643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 1 InterviewSolution

1.	If A,B,C are the angles of a non right angled triangle ABC. Then find the value of:`\|[tanA,1,1],[1,tanB,1],[1,1,tanC]\|`
Answer» Since the determinant is sysmmetrical it is easier to expand by Sarrus rule. Hence. `\|{:(tanA,,1,,1),(1,,tanB,,1),(1,,1,,tanC):}\|` = tan A tan B tan C +1+1 - tan A -tan B-tan C =2 [as in a triangle tan A+ tan B + tan C= tan A tan B tan C)

Discussion

2.	If p + q + r = a + b + c = 0, then the determinant `\|{:(pa,qb,rc),(qc,ra,pb),(rb,pc,qa):}\|` equals
Answer» Correct Answer - A `\|{:(pa,,qb,,rc),(qc,,ra,,pb),(rb,,pc,,qa):}\|` `=pqr(a^(3)+b^(3)+c^(3)-3abc)-(abc(p^(3)+q^(3)+r^(3)-3pqr)` `=pqr(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)` `-abc (p+q+r)(p^(2)+q^(2)+r^(2)-pq-qr-pr)` `=0`

Discussion

3.	If `\|(a,a +d,a +2d),(a^(2),(a + d)^(2),(a + 2d)^(2)),(2a + 3d,2 (a +d),2a +d)\| = 0`, thenA. `d = 0`B. `a +d = 0`C. `d = 0 ro a +d = 0`D. none of these
Answer» Correct Answer - C

Discussion

4.	What is the value of the following determinant?
Answer» \(\Delta = \begin{vmatrix} 4 &a &b+c \\[0.3em] 4 & b & c+a \\[0.3em] 4 & c &a+b \end{vmatrix}\)

Discussion

5.	Evaluate `\|(cosalphacosbeta,cosalphas inbeta,-sinalpha),(-sinbeta,cosbeta,0),(sinalphacosbeta,sinalphasinbeta,cosalpha)\|`
Answer» Let the given determinant be `Delta`. Then, `Delta = \|["cos"alpha"cos"beta, "cos" alpha "sin"beta, -"sin"beta],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos" beta, "sin"alpha "sin"beta, "cos" alpha]\|` `= (1)/("sin"alpha"cos"alpha)\|["sin"alpha"cos"alpha"cos"beta, "sin" alpha "cos"alpha"sin"beta, -"sin"^(2)beta],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta, "cos"^(2) alpha]\|` `[R_(1) to ("sin" alpha)R_(1), ("cos"alpha)R_(3) " and dividing "Delta "by "("sin" alpha "cos" alpha)]` `= (1)/("sin"alpha"cos"alpha)\|[0, 0, -1],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta, "cos"^(2) alpha]\|[R_(1) to (R_(1) -R_(3))]` `= (1)/("sin"alpha"cos"alpha)(-1)\|[-"sin" beta, "cos" beta],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta]\|` `= (("sin"alpha"cos"alpha))/(("sin" alpha "cos"alpha))(-1)\|[-"sin" beta, "cos" beta],["cos"beta, "sin"beta]\|["taking sin"alpha "cos"alpha "common from"R_(2)]` `= (-1) *[-"sin"^(2)beta-"cos"^(2) beta]=("sin"^(2) beta + "cos"^(2) beta) =1.` Hence, `Delta = 1.`

Discussion

6.	Using the property of determinants and without expanding in questions 1 to 7 prove that , `\|{:(a-b,b-c,c-a),(b-c,c-a,a-b),(c-a,a-b,b-c):}\|=0`
Answer» `L.H.S.=\|{:(a-b,b-c,c-a),(b-c,c-a,a-b),(c-a,a-b,b-c):}\|=\|{:(0,b-c,c-a),(0,c-a,a-b),(0,a-b,b-c):}\|` `((C_(1)toC_(1)+C_(2)+C_(3))` `=0 " "(because" all elements of "C_(1)" are zero)` =R.H.S.

Discussion

7.	Using the property of determinants and without expanding, prove that:`\|0a-b-a0-c b c0\|=0`
Answer» `Let Delta=\|{:(0,a,-b),(-a,0,-c),(b,c,0):}\|=0` `rArr" " Delta=(-1)(-1)(-1)\|{:(0,a,-b),(-a,0,-c),(b,c,0):}\|=0` `rArr" " Delta=-\|{:(0,a,-b),(-a,0,-c),(b,c,0):}\|=0" "(CharrR)` `rArr" "Delta=-Delta` `rArr" "2Delta=0rArr" "Delta=0` Hence Proved.

Discussion

8.	Value of \(\begin{vmatrix}cos50^o &sin10^o \\[0.3em]sin50^o & cos10^o \\[0.3em]\end{vmatrix}\) is :(a) 0(b) 1(c) 1/2(d) – 1/2
Answer» Answer is (c) \(\begin{vmatrix}cos50^o &sin10^o \\[0.3em]sin50^o & cos10^o \\[0.3em]\end{vmatrix}\) = cos 50° cos 10° – sin 50° sin 10° = cos (50° + 10°) = cos 60° = 1/2

8.

Value of \(\begin{vmatrix}cos50^o &sin10^o \\[0.3em]sin50^o & cos10^o \\[0.3em]\end{vmatrix}\) is :(a) 0(b) 1(c) 1/2(d) – 1/2

Answer»

Answer is (c)

\(\begin{vmatrix}cos50^o &sin10^o \\[0.3em]sin50^o & cos10^o \\[0.3em]\end{vmatrix}\)

= cos 50° cos 10° – sin 50° sin 10°

= cos (50° + 10°)

= cos 60° = 1/2

Discussion

9.	If `\|(x, x^2, x^3 +1), (y, y^2, y^3+1), (z, z^2, z^3+1)\|` = 0 and x ,y and z are not equal to any other, prove that, xyz = -1
Answer» Let the given determinant be `Delta`. Then, `Delta = \|{:(x, x^(2), 1+x^(3)),(y, y^(2), 1+y^(3)), (z, z^(2), 1+z^(3)):}\|= \|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}\| + \|{:(x, x^(2), x^(3)),(y, y^(2), y^(3)), (z, z^(2), z^(3)):}\|` `=\|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}\| +(xyz)* \|{:(1, x, x^(2)),(1, y, y^(2)), (1, z, z^(2)):}\|` `=\|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}\| +(xyz)(-1)^(2)* \|{:(x, x^(2),1 ),(y, y^(2), 1), (z, z^(2), 1):}\| ["interchanging the columns of the 2nd det. twice"]` `=(1+xyz)\|{:(x, x^(2), 1),(y, y^(2), 1), (z, z^(2), 1):}\| ` `=(1+xyz)\|{:(" "x, " "x^(2), 1),((y-x), (y^(2)-x^(2)), 0), ((z-x), (z^(2)-x^(2)), 0):}\| [R_(2) to (R_(2)-R_(1)), R_(3) to (R_(3) -R_(1))]` `=(1+xyz)(y-x)(z-x)\|{:(x, " "x^(2), 1),(1, y+z, 0), (1, z+x, 0):}\|` `=(1+xyz)(y-x)(z-x)1\|{:(1, y+x),(1, z+x):}\| ["expanding by"C_(3)]` `=(1+xyz)(y-x)(z-x)(z-y).` `therefore Delta = 0 rArr (1+xyz)(y-x)(z-x)(z-y) =0` `rArr (1+xyz)=0 [because (y-x) ne 0, (z-x) ne 0, (z-y)ne 0]` `rArr xyz = -1` Hence, xyz =-1

Discussion

10.	Prove that `\|[1+a,1,1],[1,1+b,1],[1,1,1+c]\|=(abc)(1/a+1/b+1/c+1)=(bc+ca+ab+abc)`
Answer» The given determinant `=\|{:(1+a, " "1, " "1),(" "1, 1+b, " "1),(" "1, " "1, 1+c):}\|` `=(abc)\|{:((1)/(a)+1, (1)/(a), (1)/(a)),((1)/(b), (1)/(b)+1, (1)/(b)),((1)/(c), (1)/(c), (1)/(c)+1):}\|` `["taking a, b, c common from"R_(1), R_(2) " and " R_(3) "respectively"]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)\|{:(1, 1, 1),((1)/(b), (1)/(b)+1, (1)/(b)),((1)/(c), (1)/(c), (1)/(c)+1):}\|` `["applying "R_(1) to R_(1) + R_(2) + R_(3) " and taking out" ((1)/(a) + (1)/(b) + (1)/(c) +1) "common from"R_(1)]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)\|{:(0, 0, 1),(0, 1, (1)/(b)),(-1, -1, (1)/(c)+1):}\|["applying"C_(1) to (C_(1) -C_(3))"and"C_(2) to (C_(2) -C_(3))]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)(1)\|{:(0, 1),(-1, -1):}\|["expanding by 1st row"]` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1)1` `=(abc)((1)/(a) + (1)/(b)+(1)/(c)+1) = (bc+ca+ab+abc)` Hence, the result follows.

Discussion

11.	Prove that `[[x, x^2 , 1+px^3], [y, y^2, 1+py^3] ,[z, z^2, 1+pz^3]] = (1+pxyz)(x-y)(y-z)(z-x)`
Answer» Let the given determinant be `Delta`. Then, `Delta = \|{:(x, x^(2), 1+px^(3)),(y, y^(2), 1+py^(3)),(z, z^(2), 1+pz^(3)):}\|` `= \|{:(x, x^(2), 1),(y, y^(2), 1),(z, z^(2), 1):}\| + \|{:(x, x^(2), px^(3)),(y, y^(2), py^(3)),(z, z^(2), pz^(3)):}\|" "["say"Delta_(1) + Delta_(2)]` `=(-1)* \|{:(x, 1, x^(2)),(y, 1, y^(2)),(z, 1, z^(2)):}\| + p\|{:(x, x^(2), x^(3)),(y, y^(2), y^(3)),(z, z^(2), z^(3)):}\|["by"C_(2) harr C_(3) "in"Delta_(1) ["taking p common from"C_(3) "in"Delta_(2)]` `=(-1)(-1) \|{:(1, x, x^(2)),(1, y, y^(2)),(1, z, z^(2)):}\| + (pxyz)\|{:(1, x, x^(2)),(1, y, y^(2)),(1, z, z^(2)):}\|["by"C_(1) harr C_(2)] ["taking x, y, z common from"R_(1), R_(2), R_(3)"resp."]` ` =(1+pxyz)\|{:(1, x, x^(2)), (1, y, y^(2)),(1, z, z^(2)):}\|` `=(1+pxyz) \|{:(1, x, x^(2)),(0, y-x, y^(2)-x^(2)),(0, z-x, z^(2)-x^(2)):}\| ["by"R_(2) to R_(2) - R_(1) "and"R_(3) to R_(3)-R_(1) ]` `=(1+pxyz)(y-x)(z-x) \|{:(1, x, x^(2)),(0, 1, y+x),(0, 1, z+x):}\| ["taking (y-x) common from"R_(2) "and (z-x) common from"R_(3)]` `=(1+pxyz)(y-x)(z-x)1\|{:(1, y+x),(1, z+x):}\|` `=(1+pxyz)(y-x)(z-x)*{(z+y)-(y-x)}` `= (1+pxyz)(y-x)(z-x)(z-y)` `=(1+pxyz)(x-y)(y-z)(z-x)=RHS. `therefore` LHS = RHS.

Discussion

12.	Mark the tick against the correct answer in the following:\(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\) = ?A. 1B. 0C. cos 50°D. sin 50°
Answer» Correct answer B. 0 To find: value of \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\) Formula used: (i) cos\(\theta\) = sin(90 - \(\theta\)) We have \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\) On expanding the above, ⇒ {cos 70°} {cos 20°} – {sin 70°} {sin 20°} On applying formula cos\(\theta\) = sin(90 - \(\theta\) ) ⇒ {sin (90 – 70)} {sin (90 – 20)} - {sin 70°} {sin 20°} ⇒ {sin 20°} {sin 70°} - {sin 70°} {sin 20°} = 0

12.

Mark the tick against the correct answer in the following:\(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\) = ?A. 1B. 0C. cos 50°D. sin 50°

Answer»

Correct answer B. 0

To find: value of \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\)

Formula used: (i) cos\(\theta\) = sin(90 - \(\theta\))

We have \(\begin{vmatrix} cos 70^\circ & sin 20^\circ\\[0.3em] sin 70^\circ & cos 20^\circ \\[0.3em] \end{vmatrix}\)

On expanding the above,

⇒ {cos 70°} {cos 20°} – {sin 70°} {sin 20°}

On applying formula cos\(\theta\) = sin(90 - \(\theta\) )

⇒ {sin (90 – 70)} {sin (90 – 20)} - {sin 70°} {sin 20°}

⇒ {sin 20°} {sin 70°} - {sin 70°} {sin 20°}

= 0

Discussion

13.	Let A be a square matrix of order 3 x 3 then \|KA\| is equal to (A) K \|A\| (B) K2 \|A\| (C) K3 \|A\| (D) 3K \|A\|
Answer» Answer is (C) \|KA\| = K³ \|A\|

13.

Let A be a square matrix of order 3 x 3 then |KA| is equal to (A) K |A| (B) K2 |A| (C) K3 |A| (D) 3K |A|

Answer»

Answer is (C)

|KA| = K³ |A|

Discussion

14.	If \(\begin{vmatrix}3 & x \\x & x \\\end{vmatrix}\) = \(\begin{vmatrix}-2 & 2 \\4 & 1 \\\end{vmatrix}\), find the value of x
Answer» \(\begin{vmatrix}3 & x \\x & x \\\end{vmatrix}\) = \(\begin{vmatrix}-2 & 2 \\4 & 1 \\\end{vmatrix}\) ⇒ 3x – x² = – 2 – 8 ⇒ x² – 3x – 10 = 0 ⇒ x = 5, -2.

14.

If \(\begin{vmatrix}3 & x \\x & x \\\end{vmatrix}\) = \(\begin{vmatrix}-2 & 2 \\4 & 1 \\\end{vmatrix}\), find the value of x

Answer»

\(\begin{vmatrix}3 & x \\x & x \\\end{vmatrix}\) = \(\begin{vmatrix}-2 & 2 \\4 & 1 \\\end{vmatrix}\)

⇒ 3x – x² = – 2 – 8
⇒ x² – 3x – 10 = 0
⇒ x = 5, -2.

Discussion

15.	Varify A (adjA)=(adjA)A `[{:(1,-1,2),(3,0,-2),(1,0,3):}]=`
Answer» `therefore" "\|A\|=[{:(1,-1,2),(3,0,-2),(1,0,3):}]=0` =1(0-0)-(-1)(9+2)+2(0-0)=11 `rArr" "\|A\|.I_(3)=11[{:(1,0,0),(0,1,0),(0,0,1):}]=[{:(11,0,0),(0,11,0),(0,0,11):}]` `A_(11)=(-1)^(1+1)\|{:(0,-2),(0,3):}\|` `A_(12)=(-1)^(1+2)[{:(3,-2),(1,3):}]=-11,` `A_(13)=(-1)^(1+3)[{:(3,0),(1,3):}]=0,` `A_(21)=(-1)^(2+1)[{:(-1,2),(0,3):}]=1,` `A_(22)=(-1)^(2+2)[{:(1,2),(1,3):}]=1,` `A_(23)=(-1)^(2+3)[{:(1,-1),(1,0):}]=2,` `A_(31)=(-1)^(3+1)[{:(-1,2),(0,-2):}]=8,` `A_(32)=(-1)^(3+2)[{:(1,2),(3,-2):}]=8,` `A_(33)=(-1)^(3+3)[{:(1,-1),(3,0):}]=3,` `therefore" adj A"=[{:(0,-11,0),(3,1,-1),(2,8,3):}]=[{:(0,3,2),(-11,1,8),(0,-1,3):}]` `"Now A. (adj A)"=[{:(0,-1,2),(3,1,-2),(1,0,3):}]=[{:(0,3,2),(-11,1,8),(0,-1,3):}]` `=[{:(0+11+0,3-1-2,2-8+6),(0+0+0,9+0+2,6+0-6),(0+0+0,3+0=3,2+0+9):}]` `=[{:(11,0,0),(0,11,0),(0,0,11):}]` `"(adj A).A"=[{:(0,3,2),(-11,1,8),(0,-1,3):}]=[{:(1,-1,2),(3,0,-2),(1,0,3):}]` `=[{:(0+9+2,0+0+0,0-6+6),(-11+3+8,11+0+0,-22-2+24),(0-3+3,0+0+0,0+2+9):}]` `=[{:(11,0,0),(0,11,0),(0,0,11):}]` `therefore`A.(adj A)=(adjA)A=\|A\|.`I_(3)`

Discussion

16.	Let `A=[{:(1,-2,1),(-2,3,1),(1,1,5):}]`. Verify that ltbtgt (i) `[adjA]^(-1)=adj (A^(-1))` (ii) `(A^(-1)^(-1)=A`
Answer» `A=[{:(1,-2,1),(-2,3,1),(1,1,5):}]` `therefore" "\|A\|=[{:(1,-2,1),(-2,3,1),(1,1,5):}]` `=1(15-1)-(-2)(-10-1)+1(-2-3) ` `=14-22-5-13ne0` `A_(11)=14, A_(12)=11, A_(13)=-5` `A_(21)=11, A_(22)=4, A_(23)=-3` `A_(31)=-5, A_(32)=-3, A_(33)=-1` `therefore" "A=[{:(14,11,-5),(11,4,-3),(-5,-3,-1):}]=[{:(14,11,-5),(11,4,-3),(-5,-3,-1):}]` `"and A"^(-1)=1/\|A\|="adj A"=-1/13[{:(14,11,-5),(11,4,-3),(-5,-3,-1):}]=[{:(-14/13,-11/13,5/13),(122/12,-4/13,3/13),(5/13,3/13,1/13):}]` (i) `Let B = adj A=[{:(14,11,-5),(11,4,-3),(-5,-3,-1):}]rArr\|B\|=[{:(14,11,-5),(11,4,-3),(-5,-3,-1):}]` =14(-4-9)=11(-11-15)-5(-33+20) `=-182+286+65=169ne0` `B_(11)=-13, B_(12)=26, B_(13)=-13` `B_(21)=26, B_(22)=-39, B_(23)=-13` `B_(31)=-13, B_(32)=-13, B_(33)=-65` `therefore" adj B"=[{:(-13,26,-13),(26,-39,-13),(-13,-13,-65):}]=[{:(-13,26,-13),(26,-39,-13),(-13,-13,-65):}]` `"andB"^(-1)=1/\|B\|"adj B="1/169[{:(-13,26,-13),(26,-39,-13),(-13,-13,-65):}]=1/13[{:(-1,2,-1),(1,-3,-1),(-1,-1,-5):}]` `"Let C"=A^(-1)=[{:(-14/13,-11/13,5/13),(-11/13,-4/13,3/13),(5/13,3/13,1/13):}]` `C_(11)=-13/169=-1/13,C_(12)=26/169=2/13, C_(13)=-13/169=-1/13` `C_(21)=26/169=2/13, C_(22)=39/169=-3/13, C_(23)=-13/169=-1/13` `C_(31)=-13/169=1/13, C_(32)=-13/169=-1/13, C_(33)=-65/169=-5/13` `:."adjC=adjA"^(-1)=[{:(-1/13,2/13,-1/13),(2/13,-3/13,-1/13),(-1/13,-1/13,-5/13):}]=[{:(-1/13,2/13,-1/13),(2/13,-3/13,-1/13),(-1/13,-1/13,-5/13):}]` `=1/13[{:(-1,2,-1),(2,-3,-1),(-1,-1,-5):}]` (ii) `\|A^(-1)\|=14/13(=4/169-9/169)+11/13(-11/169-15/169)+5/13(-33/169+20/169)=1-1/13` `therefore (A^(-1))^(-1)=1/\|A^(-1)\|"adj"(A^(-1))` `1/(-1/13)-1/13[{:(-1,2,-1),(2,-3,-1),(-1,-1,-5):}]=[{:(1,-2,1),(2,3,1),(-1,1,5):}]=A`

Discussion

17.	If `A=((2,-3,5),(3,2,-4),(1,1,-2))` find `A^(-1)`. Use it to solve the system of equations `2x-3y+5z=11` , `3x+2y-4z=-5` and `x+y-2z=-3`
Answer» ` A=[{:(2,-3,5),(3,2,-4),(1,1,-2):}]` `rArr" " \|A\|=[{:(2,-3,5),(3,2,-4),(1,1,-2):}]` =2(-4+4)-(-23)(-6+4)+5(3-2) `=-6+5=-1ne0` `therefore` A is invertible `"Now," A_(11)=0, A_(12)=2, A_(13)=1` `A_(21)=-1, A_(22)=-9, A_(23)=-5` `A_(31)=2, A_(32)=23, A_(33)=13` `therefore" adjA"=[{:(0,2,1),(-1,-9,-5),(2,23,13):}]=[{:(0,-1,2),(2,-9,23),(1,-5,13):}]` `"and A"^(-1)=1/\|A\|"adj A"=1/-1"=[{:(0,-1,2),(2,-9,23),(1,-5,13):}]=[{:(0,1,-2),(2,-9,23),(-1,5,-13):}]` Given system of equations, 2x-3y+5y=11 3x+2y-4z=-5 x+y-2z=-3 `"=[{:(2,-3,5),(3,2,-4),(1,1,12):}][{:(x),(y),(z):}]=[{:(11),(-5),(-3):}]` `rArr" "AX=B=rArrX=A^(-1)B` `[{:(x),(y),(z):}]=[{:(0,1,-2),(-2,9,-23),(-1,5,-13):}][{:(11),(-5),(-3):}]` `=[{:(0-5+6),(-22-45+69),(-11-25+39):}][{:(1),(2),(3):}]` `rArr " "x=1, y=2, z=3`

Discussion

18.	Solve system of linear equations, using matrix method, in questions 7 to 14. x-y+2z=7, 3x+4y-5z=-5, 2x-y+3z=12
Answer» Given system of equations, x-y+2z=7, 3x+4y-5z=-5, 2x-y+3z=12 `=[{:(1,-1,2),(3,4,-5),(2,-1,3):}][{:(x),(y),(z):}]=[{:(7),(-5),(12):}]rArrAX=B` `therefore" "A=[{:(1,-1,2),(3,4,-5),(2,-1,3):}]` `rArr" "\|A\|=[{:(1,-1,2),(3,4,-5),(2,-1,3):}]` =1(12-5)-(-1)(9+10)+2(-3-8) `=7+19-22=4ne0` `therefore` A is invertible. `"Now, "A_(11)=7, A_(12)=-19, A_(13)=-11` `A_(21)=1, A_(22)=-1, A_(23)=-1` `A_(31)=-3, A_(32)=11, A_(33)=7` `therefore" adjA"=[{:(7,-19,-11),(1,-10,-1),(-3,11,7):}]=[{:(7,1,-3),(-19,-1,11),(-11,-1,7):}]` `"and A"^(-1)=1/\|A\|"adj A"=1/4[{:(7,1,-3),(-19,-1,11),(-11,-1,7):}]` `"=1/4[{:(49-5-36),(-133+5+132),(-775+5+84):}]=[{:(2),(1),(3):}]` `therefore"x=2, y=1, z=3`

Discussion

19.	Solve system of linear equations, using matrixmethod,`x y" "+" "2z" "=" 7"``3x" "+" "4y" "" "5z" "=" "" "5``2x y" "+" "3" "z=" "12`
Answer» `A=[{:(3,2),(1,1):}],` `therefore" "A^(2)=A.A=[{:(3,2),(1,1):}][{:(3,2),(1,1):}]` `[{:(9+2,6+2),(3+1,2+1):}]=[{:(11,8),(4,3):}]` `"given that, "A^(2)+aA+bI=O` `rArr" "[{:(11,8),(4,3):}]+a[{:(3,2),(1,1):}]+b[{:(1,0),(0,1):}]=0` `rArr" "[{:(11+3a+b,8+2a),(4+a,3+a+b):}]=[{:(0,0),(0,0):}]` `"On comparing,"4+a=0 rArr a=-4` 3+a+b=0 `rArr" "3-4+b=0` `rArr" "b=1` `therefore" "a=-4, b=1`

Discussion

20.	Solve system of linear equations, using matrix method,`5x + 2y = 4``7x + 3 y = 5`
Answer» Given system of equations, 5x+2y=4 7x+3y = 4 `rArr" "[{:(5,2),(7,3):}][{:(x),(y):}]=[{:(4),(5):}]rArrAX=B` `"Here "A=[{:(5,2),(7,3):}]` `rArr" "A=[{:(5,2),(7,3):}]=15-14=1ne0` `therefore` A is invertible. `"Now "A_(11)=3, A_(12)=-7, A_(21)=-2,A_(22)=5` `therefore "adj A"=[{:(3,-7),(-2,5):}][{:(3,-2),(-7,5):}]` `"and "A^(-1)=1/\|A\|"adj A"=1/1[{:(3,-7),(-2,5):}][{:(3,-2),(-7,5):}]` From equation (1) , `AX=B rArr X=A^(-1)B` `rArr" "[{:(x),(y):}]=[{:(3,-2),(-7,5):}]=[{:(12-10),(-28+25):}]=[{:(2),(-3):}]` `therefore" "x=2, y=-3`

Discussion

21.	Solve system of linear equations, using matrix method,`2x+y+z=1``x-2y-z=3/2``3y-5z=9`
Answer» Given system of equations, 2x+y+z=1 `x-2y-z=3/2` 3y-5z=9 `rArr[{:(2,1,1),(1,-2,-1),(0,3,-5):}][{:(x),(y),(z):}]-[{:(1),(3/2),(z):}][{:(1),(3/2),(9):}]rArr AX=B` `A=[{:(2,1,1),(1,-2,-1),(0,3,-5):}] ` `rArr" "\|A\|=[{:(2,1,1),(1,-2,-1),(0,3,-5):}]` `therefore` A is invertible `"Now "A_(11)=13, A_(12)=5, A_(13)=3` `A_(21)=8, A_(22)=-10, A_(23)=-6` `A_(31)=1, A_(32)=3, A_(33)=-5` `therefore" adj A"=[{:(13,5,3),(8,-1,-6),(3,-6,-5):}]` From equation (1) `AX=B rArr X=A^(-1)B` `[{:(x),(y),(z):}]=1/34[{:(13,5,3),(8,-1,-6),(3,-6,-5):}][{:(1),(3/2),(9):}]` `1/34[{:(13,+12,+9),(5,-10,3),(3,-6,-5):}][{:(1),(1/2),(-3/2):}]` `therefore " "x=1, y=1/2, z=-3/2`

Discussion

22.	Solve system of linear equations, using matrix method,`5x + 2y = 3" "``3x + 2y = 5`
Answer» `A = [(5,2),(3,2)]` `X= [(x),(y)]` `b= [(3),(5)]` `\|A\|= \|(5,2),(3,2)\|` `=10-6=4 cancel(=) 0` `A^-1 = 1/4 [(2,-2),(-3,5)]` `X= A^-1B` `[(x), (y)] = [(4/2, -1/2),(-3/4, 5/4)][(3),(5)]` `= [(3/2,-5/2), (-9/4, 25/4)]` answer

Discussion

23.	Find the equation of the line joining A( 1,3) and B (0,0) using determinants and find k if D(k, 0) is a point such that area of triangle ABD is 3sq units.
Answer» Area of a triangle is given by, `Ar = 1/2\|[x_1,y_1,1],[x_2,y_2,1],[x_3,y_3,1]\|` For the given triangle, `3 = 1/2\|[1,3,1],[0,0,1],[k,0,1]\|` `=>\|[1,3,1],[0,0,1],[k,0,1]\| = 6` `=>[1(0-0)-3(0-k)+1(0-0)] = +-6` `=>-3k = 6 and -3k = -6` `=>k = -2 and k = 2` So, required value of `k` is `+-2`.

Discussion

24.	Find values of k if area of triangle is 4 sq. units and vertices are (i) `(k, 0), (4, 0), (0, 2)`(ii) `(–2, 0), (0, 4), (0, k)`
Answer» Area of a triangle is given by, `Ar = 1/2\|[x_1,y_1,1],[x_2,y_2,1],[x_3,y_3,1]\|` So, putting the given vertices, `Ar = 1/2\|[-2,0,1],[0,4,1],[0,k,1]\|` As Area is `4` square units, `So, 1/2\|[-2,0,1],[0,4,1],[0,k,1]\| = 4` `=>\|[-2,0,1],[0,4,1],[0,k,1]\| = 8` `=>-2(4-k)-0+1(0,0) = +-8` `=>-8+2k = +-8` `=>2k = 16 and 2k = 0` `=>k = 8 and k =0`

Discussion

25.	if\(\begin{vmatrix}6i&-3i&1\\4&3i&-1\\20&3&i\end{vmatrix}\)= x + iy, then[(6i, 3i, 1) (4, 3i, -1) (20, 3, i)](a) x = 3, y = 1 (b) x = 1, y = 3(c) x = 0, y = 3 (d) x = 0, y = 0
Answer» Correct option is (d) x = 0, y = 0

Discussion

26.	if\(\begin{vmatrix}6i&-3i&1\\4&3i&-1\\20&3&i\end{vmatrix}\)= x + iy, then[(6i, 3i, 1) (4, 3i, -1) (20, 3, i)](a) x = 3, y = 1 (b) x = 1, y = 3(c) x = 0, y = 3 (d) x = 0, y = 0
Answer» Correct option is (d) x = 0, y = 0

Discussion

27.	Which of the following is not correct in a given determinant of `A ,`where `A=([a_(i j)])_(3x3)`A. Order of minor is less than order of the det(A)B.Minor of an element can never be equal to cofactor of the same elementC. Value of a determinant is obtained by multiplying elements of a row orcolumn by corresponding cofactorsD. Order of minors and cofactors of elements of A is same
Answer» `A= [a_(ij) ]_(3xx3)` minor=`2 xx2` `[(1,0,0),(0,1,0),(0,0,1)] ` `c_(11) = 1` minor can be equal to cofactor

Discussion

28.	Statement -1 : Determinant of a skew-symmetric matrix of order 3 is zero. Statement -2 : For any matrix A, Det `(A) = "Det"(A^(T)) and "Det" (-A) = - "Det" (A)` where Det (B) denotes the determinant of matrix B. Then,A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 6B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 6C. Statement 1 is true, Statement 2 is FalseD. Statement 1 is False, Statement 2 is true
Answer» Correct Answer - C Let A be a skew - symmetric matrix of order 3 Then, `A^(T) = -A` `rArr Det (A^(T)) = Det (-A)` `rArr Det (A) = (-1)^(3) Det (A)` `rArr Det(A) = - Det(A)` `rArr 2Det(A) = 0` `rArr Det (A) = 0` So, statement -1 is true. For any square matrix of order n, we have `Det(A^(T)) = Det(A) and Det(-A) = (-1)^(n) Det(A)` So, statement -2 is not true

Discussion

29.	Evaluate `\|{:(265, 240, 219),(240, 225, 198), (219, 198, 181):}\|`
Answer» Let the given determinant be `Delta`. Then, `Delta =\|{:(265, 240, 219),(240, 225, 198), (219, 198, 181):}\|` `=\|{:(46, 21, 219),(42, 27, 198), (38, 17, 181):}\|" "[C_(1) to (C_(1) -C_(3)) " and " C_(2) to (C_(2)-C_(3))]` `=\|{:(4, 21, " "9),(-12, 27, -72),(4, 17, " "11):}\|" "[C_(1) to (C_(1) -2C_(2)) " and " C_(3) to (C_(3)-10C_(2))]` `=\|{:(0, 4, " "-2),(0, 78, -39),(4, 17, " "11):}\|" "[R_(1) to (R_(1) -R_(3)) " and " R_(2) to (R_(2)+3R_(3))]` `=2(39)\|{:(0, 2, -1),(0, 2, -1),(4, 17, 11):}\| " "["taking 2 common from "R_(1) " and 39 common from "R_(2)]` `=(78 xx 0) =0 " "[because R_(1) " and"R_(2) " are identical"].`

Discussion

30.	If `A(x_1, y_1),B(x_2, y_2) ` and `C(x_3,y_3)`are vertices of an equilateral triangle whose each side is equal to `a`, then prove that`\|[x_1,y_1, 2],[x_2,y_2, 2],[x_3,y_3, 2]\|^2=3a^4`
Answer» Area of an equilateral triangle=`(sqrt3 a^2)/4 ` area= `1/2\|(x_1, y_1, 1),(x_2, y_2, 1),(x_3, y_3, 1)\|` `sqrt3 a^2/4 = 1/2 \|(x_1, y_1,1),(x_2,y_2,1),(x_3,y_3,1)\|` `sqrt3 a^2 = 2\|(x_1 , y_1, 1),(x_2, y_2, 1),(x_3, y_3,1)\|` `sqrt3 a^2 = \|(x_1, y_1, 2),(x_2,y_2,2),(x_3,y_3,2)\|` `3a^4 = \|(x_1, y_1, 2),(x_2,y_2, 2),(x_3,y_3,2)\|^2` hence proved

Discussion

31.	Show that \(\begin{vmatrix}sin\,10° & -cos\,10° \\[0.3em]sin\,80° &cos\,80°\end{vmatrix}\) = 1
Answer» Using, sin(A+B) = sin A × cos B + cos A × sin B ⇒ \|A\| = sin10° × cos80° + cos10° x sin80° \|A\| = sin(10 + 80)° \|A\| = sin90° \|A\| = 1 Hence Proved.

31.

Show that \(\begin{vmatrix}sin\,10° & -cos\,10° \\[0.3em]sin\,80° &cos\,80°\end{vmatrix}\) = 1

Answer»

Using,

sin(A+B) = sin A × cos B + cos A × sin B

⇒ |A| = sin10° × cos80° + cos10° x sin80°

|A| = sin(10 + 80)°

|A| = sin90°

|A| = 1

Hence Proved.

Discussion

32.	Usingproperties of determinants, prove the following:`\|3"a"-"a"+"b"-"a"+"c""a"-"b"3"b""c"-"b""a"-"c""b"-"c"3"c"\|=3("a"+"b"+"c")("a b"+"b c"+"c a")`
Answer» Let the value of the given determinant be `Delta`. Then, `Delta =\|{:(" "3a,-a+b, -a+c),(a-b, " "3b, c-b),(a-c, b-c, 3c):}\|` `=\|{:(a+b+c,-a+b, -a+c),(a+b+c, " "3b, c-b),(a+b+c, b-c, " "3c):}\| [C_(1) to (C_(1) + C_(2) + C_(3))]` `=(a+b+c)\|{:(1,-a+b, -a+c),(1, " "3b, c-b),(1, b-c, 3c):}\| ["taking (a+b+c) common from"R_(1)]` `=(a+b+c)\|{:(1,-a+b, -a+c),(1, 2b+a, a-b),(0, a-c, 2c+a):}\| [R_(2) to (R_(2) -R_(1)) "and"(R_(3) -R_(1))]` `=(a+b+c)1\|{:(2b+a, a-b),(a-c, 2c+a):}\| ["expanded by"C_(1)]` ` = (a+b+c)[(2b+a)(2c+a)-(a-c)(a-b)]` ` =(a+b+c)[(4bc+2ab+2ac+a^(2))-(a^(2)-ab-ac +bc)]` `=3(a+b+c)(ab+bc+ca).` Hence, `Delta = 3(a+b+c) (ab+bc+ca)`

Discussion

33.	Usingproperties of determinants, prove that following:`\|"a"+"b"+2"c""a""b""c""b"+"c"+2"a""b""c""a""c"+"a"+2"b"\|=2("a"+"b"+"c")^3`
Answer» Let the value of the given determinant be `Delta`. Then, `Delta = \|{:(a+b+2c, " "a, " "b),(" "c, b+c+2a, " "b),(" "c, " "a, c+a+2b):}\|` `= \|{:(a+b+c, -(a+b+c), " "0),(" "c, b+c+2a, " "b),(" "0, -(a+b+c), (a+b+c)):}\| {{:("by"R_(1) to (R_(1) -R_(2))),("and"R_(3) to (R_(3) - R_(2))):}}` `=(a+b+c)^(2) * \|{:(1, " "-1, " "0),(c, b+c+2a, " "b),(0, " "-1, " "1):}\|` `["taking(a+b+c) common from each one of "R_(1) "and"R_(3)]` `=(a+b+c)^(2) * \|{:(1, " "-1, " "0),(0, b+2c+2a, " "b),(0, " "-1, " "1):}\| " "[R_(2) to R_(2) -cR_(1)]` `=(a+b+c)^(2) 1 \|{:(b+2c+2a, b),(-1, 1):}\|["expanded by"C_(1)]` `=(a+b+c)^(2) * (b+2c+2a+b) = 2(a+b+c)^(3)` Hence, `Delta =2(a+b+c)^(3)`

Discussion

34.	Prove that `\|{:(b+c, c+a, a+b),(c+a, a+b,b+c),(a+b, b+c, c+a):}\| =2(a+b+c)(ab+bc+ca-a^(2)-b^(2)-c^(2)).`
Answer» Let the given determinant be `Delta`. Then, `Delta =\|{:(b+c, c+a, a+b),(c+a, a+b,b+c),(a+b, b+c, c+a):}\| ` `=\|{:(2(a+b+c), 2(a+b+c), 2(a+b+c)),(c+a, a+b,b+c),(a+b, b+c, c+a):}\| [R_(1) to (R_(1) + R_(2) + R_(3))] ` `=2(a+b+c)\|{:(" "1, " "1, " "1),(c+a, a+b,b+c),(a+b, b+c, c+a):}\| ["taking(a+b+c)common from"R_(1)]` `=2(a+b+c)\|{:(" "1, " "0, " "0),(c+a, b-c,b-a),(a+b, c-a, c-b):}\| [{:(C_(2) to (C_(2)-C_(1))" and"),(C_(3) to (C_(3) -C_(1))):}]` `=2(a+b+c)1\|{:(b-c, b-a), (c-a, c-b):}\| ["expanded by"R_(1)]` `2(a+b+c) *[(b-c)(c-b)-(c-a)(b-a)]` `=2(a+b+c)(ab+bc+ca-a^(2)-b^(2)-c^(2))`. Hence, `Delta =2(a+b+c)(ab+bc+ca-a^(2)-b^(2)-c^(2))`.

Discussion

35.	Solve the determinant using properties`\|[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]\| = (a+b+c)^3`
Answer» Let the value of the given determinant be `Delta`. The, `Delta = \|{:(a-b-c, " "2a, " "2a),(" "2b, b-c-a, " "2b),(" "2c, " "2c, c-a-b):}\|` `= \|{:(a+b+c, a+b+c, a+b+c),(" "2b, b-c-a, " "2b),(" "2c, " "2c, c-a-b):}\| [R_(1) to (R_(1) + R_(2) + R_(3))]` `=(a+b+c)* \|{:(1, " "1, " "1),(2b, b-c-a, " "2b),(2c, " "2c, c-a-b):}\| ["taking (a+b+c) common from"R_(1)]` `=(a+b+c)* \|{:(1, " "0, " "0),(2b, -(a+b+c), " "0),(2c, " "0, -(a+b+c)):}\| [C_(2) to (C_(2)-C_(1))"and"C_(3) to (C_(3)-C_(1))]` `=(a+b+c)1 \|{:(-(a+b+c), " "0),(" "0, -(a+b+c)):}\| ["expanded by"R_(1)]` `=(a+b+c) (a+b+c)^(2) = (a+b+c)^(3)` Hence, `Delta = (a+b+c)^(3)`.

Discussion

36.	Which of the following is correct? (a) Determinant is a square matrix (b) Determinant is number associated to matrix(c) Determinant is a number associated with a square matrix (d) None of these
Answer» Correct option is (c) Determinant is a number associated with a square matrix

Discussion

37.	Which of the following is correct(A) Determinant is a square matrix.(B) Determinant is a number associated to a matrix.(C) Determinant is a number associated to a square matrix.(D) None of these
Answer» Correct Answer - ( c) because there is a number corresponds to the determinant of a matrix.

Discussion

38.	Find area of the triangle with vertices at the point given in each of the following : (i) (1,0), (6,0), (4,3) (ii) (2,7), (1,1), (10,8)
Answer» (i) Area of `Delta` `=1/2\|{:(1,0,1),(6,0,1),(4,3,1):}\|` `=1/2\|1(0-3)-0(6-4)+1(18-0)\|` `=1/2(-3-0+18)=15/2"sq.units"` (ii) Area of triangle `=1/2\|{:(2,7,1),(1,1,1),(10,8,1):}\|` `=1/2\|{:(2,7,1),(1,1,1),(10,8,1):}\|" "({:(R_(1)toR_(1)-R_(2)),(R_(2)toR_(2)-R_(3)):})` `1/2(-7+54)" "("Expanding along"C_(3))` =23.5 sq.units (iii) `"Area of "Delta=1/2\|{:(-2,-3,1),(3,2,1),(-1,-8,1):}\|` `=1/2\|-2(2+8)+3(3+1)+1(-24+2)\|` `=1/2\|-20+12-22\|=15`sq.units

Discussion

39.	Find the values of k if area of tringle is 4 sq. units and dvertices are : (i) (k,0), (4,0), (0,2)
Answer» (i) Area of triangle = 4 sq. units `rArr" "=1/2\|{:(k,0,1),(4,0,1),(0,2,1):}\|=+-4` `rArr 1/2[k(0-2)+1(8-0)=+-4` (Expending along`R_(1)`) `rArr" "-2k+8+-8` `"Now "-2k+8=8 or -2k+8=-8` `rArr" "k=0 or k=8 ` (ii) Area of `Delta=4` sq. units `rArr" "=1/2\|{:(-2,0,1),(0,4,1),(0,k,1):}\|=4` `rArr\|-2(4-k)-0(0-0)+1(0-0)\|=8` `rArr" "\|2k-8\|=8` `rArr" "2k-8=+-8` `rArr" "2k-8=8 or 2k-8=-08` `rArr" "k=8 or k=0`

Discussion

40.	From the matrix equation AB = AC, we can conclude that B = C provided (A) A is singular (B) A is non singular (C) A is symmetric (D) A is skew symmetric
Answer» Answer is (B) AB = AC A^-1(AB) = A^-1(AC) ⇒ (A^-1A)B = (A^-1A)C ⇒ IB = IC ⇒ B = C AB = AC ⇒ B = C if A¹ exist ie,. A is non singular

40.

From the matrix equation AB = AC, we can conclude that B = C provided (A) A is singular (B) A is non singular (C) A is symmetric (D) A is skew symmetric

Answer»

Answer is (B)

AB = AC

A^-1(AB) = A^-1(AC)

⇒ (A^-1A)B = (A^-1A)C

⇒ IB = IC ⇒ B = C AB = AC

⇒ B = C if A¹ exist

ie,. A is non singular

Discussion

41.	If all the elements in a square matrix A of order 3 are equal to 1 or - 1, then `\|A\|`, isA. an odd numberB. an even numberC. an imaginary numberD. a real number
Answer» Correct Answer - B

Discussion

42.	The value of `Delta = \|(1^(2),2^(2),3^(2)),(2^(2),3^(2),4^(2)),(3^(2),4^(2),5^(2))\|`, isA. 8B. `-8`C. 400D. 1
Answer» Correct Answer - B

Discussion

43.	If `\|{:(a-b,b-c,c-a),(x-y,y-z,z-x),(p-q,q-r,r-p):}\|=m\|{:(c,a,b),(z,x,y),(r,p,q):}\|," then m"=`A. 2B. 1C. 0D. None of these
Answer» Correct Answer - C

Discussion

44.	If the lines `a x+y+1=0,x+b y+1=0a n dx+y+c=0(a ,b ,c`beingdistinct and different from `1)`areconcurrent, then prove that `1/(1-a)+1/(1-b)+1/(1-c)=1.`
Answer» if the given lines are concurrent then `\|{:(a,,1,,1),(1,,b,,1),(1,,1,,c):}\|=0` `\|{:(a,,1-a,,1-a),(1,,b-1,,0),(1,,0,,c-a):}\|=0` [Applying` C_(2) to C_(2) -C_(1) " and "C_(3) to C_(3)-C_(1)]` `a(b-1)(c-1) -(c-1)(1-a)-(b-1)(1-a)=0` `(a)/(1-a)+(1)/(1-b)+(1)/(1-c)=0` `"(Dividing by " (1-a)(1-b)(1-c))` `(1)/(1-a)+(1)/(1-b)+(1)/(1-c)=1`

Discussion

45.	`" If " Delta_(1) =\|{:(x,,b,,b),(a,,x,,b),(a,,a,,x):}\|" and " Delta_(2)= \|{:(x,,b),(a,,x):}\|` are the given determinants thenA. `Delta_(1)=3(Delta_(2))^(2)`B. `(d)/(dx) (Delta_(1)) =3Delta_(2)`C. `(d)/(dx) (Delta_(1)) =3(Delta_(2))^(2)`D. `Delta_(1)=3Delta_(2)^(3//2)`
Answer» Correct Answer - B `Delta_(1)=x(x^(2) -ab)-b(ax-ab)+b(a^(2)-ax)` `=x^(3) -3abx +ab^(2) +a^(2)b` `(d)/(dx) (Delta _(1))=3x^(2) -3ab=3(x^(2) -ab)=3Delta_(2)`

Discussion

46.	The value of `Delta = \|(a,a +b,a +2b),(a +2b,a,a +b),(a +b,a +2b,a)\|` is equal toA. `9a^(2) (a +b)`B. `9b^(2) (a +b)`C. `a^(2) (a +b)`D. `b^(2) (a +b)`
Answer» Correct Answer - B

Discussion

47.	prove that `\|{:((a-x)^(2),,(a-y)^(2),,(a-z)^(2)),((b-x)^(2),,(b-y)^(2),,(b-z)^(2)),((c-x)^(2),,(c-y)^(2),,(c-z)^(2)):}\|` `\|{:((1+ax)^(2),,(1+bx)^(2),,(1+cx)^(2)),((1+ay)^(2),,(1+by)^(2),,(1+cy)^(2)),((1+az)^(2),,(1+bx)^(2),,(1+cz)^(2)):}\|` `=2 (b-c)(c-a)(a-b)xx (y-z) (z-x)(x-y)`
Answer» ` Delta=\|{:((a-x)^(2),,(a-y)^(2),,(a-z)^(2)),((b-x)^(2),,(b-y)^(2),,(b-z)^(2)),((c-x)^(2),,(c-y)^(2),,(c-z)^(2)):}\|` `\|{:(a^(2)-2ax+x^(2),,b^(2)-2bx+x^(2),,c^(2)-2cx+x^(2)),(a^(2)-2ay+y^(2),,b^(2)-2by+y^(2),,c^(2)-2cy+y^(2)),(a^(2)-2az+z^(2),,b^(2)-2bz+z^(2),,c^(2)-2cz+z^(2)):}\|` `= \|{:(1,,x,,x^(2)),(1,,y,,y^(2)),(1,,z,,z^(2)):}\|xx\|{:(a^(2),,-2a,,1),(b^(2),,-2b,,1),(c^(2),,-2c,,1):}\|` `=- \|{:(1,,x,,x^(2)),(1,,y,,y^(2)),(1,,z,,z^(2)):}\|\|{:(a^(2),,2a,,1),(b^(2),,2b,,1),(c^(2),,2c,,1):}\|` `= \|{:(1,,x,,x^(2)),(1,,y,,y^(2)),(1,,z,,z^(2)):}\|\|{:(1,,2a,,a^(2)),(1,,2b,,b^(2)),(1,,2c,,c^(2)):}\|" "" (in second determinant)"(C_(1) hArr C_(2))` `=\|underset(1" "z" "z^(2))underset(1" "y" "y^(2))(1" "x" "x^(2))\|xx\|underset(1" "2c" "c^(2))underset(1" "2b" "b^(2))(1" "2a" "a^(2))\|` `=2(x-y)(y-z)(z-x)(a-b)(b-c) (c-a)` Applying row by row we get `\|{:(1+2ax+a^(2)x^(2),,1+2bx+b^(2)x^(2),,1+2cx+c^(2)x^(2)),(1+2ay+a^(2)y^(2),,1+2by+b^(2)y^(2),,1+2cy+c^(2)y^(2)),(1+2az+a^(2)z^(2),,1+2bz+b^(2)z^(2),,1+2cz+c^(2)z^(2)):}\|` ` =\|{:((1+ax)^(2),,(1+bx)^(2),,(1+cx)^(2)),((1+ay)^(2),,(1+by)^(2),,(1+cy)^(2)),((1+az)^(2),,(1+bx)^(2),,(1+cz)^(2)):}\|`

Discussion

48.	If `a ne b ne c`, then solution of equation `\|{:(x-a,x-b,x-c),(x-b,x-c,x-a),(x-c, x-a,x-a):}\|=0" "is:`A. `x=0`B. `x=a+b+c`C. `x=1/2(a+b+c)(d) x=1/3(a+b+c)`D. `x=1/3(a+b+c)`
Answer» Correct Answer - D

Discussion

49.	Value of `\|{:(1+x_(1),,1+x_(1)x,,1+x_(1)x^(2)),(1+x_(2),,1+x_(2)x,,1+x_(2)x^(2)),(1+x_(3),,1+x_(3)x,,1+x_(3)x^(2)):}\|` depends uponA. x onlyB. `x_(1)`onlyC. `x_(2)`onlyD. none of these
Answer» Correct Answer - D `\|{:(1+x_(1),,1+x_(1)x,,1+x_(1)x^(2)),(1+x_(2),,1+x_(2)x,,1+x_(2)x^(2)),(1+x_(3),,1+x_(3)x,,1+x_(3)x^(2)):}\|` `=\|{:(1,,x_(1),,0),(1,,x_(2),,0),(1,,x_(3),,0):}\|\|{:(1,,1,,0),(1,,x,,0),(1,,x^(2),,0):}\|`

Discussion

50.	If `\|a^2+lambda^2a b+clambdac a-blambdaa b-clambdab^2+lambda^2b c+a c a+blambdab c-alambdac^2+lambda^2\|\|lambdac-b-clambdaa b-alambda\|=(1+a^2+b^2+c^2)^3`, then he value of `lambda`is`8`b. `27`c. `1`d. `-1`A. 8B. 27C. 1D. -1
Answer» Correct Answer - C We obserive that the elements in the pre -factor are the cofactors of the corresponding elements of the post factor .Hence `\|{:(lambda,,c,,-b),(-c,,lambda,,a),(b,,-a,,lambda):}\|= [lambda (lambda^(2) +a^(2)+b^(2)+c^(2))]^(3) =(1+a^(2)+b^(2)+c^(2))^(3)` `rArr lambda=1` Alternate solution: Writing `a=0, b=0c c=0` on both sides we get `lambda^(6)lambda^(3) =1" or " lambda=1`

Discussion

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