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Finding the Inverse of a Matrix In Exercises \(13-24,\) find the inverse of the matrix (if it exists). $$\left[ \begin{array}{ll}{3} & {1} \\ {4} & {2}\end{array}\right]$$

Short Answer

Expert verified
The inverse of the matrix \(\left[ \begin{array}{ll}{3} & {1} \ {4} & {2}\end{array}\right]\) is \(\left[ \begin{array}{cc}{1} & {-0.5} \ {-2} & {1.5}\end{array}\right]\)

Step by step solution

01

Find the determinant of the matrix (ad-bc)

The determinant will be found by multiplying the values from the main diagonal and subtracting the product of the other two. So, it will be \(ad-bc = (3*2)-(1*4)=6-4=2\)
02

Create the adjugate matrix

The adjugate matrix will be found by swapping the values on the main diagonal and changing the sign of the other two values. So, it will be \(\left[ \begin{array}{cc}{2} & {-1} \ {-4} & {3}\end{array}\right]\)
03

Find the inverse of the matrix

Multiply the adjugate matrix by \(1/determinant\). The determinant calculated in step 1 was 2. So, multiplying each of the values in the matrix by \(1/2\) gives the inverse of the matrix as \(\left[ \begin{array}{cc}{1} & {-0.5} \ {-2} & {1.5}\end{array}\right]\)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Determinant of a Matrix
The determinant of a matrix is a special scalar value that serves as a mathematical tool for solving systems of linear equations, finding the inverse of a matrix, and determining if a matrix is invertible. To calculate the determinant of a 2x2 matrix, you take the product of the entries in the main diagonal and subtract the product of the other two entries.

For instance, if you have a matrix \[ \left[ \begin{array}{cc}{a} & {b} \ {c} & {d}\end{array}\right] \], its determinant is calculated as \(ad - bc\). If the determinant is zero, the matrix doesn't have an inverse; this is known as a singular matrix. However, if the determinant is non-zero, the matrix is non-singular, and an inverse exists. In our example, the determinant is \(3*2 - 1*4 = 2\), indicating that the matrix is invertible.

A matrix with more than 2 rows and columns has a more complex formula for the determinant, involving permutations and signs, which is beyond the scope of this example but vital to know for higher-dimension matrices.
Adjugate Matrix
The adjugate (or adjoint) of a matrix is another crucial concept in linear algebra, particularly when finding the inverse of a matrix. It is a matrix whose elements are the cofactors, signed minors, of the corresponding elements of the original matrix.

For a 2x2 matrix \[ \left[ \begin{array}{cc}{a} & {b} \ {c} & {d}\end{array}\right] \], the adjugate matrix is obtained by swapping the positions of \(a\) and \(d\), and changing the signs of \(b\) and \(c\). This results in the matrix \[ \left[ \begin{array}{cc}{d} & {-b} \ {-c} & {a}\end{array}\right] \]. In the provided exercise, the adjugate is created as follows: \[ \left[ \begin{array}{cc}{2} & {-1} \ {-4} & {3}\end{array}\right] \]. The signed minors reflect the influence of each element on the determinant and ultimately, the matrix inverse.
Matrix Inversion Process
Understanding the matrix inversion process is fundamental in linear algebra as it is used to solve systems of equations and is vital in many applications such as physics, engineering, and computer science. To invert a matrix, you must follow a series of steps that include checking if an inverse exists (non-zero determinant), finding the adjugate matrix, and then multiplying the adjugate by the reciprocal of the determinant.

This is mathematically represented as the inverse matrix \(A^{-1}\) of \(A\) being equal to \(\frac{1}{det(A)} \times adj(A)\), where \(det(A)\) is the determinant of \(A\), and \(adj(A)\) is the adjugate of \(A\). In the specific case of our 2x2 matrix, the inverse is found by multiplying each element of the adjugate matrix by \(\frac{1}{2}\), resulting in \[ \left[ \begin{array}{cc}{1} & {-0.5} \ {-2} & {1.5}\end{array}\right] \], which is the inverse of the original matrix.

The matrix inversion process for larger matrices is more complex as it involves finding the matrix of minors, then the matrix of cofactors, and the adjugate, followed by the multiplication by the reciprocal of the determinant.

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Most popular questions from this chapter

HOW DO YOU SEE IT? At this point in the text, you have learned several methods for finding an equation of a line that passes through two given points. Briefly describe the methods that can be used to find the equation of the line that passes through the two points shown. Discuss the advantages and disadvantages of each method.

Using a Graphing Utility, use the matrix capabilities of a graphing utility to write the augmented matrix corresponding to the system of equations in reduced row-echelon form. Then solve the system. $$\left\\{\begin{aligned} x+2 y+2 z+4 w=& 11 \\ 3 x+6 y+5 z+12 w=& 30 \\ x+3 y-3 z+2 w=&-5 \\ 6 x-y-z+\quad w=&-9 \end{aligned}\right.$$

Breeding Facility A city zoo borrowed \(\$ 2,000,000\) at simple annual interest to construct a breeding facility. Some of the money was borrowed at \(8 \%,\) some at \(9 \%,\) and some at 12\(\% .\) Use a system of linear equations to determine how much was borrowed at each rate given that the total annual interest was \(\$ 186,000\) and the amount borrowed at 8\(\%\) was twice the amount borrowed at 12\(\% .\) Solve the system of linear equations using matrices.

Using Back-Substitution, write the system of linear equations represented by the augmented matrix. Then use back-substitution to solve. (Use variables \(x, y,\) and \(z,\) if applicable.) $$\left[ \begin{array}{rrrrr}{1} & {-1} & {2} & {\vdots} & {4} \\ {0} & {1} & {-1} & {\vdots} & {2} \\ {0} & {0} & {1} & {\vdots} & {-2}\end{array}\right]$$

Gaussian Elimination with Back-Substitution, use matrices to solve the system of equations (if possible). Use Gaussian elimination with back-substitution. $$\left\\{\begin{aligned} x-4 y+3 z-2 w=& 9 \\ 3 x-2 y+z-4 w=&-13 \\\\-4 x+3 y-2 z+w=&-4 \\\\-2 x+y-4 z+3 w=&-10 \end{aligned}\right.$$

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