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Math  /  Algebra

Questionatrices AA and BB shown below, find 5B+A5 B+A. A=[624−1]B=[14−4−2]A=\left[\begin{array}{cc} 6 & 2 \\ 4 & -1 \end{array}\right] \quad B=\left[\begin{array}{cc} 1 & 4 \\ -4 & -2 \end{array}\right]

Studdy Solution

STEP 1

What is this asking? We're taking two matrices, AA and BB, multiplying matrix BB by **5**, and then adding the result to matrix AA. Watch out! Remember, we can only add matrices if they have the same dimensions.
Also, multiplying a matrix by a number means multiplying *every* entry by that number!

STEP 2

1. Multiply BB by 5
2. Add the result to AA

STEP 3

Let's **multiply** each element of matrix BB by **5**.
This is called **scalar multiplication**!
We're basically scaling up the entire matrix.

STEP 4

5⋅B=5⋅[14−4−2]=[520−20−10]5 \cdot B = 5 \cdot \begin{bmatrix} 1 & 4 \\ -4 & -2 \end{bmatrix} = \begin{bmatrix} 5 & 20 \\ -20 & -10 \end{bmatrix} So, 5B=[520−20−10]5B = \begin{bmatrix} 5 & 20 \\ -20 & -10 \end{bmatrix}.
See how each number in BB got multiplied by **5**?
Awesome!

STEP 5

Now, let's **add** the result, which we called 5B5B, to matrix AA.
Remember, we add matrices element by element – matching positions!
Since AA and BB (and therefore 5B5B) have the same dimensions, we're good to go!

STEP 6

5B+A=[520−20−10]+[624−1]=[1122−16−11]5B + A = \begin{bmatrix} 5 & 20 \\ -20 & -10 \end{bmatrix} + \begin{bmatrix} 6 & 2 \\ 4 & -1 \end{bmatrix} = \begin{bmatrix} 11 & 22 \\ -16 & -11 \end{bmatrix}

STEP 7

Let's finish the calculation: [5+620+2−20+4−10+(−1)]=[1122−16−11]\begin{bmatrix} 5+6 & 20+2 \\ -20+4 & -10+(-1) \end{bmatrix} = \begin{bmatrix} 11 & 22 \\ -16 & -11 \end{bmatrix}

STEP 8

So, 5B+A=[1122−16−11]5B + A = \begin{bmatrix} 11 & 22 \\ -16 & -11 \end{bmatrix}!
We crushed it!

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