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Please work out all of the following problems.  Credit will be given based on the progress that you make towards the final solution.  Show your work.  No calculators allowed for this page.

Problem 1 

Let

       \( A = \begin{pmatrix} 2 & 1  \\ 1 & 2    \end{pmatrix}  \)

 Find an orthogonal matrix P and a diagonal matrix D with A  =  PDP^-1.

Solution

We find

     \( | \lambda I - A|  = \begin{vmatrix} \lambda I - 2 & -1  \\ -1 & \lambda I - 2    \end{vmatrix}  \)

     \( (\lambda - 2) (\lambda - 2)  = \lambda ^2 - 4 \lambda + 3 = (\lambda - 1) (\lambda - 3)   \)

Hence the eigenvalues are 1 and 3.  Now we find the eigenvectors by finding the null space of \( A - 1I \) .  We have

     \( | 1I - A|  = \begin{vmatrix} \lambda -1 & -1  \\ -1 & -1   \end{vmatrix}  \)

since the determinant is zero, we can use the first row which corresponds with the equation

        -x - y  =  0.

Which has eigenvector

        (1,-1)

Normalizing gives

       \( ( \frac{1}{\sqrt{2}} , - \frac{1}{\sqrt{2}} )  \)

For the eigenvector corresponding with the eigenvalue of 3, we have

     \( | 3I - A|  = \begin{vmatrix} \lambda 1 & -1  \\ -1 & 1   \end{vmatrix}  \)

again, since the determinant is zero, we can use the first row which corresponds with the equation

        x - y  =  0.

Which has eigenvector

        (1,1)

Normalizing gives

       \( ( \frac{1}{\sqrt{2}} ,  \frac{1}{\sqrt{2}} )  \)

Putting this all together, gives

     \( D = \begin{pmatrix} \lambda 1 & 0  \\ 0 & 3   \end{pmatrix}  \)     \( P = \begin{pmatrix} \lambda \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ - \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}   \end{pmatrix}  \)

Problem 2

Use the permutation definition of the determinant to find the determinant of

        \(  \begin{pmatrix} 4 & 2 & 3  \\ 0 & 0 & -2 \\ 1 & 6 & 5    \end{pmatrix}  \)

Solution

The permutations on three elements and their corresponding products are

          even             odd           odd            even           even          odd
        (1,2,3)         (1,3,2)       (2,1,3)        (2,3,1)        (3,1,2)      (3,2,1)

       (4)(0)(5) - (4)(-2)(6) - (2)(0)(5) + (2)(-2)(1) + (3)(0)(6) - (3)(0)(1)

         =  48 - 4  =  44

Problem 3

Find the inverse of A if

Solution

We augment this matrix with the identity matrix and rref it.

        \( \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 1 & 0 & 0 \\ -1 & 0 & 1 & 0 & 1 & 0\\ 3 & -1 & 5 & 0 & 0 & 1    \end{array} \right) \begin{array}{ccc|lcr} R_2 & \rightarrow &  R_2 + R_1 \\ \rightarrow & \rightarrow & \rightarrow \\ R_3 & \rightarrow &  R_3 + -3 R_1  \end{array} \left( \begin{array}{ccc|ccc} 1 & 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0\\ 0 & 0 & -1 & -3 & 0 & 1    \end{array} \right)\) 

        \(\begin{array}{ccc|lcr} R_1 & \rightarrow & R_1 + (-1)R_2 \\ \rightarrow & \rightarrow & \rightarrow   \end{array} \left( \begin{array}{ccc|ccc} 1 & 0 & -1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0\\ 0 & 0 & -1 & -3 & 0 & 1    \end{array} \right) \begin{array}{ccc|lcr} R_3 & \rightarrow & (-1)R_3 \\ \rightarrow  & \rightarrow  & \rightarrow   \end{array} \left( \begin{array}{ccc|ccc} 1 & 0 & -1 & 0 & -1 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0\\ 0 & 0 & 1 & 3 & 0 & -1     \end{array} \right)\) 

         \( \begin{array}{ccc|lcr} R_1 & \rightarrow & R_1 + R_3 \\ \rightarrow & \rightarrow & \rightarrow \\ R_2 & \rightarrow & R_2 + (-1)R_3  \end{array} \left( \begin{array}{ccc|ccc} 1 & 0 & 0 & 3 & -1 & -1 \\ 0 & 1 & 0 & -2 & 1 & 1\\ 0 & 0 & 1 & 3 & 0 & -1     \end{array} \right)\) 

We conclude that

        \(  A^{-1} = \begin{pmatrix} 3 & -1 & -1  \\ -2 & 1 & 1 \\ 3 & 0 & 01  \end{pmatrix}  \)

Problem 4

Consider the matrix

           \( A = \begin{pmatrix} 5 & 10 & 0 & 3  \\ 3 & 6 & 1 & 1 \\ 2 & 4 & 0 & 1   \\ 1 & 2 & 0 & -1    \end{pmatrix}  \)

A.     Determine the rank of A.

Solution

We find

           \(  rref(A) \begin{pmatrix} 1 & 2 & 0 & 0  \\0 & 0 & 1 & 0  \\ 0 & 0 & 0 & 1    \\ 0 & 0 & 0 & 0     \end{pmatrix}  \)

Since there are three pivots, the rank is 3.
 

B.   Find a basis for the null space of A.

Solution
 

The equation form of rref(A) is

                x₁  =  -2x₂        
          x₂  =  x₂ 
          x₃  =  0
          x₄  =  0

Since the rank of A is 3 and n  =  4, the nullity is 1.  A basis for the null space is

        {(-2, 1,0,0)}

C.     Find a basis for the column space of A using columns of A.

Solution

We see that the corners are in columns 1, 3, and 4.  Hence the first, second, and third columns of A are a basis for the column space of A.  

        {(5,3,2,1), (0,1,0,0), (3,1,1,-1)}
 

Problem 5      

Let   \( L: P_2 \rightarrow R^2 \)    be defined by

        \( L(f(t)) = ( f'(1), \int_0^1 f(x) dx \) 

A.    Prove that L is a linear transformation.
 

Let 

        f(t)  =  a₁ + b₁t + c₁t²        and        g(t)  =  a₂ + b₂t + c₂t²  

Then 

     1.     L(f(t) + g(t))  =  (b₁ + 2c₁ + b₂ + 2c₂, a₁ + 1/2 b₁ + 1/3 c₁ + a₂ + 1/2 b₂ + 1/3 c₂)     

             and

            L(f(t)) + L(g(t))  =  (b₁ + 2c₁, a₁ + 1/2 b₁ + 1/3 c₁) + (b₂ + 2c₂, a₂ + 1/2 b₂ + 1/3 c₂)

            =  (b₁ + 2c₁ + b₂ + 2c₂, a₁ + 1/2 b₁ + 1/3 c₁ + a₂ + 1/2 b₂ + 1/3 c₂)     

    2.  L(cf(t))  =  (cb₁ + 2cc₁, ca₁ + 1/2 cb₁ + 1/3 cc₁)  =  c(b₁ + 2c₁, a₁ + 1/2 b₁ + 1/3 c₁)  =  cL(f(t))

hence L is a linear transformation

B.   Let S = {x² + x, x² + 1, x} and T = {(1,1), (1,2)} be bases for P₂ and R².  Find the matrix for L with the bases S and T. 

Solution

We find

           \( P_{E \leftarrow S} \begin{pmatrix} 0 & 1 & 0  \\ 1 & 0 & 1  \\ 1 & 1 & 0 \end{pmatrix}  \)     \( P_{E \leftarrow T} \begin{pmatrix} 1 & 1 \\ 1 & 2  \end{pmatrix}  \)                

    and

        L(1,0,0)  =  L(1)  =  (0,1)

        L(0,1,0)  =  L(t)  =  (1,1/2)

        L(0,0,1)  =  L(t²)  =  (2,1/3)

Hence

              \( A_L = \begin{pmatrix} 0 & 1 & 2  \\ 1 & \frac{1}{2} & \frac{1}{3}   \end{pmatrix}  \)   

Now consider the diagram

        \(S  \begin{array}{lr}  \longrightarrow \\ P_{ E \leftarrow S} \end{array} E_{P_2}  \begin{array}{lr} \longrightarrow \\ A_L \end{array} E_{R^2}  \begin{array}{lr} \longrightarrow \\ P^{-1}_{E \leftarrow T} \end{array} T  \)   

We the matrix we want is

              \( P_{ E \leftarrow S}A_L P^{-1}_{E \leftarrow T} = \begin{pmatrix} 1 & 1   \\ 1 & 2   \end{pmatrix}  \begin{pmatrix} 0 & 1 & 2  \\ 1 & \frac{1}{2} & \frac{1}{3}   \end{pmatrix}  \begin{pmatrix} 0 & 1 & 0  \\ 1 & 0 & 1    \\ 1 & 1 & 0     \end{pmatrix}^{-1} = \begin{pmatrix} 0 & 1 & 2  \\ 1 & \frac{1}{2} & \frac{1}{3}   \end{pmatrix}  \begin{pmatrix} \frac{17}{6} & \frac{7}{3} & - \frac{4}{3}  \\   \frac{8}{3} & \frac{8}{3} & - \frac{2}{3} \end{pmatrix} \)   

Problem 6 

Show that the set

     \( S = \{ \begin{pmatrix} 1 & 2  \\ 0 & 0    \end{pmatrix}  \) , \( \begin{pmatrix} 0 & 0  \\ 2 & 1    \end{pmatrix}  \) , \( \begin{pmatrix} 0 & 2  \\ 0 & 1    \end{pmatrix}  \) , \( \begin{pmatrix} 2 & 0  \\ 1 & 0    \end{pmatrix}  \) , 

is a basis for M^2x

².

Solution

Since there are four vectors in a four dimensional space, we only need to show that the vectors are linearly independent.  We let

     \( c_1  \begin{pmatrix} 1 & 2  \\ 0 & 0    \end{pmatrix}  + c_2  \begin{pmatrix} 0 & 0  \\ 2 & 1    \end{pmatrix} + c_3  \begin{pmatrix} 0 & 2  \\ 0 & 1    \end{pmatrix}  + c_4 \begin{pmatrix} 2 & 0  \\ 1 & 0    \end{pmatrix} = \begin{pmatrix} 0 & 0  \\ 0 & 0    \end{pmatrix} \) 

This gives us the four equations 

        c₁ + 2c₄  =  0
        2c₁ + 2c₃  =  0
        2c₂ + c₄  =  0
        c₂ + c₃  =  0

Putting this into the matrix form Ac  =  0 gives

     \( \begin{pmatrix} 1 & 0 & 0 & 2  \\   2 & 0 & 2 & 0  \\   0 & 2 & 0 & 1  \\   0 & 1 & 1 & 0    \end{pmatrix} \begin{pmatrix} c_1  \\   c_2  \\   c_3 \\   c_4    \end{pmatrix} =  \begin{pmatrix} 0  \\   0  \\  0 \\   0   \end{pmatrix}  \)

We calculate 

        |A|  =  -6

Since the determinant is nonzero, the above equation has only the trivial solution.  Hence the four vectors are linearly independent.  We can conclude that they form a basis for M^2x2.

Problem 7 Use matrices to find the unknown currents in the given circuit.

Solution

First we find the loop equations.

        a -->  b -->  d -->  a :    -I₁ - 60 - 2I₂ + 10  =  0

        b -->  c -->  d -->  b :    -4I₃ - 50 + 2I₂ + 60  =  0

The node equations are 

        b:  I₁ - I₂ - I₃  =  0

        d:  -I₁ + I₂ + I₃  =  0

Notice that the second node equation is redundant, so we can discard this one.  The three equations together give

        -I₁ - 2I₂  =  50

         2I₂ - 4I₃  =  -10

         I₁ - I₂ - I₃  =  0

The corresponding matrix equation AI  =  b is

     \( \begin{pmatrix} -1 & 2 & 0  \\   0 & 2 & -4  \\   1 & -1 & -1    \end{pmatrix} \begin{pmatrix} I_1  \\   I_2  \\   I_3    \end{pmatrix} =  \begin{pmatrix} 50  \\   -10  \\  0   \end{pmatrix}  \)

Hence, the solution is I  =  A^-1b.  We have

     \( A^{-1}b \begin{pmatrix} -\frac{3}{7} & -\frac{1}{7} & \frac{4}{7}  \\   -\frac{2}{7} & \frac{1}{7} & -\frac{2}{7}  \\   -\frac{1}{7} & -\frac{3}{7} & -\frac{1}{7}    \end{pmatrix} \begin{pmatrix} 50  \\   -10  \\   0   \end{pmatrix} =  \begin{pmatrix} -20  \\   -15  \\  -5   \end{pmatrix}  \)

so that

         I₁  =  -20         I₂  =  -15        I₃  =  -5

Notice that the minus signs means that the currents are traveling in the opposite direction as the arrows indicate.

Problem 8

Graph the equation and write the equation in standard form.

        4x² + 2xy + 4y²  =  15

Solution

We first write this as a matrix equation x^TAx  =  15.

 \( \begin{pmatrix} x & y    \end{pmatrix} \begin{pmatrix} 4 & 1  \\   1 & 4    \end{pmatrix}  \begin{pmatrix} x  \\   y    \end{pmatrix} = 15 \)

Next we find the eigenvalues of this matrix.  They are 3 and 5.  We used a calculator for this, but we could have done this by hand.  We have

 \( 3I - A = \begin{pmatrix} -1 & -1 \\ -1 & -1    \end{pmatrix} \)    \(  5I - A = \begin{pmatrix} 1 & -1 \\ -1 & 1    \end{pmatrix} \)   \)

The eigenvectors are

 \( v_3 = \begin{pmatrix} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}}    \end{pmatrix} \)    \(  v_5 = \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}    \end{pmatrix} \)       

The equation can be written in the form x^TPDP^Tx  =  15

 \(  \begin{pmatrix} x & y    \end{pmatrix}  \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}  \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix} \begin{pmatrix} 3 & 0  \\   0 & 5    \end{pmatrix}  \begin{pmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}  \\ -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \end{pmatrix}^T \begin{pmatrix} x  \\   y    \end{pmatrix} = 15\)   

Let

        x'  =  P^Tx 

Then the equation of the conic becomes

        x'^TDx'  =  15

or

        3x'² + 5y'²  =  15

or

            x'²        y'²  
                  +            =  15
            5         3

We see that the graph is rotated from the (x,y) plane by an angle of 

        arctan(-1)  =  -p/4

The graph is shown below.

Problem 9

One of the following is a subspace of the space of differentiable functions. 

            I.   {f | f(0) – f '(0)  =  1}          II. {f | f(1)  =  f '(1)} 

A.     Determine which is not a subspace and explain why.

Solution
 

The first one is not a subspace.  For example f(x)  =  -x is in the set, but 2f(x)  =  -2x is not.

B.     Prove that the other one is a subspace.

        If f(1)  =  f '(1) and g(1)  =  g '(1) then

        (f + g)(1)  =  f(1) + g(1)  =  f '(1) + g'(1)  =  (f + g)'(1)

        proving closure under "+".  Also

        (cf)(1)  =  c(f(1))  =  c(f '(1))  =  (cf)'(1)

        proving closure under " . "  Hence set II is a subspace of the space of differentiable functions.

Problem 10

Prove that if A is a matrix such that A² = 0 then 0 is an eigenvalue for A.

We need to show that there is a nonzero v with 

        Av  =  0v  =  0

This is true if and only if 

        det(A)  =  0

but 

        (det A)²  =  det(A²)  =  det(0)  =  0

hence det A  =  0, and we can conclude that 0 is an eigenvalue for A.

Problem 11

Answer the following true or false. If it is true, explain why.  If it is false explain why or provide a counter example.

A.    If S  =  {v₁, v₂} is a linear independent set of vectors in R³ and v₃ is not in the span of S, then  {v₁, v₂, v₃} is a basis for R³.
 

Solution

            True, since then dim(Span{v₁, v₂, v₃})  > dim(Span(S))  =  2.  Hence dim(Span{{v₁, v₂, v₃}} = 3.  Since any 3 vectors in R³ is a basis for R³.

B.     Every orthonormal set of five vectors in R⁵ is a basis for R⁵.

Solution

True, since orthonormal vectors are always linearly independent and five linearly independent vectors in R⁵ always form a basis for R⁵.
 

C.     Let A and B be matrices such that A²v = a, B²v = b, and ABv = c.  Then

(A + B)² v  =  a + b + 2c

Solution

False,

(A + B)²v  =  (A² + AB + BA + B²)v  =  A²v + ABv + BAv + B²v  =  a + b + c + BAv

So as long as ABv is not equal to BAv, the statement is false.  Remember that matrix multiplication does not enjoy the commutative property.