Problem 227
Question
Find \(\left[\begin{array}{cc}1 & 3 \\ 3 & 10\end{array}\right]^{-1}\)
Step-by-Step Solution
Verified Answer
The inverse of the given matrix is \([\begin{array}{cc}10 & -3 \ -3 & 1\end{array}]\)
1Step 1: Identify the elements of the matrix
Identify the elements a,b,c, and d of the given matrix. Here, a=1, b=3, c=3, and d=10.
2Step 2: Compute the determinant
Compute the determinant of the matrix which is ad-bc. Substituting the values, we have 1*10 - 3*3 = 10 - 9 = 1.
3Step 3: Apply the inverse formula
We know that \(A^{-1}=\frac{1}{ad-bc}[\begin{array}{cc}d & -b \ -c & a\end{array}]\) . Substitute the known values into the formula and compute the inverse. By doing this, we get: \(A^{-1}=\frac{1}{1}[\begin{array}{cc}10 & -3 \ -3 & 1\end{array}] = [\begin{array}{cc}10 & -3 \ -3 & 1\end{array}]\) .
Key Concepts
Matrix DeterminantsElementary MatricesLinear Algebra
Matrix Determinants
Understanding the concept of determinants in matrices is essential when calculating matrix inverses in linear algebra. The determinant of a matrix is a special number that can be computed from its elements and provides critical information about the matrix's properties. For a 2x2 matrix like the one in the exercise, say \[\begin{bmatrix} a & b \ c & d \end{bmatrix}\], the determinant is calculated as \(ad-bc\).
This value plays a vital role in determining whether a matrix is invertible. If the determinant is zero, the matrix does not have an inverse, making it singular. On the other hand, a non-zero determinant ensures that the matrix has an inverse, which means it is nonsingular. For the given problem, the determinant is calculated as \(1 \times 10 - 3 \times 3 = 1\), a non-zero value, signifying that the matrix is invertible.
This value plays a vital role in determining whether a matrix is invertible. If the determinant is zero, the matrix does not have an inverse, making it singular. On the other hand, a non-zero determinant ensures that the matrix has an inverse, which means it is nonsingular. For the given problem, the determinant is calculated as \(1 \times 10 - 3 \times 3 = 1\), a non-zero value, signifying that the matrix is invertible.
Elementary Matrices
Elementary matrices are fundamental tools in linear algebra, used to facilitate various matrix operations, including finding inverses. They are derived from applying elementary row operations to identity matrices. There are three main types of row operations that correspond to elementary matrices:
- Swapping rows
- Multiplying a row by a nonzero scalar
- Adding a multiple of one row to another row
Linear Algebra
Linear algebra is a branch of mathematics dealing with vector spaces and linear mappings between these spaces. It encompasses the study of lines, planes, and subspaces and is fundamental in understanding systems of linear equations, vector spaces, and matrices. Matrices are central to linear algebra as they provide a compact way to represent and solve systems of linear equations. By using operations such as addition, multiplication, and finding inverses, matrices enable various practical applications from computer graphics to engineering.
The exercise illustrates a basic application of linear algebra: calculating the inverse of a matrix. Since matrices can transform one vector into another, knowing a matrix's inverse is akin to reversing this transformation. This is particularly useful in solving linear systems where you need to "unapply" this transformation to find solutions more efficiently.
Other exercises in this chapter
Problem 225
If \(A=\left[\begin{array}{ccc}1 & -2 & 3 \\ 2 & 3 & -1 \\ -3 & 1 & 2\end{array}\right]\) and \(B=\left[\begin{array}{lll}1 & 0 & 2 \\ 0 & 1 & 2 \\\ 1 & 2 & 0\e
View solution Problem 226
If \(A=\left[\begin{array}{cc}\cos x & \sin x \\ -\sin x & \cos x\end{array}\right]\) and \(A(a d j A)=k\left[\begin{array}{cc}1 & 0 \\ 0 & 1\end{array}\right]\
View solution Problem 228
If \(A=\left[\begin{array}{ll}5 & 2 \\ 3 & 1\end{array}\right]\), then find \(A^{-1}\).
View solution Problem 229
Find \(A^{-1}\), if the matrix \(A\) is given by \(A=\left[\begin{array}{cc}0.8 & 0.6 \\ -0.6 & 0.8\end{array}\right]\)
View solution