Math 310: Hour Exam 1 so (Solutions)

Prof. S. Smith: Fri 13 October 2000

so You must SHOW WORK to receive credit.

(If you used a calculator to do some details, write ``used calculator'' there).

Make sure to CHECK your answers when possible!

Problem 1: (a) Using either Gaussian or Gauss-Jordan elimination, find all solutions of the linear equation system Ax = bdetermined by the following augmented matrix: $(A\vert b) = \left( \begin{array}{rrr\vert r} 1 &-1 &0 &-1 \\ 2 &-1 &1 & 3 \\ 4 &-3 &1 & 0 \end{array} \right)$.

SHOW the steps of the method you use.

so Use row operations (Gauss-Jordan) $\stackrel{ A^{-2 \times 1}_2\ A^{-4 \times 1}_3 }{\rightarrow} \left( \begin{a... ...rrr\vert r} 1 & 0 &1 &4 \\ 0 & 1 &1 &5 \\ 0 & 0 &0 &-1 \end{array} \right)$

The bottom row says 0 = 1, so there are NO solutions.

(b) Which two row operations will bring the matrix $A = \left( \begin{array}{rrr} 1 &-4 &0 \\ 0 & 1 &1 \\ 0 & 5 &5 \end{array} \right)$into row-reduced echelon form U ? Give elementary row matrices accomplishing those operations by left multiplication; this, give E₁ and E₂ such that E₂ E₁ A = U.

so Row operations: $A^{4 \times 2}_1$ and $A^{-5 \times 2}_3$. Matrices: $E_1 = \left( \begin{array}{rrr} 1 & 4 &0 \\ 0 & 1 &0 \\ 0 & 0 &0 \end{array} \right)$and $E_2 = \left( \begin{array}{rrr} 1 & 0 &0 \\ 0 & 1 &0 \\ 0 &-5 &0 \end{array} \right)$

Problem 2: (a) Is the matrix $E_2 = \left( \begin{array}{rrr} 1 & 2 &3 \\ 2 & 3 &4 \\ 3 & 4 &5 \end{array} \right)$row-equivalent to the identity matrix I₃ ?

Say why/why not.

so No: for example, det(A) = 0, so A is singular; so the row-reduced echelon form of A

has a row of zeros (or, only 2 pivots), and hence cannot be the identity I₃.

(b) Given the augmented matrix $(A\vert b) = \left( \begin{array}{rr\vert r} 1 &2 &4 \\ 2 &3 &7\end{array} \right)$, find the inverse (by any method) of the coefficient matrix A; use A^-1to give a solution of the corresponding linear system Ax = b.

so By adjoint method, $A^{-1}= - \left( \begin{array}{rr} 3 &-2 \\ -2 & 1 \end{array} \right) = \left( \begin{array}{rr} -3& 2 \\ 2&-1 \end{array} \right)$, so $x = A^{-1}b = \left( \begin{array}{r} 2 \\ 1 \end{array} \right)$,

Problem 3: (a) Find the determinant of the product AB, where

$A = \left( \begin{array}{rrr} 1 &0 & 0 \\ 2 &2 & 0 \\ 3 &2 & 1 \end{array} \right)$and $B = \left( \begin{array}{rrr} 1 &5 & 4 \\ 0 &6 & 5 \\ 0 &0 & 1 \end{array} \right)$.

SHOW the steps you used (``calculator'' is not sufficient for credit here).

so As A is triangular, det(A) is the product down the diagonal, namely 1.2.1 = 2.

Similarly det(B) = 1.6.1 = 6. So det(AB) = det(A) det(B) = 2.6 = 12.

(b) For $A = \left( \begin{array}{rr} 2 &4 \\ 3 &5 \end{array} \right)$, find the adjoint adj(A), and then the inverse A^-1.

so We need the transpose of the matrix of cofactors, namely $adj(A) = \left( \begin{array}{rr} 5 &-4 \\ -3 &2 \end{array} \right)$.

Then $\det(A)=2.5-3.4=-2$, $A^{-1} = \frac{1}{det(A)} adj(A) = -\frac{1}{2}\left( \begin{array}{rr} 5 &-4 \\ -3 &2 \end{array} \right)$.

Problem 4: (a) In the space ${\bf R}^{2 \times 2}$ of all 2x2 matrices, show that the set of all upper-triangular matrices forms a subspace. What is the dimension of this subspace.

so (+) Add two general upper-triangular matrices: $\left( \begin{array}{rr} a &b \\ 0 &c \end{array} \right) + \left( \begin{a... ...} \right) = \left( \begin{array}{rr} a+d &b+e \\ 0 &c+f \end{array} \right)$;

so the sum is also upper-triangular.

(sc.mult.) Multiply a scalar times a general upper-triangular matrix: $f . \left( \begin{array}{rr} a &b \\ 0 &c \end{array} \right) = \left( \begin{array}{rr} fa &fb \\ 0 &fc \end{array} \right)$;

so the product is also upper-triangular.

Dimension: the ``free variables'' are a,b,c above, so the dimension is 3.

(b) What is the dimension of the span of the columns of the matrix $A = \left( \begin{array}{rrrr} 1 &1 & 3 & 0 \\ 0 &2 & 4 & 2 \\ 1 &3 & 7 & 2 \end{array} \right)$ ? (Explain how you know this is the dimension).

so The dimension is 2. One way: The row-reduced echelon form of A is $\left( \begin{array}{rrrr} 1 &0 & 1 &-1 \\ 0 &1 & 2 & 1 \\ 0 &0 & 0 & 0 \end{array} \right)$.

Only 2 pivots, so dim(col.space) = 2. The first two columns give one basis for the col.space).

Problem 5: (a) Find a basis for the row space of the matrix $A = \left( \begin{array}{rrrr}1 &1 &2 &-1 \\ 2 &3 &1 &-4 \\ 3 &4 &3 &-5 \end{array} \right)$.

(Say why you know your answer gives a basis for the row space).

so The row-reduced echelon form of A is $\left( \begin{array}{rrrr}1 &0 &5 & 1 \\ 0 &1 &-3 &-2 \\ 0 &0 &0 & 0 \end{array} \right)$.

Two pivots--so first two rows of rref give one basis for row space. Or, first two rows of A.

(b) Find the matrix of transition from the ``old'' basis given by the standard basis of ${\bf R}^2$ (namely (1,0)<sup>T</sup> and (0,1)<sup>T</sup>) to the ``new'' basis given by (3,5)^T and (1, 2)^T.

so One way: The matrix [new]_old is given by $\left( \begin{array}{rr} 3 &1 \\ 5 &2 \end{array} \right)$,

so the transition matrix [old]_new from old to new is given by its inverse, namely $\left( \begin{array}{rr} 2 &-1 \\ -5&3 \end{array} \right)$.

so What are the coordinates of (4,6)^T in this new basis?

so One way: Multiply transition matrix by old coordinates (4,6)^T to get new coordinates (2,-2)^T.