Fall 2001  Math 2090  Exam 2

1. Consider the consistent system of linear equations, with coefficient matrix A:

\begin{aligned}x_{1} & − & 2x_{2} & + & x_{3} & − & x_{4} & = & 1 \\ 3x_{1} & − & 6x_{2} & + & x_{3} & + & x_{4} & = & 5 \end{aligned} A = [ \begin{matrix}1 & −2 & 1 & −1 \\ 3 & −6 & 1 & 1 \end{matrix} ]

(a) Find the solution set of this system of linear equations.

The solution set is {(2s − t + 2, s, 2t − 1, t) : s, t any real numbers}

(b) For this matrix A, is there a 2 × 1 constant vector b⃗ for which the system of linear equations A x⃗ = b⃗ is inconsistent? Explain.

A is 2 × 4 and rank A = 2, so for any b⃗, rank A = rank A^{#} = 2, and A x⃗ = b⃗ is consistent.

2. Consider the matrix A = [ \begin{matrix}1 & 1 & 0 & 2 \\ 2 & 3 & 1 & 5 \\ 0 & 1 & 1 & 1 \\ 2 & 4 & 2 & 6 \end{matrix} ] \begin{matrix}⟶ \\ R2→R2−2R1 \\ R4→R4−2R1 \end{matrix} [ \begin{matrix}1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 1 & 1 & 1 \\ 0 & 2 & 2 & 2 \end{matrix} ] \begin{matrix}⟶ \\ R3→R3−R2 \\ R4→R4−2R2 \end{matrix}  [ \begin{matrix}1 & 1 & 0 & 2 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{matrix} ] = U

(a) Find a row echelon matrix that is row equivalent to A, and find the rank of A.

The row echelon matrix U given above is row equivalent to A. So rank A = 2.

Answer the following questions based (only) on your calculations in part (a). Give a brief explanation with each answer.

(b) Is the matrix A non-singular?  No, since A is 4 × 4 and rank A = 2 < 4, A is singular.

(c) Is the determinant of A equal to zero, or different from zero?

det A = 0. Since A and U are row equivalent, det A = k det U for some non-zero scalar k. U is upper triangular, so det U = 1 · 1 · 0 · 0 = 0 is the product of the diagonal entries.

(d) Does the homogeneous system of linear equations A x⃗ = 0 have a unique solution?

A x⃗ = 0 has infinitely many solutions. Since rank A = 2 < 4, there are two free variables.

3. [35 points] Let A = [ \begin{matrix}0 & 2 & 1 \\ 1 & −1 & 1 \\ 1 & 0 & 1 \end{matrix} ] and b⃗ = [ \begin{matrix}7 \\ 2 \\ 4 \end{matrix} ]. The inverse of A is A^{−1} = [ \begin{matrix}−1 & −2 & 3 \\ 0 & −1 & 1 \\ 1 & 2 & −2 \end{matrix} ].

(a) Use Gauss-Jordan elimination to calculate the inverse of the matrix A by hand.

The following sequence of row operations will reduce [A⋮I] to [I⋮A^{−1}]:

R1 ↔ R3, R2 → R2 − R1, R3 → R3 + 2R2, R1 → R1 − R3, R2 → (−1)R2

(b) Use the inverse of the matrix A to solve the system of linear equations A x⃗ = b⃗. x⃗ = A^{−1}b⃗ = [ 1 2 3 ]^{T}

(c) Use cofactor expansion (on any row or column) to calculate the determinant of A.

det A = (−1)^{2+1} | \begin{matrix}2 & 1 \\ 0 & 1 \end{matrix} | + (−1)^{3+1} |\begin{matrix}2 & 1 \\ −1 & 1 \end{matrix} | = −(2 − 0) + (2 + 1) = 1 (expanding on column one)

4. [20 points] Let B = [ \begin{matrix}1 & 3 & 5 \\ 2 & 4 & 6 \end{matrix} ] and C = [ \begin{matrix}0 & 1 \\ −1 & 1 \end{matrix} ].

In parts (a), (b), and (c), either carry out the indicated calculation, or state that it is not defined.

(a)  BC is not defined  (b)  CB = [ \begin{matrix}2 & 4 & 6 \\ 1 & 1 & 1 \end{matrix} ] (c)  C^{T} + 3C = [\begin{matrix}0 & 2 \\ −2 & 4 \end{matrix} ]

(d) A square matrix A is called orthogonal if the transpose of A is equal to the inverse of A, that is, A^{T} = A^{−1}. Is the 2 × 2 matrix C above orthogonal? Explain.

C^{T} C = [ \begin{matrix}0 & −1 \\ 1 & 1 \end{matrix} ] [ \begin{matrix}0 & 1 \\ −1 & 1 \end{matrix} ] = [\begin{matrix}1 & −1 \\ −1 & 2 \end{matrix} ] ≠ [ \begin{matrix}1 & 0 \\ 0 & 1 \end{matrix} ] = I, so C^{T} ≠ C^{−1} and C is not orthogonal

Extra Credit. If A is an orthogonal matrix, what is the determinant of A?

If A^{T} = A^{−1}, then A^{T} A = I so det A^{T} det A = det I = 1. But det A^{T} = det A, so det A^{T} det A = det A det A = (det A)^{2} = 1 and det A = ±1.