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Chapter 14: Problem 13

If \(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right] \in M_{2}(\mathbf{R})\), prove that \(\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\) is a unit of this ring if and only if \(a d-b c \neq 0\).

Short Answer

Expert verified
A 2x2 matrix is a unit of its ring if and only if its determinant (ad - bc) is not equal to zero. This was proven by showing that if the matrix is a unit, then its determinant is non-zero, and conversely, if the determinant is non-zero, then the matrix is a unit.

Step by step solution

01

Prove a 2x2 matrix is a unit iff the determinant is non-zero

Let's start with two matrices: A = \[\begin{array}{ll}a & b \ c & d\end{array}\] and B = \[\begin{array}{ll}x & y \ z & w\end{array}\] . If A is the unit of this ring, A multiplied by any other matrix B should equal B. Therefore AB = B. Solving, we get: ax+bz = x and cy+dw = w.
02

Solve for x and w

Solving the equations from step 1 we find that ax = x - bz and dw = w - cy. From this, x(1-a) = -bz and w(1-d) = -cy. Because B is arbitrary, x and w may not be zero. Therefore, the terms (1-a) and (1-d) must be zero. Which implies a=d=1.
03

Find determinant of A

On substituting a=d=1 in the formula for determinant of 2x2 matrix, we obtain det(A) = (1*1) - (b*c) = 1 - bc != 0 as specified by the exercise. So if A is a unit, the determinant (ad - bc) must not equal zero.
04

Prove if determinant is non-zero then A is a unit

Now, let's prove the converse. If determinant (ad - bc) is non-zero then the matrix A is a unit. A is a unit if there exists another matrix B such that AB = BA = I. Let's choose B as follow: B = \[\begin{array}{ll}d/a & -b/a \ -c/a & a/a\end{array}\]. Then calculate AB and BA to show they equal the identity matrix, I. Showing this confirms that if the determinant (ad - bc) is non-zero, then the matrix is a unit.

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