Exercises — Cramer's rule

The coefficients of $x,y$ in each row become the columns of $A$ (column 1 = the $x$-coefficients, column 2 = the $y$-coefficients): $A=\left(\begin{array}{cc}3& 2\\ 1& -1\end{array}\right),\mathbf{b}=\left(\begin{array}{c}7\\ 1\end{array}\right).$ To find $x$ ($i=1$), cover column 1 with $\mathbf{b}$: $A_1=\left(\begin{array}{cc}7& 2\\ 1& -1\end{array}\right).$ (Column 2 stays exactly as it was.)

(a) $\det A=(4)(1)-(2)(2)=4-4=0$. Rule does not apply — no unique solution. (b) $\det A=(4)(3)-(2)(1)=12-2=10\ne 0$. Rule applies. The condition $\det A\ne 0$ is exactly the invertibility check.

$A=\left(\begin{array}{cc}2& 3\\ 5& -1\end{array}\right),\mathbf{b}=\left(\begin{array}{c}8\\ 3\end{array}\right)$. $\det A=(2)(-1)-(3)(5)=-2-15=-17\ne 0$ ✓. $x=\frac{\det \left(\begin{array}{cc}8& 3\\ 3& -1\end{array}\right)}{-17}=\frac{(8)(-1)-(3)(3)}{-17}=\frac{-17}{-17}=1.$ $y=\frac{\det \left(\begin{array}{cc}2& 8\\ 5& 3\end{array}\right)}{-17}=\frac{(2)(3)-(8)(5)}{-17}=\frac{-34}{-17}=2.$ Check: $2(1)+3(2)=8$ ✓, $5(1)-2=3$ ✓. Answer $(x,y)=(1,2)$.

$A=\left(\begin{array}{ccc}1& 2& 1\\ 2& 5& 3\\ 1& 3& 2\end{array}\right),\mathbf{b}=\left(\begin{array}{c}4\\ 13\\ 8\end{array}\right)$. $\det A$ (cofactor along row 1): $1\underset{1}{\underbrace{(5\cdot 2-3\cdot 3)}}-2\underset{1}{\underbrace{(2\cdot 2-3\cdot 1)}}+1\underset{1}{\underbrace{(2\cdot 3-5\cdot 1)}}=1-2+1=0?$ Careful — recompute each minor: $5\cdot 2-3\cdot 3=10-9=1$; $2\cdot 2-3\cdot 1=4-3=1$; $2\cdot 3-5\cdot 1=6-5=1$. So $\det A=1(1)-2(1)+1(1)=0$. $\det A=0$ → Cramer's rule does NOT apply here. Switch to Gaussian Elimination. (This is deliberate: the trap is grinding out numerators without checking the denominator.)

$A=\left(\begin{array}{ccc}1& 1& 1\\ 2& -1& 1\\ 1& 2& -1\end{array}\right),\mathbf{b}=\left(\begin{array}{c}6\\ 3\\ 2\end{array}\right)$. $\det A$ (row 1): $1(-1\cdot -1-1\cdot 2)-1(2\cdot -1-1\cdot 1)+1(2\cdot 2-(-1)\cdot 1)$ $=1(1-2)-1(-2-1)+1(4+1)=1(-1)-1(-3)+1(5)=-1+3+5=7$. $x$: $A_1=\left(\begin{array}{ccc}6& 1& 1\\ 3& -1& 1\\ 2& 2& -1\end{array}\right)$, $\det A_1=6(1-2)-1(-3-2)+1(6+2)=6(-1)-1(-5)+1(8)=-6+5+8=7$; $x=7/7=1$. $y$: $A_2=\left(\begin{array}{ccc}1& 6& 1\\ 2& 3& 1\\ 1& 2& -1\end{array}\right)$, $\det A_2=1(3\cdot -1-1\cdot 2)-6(2\cdot -1-1\cdot 1)+1(2\cdot 2-3\cdot 1)=1(-5)-6(-3)+1(1)=-5+18+1=14$; $y=14/7=2$. $z$: $A_3=\left(\begin{array}{ccc}1& 1& 6\\ 2& -1& 3\\ 1& 2& 2\end{array}\right)$, $\det A_3=1(-1\cdot 2-3\cdot 2)-1(2\cdot 2-3\cdot 1)+6(2\cdot 2-(-1)\cdot 1)=1(-8)-1(1)+6(5)=-8-1+30=21$; $z=21/7=3$. Check: $1+2+3=6$ ✓, $2-2+3=3$ ✓, $1+4-3=2$ ✓. Answer $(1,2,3)$.

$x=\frac{\det A_1}{\det A}=\frac{12}{-4}=-3$, $y=\frac{\det A_2}{\det A}=\frac{0}{-4}=0$. Sign logic: numerator and denominator have opposite signs ($+$ over $-$), so the ratio is negative. A determinant is a signed volume — the sign flips when the column-box's orientation flips. $\det A_2=0$ just means covering column 2 with $\mathbf{b}$ collapses that box flat, giving $y=0$; the original box $\det A\ne 0$, so the system is still fine.

$A=\left(\begin{array}{cc}k& 1\\ 1& k\end{array}\right)$, $\det A=k^2-1=(k-1)(k+1)$. Fails when $\det A=0\Rightarrow k=1$ or $k=-1$. When it works ($k\ne \pm 1$): $\det A_1=\det \left(\begin{array}{cc}1& 1\\ 1& k\end{array}\right)=k-1$, so $x=\frac{k-1}{k^2-1}=\frac{k-1}{(k-1)(k+1)}=\frac{1}{k+1}$. By symmetry $\det A_2=\det \left(\begin{array}{cc}k& 1\\ 1& 1\end{array}\right)=k-1$, so $y=\frac{1}{k+1}$. Answer: $x=y=\frac{1}{k+1}$ for $k\ne \pm 1$. (At $k=1$ both equations become $x+y=1$ — infinitely many solutions. At $k=-1$ they contradict — none. Cramer correctly refuses both.)

The first column of $A^{-1}$ solves $A\mathbf{x}={\mathbf{e}}_1=\left(\begin{array}{c}1\\ 0\end{array}\right)$ (because $A{A}^{-1}=I$, so $A\times (\text{col }1)=\text{col }1\text{ of }I$). $\det A=(2)(2)-(1)(3)=1$. $x=\frac{\det \left(\begin{array}{cc}1& 1\\ 0& 2\end{array}\right)}{1}=\frac{2}{1}=2$, $y=\frac{\det \left(\begin{array}{cc}2& 1\\ 3& 0\end{array}\right)}{1}=\frac{-3}{1}=-3$. First column of $A^{-1}$ is $\left(\begin{array}{c}2\\ -3\end{array}\right)$. See Matrix Inverse.

As $k\to -1$, $\det A=(k-1)(k+1)\to 0$, so we're dividing by a vanishing volume.

• $k\to -1^{+}$ (e.g. $k=-0.9$): $x=\frac{1}{0.1}=+10$, growing to $+\infty$.
• $k\to -1^{-}$ (e.g. $k=-1.1$): $x=\frac{1}{-0.1}=-10$, going to $-\infty$. The solution blows up, exactly the numerical instability the parent warns about: near-zero $\det A$ makes $x_i$ enormous and sign-flippy. Geometrically, the two column-vectors become nearly parallel, so the box is nearly flat and a small change in $\mathbf{b}$ demands a huge mix.

$A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$, $\det A=ad-bc$. $x=\frac{\det \left(\begin{array}{cc}e& b\\ f& d\end{array}\right)}{ad-bc}=\frac{ed-bf}{ad-bc},y=\frac{\det \left(\begin{array}{cc}a& e\\ c& f\end{array}\right)}{ad-bc}=\frac{af-ce}{ad-bc}.$ Verify equation 1: $ax+by=\frac{a(ed-bf)+b(af-ce)}{ad-bc}=\frac{aed-abf+abf-bce}{ad-bc}=\frac{e(ad-bc)}{ad-bc}=e$ ✓. Verify equation 2: $cx+dy=\frac{c(ed-bf)+d(af-ce)}{ad-bc}=\frac{ced-bcf+adf-cde}{ad-bc}=\frac{f(ad-bc)}{ad-bc}=f$ ✓. Both hold identically — this is the parent's $\det A\cdot x_i=\det A_i$ made concrete for $n=2$. This alternating cancellation is the fingerprint of Multilinear and Alternating Maps.

$A=\left(\begin{array}{ccc}1& 2& 3\\ 2& -1& 1\\ 3& 1& -1\end{array}\right),\mathbf{b}=\left(\begin{array}{c}14\\ 3\\ 4\end{array}\right)$. $\det A$ (row 1): $1(-1\cdot -1-1\cdot 1)-2(2\cdot -1-1\cdot 3)+3(2\cdot 1-(-1)\cdot 3)$ $=1(1-1)-2(-2-3)+3(2+3)=0-2(-5)+3(5)=0+10+15=25$. $x$: $A_1=\left(\begin{array}{ccc}14& 2& 3\\ 3& -1& 1\\ 4& 1& -1\end{array}\right)$, $\det A_1=14(1-1)-2(-3-4)+3(3+4)=0-2(-7)+3(7)=14+21=35$; $x=35/25=7/5$. $y$: $A_2=\left(\begin{array}{ccc}1& 14& 3\\ 2& 3& 1\\ 3& 4& -1\end{array}\right)$, $\det A_2=1(3\cdot -1-1\cdot 4)-14(2\cdot -1-1\cdot 3)+3(2\cdot 4-3\cdot 3)=1(-7)-14(-5)+3(-1)=-7+70-3=60$; $y=60/25=12/5$. $z$: $A_3=\left(\begin{array}{ccc}1& 2& 14\\ 2& -1& 3\\ 3& 1& 4\end{array}\right)$, $\det A_3=1(-1\cdot 4-3\cdot 1)-2(2\cdot 4-3\cdot 3)+14(2\cdot 1-(-1)\cdot 3)=1(-7)-2(-1)+14(5)=-7+2+70=65$; $z=65/25=13/5$. Check eq 1: $\frac{7}{5}+2\cdot \frac{12}{5}+3\cdot \frac{13}{5}=\frac{7+24+39}{5}=\frac{70}{5}=14$ ✓. Answer: $(x,y,z)=\left(\frac{7}{5},\frac{12}{5},\frac{13}{5}\right)$.