Show that a two-rowed nonsingular matrix A=[ a11 a12 a21 a22 ] has the inverse

Show that a two-rowed nonsingular matrix A=[ a11 a12 ...

Key Concepts

Matrix Inversion Formula

For a square matrix that is nonsingular, its inverse can be explicitly computed in terms of its entries and its determinant. This formula provides a direct method to compute the inverse, using the cofactor (or adjugate) matrix and normalizing by the determinant, which is critical for both theoretical and practical applications in linear algebra.

Eigenvalues of an Inverse Matrix

A fundamental property in linear algebra is that if a matrix has an eigenvalue ? with a corresponding eigenvector, then its inverse (when it exists) has the eigenvalue 1/? for the same eigenvector. This relationship emerges naturally from the eigenvalue equation and ensures that applying the inverse undoes the scaling effect of the original matrix.

Characteristic Polynomial

The characteristic polynomial of a matrix encapsulates its eigenvalues as the roots of the polynomial. By examining the characteristic polynomial of the inverse matrix, one can show that its roots are the reciprocals of the roots of the original matrix's characteristic polynomial, connecting the spectral properties of a matrix and its inverse.

Spectral Mapping Theorem

The spectral mapping theorem is a powerful tool in linear algebra that relates the eigenvalues of a function of a matrix to the function of the eigenvalues of the matrix. When the function is the reciprocal function, this theorem supports the conclusion that the eigenvalues of the inverse matrix are the reciprocals of the eigenvalues of the original matrix.

Transcript

00:02 We want to derive the following formula for the inverse of a 2x2 matrix.

00:08 If we have an invertible 2x2 matrix, a, b, c, d, we want to prove that its inverse is 1 over the determinant, ad minus b, c.

00:22 And this fraction makes sense because if it's invertible, that means that determinant is non -zero, so we are not dividing by 0 here.

00:31 Multiplied by the matrix d minus b minus c .a.

00:39 So the entries on the main diagonal are swapped and the other entries are negated.

00:45 So let's fix an invertible matrix a, b, c, d.

00:58 And if it's an invertible matrix, it has an inverse.

01:01 Let's call it capital a, b, c, d.

01:06 And by matrix multiplication we get that this matrix is equal to a a plus bc a a plus b d c a plus b d c a plus d c a plus d but we also know that this matrix because it is the product of a matrix for this inverse is the identity matrix, one zero, zero, one.

01:49 So this gets us four equations by equating the entries of these two matrices farthest on the right.

01:58 We get that.

02:00 A, a plus b, c is one.

02:07 C, b plus d, d is one.

02:15 A, b plus b, d, d.

02:21 Is 0 and ca plus d c is 0.

02:35 So let's start by trying to solve for capital a, our first entry.

02:45 Well, from this bottom equation, we get the fact that dc equals minus c a.

03:06 So i'm going to try to convert this c in the first equation capital c into some a, but possibly with some extra factors.

03:20 So the first thing i'm going to do is i'm going to multiply this first equation by d because i have a conversion for dc.

03:29 So that gets me ad capital a plus b, d, c equals d.

03:43 Now, since dc is minus c a, i can rewrite this as a -d -a plus b minus c -a.

03:59 So minus b, c, a equals d.

04:08 I can factor out capital a, ad minus b c equals d.

04:18 And finally, i can solve for a equals d over ad minus b c.

04:28 And i can in fact divide by this.

04:31 I know it's non -zero because it's the determinant.

04:33 Of our original matrix, lowercase a, b, c, d.

04:37 And because it's invertible, the determinant is non -zero.

04:41 So we found that capital a is d divided by the determinant, which is looking good.

04:49 If we get that every capital letter here is equal to the lowercase letter on the right -hand side of the equation we're trying to prove, but each one divided by the determinant, then we will have derived this expression.

05:09 So a checks out.

05:11 Let's try capital b.

05:19 So for capital b, i'm going to use this third equation to first find a way to convert some form of capital d into some form of capital b.

05:29 So i have that from the third equation, bd equals minus a, b.

05:42 And from this, i can now do the same thing of substituting this expression into the second equation.

06:00 But to do that, i yet again have to multiply the second equation by, let's see, we want to convert a bd the second equation does not have a b next to its d.

06:16 So i'm going to multiply the second equation by b...

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