643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 8 InterviewSolution

351.	One root of the equation `\|(3x-8, 3, 3),(3,3x-8, 3),(3,3,3x-8)\|=0 si (A) `8/3` (B) `2/3` (C) `1/3` (D) `16/3`
Answer» Apply `C_(1) to C_(1) + C_(2) + C_(3)` and take (3x-2) common from `C_(1).`

Discussion

352.	One root of the equation `\|(3x-8, 3, 3),(3,3x-8, 3),(3,3,3x-8)\|=0 si (A) `8/3` (B) `2/3` (C) `1/3` (D) `16/3`A. `8//3`B. `2//3`C. `1//3`D. `16/3`
Answer» Correct Answer - B

Discussion

353.	Find the values of x, if\(\begin{vmatrix}1&2x&4x\\1&4&16\\1&1&1\end{vmatrix}=0\)[(1, 2x, 4x) (1, 4, 16) (1, 1, 1)] = 0
Answer» \(\begin{vmatrix}1&2x&4x\\1&4&16\\1&1&1\end{vmatrix}=0\) ⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0 ⇒ 1(-12) – 2x(-15) + 4x(-3) = 0 ⇒ -12 + 30x – 12x = 0 ⇒ 18x = 12 ⇒ x = 12/18 = 2/3

353.

Find the values of x, if\(\begin{vmatrix}1&2x&4x\\1&4&16\\1&1&1\end{vmatrix}=0\)[(1, 2x, 4x) (1, 4, 16) (1, 1, 1)] = 0

Answer»

\(\begin{vmatrix}1&2x&4x\\1&4&16\\1&1&1\end{vmatrix}=0\)

⇒ 1(4 – 16) – 2x(1 – 16) + 4x(1 – 4) = 0

⇒ 1(-12) – 2x(-15) + 4x(-3) = 0

⇒ -12 + 30x – 12x = 0

⇒ 18x = 12

⇒ x = 12/18 = 2/3

Discussion

354.	For what values of `k ,`thefollowing system of equations possesses a nontrival solution over the set ofrationals:`c+k y+3z=0,3c+k y-2z=02c+3y-4x=0.`Also find the solution for this value of `kdot`
Answer» The system x+ky+ 3z=0 3x+ky-2z=0 2x+3y-4z=0 as nontrivial solution (i.e., nonzero solution ) if the determinant if coefficients of x,y and z is zero , Here, `Delta=\|{:(1,,k,,3),(3,,k,,-2),(2,,3,,-4):}\|=0` `" or " 2k-33=0 " ok " k= 33//2` then the equations become `2x+33y+6z=0` `6x+33y-4z=0` `2x+3y-4z=0` Eliminating x we get from (2) and (4) 30y+ 10z=0 i.e., 3y+z=0 `" Let "y=lambda in R . " then " z=- 3lambda " and so "` `2x=-33lambda + 18lambda =- 15lambda` `:. x=-(15)/(2)lambda , y=lambda z=-3lambda , lambda in R`

Discussion

355.	If A = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix},\) find the value of \|A\| + \|B\|.
Answer» Given that A = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix},\) we have to find \|A\|+\|B\| Then, \|A\| = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and \|B\| = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix}\) \|A\| = 0 × 1 - i × i = -i² =1 (Expanding along R₁ and since i² = -1) \|B\| = 0 × 1 -1 × 1 = -1 (Expanding along R₁) \|A\| + \|B\| = 1 - 1 0

355.

If A = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix},\) find the value of |A| + |B|.

Answer»

Given that A = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix},\) we have to find |A|+|B|

Then, |A| = \(\begin{bmatrix} 0 & i \\ i & 1 \\ \end{bmatrix}\) and |B| = \(\begin{bmatrix} 0 & 1 \\ 1 & 1 \\ \end{bmatrix}\)

|A| = 0 × 1 - i × i

= -i²

=1 (Expanding along R₁ and since i² = -1)

|B| = 0 × 1 -1 × 1

= -1 (Expanding along R₁)

|A| + |B| = 1 - 1

0

Discussion

356.	If `Delta = \|(cos (alpha_(1) - beta_(1)),cos (alpha_(1) - beta_(2)),cos (alpha_(1) - beta_(3))),(cos (alpha_(2) - beta_(1)),cos (alpha_(2) - beta_(2)),cos (alpha_(2) - beta_(3))),(cos (alpha_(3) - beta_(1)),cos (alpha_(3) - beta_(2)),cos (alpha_(3) - beta_(3)))\|" then " Delta` equalsA. `cos alpha_(1) cos alpha_(2) cos alpha_(3) cos beta_(1) cos beta_(2) cos beta_(3)`B. `cos alpha_(1) + cos alpha_(2) + cos alpha_(3) + cos beta_(1) + cos beta_(2) + cos beta_(3)`C. `cos (alpha_(1) - beta_(1)) cos (alpha_(2) - beta_(2)) cos (alpha_(3) - beta_(3))`D. none of these
Answer» Correct Answer - D We have, `Delta = \|("cos" alpha_(1),"sin" alpha_(1),0),("cos" alpha_(2),"sin" alpha_(2),0),("cos" alpha_(3),"sin" alpha_(3),0)\|\|("cos" beta_(1),"sin "beta_(1),0),("cos" beta_(2),"sin" beta_(2),0),("cos"beta_(3),"sin "beta_(3),0)\|` `rArr Delta = 0 xx 0 = 0`

Discussion

357.	if `Delta= \|{:(3,,4,,5,,x),(4,,5,,6,,y),(5,,6,,7,,z),(x,,y,,z,,0):}\|=0` thenA. `x,y,z " are in A.P."`B. `x,y,z" are in G.P "`C. `x, y, z " are in H.P "`D. none of these
Answer» Correct Answer - A Applying `R_(1) to R_(1)+R_(3)-2R_(2)`we get `Delta = \|{:(0,,0,,0,,x+z-2y),(4,,5,,6,,y),(5,,6,,7,,z),(x,,y,,z,,0):}\|` `=-(x+z+2y) \|{:(4,,5,,6),(5,,6,,7),(x,,y,,z):}\|" ""[Expanding along "R_(1)"]"` `=-(x+z-2y) \|{:(0,,-1,,6),(0,,-1,,7),(x-2x+z,,y-z,,z):}\|` `"[Applying "C_(1) to C_(1)+C_(3) -2C_(2) " and " C_(2) to C_(2)-C_(3)"]"` `=-(x+z-2y)^(2) \|{:(-1,,6),(-1,,7):}\|` `=(x-2y+z)^(2)` Hence `Delta =0 rArr x,y,z " are in A.P "`

Discussion

358.	If a, b and c are all different from zero and `Delta = \|(1 +a,1,1),(1,1 +b,1),(1,1,1 +c)\| = 0`, then the value of `(1)/(a) + (1)/(b) + (1)/(c)` isA. abcB. `(1)/(abc)`C. `-a -b -c`D. `-1`
Answer» Correct Answer - D

Discussion

359.	Let `overset(to)(a_(r))=x_(r )hat(i)+y_(r )hat(j)+z_(r )hat(k),r=1,2,3` three mutually prependicular unit vectors then the value of `\|{:(x_(1),,-x_(2),,x_(3)),(y_(1),,y_(2) ,,y_(3)),(z_(1),,z_(2),,z_(3)):}\|` is equal toA. zeroB. `+-1`C. `+-2`D. none of these
Answer» Correct Answer - B `Delta =\|{:(x_(1),,x_(2),,x_(3)),(y_(1),,y_(2),,y_(3)),(z_(1),,z_(2),,z_(3)):}\|=\|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|` `:. Delta^(2) = \|{:(x_(1),,y_(1),,z_(2)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|\|{:(x_(1),,y_(1),,z_(1)),(x_(2),,y_(2),,z_(2)),(x_(3),,y_(3),,z_(3)):}\|` `=\|{:(x_(1)^(2)+y_(1)^(2)+z_(1)^(2),,x_(1)x_(2)+y_(1)y_(2)+z_(1)z_(2),,x_(1)x_(3)+y_(1)y_(3)+z_(1)z_(3)),(x_(1)x_(2)+y_(2)y_(1)+z_(2)z_(1),,x_(2)^(2)+y_(2)^(2)+z_(2)^(2),,x_(2)x_(3)+y_(2)y_(3)+z_(2)z_(3)),(x_(3)x_(1)+y_(3)y_(1)+z_(3)z_(1),,x_(2)x_(3)+y_(2)y_(3)+z_(2)z_(3),,x_(3)^(2)+y_(3)^(2)+z_(3)^(2)):}\|` `=\|{:(1,,0,,0),(0,,1,,0),(0,,0,,1):}\|=1 rArr Delta =+- 1`

Discussion

360.	Roots of the equations `\|{:(x,,m,,n,,1),(a,,x,,n,,1),(a,,b,,x,,1),(a,,b,,c,,1):}\|=0` areA. independent of m and nB. independent of a,b and cC. depend on m,n and a,b,cD. inedependent of m,n and a,b,c
Answer» Correct Answer - A `\|{:(x,,m,,n,,1),(a,,x,,n,,1),(a,,b,,x,,1),(a,,b,,c,,1):}\| " "[R_(1) to R_(1) -R_(2) ,R_(2) to R_(2)-R_(3),R_(3) to R_(3) -R_(4)]` `" or " \|{:(x-a,,m-x,,0,,0),(0,,x-b,,n-x,,0),(0,,0,,x-c,,0),(a,,b,,c,,1):}\|=0` `" or "\|{:(x-a,,m-x,,0),(0,,x-b,,n-x),(0,,0,,x-c):}\|=0` `" or " (x-a) \|{:((x-b),,n-x),(0,,(x-c)):}\|=0` `" or " (x-a) (x-b) (x-c) =0` Thus roots are independent of m and n.

Discussion

361.	If `\|(a,b-c,b+c),(a+c,b,c-a),(a-b,a+b,c)\|=0` then the line `ax+by+c=0` passes through the fixed point which isA. `(1,2)`B. `(1,1)`C. `(-2,1)`D. `(1,0)`
Answer» Correct Answer - B Applying `C_(1) to aC_(1)` and then `C_(1) to C_(1) +bC_(2) +cC_(3)` and taking `(a^(2)+b^(2)+c^(2))` common from `C_(1) ` we get ` Delta =((a^(2)+b^(2)+c^(2)))/(a) \|{:(1,,b-c,,c+b),(1,,b,,c-a),(1,,b+a,,c):}\|` `=((a^(2)+b^(2)+c^(2)))/(a) \|{:(1,,b-c,,c+b),(0,,c,,-a-b),(0,,a+c,,-b):}\|` `(R_(2) to R_(2)-R_(1),R_(3) to R_(3)-R_(1))` `=((a^(2)+b^(2)+c^(2)))/(a) (-bc+a^(2)+ab+ac+bc)` (expanding along `C_(1)`) `=(a^(2)+b^(2)+c^(2))(a+b+c)` hence `Delta =0 rArr a+b+c=0` Therefore line ax+by+c=0 passes through the fixed point (1,1)

Discussion

362.	if `D_(k) =\|{:( 1,,n,,n),(2k,,n^(2)+n+1 ,,n^(2)+n),(2k-1,,n^(2),,n^(2)+n+1):}\|" and " Sigma_(k=1)^(n) D_(k)=56` then n equalsA. 4B. 6C. 8D. 7
Answer» Correct Answer - D `overset(n)underset(k=1)(Sigma) D_(k) =56` `rArr \|{:(overset(n)underset(k=1)(Sigma)1,,n,,n),(overset(n)underset(k=1)(Sigma)2k,,n^(2)+n+1,,n^(2)+n),(overset(n)underset(k=1)(Sigma)(2k-1),,n^(2),,n^(2)+n+1):}\|=56` `" or " \|{:(n,,n,,n),(n(n+1),,n^(2)+n+1,,n^(2)+n),(n^(2),,n^(2),,n^(2)+n+1):}\|` Applying `C_(3) to C_(3) -C_(1) " and " C_(2) to C_(2)-C_(1)` we get `\|{:(n,,0,,0),(n(n+1),,1,,0),(n^(2),,0,,n+1):}\|=56` `" or " n(n+1) =56 " or " n=7`

Discussion

363.	Let `D_(r) = \|(a,2^(r),2^(16) -1),(b,3(4^(r)),2(14^(16) -1)),(c,7(8^(r)),4(8^(16) -1))\|`, then the value of `Sigma_(k=1)^(16) D_(k)`, is
Answer» Correct Answer - A `underset(k=1)overset(16)Sigma D_(k) = \|(a,underset(k=1)overset(16)Sigma 2^(k),2^(16) -1),(b,3 (underset(k=1)overset(16)Sigma 4^(k)),2(4^(16) -1)),(c,7 (underset(k=1)overset(16)Sigma 8^(k)),4(8^(16) -1))\|` `rArr underset(k=1)overset(16)Sigma D_(k) \|(a,2((2^(16) -1)/(2 -1)),2^(16) -1),(b,3xx 4 ((4^(16) -1)/(4 -1)),2(4^(16) -1)),(c,7xx 8 ((8^(16) -1)/(8 -1)),4(8^(16) -1))\|` `rArr underset(k=1)overset(16)Sigma D_(k) =2 \|(a,(2^(16) -1),(2^(16) -1)),(b,2(4^(16) 1),2(4^(16) -1)),(c,4(8^(16) -1),4(8^(16) -1))\| [("Taking 2 common"),(" from " C_(2))]` `rArr underset(k=1)overset(16)Sigma D_(k) = 2 xx 0 = 0`

Discussion

364.	the value of `Sigma_(r=2)^(n) (-2)^(r )\|{:( ""^(n-2)C_(r-2),,""^(n-2)C_(r-1),,""^(n-2)C_(r)),(-3,,1 ,,1),(2,,-1,,0):}\| (n gt 2)`A. `2n -1+(-1)^(n)`B. `2n+1+(-1)^(n-1)`C. `2n-3+(-1)^(n)`D. none of these
Answer» Correct Answer - A Applying `c_(1) to C_(1) +2C_(2)+C_(3)` we get `S=overset(n)underset(r=2)(Sigma)(-2)^(r ) \|{:(.^(n)C_(r),,.^(n-2)C_(r-1),,.^(n-2)C_(r )),(0,,1,,1),(0,,-1,,0):}\|` `=overset(n)underset(r=2)(Sigma) (-2)^(rn)C_(r ) =overset(n)underset(r=0)(Sigma) (-2)^(r " " n)C_(r )-(.^(n)C_(0)-2.^(n)C_(1))` `=(1-2)^(n)-(1-2n) =2n-1+(-1)^(n)`

Discussion

365.	If A = \(\begin{bmatrix}1 & 2 \\3 & -1 \\\end{bmatrix}\) and B = \(\begin{bmatrix}1 & 0 \\-1 & 0 \\\end{bmatrix},\) find \|AB\|.
Answer» Given that A = \(\begin{bmatrix} 1 & 2 \\ 3 & -1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 0 \\ -1 & 0 \\ \end{bmatrix},\) we have to find \|AB\| Then, \|A\| = \(\begin{bmatrix} 1 & 2 \\ 3 & -1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 0 \\ -1 & 0 \\ \end{bmatrix}\) \|A\| = 1 × -1 -2 × 3 = -7 (Expanding along R₁) \|B\| = 1 × 0 - 0 × -1 = 0 (Expanding along R₁) Since \|AB\| = \|A\|\|B\|, Therefore \|AB\| = -7 × 0 = 0

365.

If A = \(\begin{bmatrix}1 & 2 \\3 & -1 \\\end{bmatrix}\) and B = \(\begin{bmatrix}1 & 0 \\-1 & 0 \\\end{bmatrix},\) find |AB|.

Answer»

Given that A = \(\begin{bmatrix} 1 & 2 \\ 3 & -1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 0 \\ -1 & 0 \\ \end{bmatrix},\) we have to find |AB|

Then, |A| = \(\begin{bmatrix} 1 & 2 \\ 3 & -1 \\ \end{bmatrix}\) and B = \(\begin{bmatrix} 1 & 0 \\ -1 & 0 \\ \end{bmatrix}\)

|A| = 1 × -1 -2 × 3

= -7 (Expanding along R₁)

|B| = 1 × 0 - 0 × -1

= 0 (Expanding along R₁)

Since |AB| = |A||B|, Therefore |AB| = -7 × 0 = 0

Discussion

366.	If `\|(b +c,c +a,a +b),(a +b,b +c,c +a),(c +a,a +b,b +c)\| = k \|(a,b,c),(c,a,b),(b,c,a)\|`, then the value of k, isA. 1B. 2C. 3D. 4
Answer» Correct Answer - B

Discussion

367.	`f(x) = \|{:(x+c_(1),,x+a,,x+a),(x+b,,x+c_(2),,x+a),(x+b,,x+b,,x+c_(3)):}\| " and " g(x)= (C_(1) -x)(c_(3)-x)` Coefficient of x in f(x) isA. `(g(a)-f(b))/(b-a)`B. `(g(-a)-g(-b))/(b-a)`C. `(g(a)-g(b))/(b-a)`D. none of these
Answer» Correct Answer - C In given determinant ,applying `C_(2) to C_(2) -C_(1) " and " C_(3) to C_(3) -C_(2)` we get `f(x) = \|{:(x+c_(1),,a-c_(1),,0),(x+b,,c_(2)-b,,a-c_(2)),(x+b,,0,,c_(3)-b):}\|` `=x \|{:(1,,a-c_(1),,0),(1,,c_(2)-b,,a-c_(2)),(1,,0,,c_(3)-b):}\|+\|{:(c_(1),,a-c_(1),,0),(b,,c_(2)-b,,a-c_(2)),(b,,0,,c_(3)-b):}\|` So f(x) is linear Let f(x) = Px +Q. Then f(-a) =-aP+Q , f(-b) =-bP+Q` Then . f(0)=0 xxP +Q` `rArr Q= (bf (-a) -af(-b))/((b-a)) =f(0)` Also f(-a) `=\|{:(c_(1)-a,,0,,0),(b-a,,c_(2)-a,,0),(b-a,,b-a,,c_(3)-a):}\|` `=(c_(1) -a)(c_(1) -a) (c_(3) -a)` similarly `f(-b) =(c_(1)-b) (c_(2) -b) (c_(3) -b)` now `g(x) (c_(1) -x) (c_(2) -x)(c_(3) -x)` `rArr g(a) =f(-a) "and "g(b) =f(-b)` Now from (1) we get `f(0) =(bg (a)-ag(b))/((b-a))`

Discussion

368.	Consider the polynomial fucntion `f(x) = \|{:((1+x)^(2),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|` a,b being positive integers. the coefficient of x in f(x) isA. `2^(a)`B. `2^(a) -3xx 2^(b)+1`C. `0`D. none of these
Answer» Correct Answer - C Let `\|{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|=A +Bx+Cx^(2)+…….` Putting x=0 we get `A= \|{:(1,,1,,1),(1,,1,,1),(1,,1,,1):}\|=0` now differenting both sides with respect to x and putting x=0 we get `B= \|{:(a,,2b,,0),(1,,1,,1),(1,,1,,1):}\|+\|{:(1,,1,,1),(0,,a,,2b),(1,,1,,1):}\|+\|{:(1,,1,,1),(1,,1,,1),(2b,,0,,a):}\|=0` Hence ,coefficient of x is 0 .Since f(x)=0 and f(0)=0 ,x=0 is a repreating root of the equation f(x)=0

Discussion

369.	Consider the polynomial fucntion `f(x) = \|{:((1+x)^(2),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|` a,b being positive integers. Which of the following is true ?A. All the roots of the equation f(x)=0 are positiveB. All the roots of the equation f(x)=0 are negativeC. At least one of the equation f(x)=0 is repeating one .D. None of these
Answer» Correct Answer - C Let `\|{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|=A +Bx+Cx^(2)+…….` Putting x=0 we get `A= \|{:(1,,1,,1),(1,,1,,1),(1,,1,,1):}\|=0` now differenting both sides with respect to x and putting x=0 we get `B= \|{:(a,,2b,,0),(1,,1,,1),(1,,1,,1):}\|+\|{:(1,,1,,1),(0,,a,,2b),(1,,1,,1):}\|+\|{:(1,,1,,1),(1,,1,,1),(2b,,0,,a):}\|=0` Hence ,coefficient of x is 0 .Since f(x)=0 and f(0)=0 ,x=0 is a repreating root of the equation f(x)=0

Discussion

370.	If A is a square matrix of order n such that its elements are polynomila in x and its r-rows become identical for x = k, thenA. `(x -k)^(r)` is a factor of `\|A\|`B. `(x -k)^(r) -1` is a factor of `\|A\|`C. `(x -k)^(r) +1` is a factor `\|A\|`D. `(x -k)^(r)` is a factor of A
Answer» Correct Answer - A

Discussion

371.	If `omega` is a cube root of unity, then for polynomila is `\|(x + 1,omega,omega^(2)),(omega,x + omega^(2),1),(omega^(2),1,x + omega)\|`A. 1B. `omega`C. `omega^(2)`D. 0
Answer» Correct Answer - D

Discussion

372.	`f(x) = \|{:(x+c_(1),,x+a,,x+a),(x+b,,x+c_(2),,x+a),(x+b,,x+b,,x+c_(3)):}\| " and " g(x)= (C_(1) -x)(c_(3)-x)` `Which of the following is not true ?A. `f(-a) =g(a)`B. `f(-a) =g(-a)`C. `f(-b)=g(b)`D. none of these
Answer» Correct Answer - B In given determinant ,applying `C_(2) to C_(2) -C_(1) " and " C_(3) to C_(3) -C_(2)` we get `f(x) = \|{:(x+c_(1),,a-c_(1),,0),(x+b,,c_(2)-b,,a-c_(2)),(x+b,,0,,c_(3)-b):}\|` `=x \|{:(1,,a-c_(1),,0),(1,,c_(2)-b,,a-c_(2)),(1,,0,,c_(3)-b):}\|+\|{:(c_(1),,a-c_(1),,0),(b,,c_(2)-b,,a-c_(2)),(b,,0,,c_(3)-b):}\|` So f(x) is linear Let f(x) = Px +Q. Then f(-a) =-aP+Q , f(-b) =-bP+Q` Then . f(0)=0 xxP +Q` `rArr Q= (bf (-a) -af(-b))/((b-a)) =f(0)` Also f(-a) `=\|{:(c_(1)-a,,0,,0),(b-a,,c_(2)-a,,0),(b-a,,b-a,,c_(3)-a):}\|` `=(c_(1) -a)(c_(1) -a) (c_(3) -a)` similarly `f(-b) =(c_(1)-b) (c_(2) -b) (c_(3) -b)` now ` g(x) (c_(1) -x) (c_(2) -x)(c_(3) -x)` `rArr g(a) =f(-a) "and "g(b) =f(-b)` Now from `(1) ` we get `f(0) =(bg (a)-ag(b))/((b-a))`

Discussion

373.	`f(x) = \|{:(x+c_(1),,x+a,,x+a),(x+b,,x+c_(2),,x+a),(x+b,,x+b,,x+c_(3)):}\| " and " g(x)= (C_(1) -x)(c_(3)-x)` Which of the following is not a constant term in f(x) ?A. `(bg(a)-ag(b))/((b-a))`B. `(bf(a)-af(-b))/((b-a))`C. `(bf(-a)-ag(b))/((b-a))`D. none of these
Answer» Correct Answer - D In given determinant ,applying `C_(2) to C_(2) -C_(1) " and " C_(3) to C_(3) -C_(2)` we get `f(x) = \|{:(x+c_(1),,a-c_(1),,0),(x+b,,c_(2)-b,,a-c_(2)),(x+b,,0,,c_(3)-b):}\|` `=x \|{:(1,,a-c_(1),,0),(1,,c_(2)-b,,a-c_(2)),(1,,0,,c_(3)-b):}\|+\|{:(c_(1),,a-c_(1),,0),(b,,c_(2)-b,,a-c_(2)),(b,,0,,c_(3)-b):}\|` So f(x) is linear Let f(x) = Px +Q. Then f(-a) =-aP+Q , f(-b) =-bP+Q` Then . f(0)=0 xxP +Q` `rArr Q= (bf (-a) -af(-b))/((b-a)) =f(0)` Also f(-a) `=\|{:(c_(1)-a,,0,,0),(b-a,,c_(2)-a,,0),(b-a,,b-a,,c_(3)-a):}\|` `=(c_(1) -a)(c_(1) -a) (c_(3) -a)` similarly `f(-b) =(c_(1)-b) (c_(2) -b) (c_(3) -b)` now ` g(x) (c_(1) -x) (c_(2) -x)(c_(3) -x)` `rArr g(a) =f(-a) "and "g(b) =f(-b)` Now from `(1) ` we get `f(0) =(bg (a)-ag(b))/((b-a))`

Discussion

374.	if `omega!=1` is cube root of unity and x+y+z`!=`0 then `\|[x/(1+omega),y/(omega+omega^2),z/(omega^2+1)],[y/(omega+omega^2),z/(omega^2+1),x/(1+omega)],[z/(omega^2+1),x/(1+omega),y/(omega+omega^2)]\|`=0 if
Answer» As `1 + omega +omega^(2) =0 ` `D= \|{:((x )/(1+ omega),,(y)/(omega+omega^(2)),,(z)/(omega^(2)+1)),((y)/(omega+omega^(2)),,(z)/(omega^(2)+1),,(x)/(1+omega)),((z)/(omega^(2)+1),,(x)/(1+omega),,(y)/(omega+omega^(2))):}\|` `=\|{:(-(x)/(omega^(2)),,-y,,-(z)/(omega)),(-y,,-(z)/(omega),,-(x)/(omega^(2)),(-(z)/(omega),,-(x)/(omega^(2)),,-y):}\|` `=x^(3) +y^(3) +z^(3)-3xyz` `=(1)/(2) (X+Y+Z) {(x-y)^(2)+(y-z)^(2)+(z-x)^(2)` `rArr x=y=z" " "(":. x+y+z ne 0")"`

Discussion

375.	Suppose f(x) is a function satisfying the following conditions : (i) f(0)=2,f(1)=1 (ii) f has a minimum value at `x=5//2` (iii) for all `x,f (x) = \|{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}\|` Range of f(x) isA. `[7//16,oo)`B. `(-oo,15//16]`C. `[3//4,oo)`D. none of these
Answer» Correct Answer - A `f(x) = \|{:(2ax,,2ax-1,,2ax+b+1),(b,,b+1,,-1),(2(ax+b),,2ax+2b+1,,2ax+b):}\|` Applying `(C_(1) to C_(1)-C_(3) ,C_(2) to C_(2) -C_(3)` `f(x) =\|{:(-(b+1),,-(b+2),,2ax+b+1),((b+1),,(b+2),,-1),(b,,b+1,,2ax+b):}\|` Applying `R_(1) to R_(1)+R_(2)" and " R_(3) to R_(3)-R_(2)` we get `f(x) = \|{:(0,,0,,2ax+b),(b+1,,b+2,,-1),(-1,,-1,,2ax+b+1):}\|` `=(2ax +b) [-b-1+b+2]` `:. f(x) =2ax+b` `:. f(x) = ax^(2) +bx +c` `f(0) =2 rArr c=2` `f(1) =1rArr a+b+2=1rArr a+b =1` `f(5//2) =0 rArr 5a+ b=0` `rArr a=1//4 ,b=-5//4` hence `f(x) =(1)/(4) x^(2) -(5)/(4)x+2` Clearly . discriminant (D) of the equation f(x) =0 is less than 0. hence f(x)=0 has imaginary roots .Also f(2) `=1//2` .and minimum value of f(x) is `-((25)/(16)-4.(1)/(4) (2))/(4.(1)/(4)) =(7)/(16)` Hence range of the f(x) is `[(7)/(16),oo)`

Discussion

376.	Consider the polynomial fucntion `f(x) = \|{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|` a,b being positive integers. The constant term in f(x) isA. 2B. 1C. -1D. 0
Answer» Correct Answer - D Let `\|{:((1+x)^(a),,(1+2x)^(b),,1),(1,,(1+x)^(a),,(1+2x)^(b)),((1+2x)^(b),,1,,(1+x)^(a)):}\|=A +Bx+Cx^(2)+…….` Putting x=0 we get `A= \|{:(1,,1,,1),(1,,1,,1),(1,,1,,1):}\|=0` now differenting both sides with respect to x and putting x=0 we get `B= \|{:(a,,2b,,0),(1,,1,,1),(1,,1,,1):}\|+\|{:(1,,1,,1),(0,,a,,2b),(1,,1,,1):}\|+\|{:(1,,1,,1),(1,,1,,1),(2b,,0,,a):}\|=0` Hence ,coefficient of x is 0 .Since f(x)=0 and f(0)=0 ,x=0 is a repreating root of the equation f(x)=0

Discussion

377.	Let `A=\|1sintheta1-sintheta1sintheta-1-sintheta1\|,`where `0lt=thetalt=2pidot`Then`D e t(A)=0`(b) `D e t(A) in (2,oo)``D e t(A) in (2,4)`(d) `D e t(A) in [2,4]`
Answer» Correct Answer - (d) `:." "A=[{:(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1):}]` `rArr" "\|A\|=[{:(1,sintheta,1),(-sintheta,1,sintheta),(-1,-sintheta,1):}]` `=1(1+sin^(2)theta)-sintheta(-sintheta+sintheta)+1(sin^(2)theta+1)=2+2sin^(2)theta` `"For" 0 lethetale2pi` `-1lesin thetale1 rArr0lesin^(2)thetale1` `rArr" "1le1+sin^(2)thetale2` `rArr" "2le2+sin^(2)thetale4rArr2le\|A\|le4` `therefore" "detAin[2,4]`

Discussion

378.	Choose the correct answer in questions 17 to 19: If x, y, z are nonzero real numbers then the inverse of metrix `A=[{:(x,0,0),(0,y,0),(0,0,z):}]`is : (a) `[{:(x^(-1),0,0),(0,y^(1),0),(0,0,z^(1)):}]` (b) `xyz[{:(x^(-1),0,0),(0,y^(1),0),(0,0,z^(1)):}]` ( c) `1/(xyz)[{:(x,0,0),(0,y,0),(0,0,z):}]` (d) `1/(xyz)[{:(1,0,0),(0,1,0),(0,0,1):}]`
Answer» Correct Answer - (a) `:." "A=[{:(x,0,0),(0,y,0),(0,0,z):}]` `rArr" "\|A\|=[{:(x,0,0),(0,y,0),(0,0,z):}]=xyzne0` `A_(11)=yz, A_(12)=0, A_(13)=0` `A_(21)=0, A_(22)=zx, A_(23)=0` `A_(31)=0, A_(32)=0, A_(32)=0, A_(33)=xy` `:." adj A"=[{:(yz,0,0),(0,zx,0),(0,0,xy):}]=[{:(yz,0,0),(0,zx,0),(0,0,xy):}]` `"and A"^(-1)=1/\|A\|".adj A"=1/(xyz)[{:(yz,0,0),(0,zx,0),(0,0,xy):}]` `=[{:(1/x,0,0),(0,1/y,0),(0,0,1/z):}]` `=[{:(x^(-1),0,0),(0,y^(-1),0),(0,0,z^(-1)):}]`

Discussion

379.	Show that each of the following systems of linear equations is inconsistent : 3x + y = 5 – 6x – 2y = 9
Answer» Given : - Two equation 3x + y = 5 and – 6x – 2y = 9 Tip : - We know that For a system of 2 simultaneous linear equation with 2 unknowns (i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) (ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution. (iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent. Now, We have, 3x + y = 5 – 6x – 2y = 9 Lets find D ⇒ D = \(\begin{vmatrix} 3& 1 \\[0.3em] -6 &-2 \\[0.3em] \end{vmatrix}\) ⇒ D = – 6 – 6 ⇒ D = 0 Again, D₁ by replacing 1^st column by B Here, B = \(\begin{vmatrix} 5 \\[0.3em] 9\\[0.3em] \end{vmatrix}\) ⇒ D₁ = \(\begin{vmatrix} 5& 1 \\[0.3em] 9 &-2 \\[0.3em] \end{vmatrix}\) ⇒ D₁ = – 10 – 9 ⇒ D₁ = – 19 And, D₂ by replacing 2^nd column by B Here, B = \(\begin{vmatrix} 5 \\[0.3em] 9\\[0.3em] \end{vmatrix}\) ⇒ D₂ = \(\begin{vmatrix} 3& 5 \\[0.3em] -6 &9 \\[0.3em] \end{vmatrix}\) ⇒ D₂= 27 + 30 ⇒ D₂ = 57 So, here we can see that D = 0 and D₁ and D₂ are non – zero Hence the given system of equation is inconsistent.

379.

Show that each of the following systems of linear equations is inconsistent : 3x + y = 5 – 6x – 2y = 9

Answer»

Given : -

Two equation 3x + y = 5 and – 6x – 2y = 9

Tip : - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

(i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by

x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\)

(ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution.

(iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent.

Now,

We have,

3x + y = 5

– 6x – 2y = 9

Lets find D

⇒ D = \(\begin{vmatrix} 3& 1 \\[0.3em] -6 &-2 \\[0.3em] \end{vmatrix}\)

⇒ D = – 6 – 6

⇒ D = 0

Again,

D₁ by replacing 1^st column by B

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 9\\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 5& 1 \\[0.3em] 9 &-2 \\[0.3em] \end{vmatrix}\)

⇒ D₁ = – 10 – 9

⇒ D₁ = – 19

And,

D₂ by replacing 2^nd column by B

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 9\\[0.3em] \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 3& 5 \\[0.3em] -6 &9 \\[0.3em] \end{vmatrix}\)

⇒ D₂= 27 + 30

⇒ D₂ = 57

So, here we can see that

D = 0 and D₁ and D₂ are non – zero

Hence the given system of equation is inconsistent.

Discussion

380.	A matrix A of order 3 × 3 has determinant 5. What is the value of \|3A\|?
Answer» If the determinant of a matrix A of order m is ∆, then the determinant of matrix λA, where λ is a scalar, is λ^m∆. In this question, ∆ = 5, λ = 3 and m = 3. \|λA\| = 3³×5 = 135

380.

A matrix A of order 3 × 3 has determinant 5. What is the value of |3A|?

Answer»

If the determinant of a matrix A of order m is ∆, then the determinant of matrix λA, where λ is a scalar, is λ^m∆.

In this question, ∆ = 5, λ = 3 and m = 3.

|λA| = 3³×5

= 135

Discussion

381.	If `\|{:(a,b,0),(0,a,b),(b,0,a):}\|=0`, then which one of the following is correct ?A. a/b is one of the cube roots of unityB. a is one of the cube roots of unityC. b is one of the cube roots of unityD. a/b is one of the cube roots of `-1`
Answer» Correct Answer - D We have, `\|(a,b,0),(0,a,b),(b,0,a)\| =0` `rArr a^(3) + b^(3) = 0` [On expanding the determinant on LHS] `rArr (a//b)^(3) = -1` `rArr a//b` is one of the cube roots of `-1`

Discussion

382.	The largest value of a third order determinant whose elements are equal to `1 or 0` isA. 1B. 0C. 2D. 3
Answer» Correct Answer - C Let `Delta = \|(a_(1),b_(1),c_(1)),(a_(2),b_(2),c_(2)),(a_(3),b_(3),c_(3))\|` be a determinant of order 3. Then, `Delta = a_(1) b_(2) c_(3) + a_(3) b_(1) c_(2) + a_(2) b_(3) c_(1) - a_(1) b_(3) c_(2) - a_(2) b_(1) c_(3) - a_(3) b_(2) c_(1)` `Delta = (a_(1) b_(2) c_(3) + a_(3) b_(1) c_(2) + a_(2) b_(3) c_(1)) - (a_(1) b_(3) c_(2) + a_(2) b_(1) c_(3) + a_(3) b_(2) c_(1))` Since each element of `Delta` is either 1 or 0. Therefore, the value of the determinant cannot exceed 3. Clearly, the value of `Delta` is maximum when the value of each term in first bracket is 1 and the value of each term in the second bracket is zero. But `a_(1) b_(2) c_(3) = a_(3) b_(1) c_(2) = a_(2) b_(3) c_(1) = 1` implies that every element of the determinant `Delta` is 1 and in that case `Delta = 0`. Thus, we may have `Delta = \|(0,1,1),(1,0,1),(1,1,0)\| =2`

Discussion

383.	If `plambda^4+qlambda^3+rlambda^2+slambda+t=\|[lambda^2+3lambda, lambda-1, lambda+3] , [lambda^2+1, 2-lambda, lambda-3] , [lambda^2-3, lambda+4, 3lambda]\|` then `t=`
Answer» Since given relation is an identity it is true for all real values of `lambda` ` \|{:(0,,-1,,3),(1 ,,0,,-4),(-3,,4,,0):}\|` which is skew- symmetric determinant. So t=0

Discussion

384.	The largest value of a third order determinant whose elements are equal to `1 or 0` is
Answer» Let ` Delta=\|{:(a_(1),,b_(1),,c_(1)),(a_(2) ,,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}\|` be a determinant of order 3. Then `Delta =(a_(1) b_(2) c_(3) +a_(3)b_(1)c_(2) + a_(2)b_(3)c_(1))-(a_(1)b_(3)c_(2)+a_(2)b_(1)c_(3)+a_(3)b_(2)c_(1))` since each element of `Delta` is maximum when the value of each term in the first bracket is 1 and the value of each term in the second bracket is zero. But `a_(1)b_(2)c_(3) = a_(3)b_(1)c_(1)= 1` implies that every element of the determinant `Delta` is 1 and in that case `Delta =0` Thus we may have ` Delta=\|{:(0,,1,,1),(1 ,,0,,1),(1,,1,,0):}\|`

Discussion

385.	Prove that the value of the determinant ` \|{:(-7,,5+3i,,(2)/(3)-4i),(5-3i ,,8,,4+5i),((2)/(3) +4i,,4-5i,,9):}\|" is real "`
Answer» Let ` z=\|{:(-7,,5+3i,,(2)/(3)-4i),(5-3i ,,8,,4+5i),((2)/(3) +4i,,4-5i,,9):}\|` To Prove that this number (z) is real we have to prove that `hatz=z`. Now we know that conjugate of complex number is distributive over all algebraic operations . Hence to take conjugate of z in (1) we need not to expand determinant. to get the conjugate of z we can take conjugate of each element of determinant . Thus ` hatz=\|{:(-7,,5+3i,,(2)/(3)-4i),(5-3i ,,8,,4+5i),((2)/(3) +4i,,4-5i,,9):}\|" "(2)` Now interchanging rows into columns (taking transpose) in (2) ` " we have "hatz=\|{:(-7,,5+3i,,(2)/(3)-4i),(5-3i ,,8,,4+5i),((2)/(3) +4i,,4-5i,,9):}\|" "(3)` `" or " harz =z" " ["from (1) and (3)"] (4)` Hence z is purely real.

Discussion

386.	Solve the following equations:\(\begin{vmatrix}\text{x} & -6 & -1 \\[0.3em]2 & -3\text{x} & \text {x} - 3 \\[0.3em]-3 &2\text{x} & \text{x}+ 2 \end{vmatrix}\) = 0\|(x, -6, -1)(2, -3x, x-3)(-3, 2x, x + 2)\| = 0
Answer» Expanding with R₁ 0= x(-3x²- 6x - 2x²+ 6x) + 6(2x+4+3x-9) -1(4x - 9x) 0=x (-5x²) + 6(5x - 5) -1(-5x) 0= -5x³+ 30x - 30 + 5x 0= -5x³+ 35x - 30 x³- 7x + 6 = 0 x³- x - 6x + 6 = 0 x(x²-1) - 6(x - 1) = 0 x(x-1)(x+1) - 6(x - 1) = 0 (x - 1)(x²+ x - 6) = 0 (x-1)(x²+ 3x - 2x - 6) = 0 (x-1)(x(x + 3) - 2(x + 3) =0 (x-1)(x+3)(x-2) = 0 Either x - 1 = 0 or x + 3 = 0 or x - 2 = 0 ∴ x = 1 or x = -3 or x = 2

386.

Solve the following equations:\(\begin{vmatrix}\text{x} & -6 & -1 \\[0.3em]2 & -3\text{x} & \text {x} - 3 \\[0.3em]-3 &2\text{x} & \text{x}+ 2 \end{vmatrix}\) = 0|(x, -6, -1)(2, -3x, x-3)(-3, 2x, x + 2)| = 0

Answer»

Expanding with R₁

0= x(-3x²- 6x - 2x²+ 6x) + 6(2x+4+3x-9) -1(4x - 9x)

0=x (-5x²) + 6(5x - 5) -1(-5x)

0= -5x³+ 30x - 30 + 5x

0= -5x³+ 35x - 30

x³- 7x + 6 = 0

x³- x - 6x + 6 = 0

x(x²-1) - 6(x - 1) = 0

x(x-1)(x+1) - 6(x - 1) = 0

(x - 1)(x²+ x - 6) = 0

(x-1)(x²+ 3x - 2x - 6) = 0

(x-1)(x(x + 3) - 2(x + 3) =0

(x-1)(x+3)(x-2) = 0

Either x - 1 = 0 or x + 3 = 0 or x - 2 = 0

∴ x = 1 or x = -3 or x = 2

Discussion

387.	Find the area of the triangle whose vertices are: A(-2, 4), B(2, -6) and C(5, 4)
Answer» Area of triangle = \(\frac{1}{2}\)\(\begin{bmatrix}x_1 & y_1 & 1 \\[0.3em]x_2 & y_2 & 1 \\[0.3em]x_3 &y_3 & 1\end{bmatrix}\) = \(\frac{1}{2}\)\(\begin{bmatrix}-2 & 4& 1 \\[0.3em]2 & -6 & 1 \\[0.3em]5 &4& 1\end{bmatrix}\) Expanding with C₃ = \(\frac{1}{2}\)[(8 + 30) - (-8 - 20) + (12 - 8)] = \(\frac{1}{2}\) [38 + 28 + 4] = \(\frac{68}{2}\) = 34 sq. units

387.

Find the area of the triangle whose vertices are: A(-2, 4), B(2, -6) and C(5, 4)

Answer»

Area of triangle = \(\frac{1}{2}\)\(\begin{bmatrix}x_1 & y_1 & 1 \\[0.3em]x_2 & y_2 & 1 \\[0.3em]x_3 &y_3 & 1\end{bmatrix}\)

= \(\frac{1}{2}\)\(\begin{bmatrix}-2 & 4& 1 \\[0.3em]2 & -6 & 1 \\[0.3em]5 &4& 1\end{bmatrix}\)

Expanding with C₃

= \(\frac{1}{2}\)[(8 + 30) - (-8 - 20) + (12 - 8)]

= \(\frac{1}{2}\) [38 + 28 + 4]

= \(\frac{68}{2}\)

= 34 sq. units

Discussion

388.	Find the area of the triangle whose vertices are: P(0, 0), Q(6, 0) and R(4, 3)
Answer» Area of a triangle = \(\frac{1}{2}\)\(\begin{bmatrix} x_1 & y_1 & 1 \\[0.3em] x_2 & y_2 & 1 \\[0.3em] x_3 &y_3 & 1 \end{bmatrix}\) = \(\frac{1}{2}\)\(\begin{bmatrix} 0 & 0& 1 \\[0.3em] 6 & 0 & 1 \\[0.3em] 4 &3& 1 \end{bmatrix}\) Expanding with R₁ = \(\frac{1}{2}[18]\) = 9 sq. units

388.

Find the area of the triangle whose vertices are: P(0, 0), Q(6, 0) and R(4, 3)

Answer»

Area of a triangle = \(\frac{1}{2}\)\(\begin{bmatrix} x_1 & y_1 & 1 \\[0.3em] x_2 & y_2 & 1 \\[0.3em] x_3 &y_3 & 1 \end{bmatrix}\)

= \(\frac{1}{2}\)\(\begin{bmatrix} 0 & 0& 1 \\[0.3em] 6 & 0 & 1 \\[0.3em] 4 &3& 1 \end{bmatrix}\)

Expanding with R₁

= \(\frac{1}{2}[18]\)

= 9 sq. units

Discussion

389.	The determinant D = \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)= 0, if[(a, b, a+b) (b, c, b+c) (a+b, b+c, 0)](a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) α is a root of ax2 + 2bx + c = 0
Answer» Correct option is : (b) a, b, c are in G.P. Applying R₃ → R₃ – (R₁ + R₂ ), we get \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)=0 ∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0 ∴ (-ac + b²) (a + 2b + c) = 0 ∴ -ac + b² = 0 or a + 2b + c = 0 ∴ b² = ac ∴ a, b, c are in G.P.

389.

The determinant D = \(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)= 0, if[(a, b, a+b) (b, c, b+c) (a+b, b+c, 0)](a) a, b, c are in A.P. (b) a, b, c are in G.P. (c) a, b, c are in H.P. (d) α is a root of ax2 + 2bx + c = 0

Answer»

Correct option is : (b) a, b, c are in G.P.

Applying R₃ → R₃ – (R₁ + R₂ ), we get

\(\begin{vmatrix}1&b&a+b\\b&c&b+c\\a+b&b+c&0\end{vmatrix}\)=0

∴ a[-c(a + 2b + c) – 0] – b[-b(a + 2b + c) – 0] + (a + b) (0 – 0) = 0

∴ (-ac + b²) (a + 2b + c) = 0

∴ -ac + b² = 0 or a + 2b + c = 0

∴ b² = ac

∴ a, b, c are in G.P.

Discussion

390.	Evaluate \(\begin{vmatrix}X^2 -X +1& X-1 \\[0.3em]X+1& X+1 \\[0.3em]\end{vmatrix}\)
Answer» Theorem: This evaluation can be done in two different ways either by taking out the common things and then calculating the determinants or simply take determinant. I will prefer first method because with that chances of silly mistakes reduces. Take out x+1 from second row. \(\begin{vmatrix} X^2 -X +1& X-1 \\[0.3em] X+1& X+1 \\[0.3em] \end{vmatrix} \) ⇒(x+1) × (x²-x+1-(x-1)) ⇒ (x+1) × (x²-2x+2) ⇒ x³-2x²+2x+x²-2x+2 ⇒ x³-x²+2 .

390.

Evaluate \(\begin{vmatrix}X^2 -X +1& X-1 \\[0.3em]X+1& X+1 \\[0.3em]\end{vmatrix}\)

Answer»

Theorem: This evaluation can be done in two different ways either by taking out the common things and then calculating the determinants or simply take determinant.

I will prefer first method because with that chances of silly mistakes reduces.

Take out x+1 from second row.

\(\begin{vmatrix} X^2 -X +1& X-1 \\[0.3em] X+1& X+1 \\[0.3em] \end{vmatrix} \)

⇒(x+1) × (x²-x+1-(x-1))

⇒ (x+1) × (x²-2x+2)

⇒ x³-2x²+2x+x²-2x+2

⇒ x³-x²+2 .

Discussion

391.	If Aij is the cofactor of the element aij of \(\begin{bmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 & 5 & -7\end{bmatrix}\) then write the value of (a32 A32).
Answer» Theorem: A_ij is found by deleting i^th rowand j^th column, the determinant of left matrix is called cofactor with multiplied by (-1) (i+j). Given: i = 3 and j = 2. A₃₂=(-1)⁽³⁺²⁾(2 × 4-6 × 5) =-1 × (-22) =22 a₃₂=5 a₃₂A₃₂= 5 × 22 =110

391.

If Aij is the cofactor of the element aij of \(\begin{bmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 & 5 & -7\end{bmatrix}\) then write the value of (a32 A32).

Answer»

Theorem: A_ij is found by deleting i^th rowand j^th column, the determinant of left matrix is called cofactor with multiplied by (-1) (i+j).

Given: i = 3 and j = 2.

A₃₂=(-1)⁽³⁺²⁾(2 × 4-6 × 5)

=-1 × (-22)

=22

a₃₂=5

a₃₂A₃₂= 5 × 22

=110

Discussion

392.	Find the value of `aa n db`if thesystem of equation `a^2x-b y=a^2-ba n db x=b^2y=2+4b`(i)posses uniquesolution(ii)infinitesolutions
Answer» System of equations is `a^(2)x-by=a^(2)-b" and " bx-b^(2)y =2 +4b` (i) if system has unique solution then lines must be non- parallel or `Delta ne 0` Hence, `\|{:( a^(2),,-b),(b,,-b^(2)):}\| ne0` `" or " -a^(2)b^(2)+b^(2)ne0` ` " or " b^(2) (1-a^(2)) ne0` ` rArr bne " and "ane=1` (ii) if system has infinite solutions then lines must be coincident. Hence `(a^(2))/(b)=(b)/(b^(2)) =(a^(2)-b)/(2+4b)` `rArr b=0 " or " a = ne 1` `" if " a=1 " then " 2 + 4b =b-b^(2)` `" or " b^(2)+3b+2=0` `rArr b=-2 " or " -1` `" if " a=-1 , " then " b=-2 " or " -1` b=0 is not possible Then ordered pairs (a,b) for which system has infintie solutions are `( 1,-2) (1,-1)(-1,-2)(-1,-1)`

Discussion

393.	If a system of three linear equations `x+4a y+a=0,x+3b y+b=0,a n dx+2c y+c=0`isconsistent, then prove that `a ,b , c`are in H.P.
Answer» Since system of equation is consistent we have `Delta = \|{:(1,,4a,,a),(1,,3b,,b),(1,,2c,,c):}\|=0` `" or " 1(3bc -2bc ) -4a(c-b)+a(2c-3b)=0` `" or " bc-4ac+4ab+2ac -3ab=0` " or " bc-2ac+ =2ac` " or " bc+ ab=2ac` `" or "(1)/(a)+(1)/(c)=(2)/(b).` Thus a,b,c are in H.P.

Discussion

394.	The value of `\|alpha\|` for which the system of equation `alphax+y+z=alpha-1` `x+alphay+z=alpha-1` `x+y+alphaz=alpha-1` has no solution , is `"____"`
Answer» Correct Answer - 2 System of equation `alpha x +y+ z=alpha-1` `x+alpha y+z = alpha-1 ` `x+y +alpha z =alpha -1` Since system has no solution Therefore (1) `Delta =0` and (2) `alpha -1 ne 0` `\|{:(alpha,,1,,1),(1,,alpha,,1),(1,,1,,alpha):}\|= 0, alpha ne 1` `R_(1) to R_(1) -R_(3),R_(2) to R_(2)-R_(3)` ` \|{:(alpha-1,,0,,1-alpha),(0,,alpha-1,,1-alpha),(1,,1,,alpha):}\|=0` `" or " (alpha-1 )[alpha(alpha-1)-(1-alpha)]+(-alpha)[-(alpha-1)]=0` `" or " (alpha-1 )[alpha(alpha-1)+(alpha-1)]+(alpha-1)^(2)=0` `" or " (alpha-1)^(2)[(alpha+1)+1]=0` `" or " alpha=1 ,1, -2 " or " alpha =1,-2` Sicne system has no solution `alpha ne 1` `:. alpha=-2`

Discussion

395.	The value of `\|alpha\|` for which the system of equation `alphax+y+z=alpha-1` `x+alphay+z=alpha-1` `x+y+alphaz=alpha-1` has no solution , is `"____"`A. either -2 or 1B. -2C. 1D. not-2
Answer» Correct Answer - B for no solution or infinitely many solutions `\|{:(alpha ,,-1,,-1),(1,,-alpha,,-1),(1,,-1,,-alpha):}\|=0` `" or " alpha(alpha^(2) -1)-1(alpha-1)+(1-alpha)=0` `" or " alpha(alpha^(2)-1) -2alpha +2=0` `" or " alpha(alpha -1) (alpha+1) -2(alpha -1)=0` `" or " (alpha-1)(alpha^(2) +alpha -2) =0` `" or " (alpha -1) (alpha+2)(alpha -1) =0` `" or " (alpha-1)^(2)(alpha+2)=0` `" or " alpha=1 ,1,-2` But for `alpha =1` there are infinite solution. when `alpha=-2` we have `-2x -y-z=-3` `x+2y -z=-3` `x-y +2z=-3` Adding we get 0=-9 which is not true . Hence there is no solution.

Discussion

396.	If `c
Answer» Correct Answer - C the given system is consistent . Therefore `Delta =\|{:(1,,1,,-1),(2,,-1,,-c),(-b,,3b,,-c):}\|=0` `" or " c+bc -6b +b+ 2c+ 3bc=0` `" or " 3c+4bc-5b=0` `" or " 3c+ 4bc -5b =0` `" or " c=(5b)/(4b+3)` Now `c lt 1` `rArr (5 b)/(4b+3) lt 1 ` `" or " .(5b)/(4b +3) -1 lt 0` `rArr b in (-(3)/(4),3)`

Discussion

397.	If `a ,b ,c`are in G.P. with common ratio `r_1a n dalpha,beta,gamma`are in G.P. with common ratio `r_2`and equations `a x+alphay+z=0,b x+betay+z=0,c x+gammay+z=0`have only zero solution, then which of the following is not true?`a+b+c`b. `a b c`c.`1`d. none of theseA. `r_(1) ne 1`B. `r_(2) ne 1`C. `r_(1) ne r_(2)`D. none of these
Answer» Correct Answer - D As a,b,c are in G.P. with common ration `r_(1) ` and `alpha , beta , gamma` are in G.P having common ratio `r_(2) , a ne 0 , alpha ne 0,b =ar_(1) c=ar_(1)^(2)` `Beta =ar_(2) ,gamma =ar_(2)^(2)` Also the system of equations has only zero (trivial) solution `Delta \|{:(a,,alpha,,1),(b,,beta,,1),(c,,gamma,,1):}\|ne 0` `rArr aalpha \|{:(1,,1,,1),(r_(1),,r_(2),,1),(r_(1)^(2),,r_(2)^ (2),,1):}\| ne 0` `rArr aalpha (r_(1)-1) (r_(2) -1) (r_(1) -r_(2)) ne 0` `rArr r_(1) ne 1 , r_(2) ne 1 " and " r_(1) ne r_(2)`

Discussion

398.	Examine the consistency of the system of equations`x + 2 y = 2``2x + 3y = 3`
Answer» Given system of equations, x+2y=2 2x+3y=3 `rArr" "[{:(1,2),(2,3):}][{:(x),(y):}]=[{:(1),(2):}]rArrAX=B` `therefore" "A=[{:(1,2),(2,3):}]` `rArr" "\|A\|=[{:(1,2),(2,3):}]=3-4=-1ne0` `therefore` A is invertible. `rArr` Given system of equations is consistent.

Discussion

399.	Examine the consistency of the system of equations `x+3y=5 , 2x+6y=8`
Answer» Given system of equations x+3y=5 2x+6y =8 `rArr" "\|A\|[{:(2,-1),(1,1):}][{:(x),(y):}]=[{:(5),(8):}]rArr AX=B` `therefore" "A[{:(1,3),(2,6):}]` `rArr" "\|A\|=[{:(1,3),(2,6):}]=6-6=0` `therefore` A is non-invertible. `"Now"" adj A="[{:(6,-3),(-2,1):}]` `"Now adj A"=[{:(6,-3),(-2,1):}]` `therefore" (adj A)B"=[{:(6,-3),(-2,1):}][{:(5),(8):}]` `=[{:(30,-24),(-10,+8):}]=[{:(6),(-2):}]ne0` `rArr` Given system of equations is inconsistent.

Discussion

400.	Examine the consistency of the system of equations`x" "+" "y" "+" "z" "=" "1``2x" "+" "3y" "+" "2z" "=" "2``a x" "+" "a y" "+" "2a" "z" "=" "4`
Answer» Given system of equations, x+y+z=1 2x+3y+2z=1 ax+ay+2az = 4 `rArr" "[{:(1,1,1),(2,3,2),(a,a,2a):}][{:(x),(y),(z):}]=[{:(1),(1),(4):}]rArrAX=B` `therefore" "A=[{:(1,1,1),(2,3,2),(a,a,2a):}]` `rArr" "\|A\|=[{:(1,1,1),(2,3,2),(a,a,2a):}]` =1(6a-2a)-1(4a-2a)+1(2a-3a) `= 4a-2a-a=ne0` `therefore` A is invertible. `rArr` Given system of equations is consistent.

Discussion

Explore topic-wise InterviewSolutions in .

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If `c

Examine the consistency of the system of equations`x + 2 y = 2``2x + 3y = 3`

Examine the consistency of the system of equations `x+3y=5 , 2x+6y=8`

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If Aij is the cofactor of the element aij of \(\begin{bmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 & 5 & -7\end{bmatrix}\) then write the value of (a32 A32).