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QuestionMultiply the matrices S=(41−1−2)S=\begin{pmatrix} 4 & 1 \\ -1 & -2 \end{pmatrix} and T=(−50−43)T=\begin{pmatrix} -5 & 0 \\ -4 & 3 \end{pmatrix} to show ST≠TSS T \neq T S. Fill in the boxes: ST=____TS=____S T = \_\_\_\_ \quad T S = \_\_\_\_.

Studdy Solution

STEP 1

Assumptions1. The matrices and are given as=[41−1−] and =[−50−43]=\left[\begin{array}{cc} 4 &1 \\ -1 & -\end{array}\right] \text { and }=\left[\begin{array}{cc} -5 &0 \\ -4 &3\end{array}\right] . We need to find the product matrices $$ and $TS$.

STEP 2

First, let's calculate the product $$. The product of two matrices is calculated by taking the dot product of the rows of the first matrix with the columns of the second matrix.
=[41−1−2]×[−50−4] = \left[\begin{array}{cc} 4 &1 \\ -1 & -2\end{array}\right] \times \left[\begin{array}{cc} -5 &0 \\ -4 &\end{array}\right]

STEP 3

The first element of the product matrix isobtainedbymultiplyingtheelementsofthefirstrowof is obtained by multiplying the elements of the first row of with the elements of the first column of $$ and adding the results.
11=11⋅11+12⋅21=⋅−5+1⋅−_{11} =_{11} \cdot_{11} +_{12} \cdot_{21} = \cdot -5 +1 \cdot -

STEP 4

Calculate the value of 11_{11}.
11=4⋅−+1⋅−4=−20−4=−24_{11} =4 \cdot - +1 \cdot -4 = -20 -4 = -24

STEP 5

Similarly, calculate the rest of the elements of the product matrix $$.
12=11⋅12+12⋅22=4⋅0+1⋅3=3_{12} =_{11} \cdot_{12} +_{12} \cdot_{22} =4 \cdot0 +1 \cdot3 =321=21⋅11+22⋅21=−1⋅−5+−2⋅−4=5+8=13_{21} =_{21} \cdot_{11} +_{22} \cdot_{21} = -1 \cdot -5 + -2 \cdot -4 =5 +8 =1322=21⋅12+22⋅22=−1⋅0+−2⋅3=−_{22} =_{21} \cdot_{12} +_{22} \cdot_{22} = -1 \cdot0 + -2 \cdot3 = -

STEP 6

Now we have the product matrix $$.
=[−24313−6] = \left[\begin{array}{cc} -24 &3 \\ 13 & -6\end{array}\right]

STEP 7

Next, let's calculate the product TSTS. The product of two matrices is calculated by taking the dot product of the rows of the first matrix with the columns of the second matrix.
TS=[−50−43]×[41−1−2]TS = \left[\begin{array}{cc} -5 &0 \\ -4 &3\end{array}\right] \times \left[\begin{array}{cc} 4 &1 \\ -1 & -2\end{array}\right]

STEP 8

The first element of the product matrix TSTS is obtained by multiplying the elements of the first row of withtheelementsofthefirstcolumnof with the elements of the first column of and adding the results.
TS11=11⋅11+12⋅21=−5⋅4+0⋅−1=−20TS_{11} =_{11} \cdot_{11} +_{12} \cdot_{21} = -5 \cdot4 +0 \cdot -1 = -20

STEP 9

Similarly, calculate the rest of the elements of the product matrix TSTS.
TS12=11⋅12+12⋅22=−5⋅+⋅−2=−5TS_{12} =_{11} \cdot_{12} +_{12} \cdot_{22} = -5 \cdot + \cdot -2 = -5TS21=21⋅11+22⋅21=−4⋅4+3⋅−=−16−3=−19TS_{21} =_{21} \cdot_{11} +_{22} \cdot_{21} = -4 \cdot4 +3 \cdot - = -16 -3 = -19TS22=21⋅12+22⋅22=−4⋅+3⋅−2=−4−6=−TS_{22} =_{21} \cdot_{12} +_{22} \cdot_{22} = -4 \cdot +3 \cdot -2 = -4 -6 = -

STEP 10

Now we have the product matrix TSTS.
TS=[−20−5−19−10]TS = \left[\begin{array}{cc} -20 & -5 \\ -19 & -10\end{array}\right]

STEP 11

The product matrices and $TS$ are = \left[\begin{array}{cc} -24 &3 \\ 13 & -6\end{array}\right] \text { and } TS = \left[\begin{array}{cc} -20 & -5 \\ -19 & -10\end{array}\right] SinceSince is not equal to TSTS, we have shown that multiplication of matrices is not commutative.

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