Determinants and inverses of 2×2 matrices

Worked solution

1. 1

   Calculate the determinant.

   $detA=2\left(2\right)-1\left(3\right)=1$

   The determinant is the main-diagonal product minus the other diagonal product.

2. 2

   Swap diagonal entries and negate the others.

   $\mathrm{adj}\left(A\right)=\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)$

   This is the numerator matrix in the $2\times 2$ inverse formula.

3. 3

   Divide by the non-zero determinant.

   $A^{-1}=\frac{1}{1}\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)=\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)$

   Since the determinant is 1, this scalar leaves the entries unchanged.

4. 4

   Verify the inverse product.

   $A{A}^{-1}=\left(\begin{array}{cc}4-3& -2+2\\ 6-6& -3+4\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)$

   The product is the identity, confirming that the candidate undoes $A$.

Worked solution

1. 1

   Find the determinant before dividing.

   $detB=4\left(2\right)-0\left(1\right)=8\ne 0$

   A non-zero determinant guarantees that the inverse formula is valid.

2. 2

   Form the numerator matrix.

   $\mathrm{adj}\left(B\right)=\left(\begin{array}{cc}2& 0\\ -1& 4\end{array}\right)$

   Swap 4 and 2, and negate both off-diagonal entries.

3. 3

   Apply the scalar to every entry.

   $\begin{array}{rl}{B}^{-1}& =\frac{1}{8}\left(\begin{array}{cc}2& 0\\ -1& 4\end{array}\right)\\ & =\left(\begin{array}{cc}\frac{1}{4}& 0\\ -\frac{1}{8}& \frac{1}{2}\end{array}\right)\end{array}$

   Reducing each fraction preserves the common factor $\frac{1}{8}$.

4. 4

   Check by multiplication.

   $\begin{array}{rl}B{B}^{-1}& =\left(\begin{array}{cc}4\left(\frac{1}{4}\right)& 0\\ \frac{1}{4}+2\left(-\frac{1}{8}\right)& 2\left(\frac{1}{2}\right)\end{array}\right)\\ & =\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}$

   The lower-left cancellation verifies the sign on the fractional entry.