Whole row/column zero

Chapter 4 Class 12 Determinants
Concept wise

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Misc 17 (Method 1) Choose the correct answer. If a, b, c, are in A.P., then the determinant |β 8(π₯+2&π₯+3&π₯+2π@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then, b β a = c β b b β a β c + b = 0 2b β a β c = 0 (Common difference is equal) β¦(1) Solving |β 8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying and dividing 2 = 2/2 |β 8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| Multiplying R2 by 2 = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@π(π₯+3)&π(π₯+4)&π(π₯+2π)@π₯+4&π₯+5&π₯+2π)| = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@2π₯+6&2π₯+8&2π₯+4π@π₯+4&π₯+5&π₯+2π)| Applying R2 β R2 β R1 β R3 = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@2π₯+6β(π₯+2)β(π₯+4 )&2π₯+8β(π₯+3)β(π₯+5)&2π₯+4πβ(π₯+2π)β(π₯+2π)@π₯+4&π₯+5&π₯+2π)| = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@2π₯+6βπ₯β2βπ₯β4&2π₯+8βπ₯β3βπ₯β5&2π₯+4πβπ₯β2πβπ₯β2π@π₯+4&π₯+5&π₯+2π)| = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@0&0&4πβ2πβ2π@π₯+4&π₯+5&π₯+2π)| = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@0&0&2(ππβπβπ)@π₯+4&π₯+5&π₯+2π)| = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@0&0&2(π)@π₯+4&π₯+5&π₯+2π)| (From (1): 2b β b β c = 0) = 1/2 |β 8(π₯+2&π₯+3&π₯+2π@0&0&0@π₯+4&π₯+5&π₯+2π)| If any row or column of determinant are zero, then value of determinant is also zero. = 1/2 Γ 0 = 0 Thus, the value of determinant is 0 Correct Answer is A Misc 17 (Method 2) Choose the correct answer. If a, b, c, are in A.P., then the determinant |β 8(x+2&x+3&x+2a@x+3&x+4&x+2b@x+4&x+5&x+2c)| is A. 0 B. 1 C. x D. 2x Since a, b & c are in A.P Then a β b = c β b b + b = c + a 2b = a + c (Common difference is equal) β¦(1) Consider |β 8(π₯+2&π₯+3&π₯+2π@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| Applying R1 βR1 + R3 β 2R2 = |β 8((π₯+2)+(π₯+4)β2(π₯+3)&(π₯+3)+(π₯+5)β2(π₯+4)&(π₯+2π)+(π₯+2π)β2(π₯+2π)@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| = |β 8(π₯+2+π₯+4β2π₯β6&π₯+3+π₯+5β2π₯β8&π₯+2π+π₯+2πβ2π₯β4π@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| = |β 8(2π₯β2π₯+6β6&2π₯β2π₯+8β8&2π₯β2π₯+2π+2πβ4π@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| = |β 8(0&0&0+2(π+πβ2π)@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| = |β 8(0&0&2(ππβ2π)@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| = |β 8(0&0&0@π₯+3&π₯+4&π₯+2π@π₯+4&π₯+5&π₯+2π)| If any row or column of determinant are zero, then value of determinant is also zero. = 0 Hence, value of determinant is 0 Correct Answer is A (From (1): 2b = a + c)