CS223 // Major Exam 1 // Term 251
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Major 1 (Term 251)
Echelon forms, linear combinations, parametric solutions, independence, linear transformations, inverses and LU factorization.
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Question 1: Choose the correct answer 4 PTS
Consider the augmented matrix$$\left[\begin{array}{ccccc|c}2 & 5 & 3 & 8 & 7 & 6 \\ 0 & 5 & 7 & 4 & 2 & 3 \\ 0 & 0 & 0 & 8 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0\end{array}\right]$$
1
Which statement describes the matrix?
1 pt
Worked solution
Zero rows are at the bottom and each leading entry sits to the right of the one above (columns 1, 2, 4).
It is not reduced : the leading entries are 2, 5, 8 rather than 1, and there are nonzero entries above them.
Answer: Echelon form (b).
2
About the variables of the system the matrix represents:
1 pt
Worked solution
Five variable columns; pivots in columns 1, 2 and 4.
Basic: $x_1, x_2, x_4$ (three of them). Free: $x_3, x_5$ (two of them). None of (a) to (c) is true.
Answer: None of the above (d).
3
About the solutions:
1 pt
Worked solution
No row of the form $[0 \ \cdots \ 0 \mid c]$ with $c \neq 0$, so the system is consistent.
Consistent with free variables means infinitely many solutions.
Answer: Infinitely many solutions (c).
4
About the constants:
1 pt
Worked solution
The last column (constants) is $(6, 3, 0, 0)$, which is not all zero.
Answer: Non-homogeneous (b).
Question 2 8 PTS
1
Let $\mathbf u = \begin{bmatrix}1 \\ 2 \\ 0\end{bmatrix}$, $\mathbf v = \begin{bmatrix}-1 \\ 1 \\ 1\end{bmatrix}$ and $\mathbf w = \begin{bmatrix}-1 \\ 7 \\ 3\end{bmatrix}$. Find scalars $a$ and $b$ such that $\mathbf w = a\mathbf u + b\mathbf v$.
2 pts
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Worked solution
Solve $\begin{bmatrix}\mathbf u & \mathbf v\end{bmatrix}\begin{bmatrix}a\\b\end{bmatrix} = \mathbf w$: $$\left[\begin{array}{cc|c}1 & -1 & -1 \\ 2 & 1 & 7 \\ 0 & 1 & 3\end{array}\right]$$
Row 3 gives $b = 3$. Row 1: $a - 3 = -1$, so $a = 2$. Row 2 checks: $2(2) + 3 = 7$.
Answer: $\mathbf w = 2\mathbf u + 3\mathbf v$
2
Describe the solutions of the system in parametric vector form and give one solution.$$\begin{cases}x + y + 12z = 1 \\ x + 2y + 9z = -1\end{cases}$$
2.5 pts
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Worked solution
$R_2 \leftarrow R_2 - R_1$: $(0, 1, -3 \mid -2)$. Then $R_1 \leftarrow R_1 - R_2$: $$\left[\begin{array}{ccc|c}1 & 0 & 15 & 3 \\ 0 & 1 & -3 & -2\end{array}\right]$$
$z$ is free: $x = 3 - 15z$, $y = -2 + 3z$.
$$\mathbf x = \begin{bmatrix}3 \\ -2 \\ 0\end{bmatrix} + z\begin{bmatrix}-15 \\ 3 \\ 1\end{bmatrix},\quad z \in \mathbb R$$
Answer: One solution ($z = 0$): $(3, -2, 0)$. Another ($z = 1$): $(-12, 1, 1)$.
3
Given that$$\left[\begin{array}{ccc|c}1 & 2 & 3 & 0 \\ 5 & 2 & 8 & 0 \\ 1 & 1 & 9 & 0\end{array}\right]\sim\left[\begin{array}{ccc|c}1 & 2 & 3 & 0 \\ 0 & -8 & -7 & 0 \\ 0 & 0 & 55 & 0\end{array}\right]$$is the set $\left\{\begin{bmatrix}1 \\ 5 \\ 1\end{bmatrix},\begin{bmatrix}2 \\ 2 \\ 1\end{bmatrix},\begin{bmatrix}3 \\ 8 \\ 9\end{bmatrix}\right\}$ linearly independent?
1.5 pts
Worked solution
The echelon form has a pivot in every one of the three columns, so there is no free variable.
The homogeneous equation $A\mathbf x = \mathbf 0$ has only the trivial solution.
Answer: Yes, the vectors are linearly independent.
4a
$T:\mathbb R^2 \to \mathbb R^2$ has standard matrix $\begin{bmatrix}1 & 2 \\ 1 & h\end{bmatrix}$. Find $h$ so that $T\!\left(\begin{bmatrix}2 \\ 3\end{bmatrix}\right) = \begin{bmatrix}8 \\ 11\end{bmatrix}$.
1 pt
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Worked solution
$\begin{bmatrix}1 & 2 \\ 1 & h\end{bmatrix}\begin{bmatrix}2 \\ 3\end{bmatrix} = \begin{bmatrix}2 + 6\\ 2 + 3h\end{bmatrix}$.
The first entry is 8 for any $h$. The second needs $2 + 3h = 11$, so $h = 3$.
Answer: $h = 3$
4b
For the same $T$, find the value(s) of $h$ for which $T$ maps $\mathbb R^2$ onto $\mathbb R^2$.
1 pt
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Worked solution
$T$ is onto exactly when its standard matrix has a pivot in every row, i.e. $\det \neq 0$.
$\det\begin{bmatrix}1 & 2 \\ 1 & h\end{bmatrix} = h - 2$.
Answer: Onto for every $h \neq 2$.
Question 3 8 PTS
1
Find the inverse of $A = \begin{bmatrix}1 & 2 \\ 3 & 7\end{bmatrix}$.
1.5 pts
Four numbers, e.g. a, b, c, d for the matrix [a b; c d].
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Worked solution
$\det A = 1\cdot 7 - 2 \cdot 3 = 1$.
$A^{-1} = \dfrac{1}{\det A}\begin{bmatrix}7 & -2 \\ -3 & 1\end{bmatrix} = \begin{bmatrix}7 & -2 \\ -3 & 1\end{bmatrix}$.
Answer: $A^{-1} = \begin{bmatrix}7 & -2 \\ -3 & 1\end{bmatrix}$
2
Use the inverse from part 1 to solve$$\begin{cases}x + 2y = 5 \\ 3x + 7y = 12\end{cases}$$
1.5 pts
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Worked solution
$\mathbf x = A^{-1}\mathbf b = \begin{bmatrix}7 & -2 \\ -3 & 1\end{bmatrix}\begin{bmatrix}5 \\ 12\end{bmatrix} = \begin{bmatrix}35-24 \\ -15+12\end{bmatrix} = \begin{bmatrix}11 \\ -3\end{bmatrix}$.
Answer: $x = 11$, $y = -3$
3
Given the LU factorization of $A = \begin{bmatrix}2 & 3 \\ 4 & 5\end{bmatrix}$ with $L = \begin{bmatrix}1 & 0 \\ 2 & 1\end{bmatrix}$ and $U = \begin{bmatrix}2 & 3 \\ 0 & -1\end{bmatrix}$, use it to solve $A\mathbf x = \mathbf b$ where $\mathbf b = \begin{bmatrix}5 \\ 11\end{bmatrix}$.
2 pts
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Worked solution
Forward substitution, $L\mathbf y = \mathbf b$: $y_1 = 5$, then $2(5) + y_2 = 11$, so $\mathbf y = (5, 1)$.
Back substitution, $U\mathbf x = \mathbf y$: $-x_2 = 1 \Rightarrow x_2 = -1$; $2x_1 + 3(-1) = 5 \Rightarrow x_1 = 4$.
Check: $A\mathbf x = (8 - 3,\ 16 - 5) = (5, 11)$.
Answer: $\mathbf x = (4, -1)$
4a
Let $A = \begin{bmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{bmatrix}$. Find $A^TA$.
1 pt
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Worked solution
$A^TA = \begin{bmatrix}1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}\begin{bmatrix}1 & 0 \\ 1 & 1 \\ 0 & 1\end{bmatrix} = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$.
Answer: $A^TA = \begin{bmatrix}2 & 1 \\ 1 & 2\end{bmatrix}$
4b
With $B = \begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix}$, find the matrix $X$ such that $A^TA\,X = B$.
2 pts
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Worked solution
$\det(A^TA) = 4 - 1 = 3$, so $(A^TA)^{-1} = \tfrac13\begin{bmatrix}2 & -1 \\ -1 & 2\end{bmatrix}$.
$X = (A^TA)^{-1}B = \tfrac13\begin{bmatrix}2 & -1 \\ -1 & 2\end{bmatrix}\begin{bmatrix}2 & 3 \\ 1 & 2\end{bmatrix} = \tfrac13\begin{bmatrix}3 & 4 \\ 0 & 1\end{bmatrix}$.
Answer: $X = \begin{bmatrix}1 & \tfrac43 \\ 0 & \tfrac13\end{bmatrix}$
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