DeterminantsसारणिकDeterminants
Class 12 Maths · Chapter 4 · UP Boardकक्षा 12 गणित · अध्याय 4 · यूपी बोर्डClass 12 Maths · Chapter 4 · UP Board
Built by analysing many years of real UP Board papers, and updated after every board exam. Beginner or topper, it is for everyone.कई वर्षों के असली यूपी बोर्ड पेपरों के विश्लेषण से बना, हर बोर्ड परीक्षा के बाद अपडेट। शुरुआत करने वाले हों या टॉपर, सबके लिए।Kai saalon ke asli UP Board papers ko analyse karke bana, har board exam ke baad update. Beginner ho ya topper, sabke liye.
📊 Based on 7 years (2019–2026) of UP Board papers, current syllabus: 38 sets readवर्तमान सिलेबस के 7 वर्षों (2019–2026) के यूपी बोर्ड पेपरों पर आधारित: 38 सेट पढ़े गएCurrent syllabus ke 7 saal (2019–2026) ke UP Board papers par based: 38 sets padhe (2019, 2020, 2022, 2023, 2024, 2025, 2026)
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🔥 Most asked in this chapterइस अध्याय में सबसे ज़्यादा पूछे गएIs chapter mein sabse zyada aaye
Ranked by how many years each topic came in the paper. Tap a topic to jump to its notes.हर टॉपिक कितने वर्ष पेपर में आया, उसी क्रम में। टॉपिक पर टैप करें, नोट्स पर पहुँचें।Har topic kitne saal paper mein aaya, usi order mein. Topic par tap kijiye, notes par pahunchiye.
Show all 8 topicsसभी 8 टॉपिक देखेंSaare 8 topics dekhiye
Haven't studied anything? Read just this much and you will pass.कुछ नहीं पढ़ा? बस इतना पढ़ लीजिए, पास हो जाएँगे।Kuch nahi padha? Bas itna padh lijiye, pass ho jayenge.
This chapter carries one of the surest big questions of the whole paper: 'solve the system of equations by the matrix method' comes for 8 marks almost every time, and its first half is exactly 'find A⁻¹ by the adjoint'. So learn one method well: |A| → 9 cofactors → adj A (transpose of cofactors) → A⁻¹ = adj A/|A| → X = A⁻¹B. Then learn the area-of-triangle formula and the |kA|, |adj A| results for quick MCQ marks.यह अध्याय पूरे प्रश्नपत्र के सबसे पक्के बड़े प्रश्नों में से एक देता है: 'आव्यूह विधि से समीकरण निकाय हल कीजिए' लगभग हर बार 8 अंक के लिए आता है, और इसका पहला आधा भाग ठीक 'सहखंडज से A⁻¹ ज्ञात कीजिए' है। इसलिए एक विधि अच्छी तरह सीखिए: |A| → 9 सहखंड → adj A (सहखंडों का परिवर्त) → A⁻¹ = adj A/|A| → X = A⁻¹B। फिर त्रिभुज के क्षेत्रफल का सूत्र तथा |kA|, |adj A| के परिणाम MCQ के लिए सीखिए।Yeh chapter poore paper ke sabse pakke bade sawaalon mein se ek deta hai: 'matrix method se equations ka system solve kijiye' lagbhag har baar 8 marks ke liye aata hai, aur iska pehla aadha hissa theek 'adjoint se A⁻¹ nikaliye' hai. Isliye ek method achchhe se seekhiye: |A| → 9 cofactors → adj A (cofactors ka transpose) → A⁻¹ = adj A/|A| → X = A⁻¹B. Phir triangle ke area ka formula aur |kA|, |adj A| ke results MCQ ke liye seekhiye.
🏅 Pass-guarantee minimumपास होने के लिए न्यूनतमPass hone ke liye minimum
- Do 5 'solve by matrix method' questions with your own hand. This one method gives the 8-mark question and the 'find A⁻¹' question together.
- Learn the cofactor sign pattern [+ − +; − + −; + − +] and adj A = transpose of cofactors.
- Learn the area formula ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| and the ± trick for k.
- Learn |kA| = kⁿ|A|, |adj A| = |A|^(n−1), |AB| = |A||B| for MCQs.
- 5 'आव्यूह विधि से हल कीजिए' प्रश्न स्वयं हल कीजिए। यही एक विधि 8 अंक का प्रश्न और 'A⁻¹ ज्ञात कीजिए' दोनों देती है।
- सहखंड चिह्न प्रतिरूप [+ − +; − + −; + − +] तथा adj A = सहखंडों का परिवर्त याद कीजिए।
- क्षेत्रफल सूत्र ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| तथा k के लिए ± विधि सीखिए।
- MCQ के लिए |kA| = kⁿ|A|, |adj A| = |A|^(n−1), |AB| = |A||B| याद कीजिए।
- 5 'matrix method se solve kijiye' sawaal khud kijiye. Yahi ek method 8 marks ka sawaal aur 'A⁻¹ nikaliye' dono deta hai.
- Cofactor sign pattern [+ − +; − + −; + − +] aur adj A = cofactors ka transpose yaad kijiye.
- Area formula ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| aur k ke liye ± trick seekhiye.
- MCQ ke liye |kA| = kⁿ|A|, |adj A| = |A|^(n−1), |AB| = |A||B| yaad kijiye.
🎯 1-mark facts that keep comingबार-बार आने वाले 1 अंक के तथ्यBaar baar aane wale 1 mark facts
- A determinant is a number associated with a square matrix.
- For a 2×2 matrix, |2A| = 4|A|; for 3×3, |2A| = 8|A|.
- For a 3×3 matrix, |adj A| = |A|².
- A is singular if |A| = 0.
- If two rows are identical, the determinant is 0.
- Points (2, −6), (5, 4), (k, 4) form area 35 ⇒ k = 12 or −2.
- A(adj A) = |A| I.
- सारणिक वर्ग आव्यूह से संबद्ध एक संख्या है।
- 2×2 आव्यूह के लिए |2A| = 4|A|; 3×3 के लिए |2A| = 8|A|।
- 3×3 आव्यूह के लिए |adj A| = |A|²।
- यदि |A| = 0 तो A अव्युत्क्रमणीय है।
- दो पंक्तियाँ समान हों तो सारणिक 0।
- (2, −6), (5, 4), (k, 4) से क्षेत्रफल 35 ⇒ k = 12 या −2।
- A(adj A) = |A| I।
- Determinant square matrix se juda ek number hai.
- 2×2 matrix ke liye |2A| = 4|A|; 3×3 ke liye |2A| = 8|A|.
- 3×3 matrix ke liye |adj A| = |A|².
- Agar |A| = 0 to A singular hai.
- Do rows same hon to determinant 0.
- (2, −6), (5, 4), (k, 4) se area 35 ⇒ k = 12 ya −2.
- A(adj A) = |A| I.
The questions that come every year, with model answers. Most asked first.हर साल आने वाले प्रश्न, आदर्श उत्तरों के साथ। सबसे ज़्यादा पूछे गए पहले।Har saal aane wale sawaal, model answers ke saath. Sabse zyada aaye pehle.
Solve a system of linear equations by the matrix method (8 marks)आव्यूह विधि से रैखिक समीकरण निकाय हल करना (8 अंक)Matrix method se linear equations solve karna (8 marks)
🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important- Write the system as AX = B: A = coefficient matrix, X = [x; y; z], B = constants.
- Step 1: find |A|. If |A| ≠ 0, A⁻¹ exists and the system has a unique solution.
- Step 2: find all 9 cofactors Cᵢⱼ = (−1)^(i+j) Mᵢⱼ.
- Step 3: adj A = transpose of the cofactor matrix (cofactors of row 1 go into column 1).
- Step 4: A⁻¹ = (1/|A|) adj A.
- Step 5: X = A⁻¹B. Multiply, then read x, y, z.
- Step 6: check by putting x, y, z into any one equation; write the answer clearly.
- Reciprocal type 2/x + 3/y + 10/z = 4 …: put u = 1/x, v = 1/y, w = 1/z, solve for u, v, w, then invert.
- Common variant: 'if A = […], find A⁻¹ and hence solve …' — the same steps, using the given A.
- निकाय को AX = B रूप में लिखिए: A = गुणांक आव्यूह, X = [x; y; z], B = अचर।
- चरण 1: |A| ज्ञात कीजिए। यदि |A| ≠ 0, तो A⁻¹ है और निकाय का अद्वितीय हल है।
- चरण 2: सभी 9 सहखंड Cᵢⱼ = (−1)^(i+j) Mᵢⱼ ज्ञात कीजिए।
- चरण 3: adj A = सहखंड आव्यूह का परिवर्त (पंक्ति 1 के सहखंड स्तंभ 1 में)।
- चरण 4: A⁻¹ = (1/|A|) adj A।
- चरण 5: X = A⁻¹B। गुणा करके x, y, z पढ़िए।
- चरण 6: x, y, z को किसी एक समीकरण में रखकर जाँचिए; उत्तर स्पष्ट लिखिए।
- व्युत्क्रम प्रकार 2/x + 3/y + 10/z = 4 …: u = 1/x, v = 1/y, w = 1/z रखकर u, v, w ज्ञात कीजिए, फिर उलटिए।
- सामान्य रूप: 'यदि A = […], तो A⁻¹ ज्ञात कीजिए तथा इसकी सहायता से हल कीजिए …' — वही चरण, दिए गए A से।
- System ko AX = B roop mein likhiye: A = coefficient matrix, X = [x; y; z], B = constants.
- Step 1: |A| nikaliye. Agar |A| ≠ 0, to A⁻¹ hai aur system ka unique solution hai.
- Step 2: saare 9 cofactors Cᵢⱼ = (−1)^(i+j) Mᵢⱼ nikaliye.
- Step 3: adj A = cofactor matrix ka transpose (row 1 ke cofactors column 1 mein).
- Step 4: A⁻¹ = (1/|A|) adj A.
- Step 5: X = A⁻¹B. Multiply karke x, y, z padhiye.
- Step 6: x, y, z ko kisi ek equation mein rakh kar check kijiye; answer saaf likhiye.
- Reciprocal type 2/x + 3/y + 10/z = 4 …: u = 1/x, v = 1/y, w = 1/z rakh kar u, v, w nikaliye, phir ulta kijiye.
- Common variant: 'agar A = […], to A⁻¹ nikaliye aur isse solve kijiye …' — wahi steps, diye gaye A se.
Worked example: Solve by the matrix method: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.हल किया उदाहरण: आव्यूह विधि से हल कीजिए: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2।Solved example: matrix method se solve kijiye: x + y + z = 6, 2x − y + z = 3, x + 2y − z = 2.
- A = [1 1 1; 2 −1 1; 1 2 −1], X = [x; y; z], B = [6; 3; 2]; AX = B.
- |A| = 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = −1 + 3 + 5 = 7 ≠ 0, so a unique solution exists.
- Cofactors: C₁₁ = −1, C₁₂ = 3, C₁₃ = 5; C₂₁ = 3, C₂₂ = −2, C₂₃ = −1; C₃₁ = 2, C₃₂ = 1, C₃₃ = −3.
- adj A = [−1 3 2; 3 −2 1; 5 −1 −3].
- A⁻¹ = (1/7)[−1 3 2; 3 −2 1; 5 −1 −3].
- X = A⁻¹B = (1/7)[−6 + 9 + 4; 18 − 6 + 2; 30 − 3 − 6] = (1/7)[7; 14; 21] = [1; 2; 3].
- x = 1, y = 2, z = 3. Check: 1 + 2 + 3 = 6 ✓.
- A = [1 1 1; 2 −1 1; 1 2 −1], X = [x; y; z], B = [6; 3; 2]; AX = B।
- |A| = 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = −1 + 3 + 5 = 7 ≠ 0, अतः अद्वितीय हल है।
- सहखंड: C₁₁ = −1, C₁₂ = 3, C₁₃ = 5; C₂₁ = 3, C₂₂ = −2, C₂₃ = −1; C₃₁ = 2, C₃₂ = 1, C₃₃ = −3।
- adj A = [−1 3 2; 3 −2 1; 5 −1 −3]।
- A⁻¹ = (1/7)[−1 3 2; 3 −2 1; 5 −1 −3]।
- X = A⁻¹B = (1/7)[−6 + 9 + 4; 18 − 6 + 2; 30 − 3 − 6] = (1/7)[7; 14; 21] = [1; 2; 3]।
- x = 1, y = 2, z = 3। जाँच: 1 + 2 + 3 = 6 ✓।
- A = [1 1 1; 2 −1 1; 1 2 −1], X = [x; y; z], B = [6; 3; 2]; AX = B.
- |A| = 1(1 − 2) − 1(−2 − 1) + 1(4 + 1) = −1 + 3 + 5 = 7 ≠ 0, isliye unique solution hai.
- Cofactors: C₁₁ = −1, C₁₂ = 3, C₁₃ = 5; C₂₁ = 3, C₂₂ = −2, C₂₃ = −1; C₃₁ = 2, C₃₂ = 1, C₃₃ = −3.
- adj A = [−1 3 2; 3 −2 1; 5 −1 −3].
- A⁻¹ = (1/7)[−1 3 2; 3 −2 1; 5 −1 −3].
- X = A⁻¹B = (1/7)[−6 + 9 + 4; 18 − 6 + 2; 30 − 3 − 6] = (1/7)[7; 14; 21] = [1; 2; 3].
- x = 1, y = 2, z = 3. Check: 1 + 2 + 3 = 6 ✓.
Solve by the matrix method: 2/x + 3/y + 10/z = 4, 4/x − 6/y + 5/z = 1, 6/x + 9/y − 20/z = 2.आव्यूह विधि से हल कीजिए: 2/x + 3/y + 10/z = 4, 4/x − 6/y + 5/z = 1, 6/x + 9/y − 20/z = 2।Matrix method se solve kijiye: 2/x + 3/y + 10/z = 4, 4/x − 6/y + 5/z = 1, 6/x + 9/y − 20/z = 2.
- Put 1/x = u, 1/y = v, 1/z = w: 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2.
- A = [2 3 10; 4 −6 5; 6 9 −20]; |A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200.
- Cofactors: C₁₁ = 75, C₁₂ = 110, C₁₃ = 72; C₂₁ = 150, C₂₂ = −100, C₂₃ = 0; C₃₁ = 75, C₃₂ = 30, C₃₃ = −24.
- adj A = [75 150 75; 110 −100 30; 72 0 −24]; A⁻¹ = adj A/1200.
- [u; v; w] = (1/1200)[300 + 150 + 150; 440 − 100 + 60; 288 + 0 − 48] = (1/1200)[600; 400; 240] = [1/2; 1/3; 1/5].
- So x = 2, y = 3, z = 5.
- 1/x = u, 1/y = v, 1/z = w रखिए: 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2।
- A = [2 3 10; 4 −6 5; 6 9 −20]; |A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200।
- सहखंड: C₁₁ = 75, C₁₂ = 110, C₁₃ = 72; C₂₁ = 150, C₂₂ = −100, C₂₃ = 0; C₃₁ = 75, C₃₂ = 30, C₃₃ = −24।
- adj A = [75 150 75; 110 −100 30; 72 0 −24]; A⁻¹ = adj A/1200।
- [u; v; w] = (1/1200)[300 + 150 + 150; 440 − 100 + 60; 288 + 0 − 48] = (1/1200)[600; 400; 240] = [1/2; 1/3; 1/5]।
- अतः x = 2, y = 3, z = 5।
- 1/x = u, 1/y = v, 1/z = w rakhiye: 2u + 3v + 10w = 4, 4u − 6v + 5w = 1, 6u + 9v − 20w = 2.
- A = [2 3 10; 4 −6 5; 6 9 −20]; |A| = 2(120 − 45) − 3(−80 − 30) + 10(36 + 36) = 150 + 330 + 720 = 1200.
- Cofactors: C₁₁ = 75, C₁₂ = 110, C₁₃ = 72; C₂₁ = 150, C₂₂ = −100, C₂₃ = 0; C₃₁ = 75, C₃₂ = 30, C₃₃ = −24.
- adj A = [75 150 75; 110 −100 30; 72 0 −24]; A⁻¹ = adj A/1200.
- [u; v; w] = (1/1200)[300 + 150 + 150; 440 − 100 + 60; 288 + 0 − 48] = (1/1200)[600; 400; 240] = [1/2; 1/3; 1/5].
- To x = 2, y = 3, z = 5.
💡 Most marks are lost in one cofactor sign. Use the sign pattern [+ − +; − + −; + − +] and verify A·(adj A) = |A| I for one row before using A⁻¹.अधिकतर अंक एक सहखंड के चिह्न में कटते हैं। चिह्न प्रतिरूप [+ − +; − + −; + − +] अपनाइए और A⁻¹ लगाने से पहले एक पंक्ति के लिए A·(adj A) = |A| I जाँचिए।Zyadatar marks ek cofactor ke sign mein katte hain. Sign pattern [+ − +; − + −; + − +] use kijiye aur A⁻¹ lagane se pehle ek row ke liye A·(adj A) = |A| I check kijiye.
Adjoint and inverse of a matrixआव्यूह का सहखंडज तथा प्रतिलोमMatrix ka adjoint aur inverse
🔥 Asked in 7 of 7 years🔥 7 में से 7 वर्ष पूछा गया🔥 Aaya hai 7/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important- A is invertible (non-singular) ⇔ |A| ≠ 0. If |A| = 0, A is singular and A⁻¹ does not exist.
- adj A = transpose of the matrix of cofactors.
- Key property: A(adj A) = (adj A)A = |A| I.
- 2×2 shortcut: for A = [a b; c d], adj A = [d −b; −c a] (swap the diagonal, change the sign of the other two).
- A⁻¹ = adj A/|A|; check: A·A⁻¹ = I.
- (AB)⁻¹ = B⁻¹A⁻¹; (A⁻¹)⁻¹ = A; (A′)⁻¹ = (A⁻¹)′.
- A व्युत्क्रमणीय (व्युत्क्रमणीय/अव्युत्क्रमणीय नहीं) ⇔ |A| ≠ 0। यदि |A| = 0, तो A अव्युत्क्रमणीय है और A⁻¹ नहीं है।
- adj A = सहखंडों के आव्यूह का परिवर्त।
- मुख्य गुण: A(adj A) = (adj A)A = |A| I।
- 2×2 संक्षिप्त विधि: A = [a b; c d] के लिए adj A = [d −b; −c a] (विकर्ण बदलिए, शेष दो का चिह्न बदलिए)।
- A⁻¹ = adj A/|A|; जाँच: A·A⁻¹ = I।
- (AB)⁻¹ = B⁻¹A⁻¹; (A⁻¹)⁻¹ = A; (A′)⁻¹ = (A⁻¹)′।
- A invertible (non-singular) ⇔ |A| ≠ 0. Agar |A| = 0, to A singular hai aur A⁻¹ nahi hota.
- adj A = cofactors ki matrix ka transpose.
- Main property: A(adj A) = (adj A)A = |A| I.
- 2×2 shortcut: A = [a b; c d] ke liye adj A = [d −b; −c a] (diagonal swap, baaki do ka sign badliye).
- A⁻¹ = adj A/|A|; check: A·A⁻¹ = I.
- (AB)⁻¹ = B⁻¹A⁻¹; (A⁻¹)⁻¹ = A; (A′)⁻¹ = (A⁻¹)′.
Worked example: Find the inverse of A = [1 2 0; 0 1 1; 1 0 2] and verify A(adj A) = |A| I.हल किया उदाहरण: A = [1 2 0; 0 1 1; 1 0 2] का प्रतिलोम ज्ञात कीजिए तथा A(adj A) = |A| I सत्यापित कीजिए।Solved example: A = [1 2 0; 0 1 1; 1 0 2] ka inverse nikaliye aur A(adj A) = |A| I verify kijiye.
- |A| = 1(2 − 0) − 2(0 − 1) + 0 = 2 + 2 = 4 ≠ 0, so A⁻¹ exists.
- Cofactors: C₁₁ = 2, C₁₂ = 1, C₁₃ = −1; C₂₁ = −4, C₂₂ = 2, C₂₃ = 2; C₃₁ = 2, C₃₂ = −1, C₃₃ = 1.
- adj A = [2 −4 2; 1 2 −1; −1 2 1].
- A(adj A): row 1 of A × columns = [2 + 2 −4 + 4 2 − 2] = [4 0 0]; similarly rows 2 and 3 give [0 4 0] and [0 0 4]. So A(adj A) = 4I = |A| I ✓.
- A⁻¹ = (1/4)[2 −4 2; 1 2 −1; −1 2 1].
- |A| = 1(2 − 0) − 2(0 − 1) + 0 = 2 + 2 = 4 ≠ 0, अतः A⁻¹ है।
- सहखंड: C₁₁ = 2, C₁₂ = 1, C₁₃ = −1; C₂₁ = −4, C₂₂ = 2, C₂₃ = 2; C₃₁ = 2, C₃₂ = −1, C₃₃ = 1।
- adj A = [2 −4 2; 1 2 −1; −1 2 1]।
- A(adj A): A की पंक्ति 1 × स्तंभ = [2 + 2 −4 + 4 2 − 2] = [4 0 0]; इसी प्रकार पंक्ति 2 व 3 से [0 4 0] तथा [0 0 4]। अतः A(adj A) = 4I = |A| I ✓।
- A⁻¹ = (1/4)[2 −4 2; 1 2 −1; −1 2 1]।
- |A| = 1(2 − 0) − 2(0 − 1) + 0 = 2 + 2 = 4 ≠ 0, isliye A⁻¹ hai.
- Cofactors: C₁₁ = 2, C₁₂ = 1, C₁₃ = −1; C₂₁ = −4, C₂₂ = 2, C₂₃ = 2; C₃₁ = 2, C₃₂ = −1, C₃₃ = 1.
- adj A = [2 −4 2; 1 2 −1; −1 2 1].
- A(adj A): A ki row 1 × columns = [2 + 2 −4 + 4 2 − 2] = [4 0 0]; isi tarah row 2 aur 3 se [0 4 0] aur [0 0 4]. To A(adj A) = 4I = |A| I ✓.
- A⁻¹ = (1/4)[2 −4 2; 1 2 −1; −1 2 1].
Evaluate a determinantसारणिक का मान ज्ञात करनाDeterminant ki value nikalna
🔥 Asked in 5 of 7 years🔥 7 में से 5 वर्ष पूछा गया🔥 Aaya hai 5/7 saal🔴 Very importantबहुत महत्वपूर्णBahut important- 2×2: |a b; c d| = ad − bc.
- 3×3: expand along any row or column with signs + − +. Choose the row/column with most zeros.
- Taking a common factor from one row multiplies the determinant by that factor; |kA| = kⁿ|A| for an n×n matrix.
- If two rows (or columns) are equal or proportional, the determinant is 0.
- Interchanging two rows changes the sign; adding a multiple of one row to another changes nothing.
- Standard results: |a b c; b c a; c a b| = 3abc − a³ − b³ − c³ = −(a + b + c)(a² + b² + c² − ab − bc − ca); |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a).
- 2×2: |a b; c d| = ad − bc।
- 3×3: किसी पंक्ति या स्तंभ के अनुदिश चिह्न + − + से प्रसार कीजिए। सबसे अधिक शून्य वाली पंक्ति/स्तंभ चुनिए।
- एक पंक्ति से उभयनिष्ठ गुणनखंड निकालने पर सारणिक उस गुणनखंड से गुणा हो जाता है; n×n आव्यूह के लिए |kA| = kⁿ|A|।
- यदि दो पंक्तियाँ (या स्तंभ) समान या समानुपाती हों, तो सारणिक 0 होता है।
- दो पंक्तियाँ बदलने से चिह्न बदलता है; एक पंक्ति के गुणज को दूसरी में जोड़ने से मान नहीं बदलता।
- मानक परिणाम: |a b c; b c a; c a b| = 3abc − a³ − b³ − c³ = −(a + b + c)(a² + b² + c² − ab − bc − ca); |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a)।
- 2×2: |a b; c d| = ad − bc.
- 3×3: kisi row ya column ke along sign + − + se expand kijiye. Sabse zyada zero wali row/column chuniye.
- Ek row se common factor nikalne par determinant us factor se multiply ho jaata hai; n×n matrix ke liye |kA| = kⁿ|A|.
- Agar do rows (ya columns) same ya proportional hon, to determinant 0 hota hai.
- Do rows badalne se sign badalta hai; ek row ke multiple ko doosri mein jodne se value nahi badalti.
- Standard results: |a b c; b c a; c a b| = 3abc − a³ − b³ − c³ = −(a + b + c)(a² + b² + c² − ab − bc − ca); |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a).
Worked example: Evaluate |2 3 −1; 4 1 5; 0 −2 3|.हल किया उदाहरण: |2 3 −1; 4 1 5; 0 −2 3| का मान ज्ञात कीजिए।Solved example: |2 3 −1; 4 1 5; 0 −2 3| ki value nikaliye.
- Expand along row 1: 2(1·3 − 5·(−2)) − 3(4·3 − 5·0) + (−1)(4·(−2) − 1·0).
- = 2(3 + 10) − 3(12) − 1(−8) = 26 − 36 + 8.
- = −2.
- पंक्ति 1 के अनुदिश प्रसार: 2(1·3 − 5·(−2)) − 3(4·3 − 5·0) + (−1)(4·(−2) − 1·0)।
- = 2(3 + 10) − 3(12) − 1(−8) = 26 − 36 + 8।
- = −2।
- Row 1 ke along expand: 2(1·3 − 5·(−2)) − 3(4·3 − 5·0) + (−1)(4·(−2) − 1·0).
- = 2(3 + 10) − 3(12) − 1(−8) = 26 − 36 + 8.
- = −2.
Area of a triangle and collinearityत्रिभुज का क्षेत्रफल तथा संरेखताTriangle ka area aur collinearity
Asked in 2 of 7 years7 में से 2 वर्ष पूछा गयाAaya hai 2/7 saal- Area = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (take the absolute value).
- Points are collinear ⇔ this determinant = 0.
- If the area is given, use ± (the determinant can be positive or negative): ½ Δ = ± area. This gives two values of k.
- Example: vertices (2, −6), (5, 4), (k, 4), area 35 ⇒ ½ |50 − 10k| = 35 ⇒ k = 12 or k = −2.
- A(m, −1), B(2, 1), C(4, 5) collinear ⇒ −4m + 12 − 8 = 0 ⇒ m = 1.
- क्षेत्रफल = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (निरपेक्ष मान लीजिए)।
- बिंदु संरेख हैं ⇔ यह सारणिक = 0।
- यदि क्षेत्रफल दिया हो, तो ± लीजिए (सारणिक धनात्मक या ऋणात्मक हो सकता है): ½ Δ = ± क्षेत्रफल। इससे k के दो मान मिलते हैं।
- उदाहरण: शीर्ष (2, −6), (5, 4), (k, 4), क्षेत्रफल 35 ⇒ ½ |50 − 10k| = 35 ⇒ k = 12 या k = −2।
- A(m, −1), B(2, 1), C(4, 5) संरेख ⇒ −4m + 12 − 8 = 0 ⇒ m = 1।
- Area = ½ |x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1| (absolute value lijiye).
- Points collinear hain ⇔ yeh determinant = 0.
- Agar area diya ho, to ± lijiye (determinant positive ya negative ho sakta hai): ½ Δ = ± area. Isse k ki do values milti hain.
- Example: vertices (2, −6), (5, 4), (k, 4), area 35 ⇒ ½ |50 − 10k| = 35 ⇒ k = 12 ya k = −2.
- A(m, −1), B(2, 1), C(4, 5) collinear ⇒ −4m + 12 − 8 = 0 ⇒ m = 1.
If the area of the triangle with vertices (2, −6), (5, 4) and (k, 4) is 35 sq units, find k.यदि शीर्षों (2, −6), (5, 4) तथा (k, 4) वाले त्रिभुज का क्षेत्रफल 35 वर्ग इकाई है, तो k ज्ञात कीजिए।Agar vertices (2, −6), (5, 4) aur (k, 4) wale triangle ka area 35 sq units hai, to k nikaliye.
- Area = ½ |2 −6 1; 5 4 1; k 4 1|.
- Δ = 2(4 − 4) + 6(5 − k) + 1(20 − 4k) = 30 − 6k + 20 − 4k = 50 − 10k.
- ½(50 − 10k) = ±35 ⇒ 50 − 10k = ±70.
- 50 − 10k = 70 ⇒ k = −2; 50 − 10k = −70 ⇒ k = 12.
- k = 12 or −2.
- क्षेत्रफल = ½ |2 −6 1; 5 4 1; k 4 1|।
- Δ = 2(4 − 4) + 6(5 − k) + 1(20 − 4k) = 30 − 6k + 20 − 4k = 50 − 10k।
- ½(50 − 10k) = ±35 ⇒ 50 − 10k = ±70।
- 50 − 10k = 70 ⇒ k = −2; 50 − 10k = −70 ⇒ k = 12।
- k = 12 या −2।
- Area = ½ |2 −6 1; 5 4 1; k 4 1|.
- Δ = 2(4 − 4) + 6(5 − k) + 1(20 − 4k) = 30 − 6k + 20 − 4k = 50 − 10k.
- ½(50 − 10k) = ±35 ⇒ 50 − 10k = ±70.
- 50 − 10k = 70 ⇒ k = −2; 50 − 10k = −70 ⇒ k = 12.
- k = 12 ya −2.
|kA|, |AB|, |adj A| results (MCQ)|kA|, |AB|, |adj A| परिणाम (MCQ)|kA|, |AB|, |adj A| results (MCQ)
Asked in 1 of 7 years7 में से 1 वर्ष पूछा गयाAaya hai 1/7 saal- |kA| = kⁿ|A| for an n×n matrix: for 2×2, |2A| = 4|A|; for 3×3, |2A| = 8|A|.
- |AB| = |A|·|B|; |A′| = |A|; |A⁻¹| = 1/|A|.
- |adj A| = |A|^(n − 1): for 3×3, |adj A| = |A|².
- A determinant is a number associated with a SQUARE matrix only.
- n×n आव्यूह के लिए |kA| = kⁿ|A|: 2×2 में |2A| = 4|A|; 3×3 में |2A| = 8|A|।
- |AB| = |A|·|B|; |A′| = |A|; |A⁻¹| = 1/|A|।
- |adj A| = |A|^(n − 1): 3×3 के लिए |adj A| = |A|²।
- सारणिक केवल वर्ग आव्यूह से संबद्ध एक संख्या है।
- n×n matrix ke liye |kA| = kⁿ|A|: 2×2 mein |2A| = 4|A|; 3×3 mein |2A| = 8|A|.
- |AB| = |A|·|B|; |A′| = |A|; |A⁻¹| = 1/|A|.
- |adj A| = |A|^(n − 1): 3×3 ke liye |adj A| = |A|².
- Determinant sirf SQUARE matrix se jude ek number hai.
Minors and cofactorsउपसारणिक तथा सहखंडMinors aur cofactors
Not asked yetअभी तक नहीं पूछा गयाAbhi tak nahi aaya- Minor Mᵢⱼ = determinant left after deleting row i and column j.
- Cofactor Cᵢⱼ = (−1)^(i+j) Mᵢⱼ.
- |A| = sum of (elements of any row) × (their own cofactors).
- Elements of one row × cofactors of a DIFFERENT row add up to 0.
- उपसारणिक Mᵢⱼ = i वीं पंक्ति तथा j वाँ स्तंभ हटाने पर बचा सारणिक।
- सहखंड Cᵢⱼ = (−1)^(i+j) Mᵢⱼ।
- |A| = (किसी पंक्ति के अवयव) × (उनके अपने सहखंड) का योग।
- एक पंक्ति के अवयव × किसी अन्य पंक्ति के सहखंडों का योग 0 होता है।
- Minor Mᵢⱼ = row i aur column j hatane par bacha determinant.
- Cofactor Cᵢⱼ = (−1)^(i+j) Mᵢⱼ.
- |A| = (kisi row ke elements) × (unke apne cofactors) ka jod.
- Ek row ke elements × kisi doosri row ke cofactors ka jod 0 hota hai.
Consistency of a systemनिकाय की संगतताSystem ki consistency
Not asked yetअभी तक नहीं पूछा गयाAbhi tak nahi aaya- |A| ≠ 0: consistent with a unique solution X = A⁻¹B.
- |A| = 0 and (adj A)B ≠ O: inconsistent (no solution).
- |A| = 0 and (adj A)B = O: may have infinitely many solutions or none (check further).
- Consistent = at least one solution; inconsistent = no solution.
- |A| ≠ 0: संगत तथा अद्वितीय हल X = A⁻¹B।
- |A| = 0 तथा (adj A)B ≠ O: असंगत (कोई हल नहीं)।
- |A| = 0 तथा (adj A)B = O: अनंत हल हो सकते हैं या कोई नहीं (आगे जाँचिए)।
- संगत = कम से कम एक हल; असंगत = कोई हल नहीं।
- |A| ≠ 0: consistent aur unique solution X = A⁻¹B.
- |A| = 0 aur (adj A)B ≠ O: inconsistent (koi solution nahi).
- |A| = 0 aur (adj A)B = O: infinite solutions ho sakte hain ya koi nahi (aage check kijiye).
- Consistent = kam se kam ek solution; inconsistent = koi solution nahi.
Full derivations, tricky questions, common mistakes and quick revision.पूरी व्युत्पत्तियाँ, कठिन प्रश्न, आम गलतियाँ और त्वरित पुनरावृत्ति।Poori derivations, tricky sawaal, aam galtiyan aur quick revision.
📐 Prove |1+a 1 1; 1 1+b 1; 1 1 1+c| = abc(1 + 1/a + 1/b + 1/c)सिद्ध कीजिए |1+a 1 1; 1 1+b 1; 1 1 1+c| = abc(1 + 1/a + 1/b + 1/c)Prove kijiye |1+a 1 1; 1 1+b 1; 1 1 1+c| = abc(1 + 1/a + 1/b + 1/c)
- Take a, b, c common from R₁, R₂, R₃: Δ = abc |1/a + 1 1/a 1/a; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|.
- R₁ → R₁ + R₂ + R₃: every element of R₁ becomes 1 + 1/a + 1/b + 1/c. Take it common.
- Δ = abc(1 + 1/a + 1/b + 1/c) |1 1 1; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|.
- C₂ → C₂ − C₁, C₃ → C₃ − C₁: the determinant becomes |1 0 0; 1/b 1 0; 1/c 0 1| = 1.
- Δ = abc(1 + 1/a + 1/b + 1/c). Proved. (Check by expansion: Δ = abc + ab + bc + ca.)
- R₁, R₂, R₃ से a, b, c उभयनिष्ठ लीजिए: Δ = abc |1/a + 1 1/a 1/a; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|।
- R₁ → R₁ + R₂ + R₃: R₁ का प्रत्येक अवयव 1 + 1/a + 1/b + 1/c हो जाता है। इसे उभयनिष्ठ लीजिए।
- Δ = abc(1 + 1/a + 1/b + 1/c) |1 1 1; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|।
- C₂ → C₂ − C₁, C₃ → C₃ − C₁: सारणिक |1 0 0; 1/b 1 0; 1/c 0 1| = 1 हो जाता है।
- Δ = abc(1 + 1/a + 1/b + 1/c)। इति सिद्धम्। (प्रसार से जाँच: Δ = abc + ab + bc + ca।)
- R₁, R₂, R₃ se a, b, c common lijiye: Δ = abc |1/a + 1 1/a 1/a; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|.
- R₁ → R₁ + R₂ + R₃: R₁ ka har element 1 + 1/a + 1/b + 1/c ho jaata hai. Ise common lijiye.
- Δ = abc(1 + 1/a + 1/b + 1/c) |1 1 1; 1/b 1/b + 1 1/b; 1/c 1/c 1/c + 1|.
- C₂ → C₂ − C₁, C₃ → C₃ − C₁: determinant |1 0 0; 1/b 1 0; 1/c 0 1| = 1 ho jaata hai.
- Δ = abc(1 + 1/a + 1/b + 1/c). Proved. (Expansion se check: Δ = abc + ab + bc + ca.)
📐 Show |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a)दिखाइए |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a)Dikhaiye |a b c; a² b² c²; a³ b³ c³| = abc(a − b)(b − c)(c − a)
- Take a, b, c common from C₁, C₂, C₃: Δ = abc |1 1 1; a b c; a² b² c²|.
- C₁ → C₁ − C₂, C₂ → C₂ − C₃: Δ = abc |0 0 1; a − b b − c c; a² − b² b² − c² c²|.
- Take (a − b) from C₁ and (b − c) from C₂: Δ = abc(a − b)(b − c) |0 0 1; 1 1 c; a + b b + c c²|.
- Expand along R₁: |1 1; a + b b + c| = (b + c) − (a + b) = c − a.
- Δ = abc(a − b)(b − c)(c − a). Proved.
- C₁, C₂, C₃ से a, b, c उभयनिष्ठ: Δ = abc |1 1 1; a b c; a² b² c²|।
- C₁ → C₁ − C₂, C₂ → C₂ − C₃: Δ = abc |0 0 1; a − b b − c c; a² − b² b² − c² c²|।
- C₁ से (a − b) तथा C₂ से (b − c) उभयनिष्ठ: Δ = abc(a − b)(b − c) |0 0 1; 1 1 c; a + b b + c c²|।
- R₁ के अनुदिश प्रसार: |1 1; a + b b + c| = (b + c) − (a + b) = c − a।
- Δ = abc(a − b)(b − c)(c − a)। इति सिद्धम्।
- C₁, C₂, C₃ se a, b, c common: Δ = abc |1 1 1; a b c; a² b² c²|.
- C₁ → C₁ − C₂, C₂ → C₂ − C₃: Δ = abc |0 0 1; a − b b − c c; a² − b² b² − c² c²|.
- C₁ se (a − b) aur C₂ se (b − c) common: Δ = abc(a − b)(b − c) |0 0 1; 1 1 c; a + b b + c c²|.
- R₁ ke along expand: |1 1; a + b b + c| = (b + c) − (a + b) = c − a.
- Δ = abc(a − b)(b − c)(c − a). Proved.
Solve: 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4 (matrix method).हल कीजिए: 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4 (आव्यूह विधि)।Solve kijiye: 3x − 2y + 3z = 8, 2x + y − z = 1, 4x − 3y + 2z = 4 (matrix method).
- |A| = 3(2 − 3) + 2(4 + 4) + 3(−6 − 4) = −3 + 16 − 30 = −17 ≠ 0.
- Cofactors: C₁₁ = −1, C₁₂ = −8, C₁₃ = −10; C₂₁ = −5, C₂₂ = −6, C₂₃ = 1; C₃₁ = −1, C₃₂ = 9, C₃₃ = 7.
- adj A = [−1 −5 −1; −8 −6 9; −10 1 7].
- X = (1/−17)[−8 − 5 − 4; −64 − 6 + 36; −80 + 1 + 28] = (1/−17)[−17; −34; −51] = [1; 2; 3].
- x = 1, y = 2, z = 3.
- |A| = 3(2 − 3) + 2(4 + 4) + 3(−6 − 4) = −3 + 16 − 30 = −17 ≠ 0।
- सहखंड: C₁₁ = −1, C₁₂ = −8, C₁₃ = −10; C₂₁ = −5, C₂₂ = −6, C₂₃ = 1; C₃₁ = −1, C₃₂ = 9, C₃₃ = 7।
- adj A = [−1 −5 −1; −8 −6 9; −10 1 7]।
- X = (1/−17)[−8 − 5 − 4; −64 − 6 + 36; −80 + 1 + 28] = (1/−17)[−17; −34; −51] = [1; 2; 3]।
- x = 1, y = 2, z = 3।
- |A| = 3(2 − 3) + 2(4 + 4) + 3(−6 − 4) = −3 + 16 − 30 = −17 ≠ 0.
- Cofactors: C₁₁ = −1, C₁₂ = −8, C₁₃ = −10; C₂₁ = −5, C₂₂ = −6, C₂₃ = 1; C₃₁ = −1, C₃₂ = 9, C₃₃ = 7.
- adj A = [−1 −5 −1; −8 −6 9; −10 1 7].
- X = (1/−17)[−8 − 5 − 4; −64 − 6 + 36; −80 + 1 + 28] = (1/−17)[−17; −34; −51] = [1; 2; 3].
- x = 1, y = 2, z = 3.
Show that A(a, b + c), B(b, c + a), C(c, a + b) are collinear.दिखाइए कि A(a, b + c), B(b, c + a), C(c, a + b) संरेख हैं।Dikhaiye ki A(a, b + c), B(b, c + a), C(c, a + b) collinear hain.
- Δ = |a b + c 1; b c + a 1; c a + b 1|.
- C₂ → C₂ + C₁: C₂ becomes (a + b + c) in every row.
- Take (a + b + c) common: C₂ and C₃ are now both columns of 1s, so the determinant = 0.
- Area = 0 ⇒ the points are collinear.
- Δ = |a b + c 1; b c + a 1; c a + b 1|।
- C₂ → C₂ + C₁: प्रत्येक पंक्ति में C₂ = (a + b + c)।
- (a + b + c) उभयनिष्ठ लेने पर C₂ तथा C₃ दोनों 1 के स्तंभ हैं, अतः सारणिक = 0।
- क्षेत्रफल = 0 ⇒ बिंदु संरेख हैं।
- Δ = |a b + c 1; b c + a 1; c a + b 1|.
- C₂ → C₂ + C₁: har row mein C₂ = (a + b + c).
- (a + b + c) common lene par C₂ aur C₃ dono 1 ke columns hain, isliye determinant = 0.
- Area = 0 ⇒ points collinear hain.
⚠️ Common mistakesआम गलतियाँAam galtiyan
- Writing the cofactor matrix as adj A → adj A is its TRANSPOSE.
- Wrong sign for C₁₂, C₂₁, C₂₃, C₃₂ → these carry a minus sign.
- X = BA⁻¹ → it is X = A⁻¹B.
- Using only +35 for the area → use ±35, two values of k.
- |2A| = 2|A| → it is 2ⁿ|A|.
- सहखंड आव्यूह को ही adj A लिखना → adj A उसका परिवर्त है।
- C₁₂, C₂₁, C₂₃, C₃₂ का चिह्न गलत → इन पर ऋण चिह्न लगता है।
- X = BA⁻¹ लिखना → सही है X = A⁻¹B।
- क्षेत्रफल के लिए केवल +35 लेना → ±35 लीजिए, k के दो मान।
- |2A| = 2|A| लिखना → सही है 2ⁿ|A|।
- Cofactor matrix ko hi adj A likhna → adj A uska TRANSPOSE hai.
- C₁₂, C₂₁, C₂₃, C₃₂ ka sign galat → in par minus sign lagta hai.
- X = BA⁻¹ likhna → sahi hai X = A⁻¹B.
- Area ke liye sirf +35 lena → ±35 lijiye, k ki do values.
- |2A| = 2|A| likhna → sahi hai 2ⁿ|A|.
💎 Rarely asked but importantकम पूछा गया पर ज़रूरीKam aaya par zaroori
- |a b c; b c a; c a b| = 3abc − a³ − b³ − c³.
- |x + y + 2z x y; z y + z + 2x y; z x z + x + 2y| = 2(x + y + z)³.
- |−a² ab ac; ba −b² bc; ca cb −c²| = 4a²b²c².
- adj(adj A) = |A|^(n−2) A.
- |a b c; b c a; c a b| = 3abc − a³ − b³ − c³।
- |x + y + 2z x y; z y + z + 2x y; z x z + x + 2y| = 2(x + y + z)³।
- |−a² ab ac; ba −b² bc; ca cb −c²| = 4a²b²c²।
- adj(adj A) = |A|^(n−2) A।
- |a b c; b c a; c a b| = 3abc − a³ − b³ − c³.
- |x + y + 2z x y; z y + z + 2x y; z x z + x + 2y| = 2(x + y + z)³.
- |−a² ab ac; ba −b² bc; ca cb −c²| = 4a²b²c².
- adj(adj A) = |A|^(n−2) A.
⚡ Quick revisionत्वरित पुनरावृत्तिQuick revision
- |a b; c d| = ad − bc.
- Cᵢⱼ = (−1)^(i+j) Mᵢⱼ; sign pattern + − + / − + − / + − +.
- adj A = (cofactor matrix)′.
- A(adj A) = |A| I; A⁻¹ = adj A/|A|.
- AX = B ⇒ X = A⁻¹B when |A| ≠ 0.
- |A| = 0, (adj A)B ≠ O ⇒ no solution.
- Area = ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|; collinear ⇔ 0.
- |kA| = kⁿ|A|, |AB| = |A||B|, |A′| = |A|.
- |adj A| = |A|^(n−1), |A⁻¹| = 1/|A|.
- Two equal rows ⇒ determinant 0.
- |a b; c d| = ad − bc।
- Cᵢⱼ = (−1)^(i+j) Mᵢⱼ; चिह्न + − + / − + − / + − +।
- adj A = (सहखंड आव्यूह)′।
- A(adj A) = |A| I; A⁻¹ = adj A/|A|।
- |A| ≠ 0 हो तो AX = B ⇒ X = A⁻¹B।
- |A| = 0, (adj A)B ≠ O ⇒ कोई हल नहीं।
- क्षेत्रफल = ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|; संरेख ⇔ 0।
- |kA| = kⁿ|A|, |AB| = |A||B|, |A′| = |A|।
- |adj A| = |A|^(n−1), |A⁻¹| = 1/|A|।
- दो समान पंक्तियाँ ⇒ सारणिक 0।
- |a b; c d| = ad − bc.
- Cᵢⱼ = (−1)^(i+j) Mᵢⱼ; sign + − + / − + − / + − +.
- adj A = (cofactor matrix)′.
- A(adj A) = |A| I; A⁻¹ = adj A/|A|.
- |A| ≠ 0 ho to AX = B ⇒ X = A⁻¹B.
- |A| = 0, (adj A)B ≠ O ⇒ koi solution nahi.
- Area = ½|x₁ y₁ 1; x₂ y₂ 1; x₃ y₃ 1|; collinear ⇔ 0.
- |kA| = kⁿ|A|, |AB| = |A||B|, |A′| = |A|.
- |adj A| = |A|^(n−1), |A⁻¹| = 1/|A|.
- Do same rows ⇒ determinant 0.
🚫 Removed, skip itहटाया गया, इसे छोड़ देंHata diya gaya, ise chhod dijiye
- 'Prove using properties of determinants' (5 marks) — the properties section was removed from the rationalised NCERT book and the UPMSP 2026-27 syllabus text does not list it, yet older papers asked it often and recent papers have still asked to evaluate determinants like |a b c; a² b² c²; a³ b³ c³|. If you have time, learn the standard results in the Topper section; otherwise skip the long proofs.'सारणिक के गुणधर्मों से सिद्ध कीजिए' (5 अंक) — गुणधर्म वाला भाग संशोधित NCERT पुस्तक से हटा दिया गया है और UPMSP 2026-27 पाठ्यक्रम में नहीं है, फिर भी पुराने प्रश्नपत्रों में यह अक्सर आया और हाल के प्रश्नपत्रों में |a b c; a² b² c²; a³ b³ c³| जैसे सारणिक का मान पूछा गया। समय हो तो टॉपर भाग के मानक परिणाम सीखिए; अन्यथा लंबे प्रमाण छोड़ दीजिए।'Determinant ki properties se prove kijiye' (5 marks) — properties wala hissa rationalised NCERT book se hata diya gaya hai aur UPMSP 2026-27 syllabus mein nahi hai, phir bhi purane papers mein yeh aksar aaya aur haal ke papers mein |a b c; a² b² c²; a³ b³ c³| jaise determinant ki value poochhi gayi. Time ho to Topper section ke standard results seekhiye; warna lambe proofs chhod dijiye.
Notes written by Simple Learning Hub from the current NCERT syllabus. Year counts come only from UP Board papers we actually read.नोट्स Simple Learning Hub ने वर्तमान NCERT पाठ्यक्रम से लिखे हैं। वर्षों की गिनती केवल उन यूपी बोर्ड पेपरों से है जो हमने सच में पढ़े।Notes Simple Learning Hub ne current NCERT syllabus se likhe hain. Saalon ki ginti sirf un UP Board papers se hai jo humne sach mein padhe.