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To find the values of [tex]\( m \)[/tex] and [tex]\( n \)[/tex] such that the matrices [tex]\(\begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix}\)[/tex] and [tex]\(\begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix}\)[/tex] are inverses of each other, we need to use the property of inverse matrices: the product of a matrix and its inverse is the identity matrix.
If [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are inverse matrices, then:
[tex]\[ AB = I \][/tex]
where [tex]\(I\)[/tex] is the identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
Given:
[tex]\[ A = \begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix} \][/tex]
We calculate the product [tex]\( AB \)[/tex] and set it equal to the identity matrix:
[tex]\[ AB = \begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix} \begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix} \][/tex]
Perform the multiplication:
[tex]\[ AB = \begin{pmatrix} (2m \cdot 9 + 7 \cdot (-5)) & (2m \cdot n + 7 \cdot 4) \\ (5 \cdot 9 + 9 \cdot (-5)) & (5 \cdot n + 9 \cdot 4) \end{pmatrix} \][/tex]
Simplify each element:
[tex]\[ AB = \begin{pmatrix} (18m - 35) & (2mn + 28) \\ (45 - 45) & (5n + 36) \end{pmatrix} \][/tex]
Since [tex]\( AB \)[/tex] must be equal to the identity matrix, we get:
[tex]\[ \begin{pmatrix} 18m - 35 & 2mn + 28 \\ 0 & 5n + 36 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
This gives us the following system of equations:
1. [tex]\( 18m - 35 = 1 \)[/tex]
2. [tex]\( 2mn + 28 = 0 \)[/tex]
3. [tex]\( 5n + 36 = 1 \)[/tex]
Solve the first equation for [tex]\( m \)[/tex]:
[tex]\[ 18m - 35 = 1 \][/tex]
[tex]\[ 18m = 36 \][/tex]
[tex]\[ m = 2 \][/tex]
Solve the third equation for [tex]\( n \)[/tex]:
[tex]\[ 5n + 36 = 1 \][/tex]
[tex]\[ 5n = -35 \][/tex]
[tex]\[ n = -7 \][/tex]
Verify the second equation with these values for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ 2mn + 28 = 2(2)(-7) + 28 = -28 + 28 = 0 \][/tex]
Thus, the values that satisfy all conditions are:
[tex]\[ m = 2 \][/tex]
[tex]\[ n = -7 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ m = 2, n = -7 \][/tex]
If [tex]\(A\)[/tex] and [tex]\(B\)[/tex] are inverse matrices, then:
[tex]\[ AB = I \][/tex]
where [tex]\(I\)[/tex] is the identity matrix:
[tex]\[ I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
Given:
[tex]\[ A = \begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix} \][/tex]
[tex]\[ B = \begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix} \][/tex]
We calculate the product [tex]\( AB \)[/tex] and set it equal to the identity matrix:
[tex]\[ AB = \begin{pmatrix} 2m & 7 \\ 5 & 9 \end{pmatrix} \begin{pmatrix} 9 & n \\ -5 & 4 \end{pmatrix} \][/tex]
Perform the multiplication:
[tex]\[ AB = \begin{pmatrix} (2m \cdot 9 + 7 \cdot (-5)) & (2m \cdot n + 7 \cdot 4) \\ (5 \cdot 9 + 9 \cdot (-5)) & (5 \cdot n + 9 \cdot 4) \end{pmatrix} \][/tex]
Simplify each element:
[tex]\[ AB = \begin{pmatrix} (18m - 35) & (2mn + 28) \\ (45 - 45) & (5n + 36) \end{pmatrix} \][/tex]
Since [tex]\( AB \)[/tex] must be equal to the identity matrix, we get:
[tex]\[ \begin{pmatrix} 18m - 35 & 2mn + 28 \\ 0 & 5n + 36 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \][/tex]
This gives us the following system of equations:
1. [tex]\( 18m - 35 = 1 \)[/tex]
2. [tex]\( 2mn + 28 = 0 \)[/tex]
3. [tex]\( 5n + 36 = 1 \)[/tex]
Solve the first equation for [tex]\( m \)[/tex]:
[tex]\[ 18m - 35 = 1 \][/tex]
[tex]\[ 18m = 36 \][/tex]
[tex]\[ m = 2 \][/tex]
Solve the third equation for [tex]\( n \)[/tex]:
[tex]\[ 5n + 36 = 1 \][/tex]
[tex]\[ 5n = -35 \][/tex]
[tex]\[ n = -7 \][/tex]
Verify the second equation with these values for [tex]\( m \)[/tex] and [tex]\( n \)[/tex]:
[tex]\[ 2mn + 28 = 2(2)(-7) + 28 = -28 + 28 = 0 \][/tex]
Thus, the values that satisfy all conditions are:
[tex]\[ m = 2 \][/tex]
[tex]\[ n = -7 \][/tex]
Therefore, the values of [tex]\( x \)[/tex] and [tex]\( y \)[/tex] are:
[tex]\[ m = 2, n = -7 \][/tex]
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