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Chapter 9: Problem 59

Explain how to find the multiplicative inverse for a \(2 \times 2\) invertible matrix.

Short Answer

Expert verified
The multiplicative inverse of a \(2 \times 2\) matrix \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) is found by the formula \(A^{-1} = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\), only if the determinant \(ad-bc\) is not zero.

Step by step solution

01

Identify the original \(2 \times 2\) matrix

Let's say the \(2 \times 2\) matrix is denoted by \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\). The elements \(a, b, c, d\) represent any real numbers.
02

Find the determinant of the matrix

The determinant of this \(2 \times 2\) matrix is calculated as the product of the elements on the main diagonal minus the product of the elements off the main diagonal. It is denoted as \(\det(A)\), and it is calculated as \(\det(A) = ad - bc\). The determinant must not be equal to zero for the matrix to be invertible.
03

Create the adjugate matrix

The adjugate of \(A\) is obtained by interchanging the elements on the main diagonal and multiplying the off-diagonal elements by \(-1\). This results in the matrix \(adj(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).
04

Multiply the adjugate by \(1/\)determinant

The multiplicative inverse of the matrix \(A\), denoted \(A^{-1}\), is found by multiplying each element of the adjugate matrix by \(\frac{1}{\det(A)}\). This results in \(A^{-1} = \frac{1}{\det(A)} \times adj(A) = \frac{1}{ad-bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix}\).

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Most popular questions from this chapter

Solve: \(\quad 3^{2 x-8}=27 .\) (Section \(4.4,\) Example 1 )

Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{ll} {2} & {4} \\ {3} & {1} \\ {4} & {2} \end{array}\right], \quad B=\left[\begin{array}{rrr} {3} & {2} & {0} \\ {-1} & {-3} & {5} \end{array}\right] $$

Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{l} {x+2 y=z-1} \\ {x=4+y-z} \\ {x+y-3 z=-2} \end{array}\right. $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ 2 X+5 A=B $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{lll} {6} & {2} & {-3} \end{array}\right], B=\left[\begin{array}{lll} {4} & {-2} & {3} \end{array}\right] $$
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