Problem 16
Question
( Refer to Examples 3-5.) LetA be the given matrix. Find \(A^{-1}\) without a calculator. $$ \left[\begin{array}{rr} 1 & 0 \\ 1 & -1 \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse matrix \(A^{-1}\) is \(\begin{bmatrix} 1 & 0 \\ 1 & -1 \end{bmatrix}\).
1Step 1: Identify the Matrix
The given matrix is \[A = \begin{bmatrix}1 & 0 \1 & -1\end{bmatrix}\] This is a 2x2 matrix, which is important because we will be using the formula for the inverse of a 2x2 matrix.
2Step 2: Recall the Inverse Formula for 2x2 Matrix
For a 2x2 matrix \(A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\), the inverse is given by\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]provided that \(ad - bc eq 0\). This value \(ad - bc\) is called the determinant of the matrix.
3Step 3: Calculate the Determinant
Identify the elements in the matrix: \(a = 1\), \(b = 0\), \(c = 1\), and \(d = -1\).The determinant is calculated as \(ad - bc\):\[1(-1) - 0(1) = -1.\]Since the determinant is \(-1\), which is not zero, the inverse exists.
4Step 4: Write the Expression for the Inverse
Using the inverse formula for a 2x2 matrix, substitute the values into the formula:\[A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & 0 \ -1 & 1 \end{bmatrix}\]This simplifies to:\[A^{-1} = -1 \times \begin{bmatrix} -1 & 0 \ -1 & 1 \end{bmatrix}\]
5Step 5: Multiply to Find the Inverse
Multiply each entry of the matrix by \(-1\):\[A^{-1} = \begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix}\] Check the simplified result to confirm correctness.
Key Concepts
Determinant of a MatrixInverse of 2x2 MatrixMatrix Algebra
Determinant of a Matrix
When working with matrices, particularly in finding the inverse, understanding the determinant is crucial. The determinant is a special number that can tell us if a matrix is invertible. For a 2x2 matrix, the determinant is calculated using a simple formula. Given a matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is:\[ad - bc\]For example, consider the matrix \(\begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix}\). Let's identify its elements:
- \(a = 1\)
- \(b = 0\)
- \(c = 1\)
- \(d = -1\)
Inverse of 2x2 Matrix
Finding the inverse of a 2x2 matrix is a straightforward process if you know the formula and the determinant of the matrix is not zero. The formula for the inverse of a 2x2 matrix is \[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]Using the earlier example, we have matrix \(\begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix}\) with a determinant of \(-1\). We substitute the elements and the determinant into the inverse formula:\[A^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & 0 \ -1 & 1 \end{bmatrix}\]This can be simplified by multiplying each element inside the bracket by \(-1\), which gives us:\[A^{-1} = \begin{bmatrix} 1 & 0 \ 1 & -1 \end{bmatrix}\]Understanding how to use this formula will allow you to invert any 2x2 matrix confidently.
Matrix Algebra
Matrix algebra involves operations on matrices, similar to how Algebra deals with numbers and variables. It is a cornerstone of linear algebra, a subject that is key in various applications like computer graphics, physics, and engineering.
Some basic matrix operations include:
- Matrix addition and subtraction: Adding or subtracting matrices involves adding or subtracting their corresponding elements.
- Scalar multiplication: Multiplying each element of the matrix by a scalar (a number).
- Matrix multiplication: Combining matrices in a way similar to multiplying numbers, but with special rules that involve multiplying rows by columns.
- Finding inverses: Used especially in solving system of linear equations, where the inverse matrix can help in finding the solution.
Other exercises in this chapter
Problem 16
If possible, find each of the following. (a) \(A+B\) (b) \(3 A\) (c) \(2 A-3 B\) $$A=\left[\begin{array}{rrr}1 & -2 & 5 \\\3 & -4 & -1\end{array}\right]$$ $$B=\
View solution Problem 16
Let \(A\) be the given matrix. Find det \(A\) by using the method of co factors. $$ \left[\begin{array}{rrr} 1 & 1 & 5 \\ -3 & -3 & 0 \\ 7 & 0 & 0 \end{array}\r
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The augmented matrix is in row-echelon form and represents a linear system. Solve the system by using backwand substitution, if possible. Write the solution as
View solution Problem 16
If possible, solve the system. $$ \begin{array}{lr} 2 x-y+2 z= 6 \\ -x+y+z= 0 \\ -x \quad \quad -3 z=-6 \end{array} $$
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