Exercises — Determinant of 2×2 matrix

WHAT: apply $ad-bc$ with $a=4,b=1,c=2,d=3$. WHY: direct definition, nothing else needed. $\det =(4)(3)-(1)(2)=12-2=10$ Meaning: areas are scaled by $10$; the sign is $+$, so orientation is preserved.

$\det =(5)(2)-(5)(2)=10-10=0$ A zero determinant means the matrix is singular (non-invertible). Look at the columns: $(5,2)$ and $(5,2)$ are identical, so they lie on the same line — the parallelogram is flat, area $0$. See Linear independence: the columns are dependent.

$\det =(0)(0)-(7)(-3)=0-(-21)=21$ WHY the sign flip inside: $bc=7\times (-3)=-21$, and subtracting a negative adds. This is exactly where students slip — track the sign of every entry.

WHY the determinant: area of the parallelogram from two vectors is $|ad-bc|$ — this is the whole point of the determinant (see Area of parallelogram). Put the vectors as columns: $A=\left[\begin{array}{cc}3& 1\\ 1& 4\end{array}\right],\det =(3)(4)-(1)(1)=12-1=11.$ Area $=|11|=11$. Positive sign $\Rightarrow$ $\vec{v}$ is counter-clockwise from $\vec{u}$.

Part 1: From Inverse of a 2×2 matrix, $\det (A)\det (A^{-1})=\det (I)=1$, so $\det (A^{-1})=\frac{1}{\det (A)}=\frac{1}{4}.$ Part 2: First check the actual $A$: $\det =(3)(2)-(2)(1)=6-2=4$ (matches). The inverse formula swaps $a\leftrightarrow d$, negates the off-diagonal, then divides by the determinant: $A^{-1}=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -1& 3\end{array}\right]=\left[\begin{array}{cc}\frac{1}{2}& -\frac{1}{2}\\ -\frac{1}{4}& \frac{3}{4}\end{array}\right].$

Singular means $\det =0$. WHY: zero area $\Rightarrow$ no inverse. $\det =k\cdot k-3\cdot 2=k^2-6=0\Rightarrow k=\pm \sqrt{6}.$ Two answers — a common oversight is to keep only $+\sqrt{6}$.

$\det (AB)$: multiplicativity, $\det (AB)=\det A\det B=5\cdot (-2)=-10$. $\det (3A)$: scaling a $2\times 2$ matrix by $3$ scales both columns by $3$, so area scales by $3\times 3=9$: $\det (3A)=3^2\det (A)=9\cdot 5=45.$ $\det (A^{T}B)$: $\det (A^T)=\det (A)=5$ (transpose invariance), so $\det (A^{T}B)=5\cdot (-2)=-10$.

Rotation: $\det (R)=(0)(0)-(-1)(1)=0+1=1$. A rotation moves the square rigidly — area unchanged (factor $1$) and orientation preserved (positive sign). See Transformations and scaling. Reflection: $\det =(0)(0)-(1)(1)=-1$. Area unchanged (magnitude $1$) but the sign is negative — a reflection flips orientation, turning a counter-clockwise square clockwise, like a mirror image.

Start with $\det \left[\begin{array}{cc}a& b\\ c& d\end{array}\right]=ad-bc$. Swap columns to get $\left[\begin{array}{cc}b& a\\ d& c\end{array}\right]$: $\det =b\cdot c-a\cdot d=bc-ad=-(ad-bc).$ So the determinant flips sign. Geometric reason: swapping the two spanning arrows swaps which one is "counter-clockwise," reversing the signed orientation.

Set up. Coefficient matrix $A=\left[\begin{array}{cc}2& 1\\ 1& 3\end{array}\right]$, right side $(5,10)$. Main determinant: $\det A=(2)(3)-(1)(1)=6-1=5$. Non-zero $\Rightarrow$ unique solution exists. Replace column 1 with the right side for $x$: $D_x=\det \left[\begin{array}{cc}5& 1\\ 10& 3\end{array}\right]=(5)(3)-(1)(10)=15-10=5.$ Replace column 2 for $y$: $D_y=\det \left[\begin{array}{cc}2& 5\\ 1& 10\end{array}\right]=(2)(10)-(5)(1)=20-5=15.$ Divide: $x=\frac{D_x}{\det A}=\frac{5}{5}=1,y=\frac{D_y}{\det A}=\frac{15}{5}=3.$ Check: $2(1)+3=5$ ✓, $1+3(3)=10$ ✓.

Singular $\Rightarrow$ determinant zero: $\det =(4-\lambda )(3-\lambda )-(1)(2)=0.$ Expand: $12-4\lambda -3\lambda +{\lambda }^2-2={\lambda }^2-7\lambda +10=0.$ Factor: $(\lambda -2)(\lambda -5)=0\Rightarrow \lambda =2\text{ or }\lambda =5.$ Sanity check via a property: the product of eigenvalues equals the determinant: $2\times 5=10$, and $\det \left[\begin{array}{cc}4& 1\\ 2& 3\end{array}\right]=12-2=10$ ✓.

Strategy: $\det =0$ means the columns are parallel — the second column must be a scalar multiple of the first. Pick first column $(2,3)$ and multiply by $2$ to get second column $(4,6)$: $A=\left[\begin{array}{cc}2& 4\\ 3& 6\end{array}\right],\det =(2)(6)-(4)(3)=12-12=0.$ All entries non-zero ✓, and it is not a multiple of the all-ones matrix ✓. (Infinitely many valid answers exist — any parallel-column matrix works. See Linear independence.)

Requirement: $ad-bc=-3$. Easiest is a reflection-then-scale: take $\left[\begin{array}{cc}3& 0\\ 0& -1\end{array}\right]$: $\det =(3)(-1)-(0)(0)=-3.$ Area factor $|-3|=3$ (tripled) ✓, negative sign = orientation reversed ✓.

Take $A=B=I=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$. $\det (A)+\det (B)=1+1=2.$ But $A+B=\left[\begin{array}{cc}2& 0\\ 0& 2\end{array}\right]$, so $\det (A+B)=(2)(2)-0=4\ne 2.$ Conclusion: determinant is not additive. It is multiplicative ($\det (AB)=\det A\det B$) because area-scaling factors compose by multiplication when you apply one transformation after another (Matrix multiplication) — but adding matrices is not "doing one then the other," so no such rule exists for sums.

L1 plug-in ::: apply $ad-bc$, watch signs L2 area & inverse ::: columns span area; inverse $=\frac{1}{\det }\left[\begin{array}{cc}d& -b\\ -c& a\end{array}\right]$ L3 properties ::: $\det (AB)=\det A\det B$, $\det (kA)=k^2\det A$, swap flips sign L4 systems ::: Cramer $x=D_x/\det A$; eigenvalues from $\det =0$ L5 build & prove ::: engineer $\det$ to a target; $\det$ is not additive