12th Grade
How to Use an Inverse Matrix to Solve Linear Equations?
{
"voice_prompt": "Pause briefly between single letters and variables like A, X, B, and I.",
"manuscript": {
"title": {
"text": "How to Use an Inverse Matrix to Solve Linear Equations",
"audio": "How to Use an Inverse Matrix to Solve Linear Equations"
},
"description": {
"text": "To solve a system of linear equations using matrices, you can represent the system in the form $AX = B$. If the coefficient matrix A has an inverse, you can find the solution by calculating $X = A^{-1}B$.",
"audio": "To solve a system of linear equations using matrices, you can represent the system in the form A times X equals B. If the coefficient matrix A has an inverse, you can find the solution by calculating X equals A-inverse times B."
},
"scenes": [
{
"text": "Imagine you're managing a workshop and need to solve a system of equations for your production. First, you can represent the system in a compact matrix form, A X equals B.",
"latex": "A = \\begin{bmatrix} a & b \\\\ c & d \\end{bmatrix}, \\quad X = \\begin{bmatrix} x \\\\ y \\end{bmatrix}, \\quad B = \\begin{bmatrix} e \\\\ f \\end{bmatrix}"
},
{
"text": "The goal is to find the values in the vector X. If the coefficient matrix A has an inverse, you can multiply both sides of the equation by A-inverse on the left.",
"latex": "A^{-1}(AX) = A^{-1}B"
},
{
"text": "Since A-inverse times A equals the identity matrix I, and I times X equals X, the equation simplifies to X equals A-inverse times B. This is the key formula for solving the system.",
"latex": "IX = A^{-1}B \\\\ X = A^{-1}B"
},
{
"text": "For an example, take the system 2x plus y equals 5, and 3x plus 2y equals 8. You can write this in matrix form.",
"latex": "A = \\begin{bmatrix} 2 & 1 \\\\ 3 & 2 \\end{bmatrix}, \\quad X = \\begin{bmatrix} x \\\\ y \\end{bmatrix}, \\quad B = \\begin{bmatrix} 5 \\\\ 8 \\end{bmatrix}"
},
{
"text": "Next, you need the inverse of A. First, you find the determinant, which is 2 times 2 minus 1 times 3, and that equals 1. Since the determinant is not zero, the inverse exists.",
"latex": "\\det(A) = (2)(2) - (1)(3) = 1"
},
{
"text": "Using the formula for a 2-by-2 inverse, you get the matrix two, negative one, negative three, two.",
"latex": "A^{-1} = \\begin{bmatrix} 2 & -1 \\\\ -3 & 2 \\end{bmatrix}"
},
{
"text": "Now you can find X by calculating A-inverse times B. You multiply the inverse of A by the vector B.",
"latex": "X = A^{-1}B = \\begin{bmatrix} 2 & -1 \\\\ -3 & 2 \\end{bmatrix} \\begin{bmatrix} 5 \\\\ 8 \\end{bmatrix}"
},
{
"text": "To get the result, you multiply row by column. The first element is 2 times 5 plus negative 1 times 8, which is 2. The second element is negative 3 times 5 plus 2 times 8, which is 1. The solution is x equals 2 and y equals 1.",
"latex": "X = \\begin{bmatrix} 2 \\\\ 1 \\end{bmatrix}"
},
{
"text": "You can always verify this by plugging x equals 2 and y equals 1 back into the original equations. This matrix method gives you a structured way to find solutions.",
"latex": "2(2) + 1 = 5 \\\\ 3(2) + 2(1) = 8"
},
{
"text": "Remember, if the determinant of A was zero, the inverse wouldn't exist. In that case, this method wouldn't give a unique solution.",
"latex": "\\det(A) = 0 \\implies \\text{No unique solution}"
}
],
"outro": {
"text": "So, by turning a system into AX = B and using the inverse, you can solve for X. This method is quick and powerful when the inverse exists and is widely used in many fields, especially when computers handle the calculations.",
"audio": "So, by turning a system into A times X equals B and using the inverse, you can solve for X. This method is quick and powerful when the inverse exists and is widely used in many fields, especially when computers handle the calculations."
}
}
}
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