Matrix Inverse 2 ( 2) 1 = 2 1 2

Transcription

1 Name: Matrix Inverse For Scalars, we have what is called a multiplicative identity. This means that if we have a scalar number, call it r, then r multiplied by the multiplicative identity equals r. Without thinking we use this identity all the time, it is. r r r. () There is also a multiplicative inverse. This mean that if we multiply r by it multiplicative inverse then we should get the multiplicative identity,. For scalars we write the multiplicative inverse as r. Normally we think of this as: r r r r. () So if we are told to find the multiplicative inverse for -, we are looking for the number which we can multiply by - to get. So, ( ). (3) Therefore, the multiplicative inverse for - is. For Matrices this is more complicated. There is also a multiplicative identity, we call this matrix I. I is a square matrix with ones along the diagonal and zeros everywhere else. Therefore this matrix looks like: I m,m Just like scalars, any matrix, A, can be multiplied by I to get AI IA A (within the rules of multiplication of matrices, ie I must be the correct size) So like scalars, we can find the multiplicative inverse of a matrix. However, unlike scalars only some matrices have a multiplicative inverse. First the matrix must be square. Second the det(a) can t equal but we will explain that later. If we have a matrix A, lets call its multiplicative inverse A (normally a multiplicative inverse is just called an inverse, so I will refer to multiplicative inverse simply as inverse). Since the multiplicative identity for matrices is I. Then the inverse is defined to satisfy this equation: A A AA A A A A I () If A is a x matrix, then A has components. We can solve for A several different ways. So first, we can solve it using substitution or elimination. For notational reasons, lets say A B.

2 Then we have the following equation: ( ) ( A B ) I ( ) (5) a, b, b, a, a, b, b, (6) (7) That leaves us with equations: b, + a, b, (8) b, + a, b, (9) a, b, + a, b, () a, b, + a, b, () () If we think of the elements of A as constants, then by pairing equation (7) and equation (9), we can solve for b, and b,. Likewise if we pair equation (8) and equation (), we solve for b, and b,. Lets start with find b, and b,. Solving equation (7) for b,. And substituting it to equation (9) And substituting it into equation (9) b, a,b, (3) () a, a, b, + a, b, (5) Then we can solve for b,. As you see, this can get very messy. Therefore I will introduce the next method, Gauss-Jordan Elimination. So again we start with the equation: A A I (6) To begin with, we set up our augmented matrix: a, a, a,

3 Dividing the top row by : a, a, a, Then multiplying the top row by a, and adding it to the second row: a, a,a, a + a,, We can simplify this matrix : Then mulitplying the second row by a, a,a, + a, a, a, a, + a,, we now have a, a, ( a, a, + a, ) a, a, + a, Next multiply the seconds for by a, and add it to the first: a, a, ( a, a, + a, ) + a, a, ( a, a, + a, ) ( a, a, + a, ) a, a, + a, Therefore, A a, a, ( a, a, + a, ) + a, a, ( a, a, + a, ) This looks pretty messy, but if we look just at the top left term, simplify things. By multiplying the top and bottom of becomes: or just a, ( a, a, + a, ). Then we can pull out a a, + a, a, A ( a, a, + a, ) a, a, + a, a, a, ( a, a, + a, ) + we can by ( a, a, + a, )), the top left term a, a, + a, a, + a, ( a, a, + a, ) (7) a, + a, a, from the matrix and we are left with: a, a, a, So now you should try multiplying A A and A A. Do you get I? From now on we are going to call a, + a, a, the determinant of A, which is written as det(a). Now we can use this formula for find A (when A is a x matrix), instead of using Gauss- 3

4 Jordan Elimination each time. Now notice that the det(a) could equal. If this happens this means that A does not have an inverse. So lets try some examples: A 3 6 Now to find A first we must find the determinant of A. From above we have that det(a) a, a, + a,. So, det(a) a, + a, a, (8) det(a) 3 6 (9) det(a) 8 () Now let s go back to the formula: A det(a) a, a, a, Substituting the elements from A into the formula we have: A 6 3. Now let s go back to what we know about scalars. If we were giving the equation 5x () We would solve the equation by multiplying both sides by inverse of 5 5 /5 to find that x /5. We also have equations for matrices. Let s look at: A X AX B () We will first look at the simple case. So let A be a x matrix, X be a x matrix and B be a x matrix. You should check the dimension of the matrices to make sure the A X will be a x matrix. Now let s try solving this equation like we did the scalar equation, let s multiply both sides of the

5 equation by A. Then AX B (3) A AX A B () Notice that I multiplied by equations by A on the left (A B BA ). Now if we look up at equation (3). We find that A A I. So equation (3) can be written as: IX A B (5) And I is the multiplicative identity so IX XI X. So just like in the scalar case we can solve for X. Now remember that X here is a matrix and has multiple elements. So let s try an example. Let, 5 A, B, X x, x,. We want to solve the matrix equation: AX B (6) or 5 3 x, x, First let s find A. We can find the determinant: det(a) 3 5. Now plugging into the formula for A : we have A 3 3 Now let s multiple both sides by A. Then x, x, 3 Then we can simplify the left hand side by multiplying A A: x, x, 3 5

6 Further this reduces down to by multiplying the matrices on the left: x, x, 3 Now by multiplying the matrices on the right we find: x, x, Now, if we are given a system of (scalar) equations and (scalar) unknowns we can write them as matrix equation. Say we are given: 3x + 5y (7) x + y (8) Then we can write the system of equations in 3 matrices. One matrix (which we normally call A) is a matrix of the coefficients (or numbers infront of the variables (x,y))), another matrix will be a matrix of our unknows (x,y) called X, and the third will be a matrix of constants (the numbers not multiplied by the unknowns), we will call this one B. A 3 5, B X x y Now notice that matrix equation AX B is the same as the scalar equations in (6) and (7). AX 3 5 x y 3x + 5y x + y B Now we know how to solve this equation, from equation () we have that X A B. So by finding A then multiplying it by B we can solve for X. Notice that A and B is the same matrice as the previous example so x X y 7 x matrices are nice because we have a very simple formula for their inverses. Any matrices that is bigger we can find their inverses by Gauss-Jordan Elimination. 6