QuestionLet $A=\begin{bmatrix}2 & 1 \\ -4 & -2\end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ . Show when $A\mathbf{x}=\mathbf{b}$ has no solution and describe valid $\mathbf{b}$ .

Studdy Solution

STEP 1

Assumptions1. The matrix $A$ is given by $\left[\begin{array}{rr} &1 \\ -4 & -\end{array}\right]$ . The vector $\mathbf{b}$ is given by $\left[\begin{array}{l}b_{1} \\ b_{}\end{array}\right]$
3. We need to find the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution

STEP 2

First, we need to row reduce the matrix $A$ to its row echelon form. This will allow us to see if $A$ has a pivot position in every row.
The row echelon form is a form in which all nonzero rows are above any rows of all zeros, and the leading coefficient (the first nonzero number from the left, also called the pivot) of a nonzero row is always strictly to the right of the pivot of the row above it.
$A = \left[\begin{array}{rr}2 &1 \\ -4 & -2\end{array}\right]$

STEP 3

To row reduce $A$ , we can multiply the first row by2 and add it to the second row.
$\left[\begin{array}{rr}2 &1 \\0 &0\end{array}\right]$

STEP 4

We can see that the row reduced form of $A$ does not have a pivot position in every row. This means that the system $A \mathbf{x}=\mathbf{b}$ does not have a solution for all $\mathbf{b}$ .

STEP 5

Now, let's describe the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution.
We know that a system of linear equations has a solution if and only if the rightmost column of the augmented matrix (the column of constants on the right side of the equal sign) is not a pivot column. This means that there is no leading1 in this column when the matrix is in row echelon form.

STEP 6

So, let's form the augmented matrix $[A \quad b]$ and row reduce it.
$[A \quad b] = \left[\begin{array}{rr|r}2 &1 & b_{1} \\ -4 & -2 & b_{2}\end{array}\right]$

STEP 7

Row reduce the augmented matrix by multiplying the first row by2 and adding it to the second row.
$\left[\begin{array}{rr|r}2 &1 & b_{1} \\0 &0 &2b_{1} - b_{2}\end{array}\right]$

STEP 8

From the row reduced form of the augmented matrix, we can see that the system $A \mathbf{x}=\mathbf{b}$ has a solution if and only if $2b_{1} - b_{2} =0$ . This is the condition for the rightmost column not to be a pivot column.
Therefore, the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution is given by all vectors $\mathbf{b} = \left[\begin{array}{l}b_{1} \\ b_{2}\end{array}\right]$ such that $2b_{1} - b_{2} =0$ .

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QuestionLet $A=\begin{bmatrix}2 & 1 \\ -4 & -2\end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ . Show when $A\mathbf{x}=\mathbf{b}$ has no solution and describe valid $\mathbf{b}$ .

Studdy Solution

STEP 1

Assumptions1. The matrix $A$ is given by $\left[\begin{array}{rr} &1 \\ -4 & -\end{array}\right]$ . The vector $\mathbf{b}$ is given by $\left[\begin{array}{l}b_{1} \\ b_{}\end{array}\right]$
3. We need to find the set of all $\mathbf{b}$ for which $A \mathbf{x}=\mathbf{b}$ does have a solution

STEP 2

STEP 3

To row reduce $A$ , we can multiply the first row by2 and add it to the second row.
$\left[\begin{array}{rr}2 &1 \\0 &0\end{array}\right]$

STEP 4

We can see that the row reduced form of $A$ does not have a pivot position in every row. This means that the system $A \mathbf{x}=\mathbf{b}$ does not have a solution for all $\mathbf{b}$ .

STEP 5

STEP 6

So, let's form the augmented matrix $[A \quad b]$ and row reduce it.
$[A \quad b] = \left[\begin{array}{rr|r}2 &1 & b_{1} \\ -4 & -2 & b_{2}\end{array}\right]$

STEP 7

Row reduce the augmented matrix by multiplying the first row by2 and adding it to the second row.
$\left[\begin{array}{rr|r}2 &1 & b_{1} \\0 &0 &2b_{1} - b_{2}\end{array}\right]$

STEP 8

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QuestionLet A=[21−4−2]A=\begin{bmatrix}2 & 1 \\ -4 & -2\end{bmatrix}A=[2−4​1−2​] and b=[b1b2]\mathbf{b}=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}b=[b1​b2​​]. Show when Ax=bA\mathbf{x}=\mathbf{b}Ax=b has no solution and describe valid b\mathbf{b}b.

Studdy Solution

STEP 1

STEP 2

STEP 3

To row reduce AAA, we can multiply the first row by2 and add it to the second row.[2100]\left[\begin{array}{rr}2 &1 \\0 &0\end{array}\right][20​10​]

STEP 4

We can see that the row reduced form of AAA does not have a pivot position in every row. This means that the system Ax=bA \mathbf{x}=\mathbf{b}Ax=b does not have a solution for all b\mathbf{b}b.

STEP 5

STEP 6

So, let's form the augmented matrix [Ab][A \quad b][Ab] and row reduce it.[Ab]=[21b1−4−2b2][A \quad b] = \left[\begin{array}{rr|r}2 &1 & b_{1} \\ -4 & -2 & b_{2}\end{array}\right][Ab]=[2−4​1−2​b1​b2​​]

STEP 7

Row reduce the augmented matrix by multiplying the first row by2 and adding it to the second row.[21b1002b1−b2]\left[\begin{array}{rr|r}2 &1 & b_{1} \\0 &0 &2b_{1} - b_{2}\end{array}\right][20​10​b1​2b1​−b2​​]

STEP 8

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QuestionLet $A=\begin{bmatrix}2 & 1 \\ -4 & -2\end{bmatrix}$ and $\mathbf{b}=\begin{bmatrix}b_{1} \\ b_{2}\end{bmatrix}$ . Show when $A\mathbf{x}=\mathbf{b}$ has no solution and describe valid $\mathbf{b}$ .

To row reduce $A$ , we can multiply the first row by2 and add it to the second row.
$\left[\begin{array}{rr}2 &1 \\0 &0\end{array}\right]$

We can see that the row reduced form of $A$ does not have a pivot position in every row. This means that the system $A \mathbf{x}=\mathbf{b}$ does not have a solution for all $\mathbf{b}$ .

So, let's form the augmented matrix $[A \quad b]$ and row reduce it.
$[A \quad b] = \left[\begin{array}{rr|r}2 &1 & b_{1} \\ -4 & -2 & b_{2}\end{array}\right]$

Row reduce the augmented matrix by multiplying the first row by2 and adding it to the second row.
$\left[\begin{array}{rr|r}2 &1 & b_{1} \\0 &0 &2b_{1} - b_{2}\end{array}\right]$