&0&0&\mathbf{5}\\ \end{array}\right)$  

    $\displaystyle L=\left(\begin{array}{rrrr} 1&0&0&0\\ -2&1&0&0\\ 1&-3&1&0\\ ... ...{3}&1&2&-3\\ 0&0&0&\mathbf{2}&1\\ 0&0&0&0&\mathbf{5}\\ \end{array}\right)$

Extra Credit:

Consider the vector space, $ \mathbb{M}_{2\times 2}$, of all $ 2\times 2$ real valued matrices (where the zero vector is the $ 2\times 2$ matrix of all zeros). Let $ T$ be the following map

$\displaystyle T:\left(\begin{array}{cc} a&b\\ c&d\\ \end{array}\right)\mapsto \left(\begin{array}{cc} a&0\\ 0&d\\ \end{array}\right)$

  1. Using the definition of a linear map, show that $ T$ is a linear map on the vector space of all $ 2\times 2$ real valued matrices.

    Solution:Choose any two matrices $ A$ and $ B$ and see that


    $\displaystyle T[A+B]=T\left[\left(\begin{array}{cc} a&b\\ c&d\\ \end{array}\right)+\left(\begin{array}{cc} x&y\\ w&z\\ \end{array}\right)\right]$      
      $\displaystyle =$ $\displaystyle T\left[\left(\begin{array}{cc} a+x&b+y\\ c+w&d+z\\ \end{array}\right)\right]$  
      $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} a+x&0\\ 0&d+z\\ \end{array}\right)$  
      $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} a&0\\ 0&d\\ \end{array}\right)+\left(\begin{array}{cc} x&0\\ 0&z\\ \end{array}\right)$  
      $\displaystyle =$ $\displaystyle T(A)+T(B)$  

    Also
    $\displaystyle T(rA)$ $\displaystyle =$ $\displaystyle T\left[r\left(\begin{array}{cc} a&b\\ c&d\\ \end{array}\right)\right]$  
      $\displaystyle =$ $\displaystyle \left(\begin{array}{cc} ra&\\ 0&rd\\ \end{array}\right)$  
      $\displaystyle =$ $\displaystyle r\left(\begin{array}{cc} a&\\ 0&d\\ \end{array}\right)$  
      $\displaystyle =$ $\displaystyle rT(A)$  

    So $ T$ is linear transformation.

  2. Find a matrix whose image under $ T$ is the zero matrix. Is there more than one such matrix?

    Solution:

    $\displaystyle T\left(\begin{array}{cc} 0&b\\ c&0\\ \end{array}\right)\left(\begin{array}{cc} 0&0\\ 0&0\\ \end{array}\right)$

    for all $ c,b\in\mathbb{R}$ and so there are an infinite number of matrices that get mapped to the zero matrix.


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Robert Rostermundt 2004-10-06