The inverse of a matrix A is given by \(\rm \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\) What is A equal to?
Let the given matrix inverse be denoted as \(A^{-1}\). We are given:
\(A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\)
The question asks us to find the original matrix A. A fundamental property of matrix inverses is that the inverse of the inverse of a matrix is the original matrix itself. Mathematically, this is expressed as \((A^{-1})^{-1} = A\).
Therefore, to find matrix A, we need to calculate the inverse of the given matrix \(A^{-1}\).
Finding the Original Matrix using Inverse Property
To find the inverse of a 2x2 matrix \(M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\), we use the formula:
\(M^{-1} = \frac{1}{\det(M)} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}\)
where the determinant of the matrix M is \(\det(M) = ad - bc\). The inverse exists only if \(\det(M) \neq 0\).
Let the given matrix \(A^{-1}\) be represented as \(\begin{bmatrix} a' & b' \\ c' & d' \end{bmatrix}\), where \(a' = -2\), \(b' = 1\), \(c' = \frac{3}{2}\), and \(d' = -\frac{1}{2}\).
First, we calculate the determinant of \(A^{-1}\):
\(\det(A^{-1}) = a'd' - b'c'\)
\(\det(A^{-1}) = (-2) \times \left(-\frac{1}{2}\right) - (1) \times \left(\frac{3}{2}\right)\)
\(\det(A^{-1}) = 1 - \frac{3}{2}\)
\(\det(A^{-1}) = \frac{2}{2} - \frac{3}{2}\)
\(\det(A^{-1}) = -\frac{1}{2}\)
Since the determinant \(-\frac{1}{2}\) is not zero, the inverse of \(A^{-1}\) exists.
Next, we find the adjugate of \(A^{-1}\) by swapping the diagonal elements and negating the off-diagonal elements:
Adjugate of \(A^{-1} = \begin{bmatrix} d' & -b' \\ -c' & a' \end{bmatrix} = \begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix}\)
Now, we calculate the inverse of \(A^{-1}\), which is equal to A:
\(A = (A^{-1})^{-1} = \frac{1}{\det(A^{-1})} \begin{bmatrix} d' & -b' \\ -c' & a' \end{bmatrix}\)
\(A = \frac{1}{-\frac{1}{2}} \begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix}\)
\(A = -2 \begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix}\)
Now, multiply each element inside the matrix by the scalar \(-2\):
\(A = \begin{bmatrix} (-2) \times (-\frac{1}{2}) & (-2) \times (-1) \\ (-2) \times (-\frac{3}{2}) & (-2) \times (-2) \end{bmatrix}\)
\(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
Comparing Result with Options
Let's compare our calculated matrix A with the given options:
- Option 1: \(\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\)
- Option 2: \(\begin{bmatrix} 1 & -2 \\ -3 & 4 \end{bmatrix}\)
- Option 3: \(\begin{bmatrix} 1 & 2 \\ 3 & -4 \end{bmatrix}\)
- Option 4: \(\begin{bmatrix} -1 & 2 \\ 3 & 4 \end{bmatrix}\)
Our calculated matrix \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) matches Option 1.
Matrix Inverse Calculation Summary
| Step | Description | Calculation |
|---|---|---|
| 1 | Identify the given inverse matrix \(A^{-1}\) | \(A^{-1} = \begin{bmatrix} -2 & 1 \\ \frac{3}{2} & -\frac{1}{2} \end{bmatrix}\) |
| 2 | Recall property \((A^{-1})^{-1} = A\) | Need to find the inverse of \(A^{-1}\) |
| 3 | Calculate the determinant of \(A^{-1}\) | \(\det(A^{-1}) = (-2)(-\frac{1}{2}) - (1)(\frac{3}{2}) = 1 - \frac{3}{2} = -\frac{1}{2}\) |
| 4 | Find the adjugate of \(A^{-1}\) | \(\begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix}\) |
| 5 | Calculate \(A = (A^{-1})^{-1}\) using the formula | \(A = \frac{1}{-1/2} \begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix} = -2 \begin{bmatrix} -\frac{1}{2} & -1 \\ -\frac{3}{2} & -2 \end{bmatrix}\) |
| 6 | Perform scalar multiplication | \(A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\) |
Revision Table: Key Matrix Inverse Concepts
| Concept | Description |
|---|---|
| Inverse Matrix ($A^{-1}$) | A matrix $A^{-1}$ such that $A A^{-1} = A^{-1} A = I$, where $I$ is the identity matrix. |
| Determinant ($\det(A)$) | A scalar value calculated from the elements of a square matrix. For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\det(A) = ad - bc$. |
| Singular Matrix | A square matrix whose determinant is zero. A singular matrix does not have an inverse. |
| Non-Singular Matrix | A square matrix whose determinant is non-zero. A non-singular matrix has an inverse. |
| Adjugate Matrix ($\text{adj}(A)$) | The transpose of the cofactor matrix. For a 2x2 matrix $\begin{bmatrix} a & b \\ c & d \end{bmatrix}$, $\text{adj}(A) = \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$. |
| Inverse Formula (2x2) | $A^{-1} = \frac{1}{\det(A)} \text{adj}(A)$, provided $\det(A) \neq 0$. |
| Property $(A^{-1})^{-1}$ | The inverse of the inverse of a matrix is the original matrix: $(A^{-1})^{-1} = A$. |
Additional Information: Properties of Matrix Inverses
Understanding the properties of matrix inverses is crucial for solving matrix-related problems. Here are a few important ones:
- Uniqueness of Inverse: If a matrix has an inverse, it is unique.
- Inverse of a Product: The inverse of a product of two invertible matrices is the product of their inverses in reverse order: \((AB)^{-1} = B^{-1}A^{-1}\).
- Inverse of a Transpose: The inverse of the transpose of a matrix is equal to the transpose of its inverse: \((A^T)^{-1} = (A^{-1})^T\).
- Inverse of a Scalar Multiple: If \(k\) is a non-zero scalar and \(A\) is an invertible matrix, then \((kA)^{-1} = \frac{1}{k} A^{-1}\).
- Inverse of the Inverse: As used in this problem, \((A^{-1})^{-1} = A\). This property allows us to find the original matrix A when its inverse is given.
These properties simplify many matrix calculations and are fundamental concepts in linear algebra.
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