Exercises — Inverse of 2×2 matrix

What we do: apply $\det A=ad-bc$ with $a=3,b=2,c=5,d=4$. Why: invertibility is decided entirely by the determinant — no need to build the inverse to answer the yes/no question. $\det A=(3)(4)-(2)(5)=12-10=2.$ Since $\det A=2\ne 0$, $A$ is invertible (non-singular).

What we do: compute each determinant; singular means it equals $0$.

• $\det P=(1)(4)-(2)(2)=4-4=0$ → singular (row 2 is $2\times$ row 1).
• $\det Q=(6)(2)-(3)(4)=12-12=0$ → singular (row 1 is $1.5\times$ row 2).
• $\det R=(0)(0)-(5)(5)=0-25=-25\ne 0$ → invertible (a negative determinant is fine — it just means the transform flips orientation, see Linear transformations and area). Why $R$ surprises people: two zeros on the diagonal feels degenerate, but degeneracy is about parallel columns, not about zeros sitting anywhere.

Step 1 — determinant (always first). $\det A=(4)(2)-(3)(3)=8-9=-1$. Non-zero ⟹ proceed. Why first? The whole recipe divides by $\det A$; if it were $0$ the inverse would not exist and every later step would be wasted (division by zero). Computing it first is the cheapest possible way to catch a dead end. Step 2 — adjugate. Swap the corners $4\leftrightarrow 2$, flip the middle $3\to -3$ (this is the $\mathrm{adj}A$ defined above): $\mathrm{adj}A=\left(\begin{array}{cc}2& -3\\ -3& 4\end{array}\right).$ Step 3 — divide by det. $A^{-1}=\frac{1}{-1}\left(\begin{array}{cc}2& -3\\ -3& 4\end{array}\right)=\left(\begin{array}{cc}-2& 3\\ 3& -4\end{array}\right).$ Step 4 — verify (cheap insurance): $\left(\begin{array}{cc}4& 3\\ 3& 2\end{array}\right)\left(\begin{array}{cc}-2& 3\\ 3& -4\end{array}\right)=\left(\begin{array}{cc}-8+9& 12-12\\ -6+6& 9-8\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)=I$ ✓.

Step 1 — determinant first. $\det B=(1)(5)-(2)(3)=5-6=-1$ (non-zero ⟹ inverse exists). Step 2 — adjugate then divide. $B^{-1}=\frac{1}{-1}\left(\begin{array}{cc}5& -2\\ -3& 1\end{array}\right)=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right).$ Step 3 — solve. Left-multiply $B\vec{x}=\vec{b}$ by $B^{-1}$ to get $\vec{x}=B^{-1}\vec{b}$ (see Solving linear systems): $\vec{x}=\left(\begin{array}{cc}-5& 2\\ 3& -1\end{array}\right)\left(\genfrac{}{}{0ex}{}{4}{7}\right)=\left(\genfrac{}{}{0ex}{}{(-5)(4)+2(7)}{3(4)+(-1)(7)}\right)=\left(\genfrac{}{}{0ex}{}{-20+14}{12-7}\right)=\left(\genfrac{}{}{0ex}{}{-6}{5}\right).$ Check: $1(-6)+2(5)=-6+10=4$ ✓, $3(-6)+5(5)=-18+25=7$ ✓. So $x=-6,y=5$.

$\det A=(2)(3)-(0)(1)=6$. $A^{-1}=\frac{1}{6}\left(\begin{array}{cc}3& 0\\ -1& 2\end{array}\right)=\left(\begin{array}{cc}1/2& 0\\ -1/6& 1/3\end{array}\right).$ $\det (A^{-1})=\left(\frac{1}{2}\right)\left(\frac{1}{3}\right)-\left(0\right)\left(-\frac{1}{6}\right)=\frac{1}{6}.$ Relationship: $\det (A^{-1})=\frac{1}{\det A}$. Why this must be true (area picture): $A$ scales every area by a factor $\det A=6$. The undo-machine $A^{-1}$ must scale area by the reciprocal $\frac{1}{6}$ so that doing both returns the original area — see Linear transformations and area.

Singular condition: $\det M=k\cdot k-2\cdot 8=k^2-16=0\Rightarrow k=\pm 4.$ So $M$ has no inverse exactly when $==k=4==$ or $==k=-4==$ (the two rows become proportional). Otherwise ($k\ne \pm 4$): $M^{-1}=\frac{1}{k^2-16}\left(\begin{array}{cc}k& -2\\ -8& k\end{array}\right).$ Sanity check at $k=6$: $\det =36-16=20$, $M^{-1}=\frac{1}{20}\left(\begin{array}{cc}6& -2\\ -8& 6\end{array}\right)$; multiply back to confirm $I$.

Proof. To show a matrix $X$ is the inverse of $AB$, it is enough to show $(AB)X=I$ (see Identity matrix). Take $X=B^{-1}{A}^{-1}$: $(AB)(B^{-1}{A}^{-1})=A(B{B}^{-1})A^{-1}=AI{A}^{-1}=A{A}^{-1}=I.$ We used associativity to regroup, then $B{B}^{-1}=I$ and $I$ vanishing under multiplication. So $(AB{)}^{-1}=B^{-1}{A}^{-1}$. Why the order flips (glove picture): to undo "put on socks, then shoes" you must "take off shoes, then socks" — reverse the last operation first. Numeric check. $AB=\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)\left(\begin{array}{cc}2& 0\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 0& 2\end{array}\right)$, $\det =4$, so $(AB{)}^{-1}=\frac{1}{4}\left(\begin{array}{cc}2& -2\\ 0& 2\end{array}\right)=\left(\begin{array}{cc}1/2& -1/2\\ 0& 1/2\end{array}\right).$ Separately $A^{-1}=\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)$, $B^{-1}=\left(\begin{array}{cc}1/2& 0\\ 0& 1/2\end{array}\right)$, and $B^{-1}{A}^{-1}=\left(\begin{array}{cc}1/2& 0\\ 0& 1/2\end{array}\right)\left(\begin{array}{cc}1& -1\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1/2& -1/2\\ 0& 1/2\end{array}\right).$ Match ✓.

Key idea: $(A^{-1}{)}^{-1}=A$ — undoing the undo returns the original. So just invert the given matrix. Let $C=A^{-1}$. $\det C=(2)(2)-(-1)(-3)=4-3=1.$ $A=C^{-1}=\frac{1}{1}\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right).$ Check: $A{A}^{-1}=\left(\begin{array}{cc}2& 1\\ 3& 2\end{array}\right)\left(\begin{array}{cc}2& -1\\ -3& 2\end{array}\right)=\left(\begin{array}{cc}4-3& -2+2\\ 6-6& -3+4\end{array}\right)=I$ ✓.

Step 1 — compute $A^2$ for $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$: $A^2=\left(\begin{array}{cc}{a}^2+bc& ab+bd\\ ca+dc& cb+d^2\end{array}\right)=\left(\begin{array}{cc}{a}^2+bc& b(a+d)\\ c(a+d)& d^2+bc\end{array}\right).$ Step 2 — subtract $(a+d)A$: $A^2-(a+d)A=\left(\begin{array}{cc}{a}^2+bc-a(a+d)& 0\\ 0& d^2+bc-d(a+d)\end{array}\right)=\left(\begin{array}{cc}bc-ad& 0\\ 0& bc-ad\end{array}\right)=-(\det A)I.$ Step 3 — rearrange: $A^2-(a+d)A+(\det A)I=O$ ✓ (the identity holds for every $2\times 2$ matrix). Step 4 — extract the inverse. Assume $\det A\ne 0$. Factor an $A$ out of the first two terms: $A(A-(a+d)I)=-(\det A)I\Rightarrow A\cdot \frac{(a+d)I-A}{\det A}=I.$ Therefore $A^{-1}=\frac{(a+d)I-A}{\det A}=\frac{1}{\det A}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),$ which is exactly the adjugate formula — recovered by pure algebra, no swapping-by-hand. Why this is beautiful: the same swap/negate pattern the parent note found by solving four equations falls out of one polynomial identity.

Step 1 — determinant. $\det A_{\epsilon }=(1)(1+\epsilon )-(1)(1)=1+\epsilon -1=\epsilon .$ Step 2 — inverse (swap corners, flip middle, divide by $\epsilon$): $A_{\epsilon }^{-1}=\frac{1}{\epsilon }\left(\begin{array}{cc}1+\epsilon & -1\\ -1& 1\end{array}\right)=\left(\begin{array}{cc}\frac{1+\epsilon }{\epsilon }& -\frac{1}{\epsilon }\\ -\frac{1}{\epsilon }& \frac{1}{\epsilon }\end{array}\right).$ Step 3 — limiting behaviour. As $\epsilon \to 0^{+}$ the divisor $\epsilon$ shrinks toward zero, so each entry $\frac{1}{\epsilon },\frac{1+\epsilon }{\epsilon }\to +\infty$ (and $-\frac{1}{\epsilon }\to -\infty$): the inverse blows up without bound. Step 4 — the singular limit. At exactly $\epsilon =0$ we have $A_0=\left(\begin{array}{cc}1& 1\\ 1& 1\end{array}\right)$ with $\det A_0=0$ and two identical (hence parallel) columns, so $A_0$ is singular: no inverse exists at all. The blow-up in Step 3 is the algebra warning you, as you approach that wall, that the undo-machine is about to become impossible. The story (edge picture): near-singular matrices have huge, unstable inverses — a small wobble in the input data gets amplified by the factor $1/\epsilon$. At the wall $\epsilon =0$ the area collapses to a line and the inverse ceases to exist entirely. This is why dividing by a tiny determinant is numerically dangerous — see Singular vs non-singular matrices.

The figure below plots exactly this collapse.