Answer :
To find the values of [tex]\(a\)[/tex] and [tex]\(b\)[/tex] such that the product [tex]\(XY = I\)[/tex] (the identity matrix) where:
[tex]\[ X = \begin{array}{cc} \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \end{array} \][/tex]
[tex]\[ Y = \begin{array}{cc} \begin{bmatrix} a & -5 \\ -1 & b \end{bmatrix} \end{array} \][/tex]
and
[tex]\[ I = \begin{array}{cc} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{array} \][/tex]
we need to ensure that their product yields the identity matrix:
[tex]\[ XY = I \][/tex]
Now, let's calculate the product [tex]\(XY\)[/tex]:
[tex]\[ XY = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} a & -5 \\ -1 & b \end{bmatrix} \][/tex]
We perform the matrix multiplication as follows:
- The element in the first row and first column of [tex]\(XY\)[/tex]:
[tex]\[ (3 \cdot a) + (5 \cdot -1) = 3a - 5 \][/tex]
- The element in the first row and second column of [tex]\(XY\)[/tex]:
[tex]\[ (3 \cdot -5) + (5 \cdot b) = -15 + 5b \][/tex]
- The element in the second row and first column of [tex]\(XY\)[/tex]:
[tex]\[ (1 \cdot a) + (2 \cdot -1) = a - 2 \][/tex]
- The element in the second row and second column of [tex]\(XY\)[/tex]:
[tex]\[ (1 \cdot -5) + (2 \cdot b) = -5 + 2b \][/tex]
Putting it all together:
[tex]\[ XY = \begin{bmatrix} 3a - 5 & -15 + 5b \\ a - 2 & -5 + 2b \end{bmatrix} \][/tex]
For [tex]\(XY\)[/tex] to be the identity matrix, each corresponding element must match the identity matrix:
[tex]\[ \begin{bmatrix} 3a - 5 & -15 + 5b \\ a - 2 & -5 + 2b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
This gives us the following set of equations:
1. [tex]\(3a - 5 = 1\)[/tex]
2. [tex]\(-15 + 5b = 0\)[/tex]
3. [tex]\(a - 2 = 0\)[/tex]
4. [tex]\(-5 + 2b = 1\)[/tex]
We solve these equations step by step.
From equation (1):
[tex]\[ 3a - 5 = 1 \][/tex]
[tex]\[ 3a = 6 \][/tex]
[tex]\[ a = 2 \][/tex]
From equation (2):
[tex]\[ -15 + 5b = 0 \][/tex]
[tex]\[ 5b = 15 \][/tex]
[tex]\[ b = 3 \][/tex]
We also check the other equations to ensure consistency:
From equation (3):
[tex]\[ a - 2 = 0 \][/tex]
[tex]\[ a = 2 \][/tex]
From equation (4):
[tex]\[ -5 + 2b = 1 \][/tex]
[tex]\[ 2b = 6 \][/tex]
[tex]\[ b = 3 \][/tex]
Therefore, the values that satisfy all the equations are:
[tex]\[ \boxed{a = 2 \text{ and } b = 3} \][/tex]
[tex]\[ X = \begin{array}{cc} \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \end{array} \][/tex]
[tex]\[ Y = \begin{array}{cc} \begin{bmatrix} a & -5 \\ -1 & b \end{bmatrix} \end{array} \][/tex]
and
[tex]\[ I = \begin{array}{cc} \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \end{array} \][/tex]
we need to ensure that their product yields the identity matrix:
[tex]\[ XY = I \][/tex]
Now, let's calculate the product [tex]\(XY\)[/tex]:
[tex]\[ XY = \begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} a & -5 \\ -1 & b \end{bmatrix} \][/tex]
We perform the matrix multiplication as follows:
- The element in the first row and first column of [tex]\(XY\)[/tex]:
[tex]\[ (3 \cdot a) + (5 \cdot -1) = 3a - 5 \][/tex]
- The element in the first row and second column of [tex]\(XY\)[/tex]:
[tex]\[ (3 \cdot -5) + (5 \cdot b) = -15 + 5b \][/tex]
- The element in the second row and first column of [tex]\(XY\)[/tex]:
[tex]\[ (1 \cdot a) + (2 \cdot -1) = a - 2 \][/tex]
- The element in the second row and second column of [tex]\(XY\)[/tex]:
[tex]\[ (1 \cdot -5) + (2 \cdot b) = -5 + 2b \][/tex]
Putting it all together:
[tex]\[ XY = \begin{bmatrix} 3a - 5 & -15 + 5b \\ a - 2 & -5 + 2b \end{bmatrix} \][/tex]
For [tex]\(XY\)[/tex] to be the identity matrix, each corresponding element must match the identity matrix:
[tex]\[ \begin{bmatrix} 3a - 5 & -15 + 5b \\ a - 2 & -5 + 2b \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \][/tex]
This gives us the following set of equations:
1. [tex]\(3a - 5 = 1\)[/tex]
2. [tex]\(-15 + 5b = 0\)[/tex]
3. [tex]\(a - 2 = 0\)[/tex]
4. [tex]\(-5 + 2b = 1\)[/tex]
We solve these equations step by step.
From equation (1):
[tex]\[ 3a - 5 = 1 \][/tex]
[tex]\[ 3a = 6 \][/tex]
[tex]\[ a = 2 \][/tex]
From equation (2):
[tex]\[ -15 + 5b = 0 \][/tex]
[tex]\[ 5b = 15 \][/tex]
[tex]\[ b = 3 \][/tex]
We also check the other equations to ensure consistency:
From equation (3):
[tex]\[ a - 2 = 0 \][/tex]
[tex]\[ a = 2 \][/tex]
From equation (4):
[tex]\[ -5 + 2b = 1 \][/tex]
[tex]\[ 2b = 6 \][/tex]
[tex]\[ b = 3 \][/tex]
Therefore, the values that satisfy all the equations are:
[tex]\[ \boxed{a = 2 \text{ and } b = 3} \][/tex]