Solved on Nov 25, 2023

Find values of $r$ and $s$ that satisfy $3[2r, 5s] - [s, 3r] = [-18, 9]$ . If no solution exists, state "Not Possible".

STEP 1

Assumptions1. The equation is a system of linear equations represented in matrix form. . The variables are r and s.

STEP 2

First, let's expand the left side of the equation. $\left[\begin{array}{c} 2 r \\ 5 s\end{array}\right]-\left[\begin{array}{c} s \\ r\end{array}\right]=\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ r\end{array}\right]$

STEP 3

Next, subtract the second matrix from the first one.
$\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ 3 r\end{array}\right]=\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]$

STEP 4

Now, we equate the resulting matrix to the given matrix on the right side of the equation.
$\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]=\left[\begin{array}{c} -18 \\ 9\end{array}\right]$

STEP 5

This gives us two equations1. $r - s = -18$
2. $15s -3r =9$

STEP 6

Let's solve the first equation for s $s =6r +18$

STEP 7

Now, substitute $s$ from the first equation into the second equation $15(6r +18) -3r =9$

STEP 8

implify the equation $90r +270 -3r =$

STEP 9

Combine like terms $87r +270 =9$

STEP 10

Subtract270 from both sides $87r = -261$

STEP 11

Divide both sides by87 to solve for r $r = -261 /87$

STEP 12

Calculate the value of r $r = -$

STEP 13

Now, substitute $r = -3$ into the first equation to find the value of s $s =6(-3) +18$

STEP 14

Calculate the value of s $s = -18 +18 =0$
The values of the variables that make the equation true are $r = -3$ and $s =0$ .

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Find values of $r$ and $s$ that satisfy $3[2r, 5s] - [s, 3r] = [-18, 9]$ . If no solution exists, state "Not Possible".

STEP 1

Assumptions1. The equation is a system of linear equations represented in matrix form. . The variables are r and s.

STEP 2

First, let's expand the left side of the equation. $\left[\begin{array}{c} 2 r \\ 5 s\end{array}\right]-\left[\begin{array}{c} s \\ r\end{array}\right]=\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ r\end{array}\right]$

STEP 3

Next, subtract the second matrix from the first one.
$\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ 3 r\end{array}\right]=\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]$

STEP 4

Now, we equate the resulting matrix to the given matrix on the right side of the equation.
$\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]=\left[\begin{array}{c} -18 \\ 9\end{array}\right]$

STEP 5

This gives us two equations1. $r - s = -18$
2. $15s -3r =9$

STEP 6

Let's solve the first equation for s $s =6r +18$

STEP 7

Now, substitute $s$ from the first equation into the second equation $15(6r +18) -3r =9$

STEP 8

implify the equation $90r +270 -3r =$

STEP 9

Combine like terms $87r +270 =9$

STEP 10

Subtract270 from both sides $87r = -261$

STEP 11

Divide both sides by87 to solve for r $r = -261 /87$

STEP 12

Calculate the value of r $r = -$

STEP 13

Now, substitute $r = -3$ into the first equation to find the value of s $s =6(-3) +18$

STEP 14

Calculate the value of s $s = -18 +18 =0$
The values of the variables that make the equation true are $r = -3$ and $s =0$ .

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Find values of rrr and sss that satisfy 3[2r,5s]−[s,3r]=[−18,9]3[2r, 5s] - [s, 3r] = [-18, 9]3[2r,5s]−[s,3r]=[−18,9]. If no solution exists, state "Not Possible".

STEP 1

Assumptions1. The equation is a system of linear equations represented in matrix form. . The variables are r and s.

STEP 2

STEP 3

Next, subtract the second matrix from the first one.[6r15s]−[s3r]=[6r−s15s−3r]\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ 3 r\end{array}\right]=\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right] [6r15s​]−[s3r​]=[6r−s15s−3r​]

STEP 4

Now, we equate the resulting matrix to the given matrix on the right side of the equation.[6r−s15s−3r]=[−189]\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]=\left[\begin{array}{c} -18 \\ 9\end{array}\right] [6r−s15s−3r​]=[−189​]

STEP 5

This gives us two equations1. r−s=−18r - s = -18r−s=−182. 15s−3r=915s -3r =915s−3r=9

STEP 6

Let's solve the first equation for ss=6r+18s =6r +18s=6r+18

STEP 7

Now, substitute sss from the first equation into the second equation15(6r+18)−3r=915(6r +18) -3r =915(6r+18)−3r=9

STEP 8

implify the equation90r+270−3r=90r +270 -3r =90r+270−3r=

STEP 9

Combine like terms87r+270=987r +270 =987r+270=9

STEP 10

Subtract270 from both sides87r=−26187r = -26187r=−261

STEP 11

Divide both sides by87 to solve for rr=−261/87r = -261 /87r=−261/87

STEP 12

Calculate the value of rr=−r = -r=−

STEP 13

Now, substitute r=−3r = -3r=−3 into the first equation to find the value of ss=6(−3)+18s =6(-3) +18s=6(−3)+18

STEP 14

Calculate the value of ss=−18+18=0s = -18 +18 =0s=−18+18=0The values of the variables that make the equation true are r=−3r = -3r=−3 and s=0s =0s=0.

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Find values of $r$ and $s$ that satisfy $3[2r, 5s] - [s, 3r] = [-18, 9]$ . If no solution exists, state "Not Possible".

Next, subtract the second matrix from the first one.
$\left[\begin{array}{c} 6 r \\ 15 s\end{array}\right]-\left[\begin{array}{c} s \\ 3 r\end{array}\right]=\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]$

Now, we equate the resulting matrix to the given matrix on the right side of the equation.
$\left[\begin{array}{c} 6 r - s \\ 15 s -3 r\end{array}\right]=\left[\begin{array}{c} -18 \\ 9\end{array}\right]$

This gives us two equations1. $r - s = -18$
2. $15s -3r =9$

Let's solve the first equation for s $s =6r +18$

Now, substitute $s$ from the first equation into the second equation $15(6r +18) -3r =9$

implify the equation $90r +270 -3r =$

Combine like terms $87r +270 =9$

Subtract270 from both sides $87r = -261$

Divide both sides by87 to solve for r $r = -261 /87$

Calculate the value of r $r = -$

Now, substitute $r = -3$ into the first equation to find the value of s $s =6(-3) +18$

Calculate the value of s $s = -18 +18 =0$
The values of the variables that make the equation true are $r = -3$ and $s =0$ .