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Math  /  Algebra

QuestionQuestion 4
Compute 3 A−4I23 \mathrm{~A}-4 \mathrm{I}_{2} using the following matrix, or state that the indicted operation is undefined. A=[108−5−9]A=\left[\begin{array}{cc} 10 & 8 \\ -5 & -9 \end{array}\right] 3A−4I23 A-4 I_{2} is undefined 3A−4I2=[2620−19−31]3 A-4 I_{2}=\left[\begin{array}{cc}26 & 20 \\ -19 & -31\end{array}\right] 3A−4I2=[2624−15−31]3 A-4 I_{2}=\left[\begin{array}{cc}26 & 24 \\ -15 & -31\end{array}\right] 3A−4I2=[2724−15−30]3 A-4 I_{2}=\left[\begin{array}{cc}27 & 24 \\ -15 & -30\end{array}\right]

Studdy Solution

STEP 1

What is this asking? We're taking a matrix AA, multiplying it by 3, then subtracting 4 times the 2x2 identity matrix, and seeing what we get! Watch out! Matrix multiplication and subtraction aren't always possible, so we need to make sure the dimensions match up!
Also, don't forget what the identity matrix looks like!

STEP 2

1. Define the Matrices
2. Calculate 3A3A
3. Calculate 4I24I_2
4. Calculate 3A−4I23A - 4I_2

STEP 3

We're given the matrix AA: A=[108−5−9]A = \begin{bmatrix} 10 & 8 \\ -5 & -9 \end{bmatrix} This is a **2x2 matrix**, meaning it has **2 rows** and **2 columns**.

STEP 4

The 2x2 identity matrix, I2I_2, looks like this: I2=(1001)I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} It's also a **2x2 matrix**!
This is important because we can only add or subtract matrices if they have the same dimensions.
We're in the clear!

STEP 5

To calculate 3A3A, we **multiply each element** of matrix AA by **3**.
It's like giving each number in the matrix a boost! 3A=3[108−5−9]=[3024−15−27]3A = 3 \begin{bmatrix} 10 & 8 \\ -5 & -9 \end{bmatrix} = \begin{bmatrix} 30 & 24 \\ -15 & -27 \end{bmatrix}

STEP 6

Let's do the multiplication: 3A=[3024−15−27]3A = \begin{bmatrix} 30 & 24 \\ -15 & -27 \end{bmatrix} Look at those shiny new numbers!

STEP 7

Now, let's calculate 4I24I_2.
We multiply each element of the identity matrix I2I_2 by **4**: 4I2=4[1001]=[4⋅14⋅04⋅04⋅1]4I_2 = 4 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 4 \cdot 1 & 4 \cdot 0 \\ 4 \cdot 0 & 4 \cdot 1 \end{bmatrix}

STEP 8

After performing the multiplications, we get: 4I2=[4004]4I_2 = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} The identity matrix got a makeover!

STEP 9

Finally, we **subtract** the elements of 4I24I_2 from the corresponding elements of 3A3A.
It's like a head-to-head matchup between the matrices! 3A−4I2=[3024−15−27]−[4004]=[2624−15−31]3A - 4I_2 = \begin{bmatrix} 30 & 24 \\ -15 & -27 \end{bmatrix} - \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} = \begin{bmatrix} 26 & 24 \\ -15 & -31 \end{bmatrix}

STEP 10

Let's wrap this up with the final calculations: 3A−4I2=[2624−15−31]3A - 4I_2 = \begin{bmatrix} 26 & 24 \\ -15 & -31 \end{bmatrix} Boom! There's our final matrix!

STEP 11

So, the answer is [2624−15−31] \begin{bmatrix} 26 & 24 \\ -15 & -31 \end{bmatrix} .

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