Exercises — Solving 2×2 systems using Cramer's rule

2.6.11 · D4 · Maths › Matrices & Determinants — Introduction › Solving 2×2 systems using Cramer's rule

Poore note mein, hamara system hamesha yahi hoga: $a_{1}x+b_{1}y=c_1,a_{2}x+b_{2}y=c_2,$ aur teen determinants hain (parent definitions, dobara likhe gaye taaki kuch assume na ho):

D_x=\begin{vmatrix}c_1&b_1\\c_2&b_2\end{vmatrix},\quad D_y=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}.$$ > [!definition] "Diagonal down minus diagonal up" — $2\times2$ determinant *hota kya hai* > Ek determinant $\begin{vmatrix}p&q\\r&s\end{vmatrix}$ ek single number hai: **main diagonal** > (top-left se bottom-right, $p\to s$) lo aur multiply karo, $ps$; **anti-diagonal** (top-right se > bottom-left, $q\to r$) lo aur multiply karo, $qr$; phir usi order mein **subtract** karo: $ps-qr$. Neeche figure mein arrows dekho — green arrow "plus" hai, red arrow "minus" hai. ![[deepdives/dd-maths-2.6.11-d4-s01.png]] Dekhna [[Determinant of a 2×2 matrix]] — yahan explain hai ki yeh number rows se bane parallelogram ka signed area kyun hota hai. --- ## Level 1 — Recognition *Goal: $D$, $D_x$, $D_y$ sahi sahi identify karo — abhi solve nahi karna.* > [!example] Q1. Pieces ka naam lo > System $5x - 2y = 7,\quad 3x + 4y = 1$ ke liye, $D$, $D_x$, aur $D_y$ ko **determinants ke roop mein** likho > (abhi evaluate mat karo), phir har ek evaluate karo. > [!recall]- Solution Q1 > Template se match karo. Yahan $a_1=5,\ b_1=-2,\ c_1=7$ aur $a_2=3,\ b_2=4,\ c_2=1$. > $$D=\begin{vmatrix}5&-2\\3&4\end{vmatrix},\quad > D_x=\begin{vmatrix}7&-2\\1&4\end{vmatrix},\quad > D_y=\begin{vmatrix}5&7\\3&1\end{vmatrix}.$$ > **YEH KYUN:** $D_x$ mein column 1 ($x$-column) ko constants $\binom{7}{1}$ se swap kiya gaya hai; $D_y$ mein > column 2 ($y$-column) swap hua. "Diagonal down minus diagonal up" se evaluate karo: > $$D=(5)(4)-(-2)(3)=20+6=26,$$ > $$D_x=(7)(4)-(-2)(1)=28+2=30,$$ > $$D_y=(5)(1)-(7)(3)=5-21=-16.$$ > [!example] Q2. Kaun sa column swap hua? > Kisine system $2x + 6y = 9,\ -x + 8y = 9$ ke liye $\begin{vmatrix}2&9\\-1&9\end{vmatrix}$ likha. Kya yeh $D_x$ hai ya $D_y$? Aur kya yeh sahi bhi hai? > [!recall]- Solution Q2 > Constant column yahan $\binom{9}{9}$ hai. Likhe hue determinant mein, column 2 hai $\binom{9}{9}$ > aur column 1 original $a$-column $\binom{2}{-1}$ hai. **Column 2** ko constants se swap karna > $==D_y==$ ki recipe hai. Toh yeh $D_y$ hai. > Kya yeh sahi hai? Template kehta hai $D_y=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix} > =\begin{vmatrix}2&9\\-1&9\end{vmatrix}$ — **haan, sahi hai.** Value: $(2)(9)-(9)(-1)=18+9=27$. > [!mistake] L1 trap — "constants upar hain isliye yeh $D_x$ hai." > **Kyun sahi lagta hai:** constant column clearly dikhta hai, aur students "answer numbers upar" ko $x$ ke numerator se associate karte hain. **Sachchi baat:** *kaun sa variable* — yeh decide hota hai *kaun sa column replace hua* se, sirf constants ki maujoodgi se nahi. Constants **column 1** mein → $D_x$; constants **column 2** mein → $D_y$. **Fix:** apne aap se bolo "variable = column." --- ## Level 2 — Application *Goal: ek saaf unique system poora end-to-end solve karo.* > [!example] Q3. Solve karo > $$3x + 2y = 12, \qquad x - y = 1.$$ > [!recall]- Solution Q3 > **Step 1 — pehle $D$ (KYUN: agar $0$ hai toh rukna padega).** > $D=\begin{vmatrix}3&2\\1&-1\end{vmatrix}=(3)(-1)-(2)(1)=-3-2=-5\neq0.$ Achha, unique solution hai. > **Step 2 — $D_x$** (column 1 ko $\binom{12}{1}$ se swap karo): > $D_x=\begin{vmatrix}12&2\\1&-1\end{vmatrix}=(12)(-1)-(2)(1)=-12-2=-14.$ > **Step 3 — $D_y$** (column 2 swap karo): > $D_y=\begin{vmatrix}3&12\\1&1\end{vmatrix}=(3)(1)-(12)(1)=3-12=-9.$ > **Step 4:** $x=\dfrac{D_x}{D}=\dfrac{-14}{-5}=\dfrac{14}{5},\quad y=\dfrac{D_y}{D}=\dfrac{-9}{-5}=\dfrac{9}{5}.$ > **Check** eq.(2) mein: $\tfrac{14}{5}-\tfrac{9}{5}=\tfrac{5}{5}=1.$ ✓ > [!example] Q4. Answer mein fractions > $$4x + 5y = 3, \qquad 6x - 2y = 8.$$ > [!recall]- Solution Q4 > $D=\begin{vmatrix}4&5\\6&-2\end{vmatrix}=(4)(-2)-(5)(6)=-8-30=-38.$ > $D_x=\begin{vmatrix}3&5\\8&-2\end{vmatrix}=(3)(-2)-(5)(8)=-6-40=-46.$ > $D_y=\begin{vmatrix}4&3\\6&8\end{vmatrix}=(4)(8)-(3)(6)=32-18=14.$ > $x=\dfrac{-46}{-38}=\dfrac{23}{19},\quad y=\dfrac{14}{-38}=-\dfrac{7}{19}.$ > **Check** eq.(1): $4\cdot\tfrac{23}{19}+5\cdot(-\tfrac{7}{19})=\tfrac{92-35}{19}=\tfrac{57}{19}=3.$ ✓ > [!example] Q5. Ek constant zero hai > $$2x + 7y = 0, \qquad 5x + 3y = 29.$$ > [!recall]- Solution Q5 > $D=\begin{vmatrix}2&7\\5&3\end{vmatrix}=(2)(3)-(7)(5)=6-35=-29.$ > $D_x=\begin{vmatrix}0&7\\29&3\end{vmatrix}=(0)(3)-(7)(29)=-203.$ > $D_y=\begin{vmatrix}2&0\\5&29\end{vmatrix}=(2)(29)-(0)(5)=58.$ > $x=\dfrac{-203}{-29}=7,\quad y=\dfrac{58}{-29}=-2.$ > **Note:** $0$ constant bilkul theek hai — determinant ke andar bas ek product zero ho jaata hai, kuch tootata nahi. > **Check** eq.(1): $2(7)+7(-2)=14-14=0.$ ✓ > [!mistake] L2 trap — "answer ka sign flip karo, determinant ka nahi." > **Kyun sahi lagta hai:** jab $D<0$ hota hai toh students do negatives dekh ke jaldi mein kabhi kabhi > $x=-D_x/D$ likhte hain "positive banana" ke liye. **Sachchi baat:** formula hai $x=D_x/D$ — **har determinant ka > apna sign rakho**; negative divided by negative already sahi positive value deta hai. > Q3 mein, $-14/-5=+14/5$ automatically aata hai. **Fix:** $D_x$ aur $D$ ko sahi sign ke saath compute karo, phir ek baar divide karo — extra minus signs kabhi mat daalo. --- ## Level 3 — Analysis *Goal: $D=0$ interpret karo — no-solution vs infinite-solution classify karo.* ![[deepdives/dd-maths-2.6.11-d4-s02.png]] > [!example] Q6. Classify karo > $$3x - 6y = 9, \qquad -2x + 4y = 5.$$ > [!recall]- Solution Q6 > $D=\begin{vmatrix}3&-6\\-2&4\end{vmatrix}=(3)(4)-(-6)(-2)=12-12=0.$ > $D=0$, toh **koi unique solution nahi** — investigate karo. Ek numerator compute karo: > $D_x=\begin{vmatrix}9&-6\\5&4\end{vmatrix}=(9)(4)-(-6)(5)=36+30=66\neq0.$ > Kyunki $D=0$ par $D_x\neq0$: **koi solution nahi** (parallel, alag alag lines — figure ka blue/orange left panel). > Geometrically dono lines ka slope same hai par intercepts alag hain, isliye kabhi cross nahi karte. > [!example] Q7. Classify karo > $$x + 3y = 4, \qquad 2x + 6y = 8.$$ > [!recall]- Solution Q7 > $D=\begin{vmatrix}1&3\\2&6\end{vmatrix}=6-6=0.$ Dono investigate karo: > $D_x=\begin{vmatrix}4&3\\8&6\end{vmatrix}=24-24=0,\quad D_y=\begin{vmatrix}1&4\\2&8\end{vmatrix}=8-8=0.$ > Teeno zero → **infinitely many solutions** (ek hi same line — figure ka right panel, > jahan blue aur orange perfectly overlap karte hain). Parametrize karo: eq.(1) se, $x=4-3y$, jahan $y$ free hai. > [!example] Q8. Kaun sa case hai, aur kyun? > Fully solve kiye bina, decide karo ki > $$6x - 4y = 10, \qquad -3x + 2y = -5$$ > ke kitne solutions hain. > [!recall]- Solution Q8 > $D=\begin{vmatrix}6&-4\\-3&2\end{vmatrix}=(6)(2)-(-4)(-3)=12-12=0.$ Degenerate hai. > $D_x=\begin{vmatrix}10&-4\\-5&2\end{vmatrix}=(10)(2)-(-4)(-5)=20-20=0$ check karo. > Aur $D_y=\begin{vmatrix}6&10\\-3&-5\end{vmatrix}=(6)(-5)-(10)(-3)=-30+30=0$. > Teeno zero → **infinitely many solutions.** (Sach mein eq.(2) exactly $-\tfrac12\times$ eq.(1) hai.) > [!mistake] L3 trap — "$D=0$ mila; done, system fail ho gaya." > **Kyun sahi lagta hai:** $D=0$ division block karta hai, toh lagta hai "impossible" hai. **Sachchi baat:** > $D=0$ ka sirf matlab hai *unique nahi*. Tumhe **$D_x$ (aur $D_y$) test karna padega**: agar koi nonzero hai → no solution; > agar sab zero hain → infinitely many. Q6 aur Q7 dono mein $D=0$ hai phir bhi opposite answers hain. **Fix:** $D=0$ ke baad kabhi conclude mat karo — cases split karne ke liye ek numerator compute karo. --- ## Level 4 — Synthesis *Goal: parameters, aur Cramer ko algebra ke saath combine karna.* > [!example] Q9. Woh $k$ dhundho jisse system ka **koi unique solution na ho** > $$kx + 3y = 6, \qquad 2x + y = 4.$$ > Har woh $k$ dhundho jiske liye solution *unique nahi* hai, phir har aise $k$ ko classify karo. > [!recall]- Solution Q9 > Uniqueness exactly tab fail hoti hai jab $D=0$. > $D=\begin{vmatrix}k&3\\2&1\end{vmatrix}=(k)(1)-(3)(2)=k-6.$ > $D=0\iff k=6$. **Toh $k=6$** hi ek problematic value hai. > **Ab $k=6$ classify karo.** $k=6$ ke saath: $D_x=\begin{vmatrix}6&3\\4&1\end{vmatrix}=(6)(1)-(3)(4)=6-12=-6\neq0.$ > Kyunki $D=0,\ D_x\neq0$: **$k=6$ par koi solution nahi.** Har $k\neq6$ ke liye: **unique solution.** > (Yahan infinitely many dene wali $k$ ki koi value *nahi* hai, kyunki $D_x$ hamesha $-6\neq0$ rehta hai jab $D=0$ ho.) > [!example] Q10. Parameter ke saath solve karo, phir specialize karo > $2x + y = m,\ x + 3y = 1$ ko $x,y$ mein $m$ ke terms mein solve karo, phir woh $m$ dhundho jo $x=y$ banata hai. > [!recall]- Solution Q10 > $D=\begin{vmatrix}2&1\\1&3\end{vmatrix}=6-1=5\neq0$ har $m$ ke liye, toh unique solution hamesha exist karta hai. > $D_x=\begin{vmatrix}m&1\\1&3\end{vmatrix}=3m-1,\qquad D_y=\begin{vmatrix}2&m\\1&1\end{vmatrix}=2-m.$ > $$x=\frac{3m-1}{5},\qquad y=\frac{2-m}{5}.$$ > $x=y$ set karo: $\dfrac{3m-1}{5}=\dfrac{2-m}{5}\Rightarrow 3m-1=2-m\Rightarrow 4m=3\Rightarrow m=\dfrac34.$ > **Check:** $m=\tfrac34$ par, $x=\dfrac{3(\frac34)-1}{5}=\dfrac{\frac94-1}{5}=\dfrac{5/4}{5}=\dfrac14$, > aur $y=\dfrac{2-\frac34}{5}=\dfrac{5/4}{5}=\dfrac14.$ Sach mein $x=y=\tfrac14$. ✓ > [!example] Q11. Ek saath do conditions > System $ax + 2y = 6,\ 3x + by = 9$ ka solution $x=2,\ y=1$ jaana jaata hai. $a$ aur $b$ dhundho, > phir Cramer's rule se confirm karo ki yahi sach mein *the* solution hai. > [!recall]- Solution Q11 > $x=2,\ y=1$ dono equations mein plug karo: > eq.(1): $2a+2=6\Rightarrow a=2$. eq.(2): $6+b=9\Rightarrow b=3$. > Toh system hai $2x+2y=6,\ 3x+3y=9$. **Par dhyan do:** > $D=\begin{vmatrix}2&2\\3&3\end{vmatrix}=6-6=0.$ Unique nahi! > $D_x=\begin{vmatrix}6&2\\9&3\end{vmatrix}=18-18=0,\ D_y=\begin{vmatrix}2&6\\3&9\end{vmatrix}=18-18=0.$ > **Infinitely many solutions** — $(2,1)$ unme se *ek* hai, akela nahi. (Dono equations $x+y=3$ mein reduce ho jaati hain.) **Lesson:** "solution $(2,1)$ hai" ka matlab zaroor nahi "unique hai"; Cramer degeneracy reveal karta hai. > [!mistake] L4 trap — "$D=0$ mila; bas, system fail ho gaya." > **Kyun sahi lagta hai:** Q9 mein $D=0$ set karne se hi poora problem solve ho gaya, toh students har jagah $D=0$ par ruk jaate hain. **Sachchi baat:** question abhi bhi pooch sakta hai *kaun sa* degenerate case hai (no solution vs > infinite), jiske liye $D_x,D_y$ chahiye — dekho Q11, jahan $D=0$ actually infinitely many deta hai, "no solution" nahi. **Fix:** $D=0$ ko ek fork samjho, koi endpoint nahi. --- ## Level 5 — Mastery *Goal: prove karo aur banao — machinery khud samjho.* > [!example] Q12. $y$-numerator prove karo > $a_1x+b_1y=c_1,\ a_2x+b_2y=c_2$ se shuru karke, elimination se $y=\dfrac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}$ derive karo, aur dikhao ki numerator $D_y$ ke barabar hai. > [!recall]- Solution Q12 > **$x$ eliminate karo** (KYUN: hume $y$ akela chahiye). Eq.(1) ko $a_2$ se aur eq.(2) ko $a_1$ se multiply karo: > $$a_1a_2x+a_2b_1y=a_2c_1,\qquad a_1a_2x+a_1b_2y=a_1c_2.$$ > Dono $x$-terms ab $a_1a_2x$ hain — pehle ko doosre mein se subtract karo: > $$(a_1b_2-a_2b_1)\,y = a_1c_2-a_2c_1.$$ > KYUN itna clean lagta hai: shared $a_1a_2x$ exactly cancel ho jaata hai. $D=a_1b_2-a_2b_1$ se divide karo: > $$y=\frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}.$$ > Numerator $a_1c_2-a_2c_1=\begin{vmatrix}a_1&c_1\\a_2&c_2\end{vmatrix}=D_y$: $A$ ka column 2 > constants se replace kiya gaya. $\blacksquare$ > [!example] Q13. Elimination se vs Cramer se consistency check > Dikhao ki $x + y = 2,\ 2x + 2y = 5$ inconsistent hai (no solution) **do tareekon se**: (a) elimination se, > (b) Cramer's classification se. Confirm karo ki dono verdicts agree karte hain. > [!recall]- Solution Q13 > **(a) Elimination.** Eq.(1) ko $2$ se multiply karo: $2x+2y=4$. Eq.(2) se compare karo: $2x+2y=5$. > Subtract karo: $0=1$, contradiction → **koi solution nahi.** > **(b) Cramer.** $D=\begin{vmatrix}1&1\\2&2\end{vmatrix}=2-2=0$; > $D_x=\begin{vmatrix}2&1\\5&2\end{vmatrix}=4-5=-1\neq0.$ $D=0,\ D_x\neq0\Rightarrow$ **koi solution nahi.** > Dono agree karte hain. General classification ke liye dekho [[Consistency of linear systems]]. > [!example] Q14. Geometry banao > Explain karo, "$D$ = coefficient rows se bane parallelogram ka signed area" idea use karke, ki system $4x + 6y = c_1,\ 2x + 3y = c_2$ ka kisi bhi $c_1,c_2$ ke liye *kabhi bhi* unique solution kyun nahi ho sakta. > [!recall]- Solution Q14 > Coefficient rows hain $(4,6)$ aur $(2,3)$. Dekho ki $(4,6)=2\cdot(2,3)$: yeh **same > direction** mein point karte hain (row 1 exactly row 2 ka do guna hai). Ek hi line par do arrows ek **zero area** ka parallelogram span karte hain, > isliye $D=0$ — confirm karo: $D=(4)(3)-(6)(2)=12-12=0$. > Kyunki $D=0$ $c_1,c_2$ se independent hai, lines **ya toh identical hain ya parallel**, kabhi ek point par cross nahi karte > → **kabhi bhi unique solution nahi.** > Dono mein se kaun sa? Constants par depend karta hai: $D_x=\begin{vmatrix}c_1&6\\c_2&3\end{vmatrix}=3c_1-6c_2 > =3(c_1-2c_2)$. Agar $c_1=2c_2$ ⇒ $D_x=0$ ⇒ infinitely many; warna koi solution nahi. Picture yeh hai: $D=0$ row-parallelism hi unique crossing forbid karta hai. Dekho [[Determinant of a 2×2 matrix]]. > [!mistake] L5 trap — "parallel rows sahi constants ke liye phir bhi cross ho sakte hain." > **Kyun sahi lagta hai:** $c_1,c_2$ change karne se lines shift hoti hain, toh *lagta hai* unhe ek > crossing par nudge kar sakte hain. **Sachchi baat:** constants lines shift karte hain par **unka slope nahi badal sakte**; parallel lines parallel hi rehti hain. $D$ sirf coefficients par depend karta hai, toh jab ek baar $D=0$ ho gaya toh koi constant unique solution create nahi kar sakta — sirf "no solution" aur "infinitely many" mein se choose kar sakta hai. **Fix:** yaad rakho $D$ constants ko bilkul ignore karta hai. --- > [!recall]- Answer key (quick reference) > Q1 $D=26,D_x=30,D_y=-16$. Q2 $D_y=27$, sahi. Q3 $x=\tfrac{14}{5},y=\tfrac95$. Q4 $x=\tfrac{23}{19},y=-\tfrac{7}{19}$. > Q5 $x=7,y=-2$. Q6 koi solution nahi. Q7 infinitely many. Q8 infinitely many. Q9 $k=6$ → koi solution nahi; > warna unique. Q10 $x=\tfrac{3m-1}{5},y=\tfrac{2-m}{5}$, $x=y$ at $m=\tfrac34$. Q11 $a=2,b=3$, > infinitely many. Q12 proof. Q13 koi solution nahi (dono methods). Q14 hamesha $D=0$, kabhi unique nahi. ## Connections - [[Solving 2×2 systems using Cramer's rule]] — parent note jis par yeh drill train karti hai. - [[Determinant of a 2×2 matrix]] — "diagonal down minus up" engine. - [[Consistency of linear systems]] — L3 aur Q13 ke peeche classification. - [[Matrix form of linear equations Ax=b]] — yahan ka har system $A\mathbf{x}=\mathbf{c}$ ke roop mein. - [[Inverse of a 2×2 matrix]] — wahi $1/D$ andar chhupa hua. - [[Cramer's rule for 3×3 systems]] — agla step: teen column swaps.