Problem 17
Question
Use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ A=\left[\begin{array}{rr} {10} & {-2} \\ {-5} & {1} \end{array}\right] $$
Step-by-Step Solution
Verified Answer
The inverse of the given matrix \( A = \left[ \begin{array}{rr} {10} & {-2} \ {-5} & {1} \end{array} \right] \)does not exist because the determinant of the matrix is 0. The determinant is the value needed to calculate the inverse, and when it equals 0, the inverse does not exist.
1Step 1: Define the Matrix and its components
Find the variables \(a\), \(b\), \(c\) and \(d\) from the given matrix A. In this case, \(a = 10\), \(b = -2\), \(c = -5\) and \(d = 1\) .
2Step 2: Calculate the inverse
The formula for determining the inverse of a 2x2 matrix is given by \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \ {-c} & {a}\end{array}\right]\). Plug in the values \(a = 10\), \(b = -2\), \(c = -5\) and \(d = 1\) to find \(A^{-1}\).\nSo, first calculate the denominator \(ad - bc\) as \(10*1 - (-2)*(-5) = 10 - 10 = 0\). In this case, as the determinant is 0, the inverse of the matrix A does not exist.
3Step 3: Confirm the inverse matrix
The inverse of a matrix A does not exist if the determinant, which is the denominator in the equation, is equal to zero. This is a requisite for the inverse of a matrix.
Key Concepts
Determinant of a Matrix2x2 Matrix InversionMatrix Algebra
Determinant of a Matrix
The determinant of a matrix is a scalar value that is a function of the entries of a square matrix. It provides important information about the matrix, including whether the matrix is invertible and properties of the linear transformation it represents. For a 2x2 matrix given by
\[\[\begin{align*}A &= \begin{bmatrix}a & b \c & d\end{bmatrix},\end{align*}\]\]the determinant is calculated as
\[\[\begin{align*}det(A) &= ad - bc.\end{align*}\]\]When the determinant is zero, the matrix does not have an inverse, implying that the linear transformation associated with the matrix is non-invertible, and hence is not one-to-one. This condition also indicates that the matrix compresses the space into a lower dimension. In simpler terms, if the determinant of a matrix is zero, it means that there's no unique solution to the set of linear equations represented by the matrix.
\[\[\begin{align*}A &= \begin{bmatrix}a & b \c & d\end{bmatrix},\end{align*}\]\]the determinant is calculated as
\[\[\begin{align*}det(A) &= ad - bc.\end{align*}\]\]When the determinant is zero, the matrix does not have an inverse, implying that the linear transformation associated with the matrix is non-invertible, and hence is not one-to-one. This condition also indicates that the matrix compresses the space into a lower dimension. In simpler terms, if the determinant of a matrix is zero, it means that there's no unique solution to the set of linear equations represented by the matrix.
2x2 Matrix Inversion
Inverting a 2x2 matrix is a common operation in linear algebra, often required to solve systems of equations. The inverse of a 2x2 matrix A, if it exists, is given by
\[\[\begin{align*}A^{-1} &= \frac{1}{det(A)}\begin{bmatrix}d & -b \-c & a\end{bmatrix},\end{align*}\]\]where the terms a, b, c, d are the entries of matrix A, and det(A) is the determinant. The inverse matrix, when multiplied by the original matrix A, yields the identity matrix, symbolized as I. It's crucial to ensure that the determinant (ad - bc) is not zero since division by zero is undefined, and hence the inverse, in that case, would not exist. When teaching this concept, it is important to provide a clear understanding of each term in the inversion formula and emphasize the significance of the determinant.
\[\[\begin{align*}A^{-1} &= \frac{1}{det(A)}\begin{bmatrix}d & -b \-c & a\end{bmatrix},\end{align*}\]\]where the terms a, b, c, d are the entries of matrix A, and det(A) is the determinant. The inverse matrix, when multiplied by the original matrix A, yields the identity matrix, symbolized as I. It's crucial to ensure that the determinant (ad - bc) is not zero since division by zero is undefined, and hence the inverse, in that case, would not exist. When teaching this concept, it is important to provide a clear understanding of each term in the inversion formula and emphasize the significance of the determinant.
Matrix Algebra
Matrix algebra involves various operations such as addition, subtraction, multiplication, and finding the inverse of matrices. When multiplying matrices, one must consider the order of multiplication as it can affect the result; matrix multiplication is not commutative.
For a matrix to have an inverse, it must be a square matrix, and its determinant must not be zero. Multiplicative inverse properties include:
For a matrix to have an inverse, it must be a square matrix, and its determinant must not be zero. Multiplicative inverse properties include:
- If A has an inverse, then multiplying A with its inverse in any order will result in the identity matrix:
\[\[\begin{align*}A A^{-1} &= A^{-1} A = I.\end{align*}\]\] - For a product of matrices AB to be invertible, both A and B individually must be invertible:
Other exercises in this chapter
Problem 17
Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrrr|r} {1} & {-1} & {1} & {1} & {3} \\ {0} & {1} & {-2} & {-1} & {0} \\ {2}
View solution Problem 17
Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{r}{x+2 y=3} \\\\{3 x-4 y=4}\end{array}\right.$$
View solution Problem 17
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{r} {x+2 y+3 z=5} \\ {y-5
View solution Problem 18
Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {
View solution