Math  /  Algebra

QuestionConsider the following matrix. A=[210312011]A=\left[\begin{array}{ccc} 2 & 1 & 0 \\ 3 & -1 & 2 \\ 0 & 1 & -1 \end{array}\right]
Choose the correct description of AA. Find A1A^{-1} if it exists.

Studdy Solution

STEP 1

What is this asking? We're given a matrix AA and we need to figure out if it has an *inverse*, and if it does, we need to *find* it! Watch out! Not all matrices have inverses!
If the *determinant* is zero, then no inverse exists.

STEP 2

1. Calculate the Determinant
2. Find the Inverse

STEP 3

Let's **calculate the determinant** of AA to see if an inverse exists.
Remember, if the determinant is **zero**, the matrix *doesn't* have an inverse!

STEP 4

We can use the formula for the determinant of a 3x3 matrix. det(A)=a(eifh)b(difg)+c(dheg)\det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) where A=abcdefghiA=\begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array}

STEP 5

In our case, a=2a=2, b=1b=1, c=0c=0, d=3d=3, e=1e=-1, f=2f=2, g=0g=0, h=1h=1, and i=1i=-1.
Let's plug these values into the determinant formula: det(A)=2((1)(1)(2)(1))1((3)(1)(2)(0))+0((3)(1)(1)(0))\det(A) = 2((-1)(-1) - (2)(1)) - 1((3)(-1) - (2)(0)) + 0((3)(1) - (-1)(0))

STEP 6

Now, let's simplify: det(A)=2(12)1(30)+0(30)\det(A) = 2(1 - 2) - 1(-3 - 0) + 0(3 - 0) det(A)=2(1)1(3)+0\det(A) = 2(-1) - 1(-3) + 0det(A)=2+3+0\det(A) = -2 + 3 + 0det(A)=1\det(A) = 1

STEP 7

The determinant is **1**, which is *not* zero!
This means AA *does* have an inverse.
Awesome!

STEP 8

To find the inverse, we'll use the formula: A1=1det(A)adj(A)A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) where adj(A)\text{adj}(A) is the **adjugate** of AA.

STEP 9

First, let's find the **matrix of cofactors**.
Remember, the cofactor of an element is found by taking the determinant of the smaller matrix formed by *excluding* the element's row and column, and then multiplying by (1)i+j(-1)^{i+j}, where ii is the row number and jj is the column number.

STEP 10

The matrix of cofactors is: C=((1)(1)(2)(1))((3)(1)(2)(0))((3)(1)(1)(0))((1)(1)(0)(1))((2)(1)(0)(0))((2)(1)(1)(0))((1)(2)(0)(1))((2)(2)(0)(3))((2)(1)(1)(3))C = \begin{array}{ccc} ((-1)(-1) - (2)(1)) & -((3)(-1) - (2)(0)) & ((3)(1) - (-1)(0)) \\ -((1)(-1) - (0)(1)) & ((2)(-1) - (0)(0)) & -((2)(1) - (1)(0)) \\ ((1)(2) - (0)(-1)) & -((2)(2) - (0)(3)) & ((2)(-1) - (1)(3)) \end{array}

STEP 11

Simplifying, we get: C=133122245C = \begin{array}{ccc} -1 & 3 & 3 \\ 1 & -2 & -2 \\ 2 & -4 & -5 \end{array}

STEP 12

The adjugate of AA is the **transpose** of the matrix of cofactors.
To find the transpose, we swap the rows and columns: adj(A)=112324325\text{adj}(A) = \begin{array}{ccc} -1 & 1 & 2 \\ 3 & -2 & -4 \\ 3 & -2 & -5 \end{array}

STEP 13

Now we can find A1A^{-1}: A1=[112324325]A^{-1} = \left[\begin{array}{ccc} -1 & 1 & 2 \\ 3 & -2 & -4 \\ 3 & -2 & -5 \end{array}\right]

STEP 14

So, the inverse of AA is: A1=112324325A^{-1} = \begin{array}{ccc} -1 & 1 & 2 \\ 3 & -2 & -4 \\ 3 & -2 & -5 \end{array}

STEP 15

The inverse of matrix AA is: A1=112324325A^{-1} = \begin{array}{ccc} -1 & 1 & 2 \\ 3 & -2 & -4 \\ 3 & -2 & -5 \end{array}

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