Solved on Dec 06, 2023

Solve the matrix equation for $Y$ . Simplify all elements. Find the size of $Y$ if the matrix exists, or select undefined.

STEP 1

Assumptions
1. We are given a matrix equation of the form $A = BY$ , where $A$ and $B$ are known matrices and $Y$ is the unknown matrix we need to solve for.
2. Matrix $A$ is given by $\left[\begin{array}{cc} -4 & 0 \\ -5 & -1 \end{array}\right]$ .
3. Matrix $B$ is given by $\left[\begin{array}{cc} 5 & 7 \\ -2 & -1 \end{array}\right]$ .
4. We assume that matrix $B$ is invertible, which will allow us to find $Y$ by multiplying both sides of the equation by $B^{-1}$ .

STEP 2

To solve for $Y$ , we need to isolate it on one side of the equation. This can be done by multiplying both sides of the equation by the inverse of matrix $B$ , denoted as $B^{-1}$ .
$B^{-1}A = B^{-1}BY$

STEP 3

Since $B^{-1}B$ is the identity matrix $I$ , the equation simplifies to:
$B^{-1}A = IY$

STEP 4

The identity matrix $I$ does not change matrix $Y$ , so we have:
$B^{-1}A = Y$

STEP 5

Before we can find $B^{-1}A$ , we need to calculate the inverse of matrix $B$ , $B^{-1}$ . The inverse of a $2 \times 2$ matrix $\left[\begin{array}{cc} a & b \\ c & d \end{array}\right]$ is given by:
$\frac{1}{ad - bc}\left[\begin{array}{cc} d & -b \\ -c & a \end{array}\right]$

STEP 6

First, we calculate the determinant of matrix $B$ , denoted as $det(B)$ .
$det(B) = ad - bc$

STEP 7

Plug in the values from matrix $B$ into the determinant formula.
$det(B) = (5)(-1) - (7)(-2)$

STEP 8

Calculate the determinant.
$det(B) = -5 + 14 = 9$

STEP 9

Since the determinant is not zero, matrix $B$ is invertible. Now we can find $B^{-1}$ by applying the formula from STEP_5.
$B^{-1} = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]$

STEP 10

Now we multiply $B^{-1}$ by matrix $A$ to find $Y$ .
$Y = B^{-1}A$

STEP 11

Perform the matrix multiplication.
$Y = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]\left[\begin{array}{cc} -4 & 0 \\ -5 & -1 \end{array}\right]$

STEP 12

Carry out the multiplication of the first row of $B^{-1}$ with the first column of $A$ .
$Y_{11} = \frac{1}{9}((-1)(-4) + (-7)(-5))$

STEP 13

Calculate the value of $Y_{11}$ .
$Y_{11} = \frac{1}{9}(4 + 35) = \frac{1}{9}(39)$

STEP 14

Carry out the multiplication of the first row of $B^{-1}$ with the second column of $A$ .
$Y_{12} = \frac{1}{9}((-1)(0) + (-7)(-1))$

STEP 15

Calculate the value of $Y_{12}$ .
$Y_{12} = \frac{1}{9}(0 + 7) = \frac{1}{9}(7)$

STEP 16

Carry out the multiplication of the second row of $B^{-1}$ with the first column of $A$ .
$Y_{21} = \frac{1}{9}((2)(-4) + (5)(-5))$

STEP 17

Calculate the value of $Y_{21}$ .
$Y_{21} = \frac{1}{9}(-8 - 25) = \frac{1}{9}(-33)$

STEP 18

Carry out the multiplication of the second row of $B^{-1}$ with the second column of $A$ .
$Y_{22} = \frac{1}{9}((2)(0) + (5)(-1))$

STEP 19

Calculate the value of $Y_{22}$ .
$Y_{22} = \frac{1}{9}(0 - 5) = \frac{1}{9}(-5)$

STEP 20

Now we can write down the matrix $Y$ with the calculated elements.
$Y = \left[\begin{array}{cc} \frac{39}{9} & \frac{7}{9} \\ \frac{-33}{9} & \frac{-5}{9} \end{array}\right]$

STEP 21

Simplify the elements of matrix $Y$ .
$Y = \left[\begin{array}{cc} \frac{39}{9} & \frac{7}{9} \\ \frac{-33}{9} & \frac{-5}{9} \end{array}\right] = \left[\begin{array}{cc} \frac{13}{3} & \frac{7}{9} \\ \frac{-11}{3} & \frac{-5}{9} \end{array}\right]$
The solution for matrix $Y$ is:
$Y = \left[\begin{array}{cc} \frac{13}{3} & \frac{7}{9} \\ \frac{-11}{3} & \frac{-5}{9} \end{array}\right]$

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Solve the matrix equation for YYY. Simplify all elements. Find the size of YYY if the matrix exists, or select undefined.

STEP 1

STEP 2

To solve for YYY, we need to isolate it on one side of the equation. This can be done by multiplying both sides of the equation by the inverse of matrix BBB, denoted as B−1B^{-1}B−1.B−1A=B−1BYB^{-1}A = B^{-1}BYB−1A=B−1BY

STEP 3

Since B−1BB^{-1}BB−1B is the identity matrix III, the equation simplifies to:B−1A=IYB^{-1}A = IYB−1A=IY

STEP 4

The identity matrix III does not change matrix YYY, so we have:B−1A=YB^{-1}A = YB−1A=Y

STEP 5

STEP 6

First, we calculate the determinant of matrix BBB, denoted as det(B)det(B)det(B).det(B)=ad−bcdet(B) = ad - bcdet(B)=ad−bc

STEP 7

Plug in the values from matrix BBB into the determinant formula.det(B)=(5)(−1)−(7)(−2)det(B) = (5)(-1) - (7)(-2)det(B)=(5)(−1)−(7)(−2)

STEP 8

Calculate the determinant.det(B)=−5+14=9det(B) = -5 + 14 = 9det(B)=−5+14=9

STEP 9

Since the determinant is not zero, matrix BBB is invertible. Now we can find B−1B^{-1}B−1 by applying the formula from STEP_5.B−1=19[−1−725]B^{-1} = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]B−1=91​[−12​−75​]

STEP 10

Now we multiply B−1B^{-1}B−1 by matrix AAA to find YYY.Y=B−1AY = B^{-1}AY=B−1A

STEP 11

Perform the matrix multiplication.Y=19[−1−725][−40−5−1]Y = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]\left[\begin{array}{cc} -4 & 0 \\ -5 & -1 \end{array}\right]Y=91​[−12​−75​][−4−5​0−1​]

STEP 12

Carry out the multiplication of the first row of B−1B^{-1}B−1 with the first column of AAA.Y11=19((−1)(−4)+(−7)(−5))Y_{11} = \frac{1}{9}((-1)(-4) + (-7)(-5))Y11​=91​((−1)(−4)+(−7)(−5))

STEP 13

Calculate the value of Y11Y_{11}Y11​.Y11=19(4+35)=19(39)Y_{11} = \frac{1}{9}(4 + 35) = \frac{1}{9}(39)Y11​=91​(4+35)=91​(39)

STEP 14

Carry out the multiplication of the first row of B−1B^{-1}B−1 with the second column of AAA.Y12=19((−1)(0)+(−7)(−1))Y_{12} = \frac{1}{9}((-1)(0) + (-7)(-1))Y12​=91​((−1)(0)+(−7)(−1))

STEP 15

Calculate the value of Y12Y_{12}Y12​.Y12=19(0+7)=19(7)Y_{12} = \frac{1}{9}(0 + 7) = \frac{1}{9}(7)Y12​=91​(0+7)=91​(7)

STEP 16

Carry out the multiplication of the second row of B−1B^{-1}B−1 with the first column of AAA.Y21=19((2)(−4)+(5)(−5))Y_{21} = \frac{1}{9}((2)(-4) + (5)(-5))Y21​=91​((2)(−4)+(5)(−5))

STEP 17

Calculate the value of Y21Y_{21}Y21​.Y21=19(−8−25)=19(−33)Y_{21} = \frac{1}{9}(-8 - 25) = \frac{1}{9}(-33)Y21​=91​(−8−25)=91​(−33)

STEP 18

Carry out the multiplication of the second row of B−1B^{-1}B−1 with the second column of AAA.Y22=19((2)(0)+(5)(−1))Y_{22} = \frac{1}{9}((2)(0) + (5)(-1))Y22​=91​((2)(0)+(5)(−1))

STEP 19

Calculate the value of Y22Y_{22}Y22​.Y22=19(0−5)=19(−5)Y_{22} = \frac{1}{9}(0 - 5) = \frac{1}{9}(-5)Y22​=91​(0−5)=91​(−5)

STEP 20

Now we can write down the matrix YYY with the calculated elements.Y=[39979−339−59]Y = \left[\begin{array}{cc} \frac{39}{9} & \frac{7}{9} \\ \frac{-33}{9} & \frac{-5}{9} \end{array}\right]Y=[939​9−33​​97​9−5​​]

STEP 21

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Solve the matrix equation for $Y$ . Simplify all elements. Find the size of $Y$ if the matrix exists, or select undefined.

To solve for $Y$ , we need to isolate it on one side of the equation. This can be done by multiplying both sides of the equation by the inverse of matrix $B$ , denoted as $B^{-1}$ .
$B^{-1}A = B^{-1}BY$

Since $B^{-1}B$ is the identity matrix $I$ , the equation simplifies to:
$B^{-1}A = IY$

The identity matrix $I$ does not change matrix $Y$ , so we have:
$B^{-1}A = Y$

First, we calculate the determinant of matrix $B$ , denoted as $det(B)$ .
$det(B) = ad - bc$

Plug in the values from matrix $B$ into the determinant formula.
$det(B) = (5)(-1) - (7)(-2)$

Calculate the determinant.
$det(B) = -5 + 14 = 9$

Since the determinant is not zero, matrix $B$ is invertible. Now we can find $B^{-1}$ by applying the formula from STEP_5.
$B^{-1} = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]$

Now we multiply $B^{-1}$ by matrix $A$ to find $Y$ .
$Y = B^{-1}A$

Perform the matrix multiplication.
$Y = \frac{1}{9}\left[\begin{array}{cc} -1 & -7 \\ 2 & 5 \end{array}\right]\left[\begin{array}{cc} -4 & 0 \\ -5 & -1 \end{array}\right]$

Carry out the multiplication of the first row of $B^{-1}$ with the first column of $A$ .
$Y_{11} = \frac{1}{9}((-1)(-4) + (-7)(-5))$

Calculate the value of $Y_{11}$ .
$Y_{11} = \frac{1}{9}(4 + 35) = \frac{1}{9}(39)$

Carry out the multiplication of the first row of $B^{-1}$ with the second column of $A$ .
$Y_{12} = \frac{1}{9}((-1)(0) + (-7)(-1))$

Calculate the value of $Y_{12}$ .
$Y_{12} = \frac{1}{9}(0 + 7) = \frac{1}{9}(7)$

Carry out the multiplication of the second row of $B^{-1}$ with the first column of $A$ .
$Y_{21} = \frac{1}{9}((2)(-4) + (5)(-5))$

Calculate the value of $Y_{21}$ .
$Y_{21} = \frac{1}{9}(-8 - 25) = \frac{1}{9}(-33)$

Carry out the multiplication of the second row of $B^{-1}$ with the second column of $A$ .
$Y_{22} = \frac{1}{9}((2)(0) + (5)(-1))$

Calculate the value of $Y_{22}$ .
$Y_{22} = \frac{1}{9}(0 - 5) = \frac{1}{9}(-5)$

Now we can write down the matrix $Y$ with the calculated elements.
$Y = \left[\begin{array}{cc} \frac{39}{9} & \frac{7}{9} \\ \frac{-33}{9} & \frac{-5}{9} \end{array}\right]$