Problem 17
Question
For the following exercises, find the multiplicative inverse of each matrix, if it exists. $$\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$$
Step-by-Step Solution
Verified Answer
The given matrix does not have an inverse because its determinant is zero.
1Step 1: Understanding the Multiplicative Inverse
The multiplicative inverse of a matrix \(A\) is another matrix \(B\) such that \(A \cdot B = I\), where \(I\) is the identity matrix of the same size. For a 2x2 matrix, you can find the inverse if the determinant is not zero.
2Step 2: Calculate the Determinant
For a 2x2 matrix \(\begin{bmatrix} a & b \ c & d \end{bmatrix}\), the determinant is computed as \(ad - bc\). For the matrix \(\begin{bmatrix} 1 & 1 \ 2 & 2 \end{bmatrix}\), the determinant is \(1 \times 2 - 1 \times 2 = 2 - 2 = 0\).
3Step 3: Determine if the Inverse Exists
Since the determinant of the given matrix is 0, the matrix does not have an inverse. A matrix must have a non-zero determinant to have an inverse.
Key Concepts
Determinant of a MatrixIdentity MatrixMultiplicative Inverse
Determinant of a Matrix
The determinant of a matrix is a special number that can be calculated from its elements. It plays a critical role in determining whether a matrix has an inverse.
For a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]the determinant is calculated as:\[\text{det}(A) = ad - bc\]
This formula is simple yet powerful. It helps in checking one essential condition: a matrix is invertible if and only if its determinant is not zero.
In the given exercise, the matrix was \(\begin{bmatrix} 1 & 1 \ 2 & 2 \end{bmatrix}\). By applying the formula, the determinant comes out as 0, which means, unfortunately, this matrix doesn't have an inverse. Remember, checking the determinant is always the first step in finding an inverse.
For a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]the determinant is calculated as:\[\text{det}(A) = ad - bc\]
This formula is simple yet powerful. It helps in checking one essential condition: a matrix is invertible if and only if its determinant is not zero.
In the given exercise, the matrix was \(\begin{bmatrix} 1 & 1 \ 2 & 2 \end{bmatrix}\). By applying the formula, the determinant comes out as 0, which means, unfortunately, this matrix doesn't have an inverse. Remember, checking the determinant is always the first step in finding an inverse.
Identity Matrix
An identity matrix is like a mirror in the world of matrices. It is a square matrix with ones on the diagonal and zeros elsewhere.
For a 2x2 matrix, it looks like:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
Its special property is that when you multiply any matrix by the identity matrix, the original matrix remains unchanged. This property is akin to multiplying any number by 1.
This feature is crucial when we talk about inverses. For a matrix to have an inverse, multiplying it by its inverse must yield the identity matrix. Think of it as an equation:\[A \cdot B = I\]where \(B\) is the inverse of \(A\).
Unfortunately, if a matrix doesn’t have an inverse, you can't use the identity matrix as a "checkpoint" to measure it up against. In our given exercise, since the determinant is zero, no such mirror-image matrix can fulfill the inversion condition.
For a 2x2 matrix, it looks like:\[I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}\]
Its special property is that when you multiply any matrix by the identity matrix, the original matrix remains unchanged. This property is akin to multiplying any number by 1.
This feature is crucial when we talk about inverses. For a matrix to have an inverse, multiplying it by its inverse must yield the identity matrix. Think of it as an equation:\[A \cdot B = I\]where \(B\) is the inverse of \(A\).
Unfortunately, if a matrix doesn’t have an inverse, you can't use the identity matrix as a "checkpoint" to measure it up against. In our given exercise, since the determinant is zero, no such mirror-image matrix can fulfill the inversion condition.
Multiplicative Inverse
The multiplicative inverse of a matrix is like finding the reciprocal for numbers. It is a matrix that, when multiplied with the original, gives the identity matrix.
To find the inverse of a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]you rearrange the elements, change signs, and divide by the determinant:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
But beware, if the determinant \((ad-bc)\) is zero, you can’t make an inverse. The zero determinant means division is impossible, much like how you can't divide by zero in arithmetic.
In our exercise, the determinant was zero, meaning no multiplicative inverse exists for the provided matrix. Always check the determinant before diving into calculations.
To find the inverse of a 2x2 matrix \[A = \begin{bmatrix} a & b \ c & d \end{bmatrix}\]you rearrange the elements, change signs, and divide by the determinant:\[A^{-1} = \frac{1}{ad-bc} \begin{bmatrix} d & -b \ -c & a \end{bmatrix}\]
But beware, if the determinant \((ad-bc)\) is zero, you can’t make an inverse. The zero determinant means division is impossible, much like how you can't divide by zero in arithmetic.
In our exercise, the determinant was zero, meaning no multiplicative inverse exists for the provided matrix. Always check the determinant before diving into calculations.
Other exercises in this chapter
Problem 16
For the following exercises, solve each system by substitution. $$ \begin{aligned} 4 x+6 y+9 z &=0 \\\\-5 x+2 y-6 z &=3 \\ 7 x-4 y+3 z &=-3 \end{aligned} $$
View solution Problem 17
For the following exercises, find the determinant. \(\left|\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right|\)
View solution Problem 17
Use any method to solve the system of nonlinear equations. $$ \begin{array}{r} -x^{2}+y=2 \\ -x+y=2 \end{array} $$
View solution Problem 17
Use the matrices below to perform scalar multiplication. \(A=\left[\begin{array}{rr}4 & 6 \\ 13 & 12\end{array}\right], B=\left[\begin{array}{rr}3 & 9 \\ 21 & 1
View solution