Problem 3
Question
Find the inverse of the matrix or state that the matrix is not invertible. $$ C=\left[\begin{array}{rr} 6 & 15 \\ 14 & 35 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The matrix \( C \) is not invertible because its determinant is 0.
1Step 1: Check if the matrix is square
The given matrix \( C \) is a 2x2 matrix: \( \begin{bmatrix} 6 & 15 \ 14 & 35 \end{bmatrix} \). It has two rows and two columns, so it is a square matrix.
2Step 2: Calculate the determinant
The determinant of a 2x2 matrix \( \begin{bmatrix} a & b \ c & d \end{bmatrix} \) is calculated as \( ad - bc \). For our matrix \( C \), the determinant is \( 6 \times 35 - 15 \times 14 \). Calculate each product: \( 6 \times 35 = 210 \) and \( 15 \times 14 = 210 \). Thus, the determinant is \( 210 - 210 = 0 \).
3Step 3: Determine invertibility based on determinant
A square matrix is invertible if and only if its determinant is non-zero. Since the determinant of matrix \( C \) is 0, matrix \( C \) is not invertible.
Key Concepts
Square MatrixDeterminantInvertibility2x2 Matrix
Square Matrix
A square matrix is a simple yet crucial concept in linear algebra. It is defined as a matrix with an equal number of rows and columns.
This property is essential because it forms the basis for many matrix operations, including finding an inverse. **Square matrices** can be of size 2x2, 3x3, or more, depending on the number of rows and columns.
This property is essential because it forms the basis for many matrix operations, including finding an inverse. **Square matrices** can be of size 2x2, 3x3, or more, depending on the number of rows and columns.
- For example, a **2x2 matrix** has two rows and two columns.
- A **3x3 matrix** has three rows and three columns, and so forth.
Determinant
The determinant is a special number calculated from a square matrix. It offers insight into several properties of the matrix, including invertibility. For a 2x2 matrix, the determinant is calculated using the formula:
If the determinant is zero, it indicates that the matrix squashes the space it transforms into a lower dimension, thus making it non-invertible.
- \[Determinant = ad - bc\]
- Where \( a \), \( b \), \( c \), and \( d \) are the elements of the matrix arranged as:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]
If the determinant is zero, it indicates that the matrix squashes the space it transforms into a lower dimension, thus making it non-invertible.
Invertibility
Invertibility is a property that tells us whether a square matrix has an inverse. The inverse of a matrix is like a reverse button: applying the inverse will undo the effect of the original matrix. However, not all matrices have inverses.
The key factor in determining whether a matrix is invertible is its determinant.
The key factor in determining whether a matrix is invertible is its determinant.
- A matrix is **invertible (or non-singular)** if and only if its determinant is non-zero.
- If the determinant is zero, the matrix is **non-invertible (or singular)**.
2x2 Matrix
2x2 matrices are the simplest form of square matrices, yet hold significant importance in linear algebra.
They consist of four elements arranged in two rows and two columns, which makes them easy to compute by hand compared to larger matrices.
Finding the **determinant** is straightforward, and for lattice transformations, the determinant gives an essential geometric interpretation, often relating to area or scaling factors. If the determinant is non-zero, you can compute the inverse using:\[ \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \times \frac{1}{ad-bc}\]
They consist of four elements arranged in two rows and two columns, which makes them easy to compute by hand compared to larger matrices.
- A typical 2x2 matrix looks like:\[\begin{bmatrix} a & b \ c & d \end{bmatrix}\]
Finding the **determinant** is straightforward, and for lattice transformations, the determinant gives an essential geometric interpretation, often relating to area or scaling factors. If the determinant is non-zero, you can compute the inverse using:\[ \begin{bmatrix} d & -b \ -c & a \end{bmatrix} \times \frac{1}{ad-bc}\]
Other exercises in this chapter
Problem 3
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