QuestionShow that matrix multiplication is not commutative by calculating $S T$ and $T S$ for the matrices $S=\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}$ and $T=\begin{bmatrix} 0 & 4 \\ -4 & 3 \end{bmatrix}$ .

Studdy Solution

STEP 1

Assumptions1. We are given twox matrices, and. . We are asked to show that matrix multiplication is not commutative, i.e., ≠ 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.
$= \left[\begin{array}{cc} 4 &1 \\ - & -2\end{array}\right] \left[\begin{array}{cc} 0 &4 \\ -4 &\end{array}\right]$

STEP 3

Calculate the elements of the resulting matrix. The element in the first row and first column is calculated by multiplying the elements of the first row of with the elements of the first column of and adding them together.
$()_{11} = ( \times0) + (1 \times -) = -$

STEP 4

Calculate the element in the first row and second column of the resulting matrix.
$()_{12} = (4 \times4) + (1 \times3) =19$

STEP 5

Calculate the element in the second row and first column of the resulting matrix.
$()_{21} = (-3 \times0) + (-2 \times -4) =8$

STEP 6

Calculate the element in the second row and second column of the resulting matrix.
$()_{22} = (-3 \times4) + (-2 \times3) = -18$

STEP 7

Combine the calculated elements to form the resulting matrix.
$= \left[\begin{array}{cc} -4 &19 \\ & -18\end{array}\right]$

STEP 8

Now, let's calculate the product TS.
$TS = \left[\begin{array}{cc} 0 &4 \\ -4 &3\end{array}\right] \left[\begin{array}{cc} 4 &1 \\ -3 & -2\end{array}\right]$

STEP 9

Calculate the elements of the resulting matrix. The element in the first row and first column is calculated by multiplying the elements of the first row of with the elements of the first column of and adding them together.
$(TS)_{11} = ( \times4) + (4 \times -3) = -12$

STEP 10

Calculate the element in the first row and second column of the resulting matrix.
$(TS)_{12} = (0 \times) + (4 \times -2) = -8$

STEP 11

Calculate the element in the second row and first column of the resulting matrix.
$(TS)_{21} = (-4 \times4) + (3 \times -3) = -25$

STEP 12

Calculate the element in the second row and second column of the resulting matrix.
$(TS)_{22} = (-4 \times) + ( \times -2) = -10$

STEP 13

Combine the calculated elements to form the resulting matrix TS.
$TS = \left[\begin{array}{cc} -12 & -8 \\ -25 & -10\end{array}\right]$

STEP 14

Now, compare the matrices and TS. They are not equal, so we have shown that matrix multiplication is not commutative.
$\neq TS$

Was this helpful?

QuestionShow that matrix multiplication is not commutative by calculating $S T$ and $T S$ for the matrices $S=\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}$ and $T=\begin{bmatrix} 0 & 4 \\ -4 & 3 \end{bmatrix}$ .

Studdy Solution

STEP 1

Assumptions1. We are given twox matrices, and. . We are asked to show that matrix multiplication is not commutative, i.e., ≠ 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.
$= \left[\begin{array}{cc} 4 &1 \\ - & -2\end{array}\right] \left[\begin{array}{cc} 0 &4 \\ -4 &\end{array}\right]$

STEP 3

Calculate the elements of the resulting matrix. The element in the first row and first column is calculated by multiplying the elements of the first row of with the elements of the first column of and adding them together.
$()_{11} = ( \times0) + (1 \times -) = -$

STEP 4

Calculate the element in the first row and second column of the resulting matrix.
$()_{12} = (4 \times4) + (1 \times3) =19$

STEP 5

Calculate the element in the second row and first column of the resulting matrix.
$()_{21} = (-3 \times0) + (-2 \times -4) =8$

STEP 6

Calculate the element in the second row and second column of the resulting matrix.
$()_{22} = (-3 \times4) + (-2 \times3) = -18$

STEP 7

Combine the calculated elements to form the resulting matrix.
$= \left[\begin{array}{cc} -4 &19 \\ & -18\end{array}\right]$

STEP 8

Now, let's calculate the product TS.
$TS = \left[\begin{array}{cc} 0 &4 \\ -4 &3\end{array}\right] \left[\begin{array}{cc} 4 &1 \\ -3 & -2\end{array}\right]$

STEP 9

Calculate the elements of the resulting matrix. The element in the first row and first column is calculated by multiplying the elements of the first row of with the elements of the first column of and adding them together.
$(TS)_{11} = ( \times4) + (4 \times -3) = -12$

STEP 10

Calculate the element in the first row and second column of the resulting matrix.
$(TS)_{12} = (0 \times) + (4 \times -2) = -8$

STEP 11

Calculate the element in the second row and first column of the resulting matrix.
$(TS)_{21} = (-4 \times4) + (3 \times -3) = -25$

STEP 12

Calculate the element in the second row and second column of the resulting matrix.
$(TS)_{22} = (-4 \times) + ( \times -2) = -10$

STEP 13

Combine the calculated elements to form the resulting matrix TS.
$TS = \left[\begin{array}{cc} -12 & -8 \\ -25 & -10\end{array}\right]$

STEP 14

Now, compare the matrices and TS. They are not equal, so we have shown that matrix multiplication is not commutative.
$\neq TS$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionShow that matrix multiplication is not commutative by calculating STS TST and TST STS for the matrices S=[41−3−2]S=\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}S=[4−3​1−2​] and T=[04−43]T=\begin{bmatrix} 0 & 4 \\ -4 & 3 \end{bmatrix}T=[0−4​43​].

Studdy Solution

STEP 1

Assumptions1. We are given twox matrices, and. . We are asked to show that matrix multiplication is not commutative, i.e., ≠ TS.

STEP 2

STEP 3

STEP 4

Calculate the element in the first row and second column of the resulting matrix.()12=(4×4)+(1×3)=19()_{12} = (4 \times4) + (1 \times3) =19()12​=(4×4)+(1×3)=19

STEP 5

Calculate the element in the second row and first column of the resulting matrix.()21=(−3×0)+(−2×−4)=8()_{21} = (-3 \times0) + (-2 \times -4) =8()21​=(−3×0)+(−2×−4)=8

STEP 6

Calculate the element in the second row and second column of the resulting matrix.()22=(−3×4)+(−2×3)=−18()_{22} = (-3 \times4) + (-2 \times3) = -18()22​=(−3×4)+(−2×3)=−18

STEP 7

Combine the calculated elements to form the resulting matrix.=[−419−18] = \left[\begin{array}{cc} -4 &19 \\ & -18\end{array}\right]=[−4​19−18​]

STEP 8

Now, let's calculate the product TS.TS=[04−43][41−3−2]TS = \left[\begin{array}{cc} 0 &4 \\ -4 &3\end{array}\right] \left[\begin{array}{cc} 4 &1 \\ -3 & -2\end{array}\right]TS=[0−4​43​][4−3​1−2​]

STEP 9

STEP 10

Calculate the element in the first row and second column of the resulting matrix.(TS)12=(0×)+(4×−2)=−8(TS)_{12} = (0 \times) + (4 \times -2) = -8(TS)12​=(0×)+(4×−2)=−8

STEP 11

Calculate the element in the second row and first column of the resulting matrix.(TS)21=(−4×4)+(3×−3)=−25(TS)_{21} = (-4 \times4) + (3 \times -3) = -25(TS)21​=(−4×4)+(3×−3)=−25

STEP 12

Calculate the element in the second row and second column of the resulting matrix.(TS)22=(−4×)+(×−2)=−10(TS)_{22} = (-4 \times) + ( \times -2) = -10(TS)22​=(−4×)+(×−2)=−10

STEP 13

Combine the calculated elements to form the resulting matrix TS.TS=[−12−8−25−10]TS = \left[\begin{array}{cc} -12 & -8 \\ -25 & -10\end{array}\right]TS=[−12−25​−8−10​]

STEP 14

Now, compare the matrices and TS. They are not equal, so we have shown that matrix multiplication is not commutative.≠TS \neq TS=TS

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

QuestionShow that matrix multiplication is not commutative by calculating $S T$ and $T S$ for the matrices $S=\begin{bmatrix} 4 & 1 \\ -3 & -2 \end{bmatrix}$ and $T=\begin{bmatrix} 0 & 4 \\ -4 & 3 \end{bmatrix}$ .

Calculate the element in the first row and second column of the resulting matrix.
$()_{12} = (4 \times4) + (1 \times3) =19$

Calculate the element in the second row and first column of the resulting matrix.
$()_{21} = (-3 \times0) + (-2 \times -4) =8$

Calculate the element in the second row and second column of the resulting matrix.
$()_{22} = (-3 \times4) + (-2 \times3) = -18$

Combine the calculated elements to form the resulting matrix.
$= \left[\begin{array}{cc} -4 &19 \\ & -18\end{array}\right]$

Now, let's calculate the product TS.
$TS = \left[\begin{array}{cc} 0 &4 \\ -4 &3\end{array}\right] \left[\begin{array}{cc} 4 &1 \\ -3 & -2\end{array}\right]$

Calculate the element in the first row and second column of the resulting matrix.
$(TS)_{12} = (0 \times) + (4 \times -2) = -8$

Calculate the element in the second row and first column of the resulting matrix.
$(TS)_{21} = (-4 \times4) + (3 \times -3) = -25$

Calculate the element in the second row and second column of the resulting matrix.
$(TS)_{22} = (-4 \times) + ( \times -2) = -10$

Combine the calculated elements to form the resulting matrix TS.
$TS = \left[\begin{array}{cc} -12 & -8 \\ -25 & -10\end{array}\right]$

Now, compare the matrices and TS. They are not equal, so we have shown that matrix multiplication is not commutative.
$\neq TS$