Math / Algebra

Question $\text { Let } A=\left(\begin{array}{ccc} 1 & -1 & -1 \\ -1 & 0 & 1 \end{array}\right)$
The matrix transformation associated to $A$ is $T_{A}: \mathbb{R}^{3} \rightarrow \mathbb{R}^{2}$ defined by the formula $T_{A}\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right)=\left[\begin{array}{l} \square \\ \square \end{array}\right]$
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Studdy Solution

STEP 1

What is this asking? We're asked to find the formula for a matrix transformation $T_A$ given its matrix $A$ .
Basically, we need to figure out what $T_A$ does to a generic vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ . Watch out! Matrix multiplication can be tricky!
Make sure you're multiplying the matrix $A$ by the vector correctly – rows of the matrix times the column of the vector.

STEP 2

1. Perform Matrix-Vector Multiplication

STEP 3

Alright, let's multiply our matrix $A = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 0 & 1 \end{bmatrix}$ by the vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ .
Remember, it's rows of $A$ times the column of our vector!

STEP 4

For the first entry in the resulting vector, we multiply the first row of $A$ by the vector: $(1) \cdot (x) + (-1) \cdot (y) + (-1) \cdot (z) = x - y - z.$ So, our first entry is $x - y - z$ .

STEP 5

Now, for the second entry, we multiply the second row of $A$ by the vector: $(-1) \cdot (x) + (0) \cdot (y) + (1) \cdot (z) = -x + z.$ Our second entry is $-x + z$ .

STEP 6

Putting it all together, we get: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$

STEP 7

So, the transformation $T_A$ is defined by: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$

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Studdy Solution

STEP 1

STEP 2

1. Perform Matrix-Vector Multiplication

STEP 3

Alright, let's multiply our matrix $A = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 0 & 1 \end{bmatrix}$ by the vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ .
Remember, it's rows of $A$ times the column of our vector!

STEP 4

For the first entry in the resulting vector, we multiply the first row of $A$ by the vector: $(1) \cdot (x) + (-1) \cdot (y) + (-1) \cdot (z) = x - y - z.$ So, our first entry is $x - y - z$ .

STEP 5

Now, for the second entry, we multiply the second row of $A$ by the vector: $(-1) \cdot (x) + (0) \cdot (y) + (1) \cdot (z) = -x + z.$ Our second entry is $-x + z$ .

STEP 6

Putting it all together, we get: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$

STEP 7

So, the transformation $T_A$ is defined by: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$

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Studdy Solution

STEP 1

STEP 2

1. Perform Matrix-Vector Multiplication

STEP 3

Alright, let's **multiply** our matrix A=[1−1−1−101]A = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 0 & 1 \end{bmatrix}A=[1−1​−10​−11​] by the vector [xyz]\begin{bmatrix} x \\ y \\ z \end{bmatrix}⎣⎡​xyz​⎦⎤​.Remember, it's rows of AAA times the column of our vector!

STEP 4

STEP 5

Now, for the **second entry**, we multiply the second row of AAA by the vector: (−1)⋅(x)+(0)⋅(y)+(1)⋅(z)=−x+z. (-1) \cdot (x) + (0) \cdot (y) + (1) \cdot (z) = -x + z. (−1)⋅(x)+(0)⋅(y)+(1)⋅(z)=−x+z. Our **second entry** is −x+z-x + z−x+z.

STEP 6

Putting it all together, we get: TA([xyz])=[x−y−z−x+z] T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix} TA​⎝⎛​⎣⎡​xyz​⎦⎤​⎠⎞​=[x−y−z−x+z​]

STEP 7

So, the transformation TAT_ATA​ is defined by: TA([xyz])=[x−y−z−x+z] T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix} TA​⎝⎛​⎣⎡​xyz​⎦⎤​⎠⎞​=[x−y−z−x+z​]

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Alright, let's multiply our matrix $A = \begin{bmatrix} 1 & -1 & -1 \\ -1 & 0 & 1 \end{bmatrix}$ by the vector $\begin{bmatrix} x \\ y \\ z \end{bmatrix}$ .
Remember, it's rows of $A$ times the column of our vector!

Now, for the second entry, we multiply the second row of $A$ by the vector: $(-1) \cdot (x) + (0) \cdot (y) + (1) \cdot (z) = -x + z.$ Our second entry is $-x + z$ .

Putting it all together, we get: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$

So, the transformation $T_A$ is defined by: $T_A\left(\begin{bmatrix} x \\ y \\ z \end{bmatrix}\right) = \begin{bmatrix} x - y - z \\ -x + z \end{bmatrix}$