643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 6 InterviewSolution

251.	If w is a complex cube root of unity. `\|(a,b,c),(b,c,a),(c,a,b)\| = -(a +b +c) (a + bk + ck^(2)) (a + bk^(2) + ck)`, then k equalsA. 1B. `-1`C. `omega`D. `-omega`
Answer» Correct Answer - C We have, `\|(a,b,c),(b,c,a),(c,a,b)\|` `= \|(a +b +c,b,c),(a +b +c,c,a),(a +b +c,a,b)\| " " ["Applying " C_(1) rarr C_(1) + C_(2) + C_(3)]` `= (a +b +c) \|(1,b,c),(1,c,a),(1,a,b)\|` `= (a +b +c) \|(1,b,c),(0,c -b,a -c),(0,a -b,b-c)\| " " [("Applying" R_(2) rarr R_(2) - R_(1)),(" "R_(3) rarr R_(3) - R_(1))]` `= (a +b +c) \|(c -b,a -c),(a -b,b -c)\|` `= (a +b +c) (-b^(2) -c^(2) -a^(2) + ac + ab + bc - 2bc)` `= -(a + b+c) (a^(2) +b^(2) + c^(2) -ab - bc - ca)` `= - (a +b +c) (a + bomega + c omega^(2)) (a + b omega^(2) + c omega)` `:. k = w`

Discussion

252.	If `a+b+c=0,`one root of `\|a-x c b c b-x a b a c-x\|=0`is`x=1`b. `x=2`c. `x=a^2+b^2+c^2`d. `x=0`A. `x=1`B. `x=2`C. `x=a^(2)+b^(2)+c^(2)`D. `x=0`
Answer» Correct Answer - D Operating `C_(1) to C_(1) +C_(2) +C_(3) `we get `(a+b+c-x) \|{:(1,,c,,b),(1,,b-x,,a),(1,,a,,c-x):}\|=0` `:. x=a+ b+c=0`

Discussion

253.	If `alpha, beta and gamma` are the roots of the equation `x^(3) + px + q = 0` (with `p != 0 and p != 0 and q != 0`), the value of the determinant `\|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)\|`, isA. pB. qC. `p^(2) - 2q`D. none of these
Answer» Correct Answer - D It is given that `alpha, beta, gamma` are the roots of the given equations. `:. alpha + beta + gamma = 0`. So, `Delta = \|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)\|` `Delta = \|(alpha + beta + gamma,beta,gamma),(alpha + beta + gamma,gamma,alpha),(alpha + beta + gamma,alpha,beta)\| " " [ " Using" C_(1) rarr C_(2) + C_(2) + C_(3)]` `Delta = \|(0,beta,gamma),(0,gamma,alpha),(0,alpha,beta)\| [ :. alpha + beta + gamma = 0]`

Discussion

254.	`\|[1,bc,b+c],[1,ca,c+a],[1,ab,a+b]\|=\|[1,a,a^2],[1,b,b^2],[1,c,c^2]\|`
Answer» Applying `C_(3) to C_(3) - (a + b +c)C_(1)`, we get `RHS = \|[1, bc, -a], [1, ca, -b], [1, ab, -c]\| = \|[1, c, bc],[1, b, ca],[1, c, ab]\|` ` = (1)/(abc) * \|[a, a^(2), abc], [b, b^(2), abc], [c, c^(2), abc]\| = \|[a, a^(2), 1], [b, b^(2), 1], [c, c^(2), 1]\| = \|[1, a, a^(2)], [1, b, b^(2)], [1, c, c^(2)]\| = LHS.`

Discussion

255.	if w is a complex cube root to unity then value of ` Delta =\|{:(a_(1)+b_(1)w,,a_(1)w^(2)+b_(1),,c_(1)+b_(1)bar(w)),(a_(2)+b_(2)w,,a_(2)w^(2)+b_(2),,c_(2)+b_(2)bar(w)),(a_(3)+b_(3)w,,a_(3)w^(2)+b_(3),,c_(3)+b_(3)bar(w)):}\|` is
Answer» Correct Answer - A `Delta =\|{:(a_(1)+b_(1)w,,a_(1)w^(2)+b_(1),,c_(1)+b_(1)hat(w)),(a_(2)+b_(2)w,,a_(2)w^(2)+b_(2),,c_(2)+b_(2)hat(w)),(a_(3)+b_(3)w,,a_(3)w^(2)+b_(3),,c_(3)+b_(3)hat(w)):}\|` `" Operating " C_(2) to wC_(2) ` we have `Delta =(1)/(w) \|{:(a_(1)+b_(1)w,,a_(1)w^(2)+b_(1),,c_(1)+b_(1)hat(w)),(a_(2)+b_(2)w,,a_(2)w^(2)+b_(2),,c_(2)+b_(2)hat(w)),(a_(3)+b_(3)w,,a_(3)w^(2)+b_(3),,c_(3)+b_(3)hat(w)):}\|` `=(1)/(w)\|{:(a_(1)+b_(1)w,,a_(1)+b_(1)w,,c_(1)+b_(1)hat(w)),(a_(2)+b_(2)w,,a_(2)+b_(2)w,,c_(2)+b_(2)hat(w)),(a_(3)+b_(3)w,,a_(3)+b_(3)w,,c_(3)+b_(3)hat(w)):}\|` `=0`

Discussion

256.	If `D = \|(1,1,1),(1,1 +x,1),(1,1,1 +y)\| " for " x!= 0, y != 0`, then D is divisible byA. x but not yB. y but not xC. neither x nor yD. both x and y
Answer» Correct Answer - D

Discussion

257.	Prove `\|(1+a,1,1),(1,1+b,1),(1,1,1+c)\|=abc(1/a+1/b+1/c+1)`
Answer» `L.H.S. = \|[1+a,1,1],[1,1+b,1],[1,1,1+c]\|` Applying `C_1->C_1-C_2` `=>\|[a,1,1],[-b,1+b,1],[0,1,1+c]\|` Applying `C_2->C_2-C_3` `=>\|[a,0,1],[-b,b,1],[0,-c,1+c]\|` `=>[a(b+bc+c)+1(bc)]` `=> ab+abc+ac+bc` `=>abc+bc+ac+ab` `=>abc(1+1/a+1/b+1/c) = R.H.S.`

Discussion

258.	Let `Delta = \|{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}\|` and the equation `px^(3) +qx^(2) +rx+s=0` has roots a,b,c where `a,b,c in R^(+)` `" if " Delta =27 " and " a^(2) +b^(2)+c^(2) =3`thenA. `3p+2q=0`B. `4p+3q=0`C. `3p+q=0`D. none of these
Answer» Correct Answer - C Multiplying `R_(1) ,R_(2),R_(3) " by " a,b,c ` respectively ,and then taking a,b,c common from` C_(1),C_(2)" and " C_(3)` we get `Delta = \|{:(-bc,,ab+ac,,ac+ab),(ab+bc,,-ac,,bc+ab),(ac+bc,,bc+ac,,-ab):}\|` Now using `C_(2) to C_(2)-C_(1)" and " C_(3) to C_(3) -C_(1) ` and then taking (ab+bc+ca) common from `C_(2) " and " C_(3)` we get `Delta =\|{:(-bc,,1,,1),(ab+bc,,-1,,0),(ac+bc,,0,,-1):}\|xx (ab +bc +ca)^(2)` Now applying `R_(2) to R_(2) +R_(1)` we get `Delta = \|{:(-bc,,1,,1),(ab,,0,,1),(ac+bc,,0,,-1):}\| (ab+bc+ca)^(2)` Expanding along `c_(2)` we get `Delta =(ab+bc+ca)^(2)[ac+bc+ca)^(2)` `=(ab+bc+ca)^(2)` `=(r//p)^(3) =r^(3)//p^(3)` Now given a,b,c are all positive then `A.M ge G.M.` `rArr (ab+bc+ac)/(3) ge (abxx bcxx ac)^(1//3)` `" or " (ab+bc+ac)^(3) ge 27a^(2)b^(2)c^(2)` `" or " (ab+bc+ca)^(3) ge 27(s^(2)//p^(2))` if `Delta =27` then ab+bc+ca =3 and given that `a^(2) +b^(2)+c^(2)=3` From `(a+b+c)^(2) =a^(2) +b^(2)+c^(2) +2 (ab+bc+ca)` we have `a+b+c = ne 3` `rArr a+b+c =3` `rArr 3p+q=0`

Discussion

259.	`\|(sin^(2) x,cos^(2) x,1),(cos^(2) x,sin^(2) x,1),(- 10,12,2)\| =`
Answer» Correct Answer - A

Discussion

260.	If `\|(x +a,b,c),(a,x +b,c),(a,b,x +c)\| = 0`, then x equalsA. `a + b +c`B. `-(a +b +c)`C. `0, a + b+ c`D. `0, -(a + b+c)`
Answer» Correct Answer - D

Discussion

261.	The value of the deterimant `\|(1,1,1,1),(1,2,3,4),(1,3,6,10),(1,4,10,20)\|` is equal to
Answer» Correct Answer - C

Discussion

262.	Prove that `\|[1,a,bc] , [1,b,ca], [1,c,ab]\|=\|[1,a,a^2] , [1,b,b^2] , [1,c,c^2]\|`
Answer» Applying `R_(1) to aR_(1), R_(2) -R_(1) "and"R_(3) to cR_(3)`, we get LHS `=(1)/(abc) * \|[a, a^(2), abc], [b, b^(2), abc], [c, c^(2), abc]\| = \|[a, a^(2), 1], [b, b^(2), 1], [c, c^(2), 1]\| = \|[1, a, a^(2)], [1, b, b^(2)], [1, c, c^(2)]\|`

Discussion

263.	If A, B and C are the angles of a triangle and `\|(1,1,1),(1 + sin A,1 + sin B,1 + sin C),(sin A + sin^(2) A,sin B + sin^(2)B,sin C + sin^(2) C)\|= 0`, then the triangle ABC isA. isoscelesB. equilateralC. right angled isoscelesD. none of these
Answer» Correct Answer - A

Discussion

264.	If `alpha, beta & gamma` are the roots of the equation `x^3+px+q=0`, then the value of the determinant `\|(alpha,beta,gamma), (beta,gamma,alpha),(gamma,alpha,beta)\|` is
Answer» `alpha[gammabeta-alpha^2]-beta[beta^2-gammaalpha]+gamma[betaalpha-gamma^2]` `3alphabetagamma-alpha^3-beta^3-gamma^3` `3alphabetagamma-(3alphabetagamma)`.

Discussion

265.	If `\|[1,x,x^2] , [x,x^2,1] , [x^2,1,x]\|=3` then find the value of `\|[x^3-1, 0, x-x^4] , [0,x-x^4, x^3-1] , [x-x^4, x^3-1, 0]\|`
Answer» ` Delta_(c)=\|{:(x^(3)-1,,0,,x-x^(4)),(0,,x-x^(4),,x^(3)-1),(x-x^(4)-1,,x^(3)-1,,0):}\|` is the determinant formed by the cofactors of determinant `Delta = \|{:(1,,x,,x^(2)),(x,,x^(2),,1),(x^(2),,1,,x):}\|` Hence `Delta_(c) =Delta^(2) =3^(2) =9`.

Discussion

266.	Prove that `\|(1,a^2+bc,a^3),(1,b^2+ca,b^3),(1,c^2+ca,c^3)\|=-(a-b)(b-c)(c-a)(a^2+b^2+c^2)`
Answer» `Delta = \|[1, a^(2), a^(3)], [1, b^(2), b^(3)], [1, c^(2), c^(3)]\| + \|[1, bc, a^(3)], [1, ca, b^(3)], [1, ab, c^(3)]\|`

Discussion

267.	Consider the system of equations `(a -1) x -y -z = 0` `x -(b -1) y +z = 0` `x + y - (c -1) z = 0` Where a, b and c are non-zero real number Statement1 : If x,y,z are not all zero, then `ab + bc + ca + abc` Statement 2 : `abc ge 27`A. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 3B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 3C. Statement 1 is true, Statement 2 is FalseD. Statement 1 is False, Statement 2 is true
Answer» Correct Answer - B It is given that x,y,z are not all zero. So, the given system of equation has a non-trivial solutions. `:. \|(a -1,-1,-1),(1,-(b -1),1),(1,1,-(c -1))\| = 0` `rArr \|(a,0,-1),(0,-b,1),(c,c,1-c)\| = 0 " " [("Applying " C_(1) rarr C_(1) - C_(3)),(C_(2) rarr C_(2) - C_(3))]` `rArr a(-b + bc -c) - (0 + bc) = 0` `rArr ab + bc + ac = abc` Using `A.M. ge G.M`., we have `(ab + bc + ca)/(3) ge (abc)^(2//3)` `rArr abc ge 3 (abc)^(2//3) rArr abc ge 27` So, both statements are true

Discussion

268.	`" If " Delta (x)=\|underset(9" "x" "-7)underset(6" "4x" "3)(1" "x^(2)" "x^(2))\|` then find the value of `overset(1)underset(0)(int)Delta (x) dx` without expanding `Delta (x)`.
Answer» Taking x common from `C_(2)` then multiplying each element of `R_(1)` with x we get `:. Delta (x)= \|{:( x,,x^(2),,x^(3)),(6,,4,,3),(9,,1,,-7):}\|` `\|{:(overset(1)underset(0)(int) xdx,,overset(1)underset(0)(int)x^(2)dx,,overset(1)underset(0)(int)x^(3)dx),(6,,4,,3),(9,,1,,-7):}\|` `\|{:([[x^(2))/(2)]_(0)^(2),,[[x^(3))/(3)]_(0)^(1),,[[x^(4))/(4)]_(0)^(1)),(6,,4,,3),(9,,1,,-7):}\|` `= \|{:(9,,1,,-7),(6,,4,,3),((1)/(2),,(1)/(3),,(1)/(4)):}\|` `=(1)/(12) \|{:(6,,4,,3),(6,,4,,3),(9,,1,,-7):}\|` (Multiplying each element of `R_(1)` with 12) `=0" "(R_(1) " and "R_(2) " are identical")`

Discussion

269.	Show that `\|{:(bc-a^(2),,ca-b^(2),,ab-c^(2)),(ca-b^(2),,ab-c^(2),,bc-a^(2)),(ab-c^(2),,bc-a^(2),,ca-b^(2)):}\|` `\|{:(a^(2),,c^(2),,2ca-b^(2)),(2ab-c^(2),,b^(2),,a^(2)),(b^(2),,2ac-a^(2),,c^(2)):}\|.`
Answer» `\|{:(bc-a^(2),,ca-b^(2),,ab-c^(2)),(ca-b^(2),,ab-c^(2),,bc-a^(2)),(ab-c^(2),,bc-a^(2),,ca-b^(2)):}\|` is formed by the cofactors of matrix `\|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|`. `rArr \|{:(bc-a^(2),,ca-b^(2),,ab-c^(2)),(ca-b^(2),,ab-c^(2),,bc-a^(2)),(ab-c^(2),,bc-a^(2),,ca-b^(2)):}\|= \|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|^(2)` `\|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|\|{:(a,,-c,,b),(b,,-a,,c),(c,,-b,,a):}\|^(2)` `\|{:(a^(2),,c^(2),,2ca-b^(2)),(2ab-c^(2),,b^(2),,a^(2)),(b^(2),,2ac-a^(2),,c^(2)):}\|.`

Discussion

270.	if `x ne 0 , y ne 0 ,z ne 0 " and " \|{:(1+x,,1,,1),(1+y,,1+2y,,1),(1+z,,1+z,,1+3z):}\|=0` then `x^(-1) +y^(-1) +z^(-1)` is equal toA. `-1`B. `-2`C. `-3`D. none of these
Answer» Correct Answer - C `\|{:(1+x,,1,,1),(1+y,,1+2y,,1),(1+z,,1+z,,3+3z):}\|` `=xyz \|{:(1+(1)/(x),,(1)/(x),,(1)/(x)),(1+(1)/(y),,2+(1)/(y),,(1)/(y)),(1+(1)/(z),,1+(1)/(z),,3+(1)/(z)):}\|` `=xyz (3+(1)/(x)+(1)/(y)+(1)/(z)) \|{:(1+(1)/(x),,(1)/(x),,(1)/(x)),(1+(1)/(y),,2+(1)/(y),,(1)/(y)),(1+(1)/(z),,1+(1)/(z),,3+(1)/(z)):}\|` `=xyz (3+(1)/(x)+(1)/(y)+(1)/(z)) \|{:(1,,0,,0),(1+(1)/(y),,1,,-1),(1+(1)/(z),,0,,2):}\|` `=2xyz (3+(1)/(x)+(1)/(y)+(1)/(z)) ` Hence the given equation gives `x^(-1) +y^(-1) +z^(-1) =-3`

Discussion

271.	the determinant `Delta=\|[a^2+x, ab, ac] , [ab, b^2+x, bc] , [ac, bc, c^2+x]\|` is divisible byA. xB. `x^(2)`C. `x^(3)`D. none of these
Answer» Correct Answer - A::B `Delta =\|{:(a^(3)+ax,,ab,,ac),(a^(2)b,,b^(2)+x,,bc),(a^(2)c,,bc,,c^(2)+x):}\|` Applying `C_(1) to C_(1) +bC_(2) +cC_(3)" and taking " a^(2)+b^(2)+c^(2)+x` common we get `Delta =(1)/(a)(a^(2) +b^(2) +c^(2)+x ) \|{:(a,,ab,,ac),(b,,b^(2)+x,,bc),(c,,bc,,c^(2)+x):}\|` Applying `C_(2) to C_(2) -bC_(1) " and " C_(3) to C_(3) -cC_(1) ` we get `Delta =(1)/(a) (a^(2) +b^(2)+c^(2)+x) \|{:(a,,0,,0),(b,,x,,0),(c,,0,,x):}\|` `=(1)/(a) (a^(2) +b^(2)+c^(2) +x) (ax^(2))` `= x^(2) (a^(2) +b^(2)+c^(2) +x)` Thus `Delta ` is divisible by x and `x^(2)`

Discussion

272.	`\|[(a+1)(a+2), a+2, 1], [(a+2)(a+3), a+3, 1], [(a+3)(a+4), a+4, 1]\| =-2`
Answer» `Delta = \|[a^(2) + 3a+2, a+2, 1], [a^(2) +5a +6, a+3, 1], [a^(2) +7a +12, a+4, 1]\|` Now, apply `R_(2) to R_(2) -R_(1) "and"R_(3) to R_(3) -R_(1)`

Discussion

273.	Let D be the determinant given by `D = \|(1,cos (beta - alpha),cos (gamma - alpha)),(cos (alpha -beta),1,cos (gamma - beta)),(cos (alpha - gamma),cos (beta - gamma),1)\|` where `alpha, beta and gamma` are real number Statement -1: The value of D is zero Statement 2: The determinant D is expressible as the product of two determinant each equal to zeroA. Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 2B. Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2C. Statement 1 is true, Statement 2 is FalseD. Statement 1 is False, Statement 2 is true
Answer» Correct Answer - A We have, `D = \|(1,cos (beta - alpha),cos (gamma - alpha)),(cos (alpha - beta),1,cos (gamma - beta)),(cos (alpha - gamma),cos (beta - gamma),1)\|` `= \|(cos^(2) alpha + sin^(2) alpha,cos beta cos alpha + sin beta sin alpha,cos gamma cos alpha + sin gamma sin alpha),(cos alpha cos beta + sin alpha sin beta,cos^(2) beta + sin^(2) beta,cos gamma cos beta + sin gamma sin beta),(cos alpha cos gamma + sin alpha sin gamma,cos beta cos gamma + sin beta sin gamma,cos^(2) gamma + sin^(2) gamma)\|` `= \|(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma ,sin gamma,0)\| \|(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)\|= 0` So, both the statement are true and statement 2 is a correct explanation for statement 1

Discussion

274.	`\|(y+z,z,y),(z,z+x,x),(y,x,x+y)\|-4xyz`
Answer» Given determinant `=\|{:(y+z, " "z, y),(" "z, z+x,x), (" "y, " "x, x+y):}\|` `=\|{:(0, -2x, -2x),(z, z+x," "x), (y, " "x, x+y):}\| [R_(1) to R_(1) -(R_(2) + R_(3))]` `=(-2x)\|{:(0, 1, 1),(z, z+x," "x), (y, " "x, x+y):}\| ["taking(-2x) common from"R_(1)]` `=(-2x)\|{:(0, 0, " "1),(z, z," "x), (y, -y, x+y):}\| [C_(2) to (C_(2) -C_(3))]` `=(-2x)1\|{:(z,z),(y,-y):}\| ["expanded by"R_(1)]`d `=(-2x)1(-yz-yz) =(-2x)(-2yz) = 4xyz.`

Discussion

275.	Prove that : `Det[[x,x^2,x^3],[y,y^2,y^3],[z,z^2,z^3]]=xyz(x-y)(y-z)(z-x)`
Answer» Take x, y and z common from `C_(1), C_(2), C_(3)` respectively

Discussion

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

Discussion

277.	Show that `\|[a,b,c],[a^2,b^2,c^2],[a^3,b^3,c^3]\|=abc(a-b)(b-c)(c-a)`
Answer» `L.H.S. = \|[a,b,c],[a^2,b^2,c^2],[a^3,b^3,c^3]\|` `= (abc)\|[1,1,1],[a,b,c],[a^2,b^2,c^2]\|` Applying `C_1->C_1-C_3 and C_2->C_2-C_3` `= (abc)\|[0,0,1],[a-b,b-c,c],[a^2-b^2,b^2-c^2,c^2]\|` `= (abc)(a-b)(b-c)\|[0,0,1],[1,1,c],[a+b,b+c,c^2]\|` `= (abc)(a-b)(b-c)[b+c-a-b]` `= (abc)(a-b)(b-c)(c-a) = R.H.S.`

Discussion

278.	Prove `\|(y+z,z+x,x+y),(z+x,x+y,y+z),(x+y,y+z,z+x)\|=2\|(x,y,z),(y,z,x),(z,x,y)\|=-2(x^3+y^3+z^3-3xyz)`
Answer» `L.H.S. = \|[y+z,z+x,x+y],[z+x,x+y,y+z],[x+y,y+z,z+x]\|` `=\|[y,z,x],[z,x,y],[x,y,z]\|+\|[z,x,y],[x,y,z],[y,z,x]\|` `=(-1)\|[y,x,z],[z,y,x],[x,z,y]\|+(-1)\|[x,z,y],[y,x,z],[z,y,x]\|` `=(-1)^2\|[x,y,z],[y,z,x],[z,x,y]\|+(-1)^2\|[x,y,z],[y,z,x],[z,x,y]\|` `=\|[x,y,z],[y,z,x],[z,x,y]\|+\|[x,y,z],[y,z,x],[z,x,y]\|` `=2\|[x,y,z],[y,z,x],[z,x,y]\|` ...(first part proved) `=2[x(zy-x^2)-y(y^2-zx)+z(yx-z^2)]` `=2[3xyz-x^3-y^3-z^3]` `=-2(x^3+y^3+z^3-3xyz) = R.H.S.`

Discussion

279.	`[[b+c,a-b,a],[c+a,b-c,b],[a+b,c-a,c]] = 3abc - a^3 - b^3 - c^3`
Answer» `L.H.S. = \|[b+c,a-b,a],[c+a,b-c,b],[a+b,c-a,c]\|` Applying `C_2->C_2-C_3` `= \|[b+c,-b,a],[c+a,-c,b],[a+b,-a,c]\|` Applying `C_1->C_+C_2` `= \|[c,-b,a],[a,-c,b],[b,-a,c]\|` `=-1 \|[c,b,a],[a,c,b],[b,a,c]\|` `= - 1[c^3-abc - bac+b^3 + a^3 - abc]` `= 3abc - a^3-b^3-c^3 = R.H.S.`

Discussion

280.	Find x, if \(\begin{vmatrix}x - 1& x - 2\\[0.3em]x & x - 3\end{vmatrix} = 0\)
Answer» \(\begin{vmatrix}x - 1& x - 2\\[0.3em]x & x - 3\end{vmatrix} = 0\) ⇒ (x – 1) (x – 3) – x(x – 2) = 0 ⇒ x² – 3x – x + 3 – x² + 2x = 0 ⇒ -2x + 3 = 0 ⇒ – 2x = -3 So, x = 3/2

280.

Find x, if \(\begin{vmatrix}x - 1& x - 2\\[0.3em]x & x - 3\end{vmatrix} = 0\)

Answer»

\(\begin{vmatrix}x - 1& x - 2\\[0.3em]x & x - 3\end{vmatrix} = 0\)

⇒ (x – 1) (x – 3) – x(x – 2) = 0

⇒ x² – 3x – x + 3 – x² + 2x = 0

⇒ -2x + 3 = 0

⇒ – 2x = -3

So, x = 3/2

Discussion

281.	Prove that : \(\begin{vmatrix}1&a&b\\[0.3em]-a&1&c\\[0.3em]-b&-c&1\end{vmatrix}\) = 1 + a2 + b2 + c2
Answer» \(\begin{vmatrix}1&a&b\\[0.3em]-a&1&c\\[0.3em]-b&-c&1\end{vmatrix}\) = 1(1 + c²) - a(- a + bc) + bc(ac + b) = 1 + c² + a² – abc + abc + b² = 1 + a² + b² + c² = R.H.S. Proved.

281.

Prove that : \(\begin{vmatrix}1&a&b\\[0.3em]-a&1&c\\[0.3em]-b&-c&1\end{vmatrix}\) = 1 + a2 + b2 + c2

Answer»

\(\begin{vmatrix}1&a&b\\[0.3em]-a&1&c\\[0.3em]-b&-c&1\end{vmatrix}\) = 1(1 + c²) - a(- a + bc) + bc(ac + b)

= 1 + c² + a² – abc + abc + b²

= 1 + a² + b² + c² = R.H.S. Proved.

Discussion

282.	If \(\begin{vmatrix}l&m\\[0.3 em]2&3\end{vmatrix} = 0\), then find l:m
Answer» \(\begin{vmatrix}l&m\\[0.3 em]2&3\end{vmatrix} = 0\) ⇒ l x 3 – 2 x m = 0 ⇒ 3l – 2m= 0 ⇒ 3l = 2m ⇒ l/m = 2/3 So, l : m = 2 : 3

282.

If \(\begin{vmatrix}l&m\\[0.3 em]2&3\end{vmatrix} = 0\), then find l:m

Answer»

\(\begin{vmatrix}l&m\\[0.3 em]2&3\end{vmatrix} = 0\)

⇒ l x 3 – 2 x m = 0

⇒ 3l – 2m= 0

⇒ 3l = 2m

⇒ l/m = 2/3

So, l : m = 2 : 3

Discussion

283.	Evaluate \( \begin{vmatrix} cos a & -sin a\\[0.3em] sin a & cos a \\[0.3em] \end{vmatrix}\) Evaluate \|(cos a, -sin a)(sin a, cos a)\|
Answer» After finding determinant we will get a trigonometric identity. cos²α +sin²α =1 ∵ sin²θ + cos²θ = 1

283.

Evaluate \( \begin{vmatrix} cos a & -sin a\\[0.3em] sin a & cos a \\[0.3em] \end{vmatrix}\) Evaluate |(cos a, -sin a)(sin a, cos a)|

Answer»

After finding determinant we will get a trigonometric identity.

cos²α +sin²α =1

∵ sin²θ + cos²θ = 1

Discussion

284.	Evaluate \( \begin{vmatrix} sin 60 ^\circ& cos 60^\circ\\[0.3em] -sin 30^\circ & cos30^\circ \\[0.3em] \end{vmatrix}\) Evaluate \|(sin60°, cos60°) (-sin 30° cos 30°)\|
Answer» After finding determinant we will get, Sin60° = \(\frac{\sqrt{3}}{2}\) = cos 30° Cos60° = \(\frac{1}{2}\) = sin30° sin 60° × cos30° + sin30° × cos60° = \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}\) = \(\frac{3}{4} + \frac{1}{4}\) = 1.

284.

Evaluate \( \begin{vmatrix} sin 60 ^\circ& cos 60^\circ\\[0.3em] -sin 30^\circ & cos30^\circ \\[0.3em] \end{vmatrix}\) Evaluate |(sin60°, cos60°) (-sin 30° cos 30°)|

Answer»

After finding determinant we will get,

Sin60° = \(\frac{\sqrt{3}}{2}\) = cos 30°

Cos60° = \(\frac{1}{2}\) = sin30°

sin 60° × cos30° + sin30° × cos60°

= \(\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2} + \frac{1}{2} \times \frac{1}{2}\)

= \(\frac{3}{4} + \frac{1}{4}\)

= 1.

Discussion

285.	Evaluate \( \begin{bmatrix} cos 65 ^\circ & sin 65^\circ\\[0.3em] sin 25^\circ & cos 25^\circ\\[0.3em] \end{bmatrix}\). Evaluate \|(cos 65° sin 65°)(sin25°, cos 25°)\|
Answer» By directly opening this determinant cos65° × cos25° -sin25° × sin65° = cos(65°+25°) ∵ cosAcosB - sinAsinB = cos(A+B) = cos90° = 0 ∵ cosAcosB - sinAsinB = cos(A+B)

285.

Evaluate \( \begin{bmatrix} cos 65 ^\circ & sin 65^\circ\\[0.3em] sin 25^\circ & cos 25^\circ\\[0.3em] \end{bmatrix}\). Evaluate |(cos 65° sin 65°)(sin25°, cos 25°)|

Answer»

By directly opening this determinant

cos65° × cos25° -sin25° × sin65° = cos(65°+25°)

∵ cosAcosB - sinAsinB = cos(A+B)

= cos90° = 0

∵ cosAcosB - sinAsinB = cos(A+B)

Discussion

286.	Evaluate the following determinants :i. \(\begin{vmatrix}x &-7 \\[0.3em]x & 5x+1 \\[0.3em]\end{vmatrix}\)ii. \(\begin{vmatrix}cos\,\theta &-sin\,\theta \\[0.3em]sin\,\theta & cos\,\theta \\[0.3em]\end{vmatrix}\)iii. \(\begin{vmatrix}cos\,15° &-sin\,15° \\[0.3em]sin\,75° & cos\,75° \\[0.3em]\end{vmatrix}\)iv. \(\begin{vmatrix}a+ib &c+id \\[0.3em]-c+id & a-ib \\[0.3em]\end{vmatrix}\)
Answer» i. Let A = \(\begin{vmatrix} x &-7 \\[0.3em] x & 5x+1 \\[0.3em] \end{vmatrix}\) ⇒ \|A\| = x(5x + 1) – (–7)x \|A\| = 5x² + 8x ii. Let A = \(\begin{vmatrix} cos\,\theta &-sin\,\theta \\[0.3em] sin\,\theta & cos\,\theta \\[0.3em] \end{vmatrix}\) ⇒ \|A\| = cosθ × cosθ – (–sinθ) x sinθ \|A\| = cos 2θ + sin 2θ \|A\| = 1 iii. Let A = \(\begin{vmatrix} cos\,15° &-sin\,15° \\[0.3em] sin\,75° & cos\,75° \\[0.3em] \end{vmatrix}\) ⇒ \|A\| = cos15° × cos75° + sin15° x sin75° \|A\| = cos(75 – 15)° \|A\| = cos60° \|A\| = 0.5. iv. A = \(\begin{vmatrix} a+ib &c+id \\[0.3em] -c+id & a-ib \\[0.3em] \end{vmatrix}\) ⇒ \|A\| = (a + ib)( a – ib) – (c + id)( –c + id) = (a + ib)( a – ib) + (c + id)( c – id) = a²– i² b² + c² – i² d²= a² – (–1)b² + c² – (–1)d² = a² + b² + c² + d²

286.

Evaluate the following determinants :i. \(\begin{vmatrix}x &-7 \\[0.3em]x & 5x+1 \\[0.3em]\end{vmatrix}\)ii. \(\begin{vmatrix}cos\,\theta &-sin\,\theta \\[0.3em]sin\,\theta & cos\,\theta \\[0.3em]\end{vmatrix}\)iii. \(\begin{vmatrix}cos\,15° &-sin\,15° \\[0.3em]sin\,75° & cos\,75° \\[0.3em]\end{vmatrix}\)iv. \(\begin{vmatrix}a+ib &c+id \\[0.3em]-c+id & a-ib \\[0.3em]\end{vmatrix}\)

Answer»

i. Let A = \(\begin{vmatrix} x &-7 \\[0.3em] x & 5x+1 \\[0.3em] \end{vmatrix}\)

⇒ |A| = x(5x + 1) – (–7)x

|A| = 5x² + 8x

ii. Let A = \(\begin{vmatrix} cos\,\theta &-sin\,\theta \\[0.3em] sin\,\theta & cos\,\theta \\[0.3em] \end{vmatrix}\)

⇒ |A| = cosθ × cosθ – (–sinθ) x sinθ

|A| = cos 2θ + sin 2θ

|A| = 1

iii. Let A = \(\begin{vmatrix} cos\,15° &-sin\,15° \\[0.3em] sin\,75° & cos\,75° \\[0.3em] \end{vmatrix}\)

⇒ |A| = cos15° × cos75° + sin15° x sin75°

|A| = cos(75 – 15)°

|A| = cos60°

|A| = 0.5.

iv. A = \(\begin{vmatrix} a+ib &c+id \\[0.3em] -c+id & a-ib \\[0.3em] \end{vmatrix}\)

⇒ |A| = (a + ib)( a – ib) – (c + id)( –c + id)

= (a + ib)( a – ib) + (c + id)( c – id)

= a²– i² b² + c² – i² d²= a² – (–1)b² + c² – (–1)d²

= a² + b² + c² + d²

Discussion

287.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case : A = \(\begin{bmatrix} 1&-3 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 3 & 5 & 2 \end{bmatrix}\)
Answer» Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij = (–1)^i+j × M_ij A = \(\begin{bmatrix} 1&-3 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 3 & 5 & 2 \end{bmatrix}\) ⇒ M₁₁ = \(\begin{bmatrix} -1&2 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}\) M₁₁ = –1 × 2 – 5 × 2 M₁₁ = –12 ⇒ M₂₁ = \(\begin{bmatrix} -3&2 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}\) M₂₁ = –3 × 2 – 5 × 2 M₂₁ = –16 ⇒ M₃₁ = \(\begin{bmatrix} -3&2 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}\) M₃₁ = –3 × 2 – (–1) × 2 M₃₁= – 4 C₁₁ = (–1)¹⁺¹× M₁₁ = 1 × –12 = –12 C₂₁ = (–1)²⁺¹× M₂₁ = –1 × –16 = 16 C₃₁ = (–1)³⁺¹ × M₃₁ = 1 × –4 = –4 Now expanding along the first column we get, \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ = 1× (–12) + 4 × 16 + 3× (–4) = –12 + 64 –12 = 40

287.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case : A = \(\begin{bmatrix} 1&-3 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 3 & 5 & 2 \end{bmatrix}\)

Answer»

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present.

Then finding the absolute value of the matrix newly formed.

Also,

C_ij = (–1)^i+j × M_ij

A = \(\begin{bmatrix} 1&-3 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 3 & 5 & 2 \end{bmatrix}\)

⇒ M₁₁ = \(\begin{bmatrix} -1&2 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}\)

M₁₁ = –1 × 2 – 5 × 2

M₁₁ = –12

⇒ M₂₁ = \(\begin{bmatrix} -3&2 \\[0.3em] 5 & 2 \\[0.3em] \end{bmatrix}\)

M₂₁ = –3 × 2 – 5 × 2

M₂₁ = –16

⇒ M₃₁ = \(\begin{bmatrix} -3&2 \\[0.3em] -1 & 2 \\[0.3em] \end{bmatrix}\)

M₃₁ = –3 × 2 – (–1) × 2

M₃₁= – 4

C₁₁ = (–1)¹⁺¹× M₁₁

= 1 × –12

= –12

C₂₁ = (–1)²⁺¹× M₂₁

= –1 × –16

= 16

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –4

= –4

Now expanding along the first column we get,

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (–12) + 4 × 16 + 3× (–4)

= –12 + 64 –12

= 40

Discussion

288.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} -1&4 \\[0.3em] 2 & 3 \end{bmatrix}\)
Answer» Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij= (–1)^i+j × M_ij A = \(\begin{bmatrix} -1&4 \\[0.3em] 2 & 3 \end{bmatrix}\) M₁₁ = 3 M₂₁= 4 C₁₁ = (–1)¹⁺¹ × M₁₁ = 1 × 3 = 3 C₂₁ = (–1)²⁺¹ × 4 = –1 × 4 = –4 Now expanding along the first column we get \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁ = –1× 3 + 2 × (–4) = –11

288.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} -1&4 \\[0.3em] 2 & 3 \end{bmatrix}\)

Answer»

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present.

Then finding the absolute value of the matrix newly formed.

Also,

C_ij= (–1)^i+j × M_ij

A = \(\begin{bmatrix} -1&4 \\[0.3em] 2 & 3 \end{bmatrix}\)

M₁₁ = 3

M₂₁= 4

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 3

= 3

C₂₁ = (–1)²⁺¹ × 4

= –1 × 4

= –4

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁

= –1× 3 + 2 × (–4)

= –11

Discussion

289.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 5&20 \\[0.3em] 0 & -1 \end{bmatrix}\)
Answer» Let M_ijand C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij = (–1)^i+j × M_ij A = \(\begin{bmatrix}5&20 \\[0.3em]0 & -1\end{bmatrix}\) M₁₁ = –1 M₂₁ = 20 C₁₁= (–1)¹⁺¹ × M₁₁ = 1 × –1 = –1 C21 = (–1)2+1 × M21 = 20 × –1 = –20 Now expanding along the first column we get, \|A\| = a₁₁ × C₁₁+ a₂₁× C₂₁ = 5× (–1) + 0 × (–20) = –5

289.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 5&20 \\[0.3em] 0 & -1 \end{bmatrix}\)

Answer»

Let M_ijand C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also,

C_ij = (–1)^i+j × M_ij

A = \(\begin{bmatrix}5&20 \\[0.3em]0 & -1\end{bmatrix}\)

M₁₁ = –1

M₂₁ = 20

C₁₁= (–1)¹⁺¹ × M₁₁

= 1 × –1 = –1

C21 = (–1)2+1 × M21 = 20 × –1 = –20

Now expanding along the first column we get,

|A| = a₁₁ × C₁₁+ a₂₁× C₂₁

= 5× (–1) + 0 × (–20)

= –5

Discussion

290.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix}2 &-1 & 0 &1 \\[0.3em]-3 & 0 & 1&-2 \\[0.3em]1 & 1 &-1&1\\[0.3em]2 & -1 &5&0\end{bmatrix}\)
Answer» Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij = (–1)^i+j× M_ij A = \(\begin{bmatrix}2 &-1 & 0 &1 \\[0.3em]-3 & 0 & 1&-2 \\[0.3em]1 & 1 &-1&1\\[0.3em]2 & -1 &5&0\end{bmatrix}\) ⇒ M₁₁ = \(\begin{bmatrix}0 &1 & -2 \\[0.3em]1 & -1 & 1 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\) M₁₁ = 0(–1×0 – 5×1) – 1(1×0 – (–1)×1) + (–2)(1×5 – (–1)×(–1)) M₁₁ = – 9 ⇒ M₂₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]1 & -1 & 1 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\) M₂₁ = –1(–1×0 – 5×1) – 0(1×0 – (–1)×1) + 1(1×5 – (–1)×(–1)) M₂₁ = 9 ⇒ M₃₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]0 & 1 & -2 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\) M₃₁ = –1(1×0 – 5×(–2)) – 0(0×0 – (–1)×(–2)) + 1(0×5 – (–1)×1) M₃₁ = – 9 ⇒ M₄₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]0 & 1 & -2 \\[0.3em]-1 & -1 &1\\[0.3em]\end{bmatrix}\) M₄₁ = –1(1×1 – (–1)×(–2)) – 0(0×1 – 1×(–2)) + 1(0×(–1) – 1×1) M₄₁ = 0 C₁₁ = (–1)¹⁺¹ × M₁₁ = 1 × (–9) = –9 C₂₁ = (–1)²⁺¹ × M₂₁ = –1 × 9 = –9 C₃₁ = (–1)³⁺¹ × M₃₁ = 1 × –9 = –9 C₄₁ = (–1)⁴⁺¹ × M₄₁ = –1 × 0 = 0 Now expanding along the first column we get \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁ = 2× (–9) + (–3) × –9 + 1× (–9) + 2× 0 = – 18 + 27 – 9 = 0

290.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix}2 &-1 & 0 &1 \\[0.3em]-3 & 0 & 1&-2 \\[0.3em]1 & 1 &-1&1\\[0.3em]2 & -1 &5&0\end{bmatrix}\)

Answer»

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present.

Then finding the absolute value of the matrix newly formed.

Also,

C_ij = (–1)^i+j× M_ij

A = \(\begin{bmatrix}2 &-1 & 0 &1 \\[0.3em]-3 & 0 & 1&-2 \\[0.3em]1 & 1 &-1&1\\[0.3em]2 & -1 &5&0\end{bmatrix}\)

⇒ M₁₁ = \(\begin{bmatrix}0 &1 & -2 \\[0.3em]1 & -1 & 1 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\)

M₁₁ = 0(–1×0 – 5×1) – 1(1×0 – (–1)×1) + (–2)(1×5 – (–1)×(–1))

M₁₁ = – 9

⇒ M₂₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]1 & -1 & 1 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\)

M₂₁ = –1(–1×0 – 5×1) – 0(1×0 – (–1)×1) + 1(1×5 – (–1)×(–1))

M₂₁ = 9

⇒ M₃₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]0 & 1 & -2 \\[0.3em]-1 & 5 &0\\[0.3em]\end{bmatrix}\)

M₃₁ = –1(1×0 – 5×(–2)) – 0(0×0 – (–1)×(–2)) + 1(0×5 – (–1)×1)

M₃₁ = – 9

⇒ M₄₁ = \(\begin{bmatrix}-1 &0 & 1 \\[0.3em]0 & 1 & -2 \\[0.3em]-1 & -1 &1\\[0.3em]\end{bmatrix}\)

M₄₁ = –1(1×1 – (–1)×(–2)) – 0(0×1 – 1×(–2)) + 1(0×(–1) – 1×1)

M₄₁ = 0

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (–9)

= –9

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × 9

= –9

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –9

= –9

C₄₁ = (–1)⁴⁺¹ × M₄₁

= –1 × 0

= 0

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ + a₄₁× C₄₁

= 2× (–9) + (–3) × –9 + 1× (–9) + 2× 0

= – 18 + 27 – 9

= 0

Discussion

291.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\)
Answer» Let M_ij and C_ijrepresents the minor and co–factor of an element, where i and j represent the row and column. The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij= (–1)^i+j × M_ij A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\) ⇒ M₁₁ = \(\begin{bmatrix} b&ca \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\) M₁₁= b × ab – c × ca M₁₁ = ab² – ac² ⇒ M₂₁ = \(\begin{bmatrix} a&bc \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\) M₂₁ = a × ab – c × bc M₂₁= a²b – c²b ⇒ M₃₁ = \(\begin{bmatrix} a&bc \\[0.3em] b & ca \\[0.3em] \end{bmatrix}\) M₃₁ = a × ca – b × bc M₃₁ = a²c – b²c C₁₁ = (–1)¹⁺¹ × M₁₁ = 1 × (ab² – ac²) = ab² – ac² C₂₁ = (–1)²⁺¹ × M₂₁ = –1 × (a²b – c²b) = c²b – a²b C₃₁ = (–1)³⁺¹ × M₃₁ = 1 × (a²c – b²c) = a²c – b²c Now expanding along the first column we get \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ = 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c) = ab² – ac² + c²b – a²b + a²c – b²c

291.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\)

Answer»

Let M_ij and C_ijrepresents the minor and co–factor of an element, where i and j represent the row and column.

The minor of the matrix can be obtained for a particular element by removing the row and column where the element is present.

Then finding the absolute value of the matrix newly formed.

Also,

C_ij= (–1)^i+j × M_ij

A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\)

⇒ M₁₁ = \(\begin{bmatrix} b&ca \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M₁₁= b × ab – c × ca

M₁₁ = ab² – ac²

⇒ M₂₁ = \(\begin{bmatrix} a&bc \\[0.3em] c & ab \\[0.3em] \end{bmatrix}\)

M₂₁ = a × ab – c × bc

M₂₁= a²b – c²b

⇒ M₃₁ = \(\begin{bmatrix} a&bc \\[0.3em] b & ca \\[0.3em] \end{bmatrix}\)

M₃₁ = a × ca – b × bc

M₃₁ = a²c – b²c

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (ab² – ac²)

= ab² – ac²

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × (a²b – c²b)

= c²b – a²b

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (a²c – b²c)

= a²c – b²c

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 1× (ab² – ac²) + 1 × (c²b – a²b) + 1× (a²c – b²c)

= ab² – ac² + c²b – a²b + a²c – b²c

Discussion

292.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 0&2 & 6 \\[0.3em] 1 & 5 & 0 \\[0.3em] 3 & 7 & 1 \end{bmatrix}\)
Answer» Let M_ijand C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij = (–1)^i+j× M_ij A = \(\begin{bmatrix} 0&2 & 6 \\[0.3em] 1 & 5 & 0 \\[0.3em] 3 & 7 & 1 \end{bmatrix}\) ⇒ M₁₁ = \(\begin{bmatrix} 5 & 0 \\[0.3em] 7 &1\\[0.3em] \end{bmatrix}\) M₁₁ = 5 × 1 – 7 × 0 M₁₁= 5 ⇒ M₂₁ = \(\begin{bmatrix}2 & 6 \\[0.3em] 7 &1\\[0.3em] \end{bmatrix}\) M₂₁= 2× 1 – 7 × 6 M₂₁ = – 40 ⇒ M₃₁ = \(\begin{bmatrix}2 & 6 \\[0.3em] 5 &0\\[0.3em] \end{bmatrix}\) M₃₁ = 2×0 – 5×6 M₃₁= –30 C₁₁ = (–1)¹⁺¹ × M₁₁ = 1 × 5 = 5 C₂₁ = (–1)²⁺¹ × M₂₁ = –1 × –40 = 40 C₃₁ = (–1)³⁺¹ × M₃₁ = 1 × –30 = –30 Now expanding along the first column we get \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ = 0× 5 + 1 × 40 + 3× (–30) = 0 + 40 – 90 = 50

292.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 0&2 & 6 \\[0.3em] 1 & 5 & 0 \\[0.3em] 3 & 7 & 1 \end{bmatrix}\)

Answer»

Let M_ijand C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed.

Also,

C_ij = (–1)^i+j× M_ij

A = \(\begin{bmatrix} 0&2 & 6 \\[0.3em] 1 & 5 & 0 \\[0.3em] 3 & 7 & 1 \end{bmatrix}\)

⇒ M₁₁ = \(\begin{bmatrix} 5 & 0 \\[0.3em] 7 &1\\[0.3em] \end{bmatrix}\)

M₁₁ = 5 × 1 – 7 × 0

M₁₁= 5

⇒ M₂₁ = \(\begin{bmatrix}2 & 6 \\[0.3em] 7 &1\\[0.3em] \end{bmatrix}\)

M₂₁= 2× 1 – 7 × 6

M₂₁ = – 40

⇒ M₃₁ = \(\begin{bmatrix}2 & 6 \\[0.3em] 5 &0\\[0.3em] \end{bmatrix}\)

M₃₁ = 2×0 – 5×6

M₃₁= –30

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × 5

= 5

C₂₁ = (–1)²⁺¹ × M₂₁

= –1 × –40

= 40

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × –30

= –30

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= 0× 5 + 1 × 40 + 3× (–30)

= 0 + 40 – 90

= 50

Discussion

293.	Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} a&h & g \\[0.3em] h & b & f \\[0.3em] g & f & c \end{bmatrix}\)
Answer» Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column. The minor of matrix can be obtained for particular element by removing the row and column where the element is present. Then finding the absolute value of the matrix newly formed. Also, C_ij= (–1)^i+j × M^ij A = \(\begin{bmatrix} a&h & g \\[0.3em] h & b & f \\[0.3em] g & f & c \end{bmatrix}\) ⇒ M₁₁ = \(\begin{bmatrix}b&f \\[0.3em] f & c\\[0.3em] \end{bmatrix}\) M₁₁= b × c – f × f M₁₁ = bc– f² ⇒ M₂₁ = \(\begin{bmatrix}h&g \\[0.3em] f & c\\[0.3em] \end{bmatrix}\) M₂₁ = h × c – f × g M₂₁ = hc – fg ⇒ M₃₁ = \(\begin{bmatrix}h&g \\[0.3em] b & f\\[0.3em] \end{bmatrix}\) M₃₁ = h × f – b × g M₃₁= hf – bg C₁₁ = (–1)¹⁺¹ × M₁₁ = 1 × (bc– f²) = bc– f² C₂₁= (–1)²⁺¹ × M₂₁ = –1 × (hc – fg) = fg – hc C₃₁ = (–1)³⁺¹ × M₃₁ = 1 × (hf – bg) = hf – bg Now expanding along the first column we get \|A\| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁ = a× (bc– f²) + h× (fg – hc) + g× (hf – bg) = abc– af² + hgf – h²c +ghf – bg²

293.

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} a&h & g \\[0.3em] h & b & f \\[0.3em] g & f & c \end{bmatrix}\)

Answer»

Let M_ij and C_ij represents the minor and co–factor of an element, where i and j represent the row and column.

The minor of matrix can be obtained for particular element by removing the row and column where the element is present.

Then finding the absolute value of the matrix newly formed.

Also,

C_ij= (–1)^i+j × M^ij

A = \(\begin{bmatrix} a&h & g \\[0.3em] h & b & f \\[0.3em] g & f & c \end{bmatrix}\)

⇒ M₁₁ = \(\begin{bmatrix}b&f \\[0.3em] f & c\\[0.3em] \end{bmatrix}\)

M₁₁= b × c – f × f

M₁₁ = bc– f²

⇒ M₂₁ = \(\begin{bmatrix}h&g \\[0.3em] f & c\\[0.3em] \end{bmatrix}\)

M₂₁ = h × c – f × g

M₂₁ = hc – fg

⇒ M₃₁ = \(\begin{bmatrix}h&g \\[0.3em] b & f\\[0.3em] \end{bmatrix}\)

M₃₁ = h × f – b × g

M₃₁= hf – bg

C₁₁ = (–1)¹⁺¹ × M₁₁

= 1 × (bc– f²)

= bc– f²

C₂₁= (–1)²⁺¹ × M₂₁

= –1 × (hc – fg)

= fg – hc

C₃₁ = (–1)³⁺¹ × M₃₁

= 1 × (hf – bg)

= hf – bg

Now expanding along the first column we get

|A| = a₁₁ × C₁₁ + a₂₁× C₂₁+ a₃₁× C₃₁

= a× (bc– f²) + h× (fg – hc) + g× (hf – bg)

= abc– af² + hgf – h²c +ghf – bg²

Discussion

294.	Find the real values of `lambda`for which the following system of linear equations has non-trivialsolutions. Also, find the non-trivial solutions.`2lambdax-2y+3z=0``x+lambday+2z=0``2x+lambdaz=0`
Answer» `\|(2 lambda ,-2, 3),(1, lamda, 2),(2,0,lamda)\|= 0` `2 lamda^3 + (-8)- 6lamda + 2lamda = 0` `2 lamda^3 - 4 lamda - 8 = 0` `lamda^3 - 2 lamda - 4 = 0` so, `lamda= 2` `(lamda- 2)(lamda^2 + 2 lamda + 2) = 0` `lamda = (-2 +- isqrt(4-0))/21 = (-2 +- 2i)/2` `lamda = -1 +- i` Answer

Discussion

295.	If A is a square matrix of order n × n such that \|A\| = λ, then write the value of \|-A\|.
Answer» Since \|kA\| = k^m\|A\| Given that k = -1, m = n and \|A\| = λ, we get \|-A\| = (-1)ⁿ×λ Hence, \|-A\| = λ if n is even and \|-A\| = -λ if n is odd.

295.

If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.

Answer»

Since |kA| = k^m|A|

Given that k = -1, m = n and |A| = λ, we get

|-A| = (-1)ⁿ×λ

Hence, |-A| = λ if n is even and |-A| = -λ if n is odd.

Discussion

296.	Write the value of \(\begin{vmatrix}sin\,20° & -cos\,20° \\[0.3em]sin\,70°& cos\,70° \\[0.3em]\end{vmatrix}\).
Answer» \(\begin{vmatrix}sin\,20° & -cos\,20° \\[0.3em]sin\,70°& cos\,70° \\[0.3em]\end{vmatrix}\) = sin 20° . cos 70° + cos 20° . sin 70° = sin (20° + 70°) = sin 90° = 1.

296.

Write the value of \(\begin{vmatrix}sin\,20° & -cos\,20° \\[0.3em]sin\,70°& cos\,70° \\[0.3em]\end{vmatrix}\).

Answer»

\(\begin{vmatrix}sin\,20° & -cos\,20° \\[0.3em]sin\,70°& cos\,70° \\[0.3em]\end{vmatrix}\)

= sin 20° . cos 70° + cos 20° . sin 70°

= sin (20° + 70°)

= sin 90°

= 1.

Discussion

297.	If A is a 3 × 3 matrix, then what will be the value of k if Det(A-1) = (Det A)k?
Answer» We are given that, Order of matrix = 3 × 3 Det(A^-1) = (Det A)^k An n-by-n square matrix A is called invertible if there exists an n-by-n square matrix B such that where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication. We know that, If A and B are square matrices of same order, then Det (AB) = Det (A).Det (B) Since, A is an invertible matrix, this means that, A has an inverse called A^-1. Then, if A and A^-1 are inverse matrices, then Det (AA^-1) = Det (A).Det (A^-1) By property of inverse matrices, AA^-1 = I ∴, Det (I) = Det (A).Det (A^-1) Since, Det (I) = 1 \(\Rightarrow\) 1 = Det (A).Det (A^-1) \(\Rightarrow\) Det (A^-1) = \(\frac{1}{Det(A)}\) \(\Rightarrow\) Det (A^-1) = Det (A)^-1 Since, according to question, Det(A^-1) = (Det A)^k \(\Rightarrow\)k = -1 Thus, the value of k is -1.

297.

If A is a 3 × 3 matrix, then what will be the value of k if Det(A-1) = (Det A)k?

Answer»

We are given that,

Order of matrix = 3 × 3

Det(A^-1) = (Det A)^k

An n-by-n square matrix A is called invertible if there exists an n-by-n square matrix B such that where In denotes the n-by-n identity matrix and the multiplication used is ordinary matrix multiplication.

We know that,

If A and B are square matrices of same order, then

Det (AB) = Det (A).Det (B)

Since, A is an invertible matrix, this means that, A has an inverse called A^-1.

Then, if A and A^-1 are inverse matrices, then

Det (AA^-1) = Det (A).Det (A^-1)

By property of inverse matrices,

AA^-1 = I

∴, Det (I) = Det (A).Det (A^-1)

Since, Det (I) = 1

\(\Rightarrow\) 1 = Det (A).Det (A^-1)

\(\Rightarrow\) Det (A^-1) = \(\frac{1}{Det(A)}\)

\(\Rightarrow\) Det (A^-1) = Det (A)^-1

Since, according to question,

Det(A^-1) = (Det A)^k

\(\Rightarrow\)k = -1

Thus, the value of k is -1.

Discussion

298.	Using properties of determinants Prove that `\|{:(a+b+c,,-c,,-b),(-c,,a+b+c,,-a),( -b,,-a,,a+b+c):}\| = 2 (a+b) (b+c) (c+a)`
Answer» `Delta =- \|{:(a+b+c,,-c,,-b),(-c,,a+b+c,,-a),(-b,,-a,,a+b+c):}\|` Applying `R_(3) to R_(3) +R_(2),R_(1) to R_(1)+R_(2)` `Delta= \|{:(a+b,,a+b,,-(a+b)),(-c,,a+b+c,,-a),(-(b+c),,b+c,,b+c):}\|` `=(a+b)(b+c) \|{:(1,,1,,-1),(-c,,a+b+c,,-a),(-1,,1,,1):}\|` Applying `R_(1) to R_(1)+R_(3)` `Delta =(a+b)(b+c) \|{:(0,,2,,0),(-c,,a+b+c,,-a),(-1,,1,,1):}\|` `= (a+b)(b+c) (-2) \|{:(-c,,-a),(-a,,1):}\|` `=2(a+b)(b+c)(c+a)`

Discussion

299.	Solve `\|{:(x^(2)-1,,x^(2)+2x+1,,2x^(2)+3x+1),(2x^(2)+x-1,,2x^(2)+5x-3,,4x^(2)+4x-3),(6x^(2)-x-2,,6x^(2)-7x+2,,12x^(2)-5x-2):}\| =0`
Answer» Correct Answer - `x=-1,(1)/(2),(2)/(3),-(2)/(3)` `\|{:(x^(2)-1,,x^(2)+2x+1,,2x^(2)+3x+1),(2x^(2)+x-1,,2x^(2)+5x-3,,4x^(2)+4x-3),((3x-2)(2x+1),,6x^(2)-7x+2,,12x^(2)-5x-2):}\|=0` `" or " \|{:((x-1)(x+1),,(x+1)^(2),,(x+1)(2x+1)),((2x-1)(x+1),,(2x-1)(x+3),,(2x-1)(2x+3)),((3x-2)(2x+1),,(3x-2)(2x+1),,(3x-2)(4x+1)):}\|=0` `" or " (x+1)(2x-1)(3x-2) \|{:(x-1,,x+1,,2x+1),(x+1,,x+3,,2x+3),(2x+1,,2x-1,,4x+1):}\|=0` `C_(2) to C_(2)-C_(1) " and "C_(3) to C_(3) -2C_(1)` `rArr (x+1) (2x-1) (3x-2) \|{:(x-1,,2,,3),(x+1,,2,,1),(2x+1,,-2,,-1):}\|=0` Applying `C_(2)to C_(2)-2C_(3)` we get `(x+1) (2x-1) (3x-2) \|{:(x-1,,-4,,3),(x+1,,0,,1),(2x+1,,0,,-1):}\|=0` `" or " 4(x+1)(2x-1)(3x-2)(3x+2)=0` `" or " x=-1, (1)/(2),(2)/(3),(2)/(3)`

Discussion

300.	Evaluate \(\begin{vmatrix} 0 & 2 & 0 \\[0.3em] 2 & 3 & 4 \\[0.3em] 4 & 5 &6 \end{vmatrix}\) \|(0,2,0),(2,3,4),(4,5,6)\|
Answer» We know that expansion of determinant with respect to first row is a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃. = 0(3 × 6-5 × 4) - 2(2 × 6-4 × 4) + 0(2 × 5-4 × 3) = 8.

300.

Evaluate \(\begin{vmatrix} 0 & 2 & 0 \\[0.3em] 2 & 3 & 4 \\[0.3em] 4 & 5 &6 \end{vmatrix}\) |(0,2,0),(2,3,4),(4,5,6)|

Answer»

We know that expansion of determinant with respect to first row is a₁₁A₁₁+a₁₂A₁₂+a₁₃A₁₃.

= 0(3 × 6-5 × 4) - 2(2 × 6-4 × 4) + 0(2 × 5-4 × 3)

= 8.

Discussion

Explore topic-wise InterviewSolutions in .

If w is a complex cube root of unity. `|(a,b,c),(b,c,a),(c,a,b)| = -(a +b +c) (a + bk + ck^(2)) (a + bk^(2) + ck)`, then k equalsA. 1B. `-1`C. `omega`D. `-omega`

If `a+b+c=0,`one root of `|a-x c b c b-x a b a c-x|=0`is`x=1`b. `x=2`c. `x=a^2+b^2+c^2`d. `x=0`A. `x=1`B. `x=2`C. `x=a^(2)+b^(2)+c^(2)`D. `x=0`

If `alpha, beta and gamma` are the roots of the equation `x^(3) + px + q = 0` (with `p != 0 and p != 0 and q != 0`), the value of the determinant `|(alpha,beta,gamma),(beta,gamma,alpha),(gamma,alpha,beta)|`, isA. pB. qC. `p^(2) - 2q`D. none of these

`|[1,bc,b+c],[1,ca,c+a],[1,ab,a+b]|=|[1,a,a^2],[1,b,b^2],[1,c,c^2]|`

if w is a complex cube root to unity then value of ` Delta =|{:(a_(1)+b_(1)w,,a_(1)w^(2)+b_(1),,c_(1)+b_(1)bar(w)),(a_(2)+b_(2)w,,a_(2)w^(2)+b_(2),,c_(2)+b_(2)bar(w)),(a_(3)+b_(3)w,,a_(3)w^(2)+b_(3),,c_(3)+b_(3)bar(w)):}|` is

If `D = |(1,1,1),(1,1 +x,1),(1,1,1 +y)| " for " x!= 0, y != 0`, then D is divisible byA. x but not yB. y but not xC. neither x nor yD. both x and y

Prove `|(1+a,1,1),(1,1+b,1),(1,1,1+c)|=abc(1/a+1/b+1/c+1)`

Let `Delta = |{:(-bc,,b^(2)+bc,,c^(2)+bc),(a^(2)+ac,,-ac,,c^(2)+ac),(a^(2)+ab,,b^(2)+ab,,-ab):}|` and the equation `px^(3) +qx^(2) +rx+s=0` has roots a,b,c where `a,b,c in R^(+)` `" if " Delta =27 " and " a^(2) +b^(2)+c^(2) =3`thenA. `3p+2q=0`B. `4p+3q=0`C. `3p+q=0`D. none of these

`|(sin^(2) x,cos^(2) x,1),(cos^(2) x,sin^(2) x,1),(- 10,12,2)| =`

If `|(x +a,b,c),(a,x +b,c),(a,b,x +c)| = 0`, then x equalsA. `a + b +c`B. `-(a +b +c)`C. `0, a + b+ c`D. `0, -(a + b+c)`

The value of the deterimant `|(1,1,1,1),(1,2,3,4),(1,3,6,10),(1,4,10,20)|` is equal to

Prove that `|[1,a,bc] , [1,b,ca], [1,c,ab]|=|[1,a,a^2] , [1,b,b^2] , [1,c,c^2]|`

If A, B and C are the angles of a triangle and `|(1,1,1),(1 + sin A,1 + sin B,1 + sin C),(sin A + sin^(2) A,sin B + sin^(2)B,sin C + sin^(2) C)|= 0`, then the triangle ABC isA. isoscelesB. equilateralC. right angled isoscelesD. none of these

If `alpha, beta & gamma` are the roots of the equation `x^3+px+q=0`, then the value of the determinant `|(alpha,beta,gamma), (beta,gamma,alpha),(gamma,alpha,beta)|` is

If `|[1,x,x^2] , [x,x^2,1] , [x^2,1,x]|=3` then find the value of `|[x^3-1, 0, x-x^4] , [0,x-x^4, x^3-1] , [x-x^4, x^3-1, 0]|`

Prove that `|(1,a^2+bc,a^3),(1,b^2+ca,b^3),(1,c^2+ca,c^3)|=-(a-b)(b-c)(c-a)(a^2+b^2+c^2)`

`" If " Delta (x)=|underset(9" "x" "-7)underset(6" "4x" "3)(1" "x^(2)" "x^(2))|` then find the value of `overset(1)underset(0)(int)Delta (x) dx` without expanding `Delta (x)`.

Show that `|{:(bc-a^(2),,ca-b^(2),,ab-c^(2)),(ca-b^(2),,ab-c^(2),,bc-a^(2)),(ab-c^(2),,bc-a^(2),,ca-b^(2)):}|` `|{:(a^(2),,c^(2),,2ca-b^(2)),(2ab-c^(2),,b^(2),,a^(2)),(b^(2),,2ac-a^(2),,c^(2)):}|.`

if `x ne 0 , y ne 0 ,z ne 0 " and " |{:(1+x,,1,,1),(1+y,,1+2y,,1),(1+z,,1+z,,1+3z):}|=0` then `x^(-1) +y^(-1) +z^(-1)` is equal toA. `-1`B. `-2`C. `-3`D. none of these

the determinant `Delta=|[a^2+x, ab, ac] , [ab, b^2+x, bc] , [ac, bc, c^2+x]|` is divisible byA. xB. `x^(2)`C. `x^(3)`D. none of these

`|[(a+1)(a+2), a+2, 1], [(a+2)(a+3), a+3, 1], [(a+3)(a+4), a+4, 1]| =-2`

`|(y+z,z,y),(z,z+x,x),(y,x,x+y)|-4xyz`

Prove that : `Det[[x,x^2,x^3],[y,y^2,y^3],[z,z^2,z^3]]=xyz(x-y)(y-z)(z-x)`

15. Using properties of determinants, prove the following `|[a,b,c],[a-b,b-c,c-a],[b+c,c+a,a+b]|`=`a^3+b^3+c^3-3abc`

Show that `|[a,b,c],[a^2,b^2,c^2],[a^3,b^3,c^3]|=abc(a-b)(b-c)(c-a)`

Prove `|(y+z,z+x,x+y),(z+x,x+y,y+z),(x+y,y+z,z+x)|=2|(x,y,z),(y,z,x),(z,x,y)|=-2(x^3+y^3+z^3-3xyz)`

`[[b+c,a-b,a],[c+a,b-c,b],[a+b,c-a,c]] = 3abc - a^3 - b^3 - c^3`

Find x, if \(\begin{vmatrix}x - 1&amp; x - 2\\[0.3em]x &amp; x - 3\end{vmatrix} = 0\)

Prove that : \(\begin{vmatrix}1&amp;a&amp;b\\[0.3em]-a&amp;1&amp;c\\[0.3em]-b&amp;-c&amp;1\end{vmatrix}\) = 1 + a2 + b2 + c2

If \(\begin{vmatrix}l&amp;m\\[0.3 em]2&amp;3\end{vmatrix} = 0\), then find l:m

Evaluate \( \begin{vmatrix} cos a &amp; -sin a\\[0.3em] sin a &amp; cos a \\[0.3em] \end{vmatrix}\) Evaluate |(cos a, -sin a)(sin a, cos a)|

Evaluate \( \begin{vmatrix} sin 60 ^\circ&amp; cos 60^\circ\\[0.3em] -sin 30^\circ &amp; cos30^\circ \\[0.3em] \end{vmatrix}\) Evaluate |(sin60°, cos60°) (-sin 30° cos 30°)|

Evaluate \( \begin{bmatrix} cos 65 ^\circ &amp; sin 65^\circ\\[0.3em] sin 25^\circ &amp; cos 25^\circ\\[0.3em] \end{bmatrix}\). Evaluate |(cos 65° sin 65°)(sin25°, cos 25°)|

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case : A = \(\begin{bmatrix} 1&amp;-3 &amp; 2 \\[0.3em] 4 &amp; -1 &amp; 2 \\[0.3em] 3 &amp; 5 &amp; 2 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} -1&amp;4 \\[0.3em] 2 &amp; 3 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 5&amp;20 \\[0.3em] 0 &amp; -1 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 1&amp;a &amp; bc \\[0.3em] 1 &amp; b &amp; ca \\[0.3em] 1 &amp; c &amp; ab \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 0&amp;2 &amp; 6 \\[0.3em] 1 &amp; 5 &amp; 0 \\[0.3em] 3 &amp; 7 &amp; 1 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} a&amp;h &amp; g \\[0.3em] h &amp; b &amp; f \\[0.3em] g &amp; f &amp; c \end{bmatrix}\)

Find the real values of `lambda`for which the following system of linear equations has non-trivialsolutions. Also, find the non-trivial solutions.`2lambdax-2y+3z=0``x+lambday+2z=0``2x+lambdaz=0`

If A is a square matrix of order n × n such that |A| = λ, then write the value of |-A|.

Write the value of \(\begin{vmatrix}sin\,20° &amp; -cos\,20° \\[0.3em]sin\,70°&amp; cos\,70° \\[0.3em]\end{vmatrix}\).

If A is a 3 × 3 matrix, then what will be the value of k if Det(A-1) = (Det A)k?

Using properties of determinants Prove that `|{:(a+b+c,,-c,,-b),(-c,,a+b+c,,-a),( -b,,-a,,a+b+c):}| = 2 (a+b) (b+c) (c+a)`

Solve `|{:(x^(2)-1,,x^(2)+2x+1,,2x^(2)+3x+1),(2x^(2)+x-1,,2x^(2)+5x-3,,4x^(2)+4x-3),(6x^(2)-x-2,,6x^(2)-7x+2,,12x^(2)-5x-2):}| =0`

Evaluate \(\begin{vmatrix} 0 &amp; 2 &amp; 0 \\[0.3em] 2 &amp; 3 &amp; 4 \\[0.3em] 4 &amp; 5 &amp;6 \end{vmatrix}\) |(0,2,0),(2,3,4),(4,5,6)|

Find x, if \(\begin{vmatrix}x - 1& x - 2\\[0.3em]x & x - 3\end{vmatrix} = 0\)

Prove that : \(\begin{vmatrix}1&a&b\\[0.3em]-a&1&c\\[0.3em]-b&-c&1\end{vmatrix}\) = 1 + a2 + b2 + c2

If \(\begin{vmatrix}l&m\\[0.3 em]2&3\end{vmatrix} = 0\), then find l:m

Evaluate \( \begin{vmatrix} cos a & -sin a\\[0.3em] sin a & cos a \\[0.3em] \end{vmatrix}\) Evaluate |(cos a, -sin a)(sin a, cos a)|

Evaluate \( \begin{vmatrix} sin 60 ^\circ& cos 60^\circ\\[0.3em] -sin 30^\circ & cos30^\circ \\[0.3em] \end{vmatrix}\) Evaluate |(sin60°, cos60°) (-sin 30° cos 30°)|

Evaluate \( \begin{bmatrix} cos 65 ^\circ & sin 65^\circ\\[0.3em] sin 25^\circ & cos 25^\circ\\[0.3em] \end{bmatrix}\). Evaluate |(cos 65° sin 65°)(sin25°, cos 25°)|

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case : A = \(\begin{bmatrix} 1&-3 & 2 \\[0.3em] 4 & -1 & 2 \\[0.3em] 3 & 5 & 2 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} -1&4 \\[0.3em] 2 & 3 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 5&20 \\[0.3em] 0 & -1 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 1&a & bc \\[0.3em] 1 & b & ca \\[0.3em] 1 & c & ab \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} 0&2 & 6 \\[0.3em] 1 & 5 & 0 \\[0.3em] 3 & 7 & 1 \end{bmatrix}\)

Write the minors and cofactors of each element of the first column of the following matrices and hence evaluate the determinant in each case :A = \(\begin{bmatrix} a&h & g \\[0.3em] h & b & f \\[0.3em] g & f & c \end{bmatrix}\)

Write the value of \(\begin{vmatrix}sin\,20° & -cos\,20° \\[0.3em]sin\,70°& cos\,70° \\[0.3em]\end{vmatrix}\).

Evaluate \(\begin{vmatrix} 0 & 2 & 0 \\[0.3em] 2 & 3 & 4 \\[0.3em] 4 & 5 &6 \end{vmatrix}\) |(0,2,0),(2,3,4),(4,5,6)|