First, apply the formulas in (9) to find A^-1. Then use A^-1 to solve the system Ax=b. ### Problem 2Consider the following matrices:Matrix **A** is a 2x2 matrix given by:\[A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}\]Matrix **b** is a 2x1 column matrix (or vector) given by:\[b = \begin{bmatrix} -1 \\ 3 \end{bmatrix}\]These matrices may be used in a variety of linear algebra problems, such as solving systems of equations, matrix multiplication, or finding determinants. **THEOREM 2: Inverses of 2 x 2 Matrices**The 2 x 2 matrix\[A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}\]is invertible if and only if $ad - bc \neq 0$, in which case\[A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.\]The above theorem describes the condition for finding the inverse of a 2 x 2 matrix. For the matrix $A$ to be invertible, the determinant $ad - bc$ must not equal zero. If this condition is met, the inverse of the matrix is calculated using the formula provided, where elements are rearranged and negated appropriately.

Answered: 2. A = 77₁b = 3 25 -1 3

Calculus: Early Transcendentals

8th Edition

ISBN:9781285741550

Author:James Stewart

Publisher:James Stewart

Chapter1: Functions And Models

Section: Chapter Questions

Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...

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Related questions

Question

First, apply the formulas in (9) to find A^-1. Then use A^-1 to solve the system Ax=b.

### Problem 2

Consider the following matrices:

Matrix **A** is a 2x2 matrix given by:

\[
A = \begin{bmatrix} 3 & 7 \\ 2 & 5 \end{bmatrix}
\]

Matrix **b** is a 2x1 column matrix (or vector) given by:

\[
b = \begin{bmatrix} -1 \\ 3 \end{bmatrix}
\]

These matrices may be used in a variety of linear algebra problems, such as solving systems of equations, matrix multiplication, or finding determinants.

**THEOREM 2: Inverses of 2 x 2 Matrices**

The 2 x 2 matrix

\[
A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}
\]

is invertible if and only if $ad - bc \neq 0$, in which case

\[
A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix}.
\]

The above theorem describes the condition for finding the inverse of a 2 x 2 matrix. For the matrix $A$ to be invertible, the determinant $ad - bc$ must not equal zero. If this condition is met, the inverse of the matrix is calculated using the formula provided, where elements are rearranged and negated appropriately.

Expert Solution

Step 1: Given formula of inverse of matrix:

Given, $A equals open square brackets table row 3 7 row 2 5 end table close square brackets space comma space space B equals open square brackets table row cell negative 1 end cell row 3 end table close square brackets$ .

If A= $open square brackets table row a b row c d end table close square brackets$ , than $A to the power of negative 1 end exponent equals fraction numerator 1 over denominator a d minus b c end fraction open square brackets table row d cell negative b end cell row cell negative c end cell a end table close square brackets$ .