10.6 The identity and inverse matrices

$5\times 1=$

$1\times \frac{2}{3}=$

$\mathrm{–\pi }\times 1=$

$1\times x=$

$2\times \frac{1}{2}=$

$\frac{-2}{3}x\frac{-3}{2}=$

Write the equation that defines two numbers $a$ and $b$ as inverses of each other.

Find the inverse of $\frac{4}{5}$ .

Find the inverse of –3.

Find the inverse of $x$ .

Is there any number that does not have an inverse, according to your definition in #7?

$\left[\begin{array}{cc}3& 8\\ -4& 12\end{array}\right]\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=$

$\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}3& 8\\ -4& 12\end{array}\right]=$

We have just seen that $\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$ acts as the multiplicative identify for a 2×2 matrix.

• A

  What is the multiplicative identity for a 3×3 matrix?

• B

  Test this identity to make sure it works.

• C

  What is the multiplicative identity for a 5×5 matrix? (I won’t make you test this one…)

• D

  What is the multiplicative identity for a 2×3 matrix?

• E

  Trick question! There isn’t one. You could write a matrix that satisfies $\mathrm{AI}=A$ , but it would not also satisfy $\mathrm{IA}=A$ —that is, it would not commute, which we said was a requirement. Don’t take my word for it, try it! The point is that only square matrices (*same number of rows as columns) have an identity matrix.

Multiply: $\left[\begin{array}{cc}2& 2\frac{1}{2}\\ -1& -1\frac{1}{2}\end{array}\right]$ $\left[\begin{array}{cc}3& 5\\ -2& -4\end{array}\right]$

Multiply: $\left[\begin{array}{cc}3& 5\\ -2& -4\end{array}\right]$ $\left[\begin{array}{cc}2& 2\frac{1}{2}\\ -1& -1\frac{1}{2}\end{array}\right]$

Find the inverse of the matrix $\left[\begin{array}{cc}3& 2\\ 5& 4\end{array}\right]$

• A

  Since we don’t know the inverse yet, we will designate it as a bunch of unknowns: $\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ will be our inverse matrix. Write down the equation that defines this unknown matrix as our inverse matrix.

• B

  Now, in your equation, you had a matrix multiplication. Go ahead and do that multiplication, and write a new equation which just sets two matrices equal to each other.

• C

  Now, remember that when we set two matrices equal to each other, every cell must be equal. So, when we set two different 2x2 matrices equal, we actually end up with four different equations. Write these four equations.

• D

  Solve for $a$ , $b$ , $c$ , and $d$ .

• E

  So, write the inverse matrix $A^{-1}$ .

• F

  Test this inverse matrix to make sure it works!