643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 9 InterviewSolution

401.	By using properties of determinants. Show that:(i) `\|[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]\|=(a+b+c)^3` (ii) `\|[x+y+2z, x, y],[ z, y+z+2x, y],[ z, x, z+x+2y]\|=2(x+y+z)^3`
Answer» (i) `L.H.S. = \|[a-b-c,2a,2a],[2b,b-c-a,2b],[2c,2c,c-a-b]\|` Applying `R_1->R_1+R_2+R_3` `=\|[a+b+c,a+b+c,a+b+c],[2b,b-c-a,2b],[2c,2c,c-a-b]\|` `=(a+b+c)\|[1,1,1],[2b,b-c-a,2b],[2c,2c,c-a-b]\|` Applying `C_2->C_2-C_1 and C_3->C_3-C_1` `=(a+b+c)\|[1,0,0],[2b,-(a+b+c),0],[2c,0,-(a+b+c)]\|` `=(a+b+c)(-(a+b+c))(-(a+b+c))\|[1,0,0],[2b,1,0],[2c,0,1]\|` `=(a+b+c)^3[1(1-0)-0+0]` `= (a+b+c)^3 = R.H.S.` Following the same steps, we can prove the second part.

Discussion

402.	Using properties of determinants, prove thefollowing:`\|xx+y x+2y x+2y xx+y x+y x+2y x\|=9y^2(x+y)`
Answer» Apply `R_(1) to R_(1) + R_(2) + R_(3) "and take 3(x+y) common from"R_(1)`

Discussion

403.	Write Minors and Cofactors of the elements of following determinants:(i) `\|2-4 0 3\|` (ii) `\|a c b d\|`
Answer» `Let A=\|{:(2,-4),(0,3):}\|` ltbrtgt Minors `M_(11)=3,M_(12)=0,M_(21)=-4,M_(22)=2` Cofactors `A_(11)=(-1)^(1+1)M_(11)=3` `A_(21)=(-1)^(2+1)M_(21)=4` `A_(22)=(-1)^(2+2)M_(22)=2` (ii) `Let A=\|{:(a,c),(b,d):}\|` Minors `M_(11)=d,M_(12)=b,` `M_(21)=c, M_(22)=a` Cofactors `A_(11) =(-1)^(1+1)M_(11)=d` `A_(12)=(-1)^(1+2)M_(21)=-c` `A_(22)=(-1)^(2+2)M_(22)=a`

Discussion

404.	By using properties of determinants. Show that:(i) `\|x+4 2x2x2xx+4 2x2x2xx+4\|=(5x-4)(4-x)^2`(ii)`\|y+k y y y y+k y y y y+k\|=k^2(2ydotk)^2`
Answer» `\|{:(x+4,2x,2x),(2x,x+4,2x),(2x,2x,x+4):}\|=\|{:(5x+4,2x,2x),(2x,x+4,2x),(2x,2x,x+4):}\|` `(R_(1)toR_(1)+R_(2)+R_(3))` `=(5x+4)\|{:(1,1,1),(2x,x+4,2x),(2x,2x,x+4):}\|` `=(5x+4)\|{:(0,0,1),(0,4-x,2x),(x-4,x-4,x+4):}\|` `(C_(1)toC_(1)-C_(3),C_(2)toC_(2)-C_(3))` `=(5x+4)(40x)(4-x)\|{:(0,0,1),(0,1,2x),(-1,-1,x+4):}\|` `=(5x+4)(4-x)^(2).1\|{:(0,1),(-1,-1):}\|` `=(5x+4)(e-x)^(2)` =R.H.S. (ii) `L.H.S.=\|{:(y+k,y,y),(y,y+k,y),(y,y,y+k):}\|` `=\|{:(3y+k,3y+k,3y+k),(y,y+k,y),(y,y,y+k):}\|` `(R_(1)toR_(1)+R_(2)+R_(2))` `=(3y+k)\|{:(1,1,1),(y,y+k,y),(y,y,y+k):}\|` `=(3y+k)\|{:(1,0,1),(0,k,y),(-k,-k,y+k):}\|` `(C_(1)toC_(1)-C_(3),C_(2)toC_(2)-C_(3))` `=(3y+k).1\|{:(0,k),(-k,-k):}\|` (Expanding along `R_(1)`) `=(3y+k)(0+K^(2))` `=K^(2)(3y+k)=R.H.S`

Discussion

405.	Using properties ofdeterminants, prove the following:`\|1xx^2x^2 1xxx^2 1\|=(1-x^3)^2`
Answer» `L.H.S.=\|{:(1,x,x^(2)),(x^(2),1,x),(x,x^(2),1):}\|=\|{:(1+x+x^(2),x,x^(2)),(x^(2)+1+x,1,x),(x+x^(2)+1,x^(2),1):}\|` `(C_(1)toC_(1)+C_(2)+C_(3))` `=(1+x+x^(2))\|{:(1,x,x^(2)),(1,1,x),(1,x^(2),1):}\|` `=(1+x+x^(2))\|{:(1,x,x^(2)),(0,1-x,x-x^(2)),(0,x^(2)-x,1-x^(2)):}\|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `=(1+x+x^(2))\|{:(1,x,x^(2)),(0,1-x,x(1-x)),(0,-x(1-x),(1-x)(1+x)):}\|` `=(1+x+x^(2))(1-x)(1-x)\|{:(1,x,x^(2)),(0,1,x),(0,-x,1+x):}\|` `=(1+x+x^(2))(1-x).1\|{:(1,x),(-x,1+x):}\|` `=(1+x+x^(2))(1-x)^(2)(1+x+x^(2))` `=[(1+x+x^(2))(1-x)^(2)]=(1-x^(3))^(2)` R.H.S.

Discussion

406.	Usingproperties of determinants, prove that`\|b+c q+r y+z c+a r+p z+x c+b p+q x+y\|=2 \|a p x b q y c r z\|`
Answer» We have, `LHS = \|[b+c,c+a, a+b],[q+r, r+p, p+q],[y+z, z+x,x+y]\|` `=2 \|[a+b+c,c+a, a+b],[p+q+r, r+p, p+q],[x+y+z, z+x,x+y]\|["applying"C_(1) to (C_(1) + C_(2) + C_(3))"and taking out 2 common from"C_(1)]` `=2 \|[a+b+c,-b, -c],[p+q+r, -q, -r],[x+y+z, -y,-z]\|[C_(2) to (C_(2) -C_(1)), C_(3) to (C_(3) -C_(1))]` `=2(-1)(-1) * \|[a+b+c,b, c],[p+q+r, q, r],[x+y+z, y,z]\|["taking out(-1) common from each one of "C_(2) "and"C_(3)]` `=2 \|[a,b, c],[p, q, r],[x, y,z]\| = RHS ["applying "C_(1) to C_(1) -(C_(2) + C_(3))]` Hence, LHS=RHS.

Discussion

407.	If `Delta_(1)=\|[a,b, c],[x, y, z],[p,q ,r]\|"and "Delta_(2) \|[q,-b, y],[-p, a, -x],[r,-c ,z]\|` then without expanding `Delta_(1) " and "Delta_(2), "prove that "Delta_(1) + Delta_(2) =0`
Answer» We have `Delta_(2)=\|[q,-b, y],[-p, a, -x],[r,-c ,z]\|` `=(-1)\|[q,-b, y],[p, -a, x],[r,-c ,z]\| ["taking (-1) common from"R_(2)]` `=(-1)(-1)\|[q,b, y],[p, a, x],[r,c ,z]\| ["taking (-1) common from"C_(2)]` `=\|[q,b, y],[p, a, x],[r,c ,z]\|=\|[q,p, r],[b, a, c],[y,x ,z]\| ["interchanging rows and columns"]` `=(-1)\|[p,q, r],[a, b, c],[x,y ,z]\|["applying"C_(1) harr C_(2)]` `=(-1)(-1)\|[a,b, c],[p, q, r],[x,y ,z]\|["applying"R_(1) harr R_(2)]` `=\|[a,b, c],[p, q, r],[x,y ,z]\|` `=(-1) \|[a,b, c],[x, y, z],[p,q ,r]\| =-Delta_(1) ["applying "R_(2) harr R_(3)]` Thus, `Delta_(2) = -Delta_(1) " and hence `Delta_(1) + Delta_(2) =0`

Discussion

408.	By using properties of determinants. Show that:(i) `\|a-b-c2a2a2bb-c-a2b2c2cc-a-b\|=(a+b+c)^3` (ii) `\|x+y+2z x y z y+z+2x y z x z+x+2y\|=2(x+y+z)^3`
Answer» `\|{:(a-b-c,2a,2a),(ab,b-c-a,2a),(2c,2c,c-a-b):}\|=\|{:(-(a+b+c),0,2a),(a+b+v,-(a+b+c),ab),(0,a+b+c,c-a-b):}\|` `(C_(1)toC_(1)-C_(2),C_(2)toC_(2)-C_(3))` `=(a+b+c)^(2)\|{:(-1,0,2a),(0,-1,2b+2a),(0,1,c-a-b):}\|` `(R_(2)toR_(2)+R_(1))` `=(a+b+c)^(2).(-1)\|{:(-1,2b+2a),(1,c-a-b):}\|` (Expanding along `C_(1)`) `=(a+b+c)^(2)(-1)(-c+a+b-2a-ab)` `=(a+b+c)^(2)(-1)(-a-b-c)` `=(a+b+c)^(2)(a+b+c)` `=(a+b+c)=R.H.S.` `\|{:(x+y+2s,x,y),(z,y+z+2z,y),(z,x,z+x+2y):}\|=\|{:(2x+2y+2z,x,y),(2x+2y+2z,y+z+2x,y),(2x+2y+2z,z,z+x+2y):}\|` `(C_(1)toC_(1)+C_(2)+C_(3))` `=(2x+2y+2z)\|{:(1,x,y),(1,y+z+2x,u),(1,x,z+x+2y):}\|` `=2(x+y+z)\|{:(1,x,y),(1,y+z+2x,0),(0,0,x+y+z):}\|` `(R_(2)toR_(2)-R_(1),R_(3)toR_(3)-R_(1))` `=2(x+y+z).1\|{:(x+y+z,0),(0,x+y+z):}\|` (Expanding along `C_(1)`) `=2(x+y+z)^(3)=R.H.S`

Discussion

409.	Prove that ` /_ \|[a+bx, c+dx, p+qx],[-ax+b, cx+d, px+q],[u,v,w]\|=(1-x^2) [[a,c,p],[b,d,q],[u,v,w]] `
Answer» We have `LHS=\|[a+bx, c+dx, p+qx],[ax+b,cx+d,px+q],[u, v, w]\|` `=\|[a-ax^(2), c-cx^(2), p-px^(2)],[ax+b,cx+d,px+q],[u, v, w]\| ["applying"R_(1) to R_(1) -xR_(2)]` `=\|[a(1-x^(2)), c(1-x^(2)), p(1-x^(2))],[ax+b,cx+d,px+q],[u, v, w]\|` `=(1-x^(2))\|[a, c, p],[ax+b,cx+d,px+q],[u, v, w]\| ["taking out "(1-x^(2))"common from"R_(1)]` `=(1-x^(2))\|[a, c, p],[b, d, q],[u, v, w]\| ["applying "R_(2) to R_(2) -xR_(1)]` =RHS. Hence, LHS = RHS.

Discussion

410.	Write minros and cofactros of the elements of the determinants: (i) `\|{:(1,0,4),(3,5,-1),(0,1,2):}\|`
Answer» (i) `Let A=\|{:(1,0,0),(0,1,0),(0,0,1):}\|` Minors `M_(11)=\|{:(1,0),(0,1):}\|=1-0=1` `M_(12)=\|{:(0,0),(0,1):}\|=0-0=0,M_(13)=\|{:(0,1),(0,0):}\|=0-0=0` `M_(21)=\|{:(0,0),(0,1):}\|=0-0=0,M_(22)=\|{:(1,0),(0,1):}\|=1-0=0,` `M_(23)=\|{:(1,0),(0,0):}\|=0-0=0` `M_(33)=\|{:(1,0),(0,1):}\|=1-0=1 Cofactors` `A_(11)=(-1)^(1+1)M_(11)=1,A_(12)=(-1)^(1+2)M_(12)=0` `A_(13)=(-1)^(1+3)M_(13)=0` `A_(21)=(-1)^(2+1)M_(21)=0,A_(22)=(-1)^(2+2)M_(22)=1` `A_(23)=(-1)^(2+3)M_(23)=0` `A_(31)=(-1)^(3+1)M_(31)=0,A_(32)=(-1)^(3+2)M_(32)=0`, `A_(33)=(-1)^(3+3)M_(33)=1` `Let A= \|{:(1,0,4),(3,5,-1),(0,1,2):}\|` Minors `=\|{:(5,-1),(1,2):}\|=10+1=11,` `M_(12)=\|{:(3,-1),(0,2):}\|=6-0=6, M_(13)=\|{:(3,5),(0,1):}\|=3-0=3` `M_(23)=\|{:(1,0),(0,1):}\|=1-0=1` `M_(31)=\|{:(0,4),(5,-1):}\|=0-20-=-20` `M_(32)=\|{:(1,4),(3,-1):}\|=-1-12=-13`, `M_(33)=\|{:(1,0),(3,5):}\|=5-0=5` Cofactors `A_(11)=(-1)^(1+1)M_(11)=11` `A_(12)=(-1)^(2+1)M_(21)=4,A_(22)=(-1)^(2+2)M_(22)=2`, `A_(23)(-1)^(3+1)M_(31)=-20,A_(32)=(-1)^(3+2)M_(32)=13,` `A_(33)=(-1)^(3+3)M_(33)=5`

Discussion

411.	Show that `Delta =Delta_(1)`, where `Delta = \|[Ax,x^(2), 1],[By, y^(2), 1],[Cz, z^(2),1]\| "and "Delta_(1) = \|[A,B, C],[x, y, z],[zy, zx,xy]\|`
Answer» We have, `Delta = \|[Ax,x^(2), 1],[By, y^(2), 1],[Cz, z^(2),1]\|` `= \|[Ax,By, Cz],[x^(2), y^(2), z^(2)],[1,1 ,1]\| ["interchanging the rows and columns"]` `=(xyz) \|[A,B, C],[x, y, z],[(1)/(x),(1)/(y) ,(1)/(z)]\| ["applying" C_(1) to (1)/(x) C_(1), C_(2) to (1)/(y) C_(2), C_(3) to (1)/(z)C_(3) " and multiplying the whole determinant by xyz"]` `=((xyz))/((xyz))* \|[A,B, C],[x, y, z],[yz,zx ,xy]\| ["applying" R_(3) to (xyz)R_(3)" and dividing the whole determinant by xyz"]` `=\|[A,B, C],[x, y, z],[zy,zx ,xy]\| =Delta_(1)` Hence, `Delta = Delta_(1)`

Discussion

412.	The sum of the products of the elements of any row of a matrix A with the corresponding cofactors of the elements of the same row is always equal toA. `\|A\|`B. `(1)/(2) \|A\|`C. 1D. 0
Answer» Correct Answer - A

Discussion

413.	Using properties of determinants prove that:\(\begin{vmatrix}1& b+c & b^2+c^2 \\[0.3em]1& c+a &c^2+a^2\\[0.3em]1 & a+b& a^2+b^2\end{vmatrix}\) = (a-b)(b-c)(c-a)\|(1,b + c,b^2 + c^2)(1,c + a,c^2 + a^2)(1, a + b, a^2 + b^2)\| = (a-b)(b-c)(c-a)
Answer» \(\begin{vmatrix} 1& b+c & b^2+c^2 \\[0.3em] 1& c+a &c^2+a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) = \(\begin{vmatrix} 0& b-c & b^2-c^2 \\[0.3em] 0& c-a &c^2-a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃] = \(\begin{vmatrix} 0& b-c & (b-a)(b+a) \\[0.3em] 0& c-a &(c-b)(c+b)\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) = (b - a)(c - b)\(\begin{vmatrix} 0& 1 & b+a \\[0.3em] 0& 1 &c+b\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) [R₁’ = R₁/(b - a) & R₂’ = R₂/(c - b)] = (b - a)(c - b)[0 + 0 + 1{(c + b) - (b + a)}][expansion by first column] = (a - b)(b - c)(c - a)

413.

Using properties of determinants prove that:\(\begin{vmatrix}1& b+c & b^2+c^2 \\[0.3em]1& c+a &c^2+a^2\\[0.3em]1 & a+b& a^2+b^2\end{vmatrix}\) = (a-b)(b-c)(c-a)|(1,b + c,b^2 + c^2)(1,c + a,c^2 + a^2)(1, a + b, a^2 + b^2)| = (a-b)(b-c)(c-a)

Answer»

\(\begin{vmatrix} 1& b+c & b^2+c^2 \\[0.3em] 1& c+a &c^2+a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)

= \(\begin{vmatrix} 0& b-c & b^2-c^2 \\[0.3em] 0& c-a &c^2-a^2\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)[R₁’ = R₁ - R₂ & R₂’ = R₂ - R₃]

= \(\begin{vmatrix} 0& b-c & (b-a)(b+a) \\[0.3em] 0& c-a &(c-b)(c+b)\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\)

= (b - a)(c - b)\(\begin{vmatrix} 0& 1 & b+a \\[0.3em] 0& 1 &c+b\\[0.3em] 1 & a+b& a^2+b^2 \end{vmatrix}\) [R₁’ = R₁/(b - a) & R₂’ = R₂/(c - b)]

= (b - a)(c - b)[0 + 0 + 1{(c + b) - (b + a)}][expansion by first column]

= (a - b)(b - c)(c - a)

Discussion

414.	Using Cofactors of elements of third column, evaluate `Delta=\|1x y z1y z x1z x y\|`
Answer» `Delta=\|{:(1,x,yz),(1,y,zx),(1,z,xy):}\|` Cofactors of the elements of third row `A_(13)=(-1)^(1+3)\|{:(1,y),(1,z):}\|=z-y` `A_(23)=(-1)^(2+3)\|{:(1,x),(1,z):}\|=(z-x)=x-z` `A_(33)=(-1)^(3+3)\|{:(1,x),(1,y):}\|=y+x` `therefore" "Delta=a_(13)A_(13)+a_(23)A_(23)+a_(33)A_(33)` `=yz(z-y)+zx(x-z)+xy(y-x)` `=yz^(2)-y^(2)z+zx^(2)-z^(2)x+xy^(2)-x^(2)y`

Discussion

415.	Using Cofactors of elements of second row, evaluate `Delta=\|5 3 8 2 0 1 1 2 3\|`
Answer» `Delta=\|{:(5,3,8),(2,0,1),(1,2,3):}\|.` Cofactors of the elements of second row `A_(21)=(-1)^(2+1)\|{:(3,8),(2,3):}\|=-(9-16)=7` `A_(22)=(-1)^(2+2)\|{:(5,8),(1,3):}\|=(15-8)=7` `A_(23)=(-1)^(2+3)\|{:(5,3),(1,2):}\|=(10-3)=7` `therefore" "Delta=a_(21)A_(21)+a_(22)A_(22)+a_(23)A_(23)` `=2xx7+0xx7+xx(-7)` =14+0-7=7

Discussion

416.	if `A_(1) ,B_(1),C_(1) …….` are respectively the cofactors of the elements `a_(1) ,b_(1),c_(1)……` of the determinant `Delta = \|{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}\|, Delta ne 0 ` then the value of `\|{:(B_(2),,C_(2)),(B_(3),,C_(3)):}\|` is equal toA. `a_(1)^(2)Delta`B. `a_(1)Delta`C. `a_(1)Delta^(2)`D. `a_(1)^(2)Delta^(2)`
Answer» Correct Answer - B `B_(2) =a_(1)c_(3) -a_(3)c_(1),C_(2)=-(a_(1)b_(3)-a_(3)b_(1))` `B_(3) =-(a_(1)C_(2) -a_(2)c_(1)) ,C_(3)=a_(1)b_(2)-a_(2)b_(1)` ` :. \|{:(B_(2),,C_(2)),(B_(3),,C_(2)):}\|= \|{:(a_(1)C_(3)-a_(3)c_(1),,-a_(1)b_(3)+a_(3)b_(1)),(-a_(1)c_(2)+a_(2)c_(1),,a_(1)b_(1)-a_(2)b_(1)):}\|` `=\|{:(a_(1)c_(3),,-a_(1)b_(3)),(-a_(1)c_(2),,a_(1)b_(2)):}\|+ \|{:(a_(1)C_(3),,a_(3)b_(1)),(-a_(1)c_(2),,-a_(2)b_(1)):}\|` `+\|{:(-a_(3)C_(1),,-a_(1)b_(3)),(-a_(1)C_(2),,a_(1)b_(2)):}\|+ \|{:(-a_(3)C_(1),,a_(3)b_(1)),(a_(2)c_(1),,-a_(2)b_(1)):}\|` ` =a_(1)^(2) \|{:(C_(3),,-b_(3)),(-c_(2),,b_(2)):}\|+a_(1)b_(1) \|{:(c_(3),,a_(3)),(-c_(2),,-a_(2)):}\|` `+a_(1)c_(1) \|{:(-a_(3),,-b_(3)),(a_(2),,b_(2)):}\|+b_(1)c_(1) \|{:(-a_(3),,a_(3)),(a_(2),,-a_(2)):}\|` `=a_(1){a_(1)(b_(2)c_(3)-b_(3)c_(2))-b_(1)(a_(2)c_(3)-a_(3)c_(2))` `+c_(1)(a_(2)b_(3)-a_(3)b_(2))}` ` =a_(1)\|{:(a_(1),,b_(1),,c_(1)),(a_(2),,b_(2),,c_(2)),(a_(3),,b_(3),,c_(3)):}\| =a_(1) Delta`

Discussion

417.	Without expanding evaluate the determinant `\|(sinalpha,cosalpha,sin(alpha+delta)),(sinbeta,cosbeta,sin(beta+delta)),(singamma,cosgamma,sin(gamma+delta))\|`
Answer» Correct Answer - A `C_(3) to C_(3) + ("sin"delta)C_(1) -("cos" delta)C_(2)` makes all zeros in `C_(3)` and hence `Delta = 0`

Discussion

418.	Using properties of determinants. Prove that `\|(sinalpha,cosalpha,cos(alpha+delta)),(sinbeta,cosbeta,cos(beta+delta)),(singamma,cosgamma,cos(gamma+delta))\|=0`
Answer» `L.H.S. = \|[sin alpha, cos alpha, cos(alpha+delta)],[sin beta, cos beta, cos(beta+delta)],[sin gamma, cos gamma, cos(gamma+delta)]\|` `= \|[sin alpha, cos alpha, cosalphacosdelta - sinalphasindelta],[sin beta, cos beta, cosbetacosdelta - sinbetasindelta],[sin gamma, cos gamma, cosgammacosdelta - singammasindelta]\|` Applying `C_3 ->C_3-cosdeltaC_2+sindeltaC_1` `= \|[sin alpha, cos alpha, 0],[sin beta, cos beta, 0],[sin gamma, cos gamma, 0]\|` If in a determinant, all elements in a row or column are `0`, then value of that determinant is `0`. Here, `C_3` is `0`. `:. \|[sin alpha, cos alpha, 0],[sin beta, cos beta, 0],[sin gamma, cos gamma, 0]\| = 0 = R.H.S.`

Discussion

419.	Using Cofactors of elements of second row, evaluate `Delta=\|(5, 3, 8),( 2, 0, 1),( 1, 2 ,3)\|`
Answer» Here, `A = \|[5,3,8],[2,0,1],[1,2,3]\|` First we will find minor of second row. `M_21 = 9-16 = -7` `M_22 = 15-8 = 7` `M_23 = 10-3 = 7` Now, we will find the cofactors using these minors. `A_21 = (-1)^(2+1)M_21 = -1(-7) = 7` `A_22 = (-1)^(2+2)M_22 = 1(7) = 7` `A_23 = (-1)^(2+3)M_23 = -1(7) = -7` Now, `a_21 = 2` `a_22 = 0` `a_23 =1` `:. Delta = a_21A_21+a_22A_22+a_23A_23` `=>Delta = 2(7)+0(7)+1(-7) = 14+0-7 = 7` `:. Delta = 7.`

Discussion

420.	Using properties of determinants. Prove that`\|[1 ,1+p,1+p+q],[2, 3+2p,4+3p+2q],[3, 6+3p, 10+6p+3q]\|=1`
Answer» `L.H.S. = \|[1,1+p,1+p+q],[2,3+2p,4+3p+2q],[3,6+3p,10+6p+3q]\|` Applying `R_2->R_2-2R_1` and `R_3->R_3-3R_1` `=\|[1,1+p,1+p+q],[0,1,2+p],[0,3,7+3p]\|` Applying `R_3->R_3-3R_2` `= \|[1,1+p,1+p+q],[0,1,2+p],[0,0,1]\|` `=[1(1-0)-0+0]` `=1 = R.H.S.`

Discussion

421.	Evaluate the following determinants:(i) \(\begin{vmatrix} x& -7 \\[0.3em] x & 5x + 1 \end{vmatrix}\)(ii) \(\begin{vmatrix} cos θ& -sin θ\\[0.3em]sin θ& cos θ\end{vmatrix}\)(iii) \(\begin{vmatrix} cos 15°& sin 15°\\[0.3em]sin 75°& cos 75°\end{vmatrix}\)(iv) \(\begin{vmatrix} a + ib & c + id\\[0.3em]-c + id& a - ib\end{vmatrix}\)
Answer» (i) \|A\| = x (5x + 1) – (–7) x \|A\| = 5x² + 8x (ii) \|A\| = cos θ × cos θ – (–sin θ) x sin θ \|A\| = cos²θ + sin²θ As we know that cos²θ + sin²θ = 1 \|A\| = 1 (iii) \|A\| = cos15° × cos75° + sin15° x sin75° As we know that cos (A – B) = cos A cos B + Sin A sin B On substituting this we get, \|A\| = cos (75 – 15)° \|A\| = cos60° \|A\| = 0.5 (iv) \|A\| = (a + ib) (a – ib) – (c + id) (–c + id) = (a + ib) (a – ib) + (c + id) (c – id) = a² – i² b² + c² – i² d² As we know that i² = -1 = a² – (–1) b² + c² – (–1) d² = a² + b² + c² + d²

421.

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

Answer»

(i) |A| = x (5x + 1) – (–7) x

|A| = 5x² + 8x

(ii) |A| = cos θ × cos θ – (–sin θ) x sin θ

|A| = cos²θ + sin²θ

As we know that cos²θ + sin²θ = 1

|A| = 1

(iii) |A| = cos15° × cos75° + sin15° x sin75°

As we know that cos (A – B) = cos A cos B + Sin A sin B

On substituting this we get, |A| = cos (75 – 15)°

|A| = cos60°

|A| = 0.5

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

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

= a² – i² b² + c² – i² d²

As we know that i² = -1

= a² – (–1) b² + c² – (–1) d²

= a² + b² + c² + d²

Discussion

422.	Evaluate the following :\(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\)
Answer» Let Δ = \(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\) = 0(0 – y³z³) – xy²(0 – x²yz³) + xz²(x²y³z – 0) = 0 + x³y³z³ + x³y³z³ = 2x³y³z³ So, Δ = 2x³y³z³

422.

Evaluate the following :\(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\)

Answer»

Let Δ = \(\begin{bmatrix}0&xy^2& xz^2 \\[0.3em]x^2y & 0 &yz^2 \\[0.3em]x^2y & zy^2& 0\end{bmatrix}\)

= 0(0 – y³z³) – xy²(0 – x²yz³) + xz²(x²y³z – 0)

= 0 + x³y³z³ + x³y³z³

= 2x³y³z³

So,

Δ = 2x³y³z³

Discussion

423.	Using co-factors of elements of 3rd columnΔ = \(\begin{vmatrix}1 & x & yz \\[0.3em]1 &y &zx \\[0.3em]1 &z &xy\end{vmatrix}\)
Answer» M₁₃ = (z – y) M₂₃ = (z-x) M₃₃ = (y - x) C₁₃ = (z - y) C₂₃ = -(z - n), C₃₃ = (y – x) Δ = yz (z – y) + zx{-(z – x)} + xy (y – x) = yz² – y²z – z²x + zx² + xy² – x²y x² (z – y) + z (y² – z²) + yz (z – y) x²(z – y) + x (y+ z) (y- z) + yz (z – y) (z – y) {x² – x (y + z) + yz} (z – y) {x² – xy – xz + yz} (z – y) {x (x – y) – z (x – y)} (z – y) (x – y) (x – z) = (z – y) (y – x) (z – x)

423.

Using co-factors of elements of 3rd columnΔ = \(\begin{vmatrix}1 & x & yz \\[0.3em]1 &y &zx \\[0.3em]1 &z &xy\end{vmatrix}\)

Answer»

M₁₃ = (z – y) M₂₃ = (z-x) M₃₃ = (y - x)

C₁₃ = (z - y) C₂₃ = -(z - n), C₃₃ = (y – x)

Δ = yz (z – y) + zx{-(z – x)} + xy (y – x)

= yz² – y²z – z²x + zx² + xy² – x²y

x² (z – y) + z (y² – z²) + yz (z – y)

x²(z – y) + x (y+ z) (y- z) + yz (z – y)

(z – y) {x² – x (y + z) + yz}

(z – y) {x² – xy – xz + yz}

(z – y) {x (x – y) – z (x – y)}

(z – y) (x – y) (x – z) = (z – y) (y – x) (z – x)

Discussion

424.	Evaluate:(i) \(\begin{vmatrix}2&3&-2\\[0.3em]3&4&0\\[0.3em]1&-1&5\end{vmatrix}\)(ii) \(\begin{vmatrix}3&-1&2\\[0.3em]1&3&-2\\[0.3em]1&4&3\end{vmatrix}\)(iii) \(\begin{vmatrix}2&0&0\\[0.3em]0&4&0\\[0.3em]0&0&5\end{vmatrix}\)(iv) \(\begin{vmatrix}2&-1&1\\[0.3em]3&1&-2\\[0.3em]1&4&-1\end{vmatrix}\)(v) \(\begin{vmatrix}1&2&-1\\[0.3em]3&1&-4\\[0.3em]1&0&-2\end{vmatrix}\)
Answer» (i) \(\begin{vmatrix}2&3&-2\\[0.3em]3&4&0\\[0.3em]1&-1&5\end{vmatrix}\) 2(20 + 0) -3 (15 – 0) -2(-3 -4) 2(20) -3(15) -2(-7) 40 – 45 + 14 = 9 (ii) \(\begin{vmatrix}3&-1&2\\[0.3em]1&3&-2\\[0.3em]1&4&3\end{vmatrix}\) = 3(9+8) + 1(3 + 2) +2(4 – 3) = 3(17) + 1(5) + 2(1) = 51 + 5 + 2 = 58 (iii) \(\begin{vmatrix}2&0&0\\[0.3em]0&4&0\\[0.3em]0&0&5\end{vmatrix}\) 2(20 – 0) – 0 + 0 = 40 (iv) \(\begin{vmatrix}2&-1&1\\[0.3em]3&1&-2\\[0.3em]1&4&-1\end{vmatrix}\) 2(-1 + 8) + 1(-3 + 2) +1(12 – 1) 2(7) + 1(-1) +11 =14 – 1 + 11 = 24 (v) \(\begin{vmatrix}1&2&-1\\[0.3em]3&1&-4\\[0.3em]1&0&-2\end{vmatrix}\) = 1(-2 + 0) -2(-6 + 4) -1(0 – 1) = -2 – 2(-2) + 1 = -2 + 4 + 1 = 3.

424.

Evaluate:(i) \(\begin{vmatrix}2&3&-2\\[0.3em]3&4&0\\[0.3em]1&-1&5\end{vmatrix}\)(ii) \(\begin{vmatrix}3&-1&2\\[0.3em]1&3&-2\\[0.3em]1&4&3\end{vmatrix}\)(iii) \(\begin{vmatrix}2&0&0\\[0.3em]0&4&0\\[0.3em]0&0&5\end{vmatrix}\)(iv) \(\begin{vmatrix}2&-1&1\\[0.3em]3&1&-2\\[0.3em]1&4&-1\end{vmatrix}\)(v) \(\begin{vmatrix}1&2&-1\\[0.3em]3&1&-4\\[0.3em]1&0&-2\end{vmatrix}\)

Answer»

(i) \(\begin{vmatrix}2&3&-2\\[0.3em]3&4&0\\[0.3em]1&-1&5\end{vmatrix}\)

2(20 + 0) -3 (15 – 0) -2(-3 -4)

2(20) -3(15) -2(-7)

40 – 45 + 14 = 9

(ii) \(\begin{vmatrix}3&-1&2\\[0.3em]1&3&-2\\[0.3em]1&4&3\end{vmatrix}\)

= 3(9+8) + 1(3 + 2) +2(4 – 3)

= 3(17) + 1(5) + 2(1)

= 51 + 5 + 2 = 58

(iii) \(\begin{vmatrix}2&0&0\\[0.3em]0&4&0\\[0.3em]0&0&5\end{vmatrix}\)

2(20 – 0) – 0 + 0 = 40

(iv) \(\begin{vmatrix}2&-1&1\\[0.3em]3&1&-2\\[0.3em]1&4&-1\end{vmatrix}\)

2(-1 + 8) + 1(-3 + 2) +1(12 – 1)

2(7) + 1(-1) +11 =14 – 1 + 11 = 24

(v) \(\begin{vmatrix}1&2&-1\\[0.3em]3&1&-4\\[0.3em]1&0&-2\end{vmatrix}\)

= 1(-2 + 0) -2(-6 + 4) -1(0 – 1)

= -2 – 2(-2) + 1 = -2 + 4 + 1 = 3.

Discussion

425.	Choose the correct answer from the following : If A is a matrix of order `3xx3" and "\|A\|=6`, then \|adj.A\|A. 36B. 216C. 729D. None of these
Answer» Correct Answer - A

Discussion

426.	Let `omega`be a complex number such that `2omega+1=z`where `z=sqrt(-3.)``If\|1 1 1 1-omega^2-1omega^2 1omega^2omega^7\|=3k ,`then`k`is equal to :`-1`(2) `1`(3) `-z`(4) `z`A. 1B. `-z`C. zD. -1
Answer» Correct Answer - 2 Here `omega ` is complex cube root of unity Applying `R_(1) to R_(1)+R_(2)+R_(3),` the given matrix reduces to ` \|{:(3,,0,,0),(1,,-omega^(2)-1,,omega^(2)),(1,,omega^(2),,omega):}\|= 3(-1-omega -omega) = -3z` `rArr k=-z`

Discussion

427.	If the system of linear equationsx+ky+3z=03x+ky-2z=02x+4y-3z=0has a non-zero solution (x,y,z) then `(xz)/(y^2)` is equal toA. 30B. -10C. 10D. -30
Answer» Correct Answer - 3 Given system of equations `x+ky+3z=0` `3x+ky -2x =0` `2x+4y -3z =0` Eliminating from first two equations we get `2x-5z=0 " or " 2x=5z` ` :. X=5lambda , z= 2lambda` from equation (iii) we have `10 lambda +4y-6lambda= 0 " or " y=-lambda` `:. (xz)/(y^(2)) =((5lambda)(2lambda))/((-lambda)^(2)) =10`

Discussion

428.	The number of values of k for which the linear equations`4x""+""k y""+""2z""=""0``k x""+""4y""+""z""=""0``2x""+""2y""+""z""=""0`posses a non-zero solution is :(1) 3 (2) 2 (3) 1 (4) zeroA. zeroB. 3C. 2D. 1
Answer» Correct Answer - 3 For non-trivial solution of the given system of linear equations `\|{:(4,,k,,2),(k,,4,,1),(2,,2,,1):}\|=0` `rArr 8+ k (2-k) + 2 (2k -8) =0` `rArr -k^(2) +6k -8 =0` `rArr k^(2) -6k +8=0` `rArr k=2,4` Clearly there exists two values of k .

Discussion

429.	For a real number `alpha,`if the system `[1alphaalpha^2alpha1alphaalpha^2alpha1][x y z]=[1-1 1]`of linear equations, has infinitely many solutions, then `1+alpha+alpha^2=`
Answer» Correct Answer - 1 Since the system of equations hasss infinitely many solutions `\|{:(1,,alpha,,alpha^(2)),(alpha,,1,,alpha),(alpha^(2),,alpha,,1):}\|=0` `rArr 1(1-alpha^(2)) -alpha(alpha-alpha^(3)) +alpha^(2)(alpha^(2)-alpha^(2))=0` `rArr (1-alpha^(2)) -alpha^(2) +alpha^(4)=0` `rArr (alpha^(2) -1)^(2)=0` `rArr alpha= +-1` For `alpha=1` we get `x+y+z=1 " and " x+y+z=-1` Hence system has no solution . Fora `alpha=-1` all three equations become x-y+z=1 which represents coincident planes. `:. 1+alpha+alpha^(2) =1`

Discussion

430.	If the determinant \(\begin{vmatrix} a & b & 2a\,\alpha+3b \\[0.3em] b& c &2b\,\alpha+3c \\[0.3em] 2a\,\alpha+3b & 2b\,\alpha+3c & 0 \end{vmatrix}\) , thenA. a, b, c are in H.P. B. α is a root of 4ax2 + 12bx + 9c = 0 or, a, b, c are in G.P. C. a, b, c are in G.P. only D. a, b, c are in A.P.
Answer» B. α is a root of 4ax² + 12bx + 9c = 0 or, a, b, c are in G.P. Expend the determinats : a[-(2bα+3c)² ]-b[-(2bα+3c)(2aα+3b)]+ (2aα+3b)[b(2bα+3c)-c(2aα+3b)] = 0 -a(2bα+3c)² + b(2bα+3c)(2aα+3b)+(2aα+3b)[2b²α+3bc-3bc-2acα] = 0 (2bα+3c) [-2abα-3ac+2abα+3b²]+ (2aα+3b)(2α)( b² -ac) = 0 (2bα+3c) [-3ac +3b² ]+ (2aα+3b)(2α)( b²- ac) = 0 (b² – ac)[4aα² + 12bα + ac] = 0 CASE 1→ (b² - ac) = 0 b² = ac {abc are in Gp} CASE 2→ (4aα² +12bα + ac)=0 {Whose one root is α}

430.

If the determinant \(\begin{vmatrix} a & b & 2a\,\alpha+3b \\[0.3em] b& c &2b\,\alpha+3c \\[0.3em] 2a\,\alpha+3b & 2b\,\alpha+3c & 0 \end{vmatrix}\) , thenA. a, b, c are in H.P. B. α is a root of 4ax2 + 12bx + 9c = 0 or, a, b, c are in G.P. C. a, b, c are in G.P. only D. a, b, c are in A.P.

Answer»

B. α is a root of 4ax² + 12bx + 9c = 0 or, a, b, c are in G.P.

Expend the determinats :

a[-(2bα+3c)² ]-b[-(2bα+3c)(2aα+3b)]+ (2aα+3b)[b(2bα+3c)-c(2aα+3b)] = 0

-a(2bα+3c)² + b(2bα+3c)(2aα+3b)+(2aα+3b)[2b²α+3bc-3bc-2acα] = 0

(2bα+3c) [-2abα-3ac+2abα+3b²]+ (2aα+3b)(2α)( b² -ac) = 0

(2bα+3c) [-3ac +3b² ]+ (2aα+3b)(2α)( b²- ac) = 0

(b² – ac)[4aα² + 12bα + ac] = 0

CASE 1→

(b² - ac) = 0

b² = ac

{abc are in Gp}

CASE 2→

(4aα² +12bα + ac)=0

{Whose one root is α}

Discussion

431.	if `alpha,beta,gamma` are non-real numbers satisfying `x^3-1=0,` then the value of `\|[lambda+1,alpha,beta],[alpha,lambda+beta,1],[beta,1,lambda+1]\|`
Answer» Correct Answer - B

Discussion

432.	If the lines `a_1x+b_1y+1=0, a_2x+b_2y+1=0 a n d a_3x+b_3y+1=0`are concurrent, show that the points `(a_1, b_1), (a_2, b_2)a n d (a_3, b_3)`are collinear.
Answer» The given lines are `a_ (1)x+b_(1)y+1=0` `a_(2)x+b_(2)y +1=0` ` a_(3) x+b_(3)y+1=0` if these lines are concurrent we must have3 `\|underset(a_(3)" "b_(3)" "1)underset(a_(2)" "b_(2)" "1)(a_(1)" "b_(1)" "1)\|=0` Which is the condition of colinearity of three points `(a_(1) b_(1)), (a_(2),b_(2)), " and " (a_(3),b_(3))` Hence if the given lies are concurrent the given points are collinear.

Discussion

433.	Let D = \(\begin{vmatrix}sinθ. cosϕ&sinθ. sinϕ&cosθ\\cosθ.cosϕ&cosθ.sinθ& -sinθ\\-sinθ.sinϕ&sinθ.cosϕ&0\end{vmatrix}\)thenD=[(sin θ ⋅ cos ϕ, sin θ ⋅ sin ϕ, cos θ) (cos θ ⋅ cos ϕ, cos θ ⋅ sin ϕ, −sin θ) (−sin θ ⋅ sin ϕ, sin θ ⋅ cos ϕ, 0)](a) D is independent of θ (b) D is independent of φ (c) D is a constant(d) dD/d at θ = π/2 is equal to 0
Answer» Correct option is (b) D is independent of φ

Discussion

434.	Evaluate: `(i) \|{:(6, -3), (7, -2):}\| " " (ii) \|{:(x^(2)-x +1, x-1),(x+1, x+1):}\|`
Answer» `(i) \|{:(6, -3), (7, -2):}\| = (-2)-7(-3) = -12 + 21 =9`. `(ii) \|{:(x^(2)-x+1, x-1), (x+1, x+1):}\| = (x^(2) -x +1)(x+1)-(x+1)(x-1)` `= (x^(3)+1)-(x^(2)-1) = (x^(3) -x^(2) +2)`.

Discussion

435.	Prove that`\|[b+c,a,b],[c+a,c,a],[a+b,b,c]\|=(a+b+c)(a-c)^2`A. `(a+b+c)(a-c)`B. `(a+b+c)(b-c)`C. `(a+b+c)(a-c)^(2)`D. `(a+b+c)(b-c)^(2)`
Answer» Correct Answer - C `Delta = \|[b, a, b], [c, c, a], [a, b, c]\| + \|[c, a, b], [a, c, a], [b, b, c]\|` Apply `R_(1) to R_(1) + R_(2) + R_(3)` in each and simplify

Discussion

436.	`\|[a, a+2b, a+2b+3c], [3a, 4a+6b, 5a+7b+9c], [6a, 9a+12b, 11a+15b+18c]\|=?`A. `a^(3)`B. `-a^(3)`C. 0D. none of these
Answer» Correct Answer - B Apply `R_(2) to R_(2) - 3R_(1) "and"R_(3) to R_(3)-6R_(1)`

Discussion

437.	`\|[a, a+2b, a+2b+3c], [3a, 4a+6b, 5a+7b+9c], [6a, 9a+12b, 11a+15b+18c]\|=?`
Answer» Apply `R_(2) to R_(2) -3R_(1) " and "R_(3) to R_(3) - 6R_(1)`

Discussion

438.	Using the property of determinants and without expanding, prove that:`\|1b c a(b+c)1c a b(c+a)1a b x(a+b)\|=0`
Answer» `L.H.S.=\|{:(1,bc,a(b+c)),(1,ca,b(c+a)),(1,ab,c(a+b)):}\|=\|{:(1,bc,ab+ca+bc),(1,ca,ba+ab+ca),(1,ab,ca+be+ab):}\|` `(C_(3)toC_(3)+C_(2))` `=(ab+bc+ca)\|{:(1,bc,1),(1,ca,1),(1,ab,1):}\|` `=(ab+bc+ca)xx0" "(because C_(1)" and "C_(3)" are same)` = 0 = R.H.S

Discussion

439.	If `f(x)`is a polynomial of degree `
Answer» we have to prove that `\|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}\| -: \|{:(1,,a,,a^(2)),( 1,,b,,b^(2)),(1,,c,,c^(2)):}\| =(f(x))/((x-a)(x-b)(x-c))` `L.H.S.=\|{:(1,,a,,(f(a))/((x-a))),( 1,,b,,(f(b))/((x-b))),(1,,c,,(f(c))/((x-c))):}\| -: [(a-b)(b-c)(c-a)]` Expanding along `C_(3)` we get `L.H.S. =(1)/((a-b)(b-c)(c-a))xx` `[(f(a)(c-b))/((x-a))+(f(b)(a-c))/((x-b))+(f(c)(b-a))/((x-c))]` Now using partial fraction method on R.H.S. we get `R.H.S. (f(x))/((x-a)(x-b)(x-c))=(A)/(x-a)+(B)/(x-b)+(C)/(x-c)` ltbr. `"(As degree of " f(x) lt 3)` `" then " A= [(f(x))/((x-b)(x-c))]_(x=a)` `=(f(a))/((b-a)(a-c))` `" Similarly "=(1)/((b-a)(b-c)) " and " C = (f(c))/((c-a)(c-b))` `:. R.H.S. =(1)/((a-b)(b-c)(c-a))xx` `[((c-b)f(a))/((x-b))+((a-c)f(b))/((x-b))+((b-a)f(c))/((x-c))]` Hence L.H.S. =R.H.S.`

Discussion

440.	By using properties of determinants. Show that:(i) `\|1a a^2 1bb^2 1cc^2\|=(a-b)(b-c)(c-a)`(ii) `\|1 1 1a b c a^3b^3c^3\|=(a-b)(b-c)(c-a)(a+b+c)`
Answer» `\|{:(1,a,a^(2)),(1,b,b^(2)),(1,c,c^(2)):}\|=\|{:(0,a-c,a^(2)-c^(2)),(0,b-c,b^(2)-c^(2)),(1,c,c^(2)):}\|" "({:(R_(1)toR_(1)-R_(3)),(R_(2)toR_(2)-R_(3)):})` `=(a-c)(b-c)\|{:(0,a-c,a^(2)-c^(2)),(0,b-c,b^(2)-c^(2)),(1,c,c^(2)):}\|` `=(a-c)(b-c)\|{:(0,a-c,a^(2)-c^(2)),(0,b-c,b^(2)-c^(2)),(1,c,c^(2)):}\|` (Expending along `C_(1)`) ` =-(c-a)(b-c)(b+c-a-c) ` `=-(b-c)(c-a)(b-a)` `=(a-b)(b-c)(c-a)=R.H.S.` (ii) `\|{:(1,1,1),(a,b,c),(a^(3),b^(3),c^(3)):}\|=\|{:(0,0,1),(a-b,b-c,c),(a^(3)-b^(3),b^(3)-c^(3),c^(3)):}\|" "({:(C_(1)toC_(1)-C_(2)),(C_(2)toC_(2)-C_(3)):})` `=(a-b)(b-c)\|{:(0,0,1),(1,1,c),(a^(2)+b^(2)+ab,b^(2)+c^(2)+cb,c^(3)):}\|` `=(a-b)(b-c).1\|{:(1,1),(a^(2)+b^(2)+ab,b^(2)+c^(2)+bc):}\|" "("Expenbding along "{R_(1))` `=(a-b)(b-c)(b^(2)+c^(2)+bc-a^(2)-b^(2)-ab)` `=(a-b)(b-c)[c^(2)-a^(2)+bc-ab]` `=(a-b)(b-c)(b-c)[(c-a)(c+a)+b(c-a)]` `=(a-b)(b-c)(c-a)(c+a+b)` `=(a-b)(b-c)(c-a)(a+b+c)` =R.H.S

Discussion

441.	Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\) Applying, C₂→C₂ + C₁ and C₃→C₃ + C₁ ⇒ Δ = \(\begin{vmatrix}a & b & c \\[0.3em]2a+2x& 2b+2y & 2c+2z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\) Taking 2 common from R₂ we get, ⇒ Δ = 2\(\begin{vmatrix}a & b & c \\[0.3em]a+x& b+y & c+z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\) As, R₂ = R₃, hence value of determinant is zero.

441.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}a & b & c \\[0.3em]a+2x& b+2y & c+2z \\[0.3em]x & y &z\end{vmatrix}\)

Applying, C₂→C₂ + C₁ and C₃→C₃ + C₁

⇒ Δ = \(\begin{vmatrix}a & b & c \\[0.3em]2a+2x& 2b+2y & 2c+2z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\)

Taking 2 common from R₂ we get,

⇒ Δ = 2\(\begin{vmatrix}a & b & c \\[0.3em]a+x& b+y & c+z \\[0.3em]a+x & b+y &c+z\end{vmatrix}\)

As,

R₂ = R₃, hence value of determinant is zero.

Discussion

442.	Using properties ofdeterminants, solve the following for x:`\|x-2 2x-3 3x-4x-4 2x-9 3x-16 x-8 2x-27 3x-64\|=0`
Answer» Let the given determinant be `Delta`. Then, `Delta = \|[x-2,2x-3, 3x-4],[x-4, 2x-9, 3x-16],[x-8,2x-27 ,3x-64]\|` `= \|[x-2,1, 2],[x-4, -1, -4],[x-8,-11 ,-40]\| [C_(1) to (C_(2) -2C_(1)), C_(3) to (C_(3) - 3C_(1))]` `= \|[x-2,1, 2],[-2, -2, -6],[-6,-12 ,-42]\| [R_(2) to (R_(2) -R_(1)), R_(3) to (R_(3) - R_(1))]` `=(-2) * (-6)* \|[x-2,1, 2],[1, 1, 3],[1, 2, 7]\|` `=12* \|[x-3,1, 2],[0, 1, 3],[-1, 2, 7]\| [C_(1) to (C_(1)-C_(2))]` `=12 * [(x-3) (7-6) -1 * (3-2)]` `=12 * (x-4)` `therefore Delta = 0 hArr 12(x-4) = 0 hArr x = 4` Hence, solution set = {4}

Discussion

443.	Using properties ofdeterminants, solve the following for `x:``\|(x-2, 2x-3, 3x-4),(x-4, 2x-9, 3x-16),( x-8, 2x-27, 3x-64)\|=0`A. {4}B. {2, 4}C. {2, 8}D. {4, 8}
Answer» Correct Answer - A Apply `C_(2) to (C_(2) - 2C_(1)) "and" C_(3) to (C_(3)-3C_(1))` Now, apply `R_(2) to (R_(2)-R_(1)) "and" R_(3) to (R_(3)-R_(1))`

Discussion

444.	Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]16 &4 & 3\end{vmatrix}\)
Answer» Let Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]16 &4 & 3\end{vmatrix}\) Applying R₃→ R₃ – R₂, we get ⇒ Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]4 &1 & -2\end{vmatrix}\) Applying R₂ → R₂ – R₁, we get ⇒ Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]4 & 1 & -2 \\[0.3em]4 &1 & -2\end{vmatrix}\) As, R₂ = R₃, Therefore the value of the determinant is zero.

444.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]16 &4 & 3\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]16 &4 & 3\end{vmatrix}\)

Applying R₃→ R₃ – R₂, we get

⇒ Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]4 &1 & -2\end{vmatrix}\)

Applying R₂ → R₂ – R₁, we get

⇒ Δ = \(\begin{vmatrix}8 &2 & 7 \\[0.3em]4 & 1 & -2 \\[0.3em]4 &1 & -2\end{vmatrix}\)

As,

R₂ = R₃,

Therefore the value of the determinant is zero.

Discussion

445.	Show that the points (1, 0), (6, 0), (0, 0) are collinear.
Answer» Since \(\begin{vmatrix} 1&0 & 1 \\[0.3em] 6& 0 & 1 \\[0.3em] 0 &0&1 \end{vmatrix}=0\) Hence (1,0),(6,0) and (0,0) are collinear.

445.

Show that the points (1, 0), (6, 0), (0, 0) are collinear.

Answer»

Since \(\begin{vmatrix} 1&0 & 1 \\[0.3em] 6& 0 & 1 \\[0.3em] 0 &0&1 \end{vmatrix}=0\)

Hence (1,0),(6,0) and (0,0) are collinear.

Discussion

446.	Write the minor and cofactor of each element of the following determinants and also evaluate the determinant in each case: ` \|[1,3,-2],[4,-5,6],[3,5,2]\| `
Answer» The minors of the elements of `Delta` are given by `M_(11) = \|{:(-1, 2), (5, 2):}\| =-12, M_(12) = \|{:(4, 2), (3, 2):}\| = 2, M_(13) = \|{:(4, -1), (3, 5):}\| =23,` `M_(21) = \|{:(-3, 2), (5, 2):}\| =-16, M_(22) = \|{:(1, 2), (3, 2):}\| = -4, M_(23) = \|{:(1, -3), (3, 5):}\| =14,` `M_(31) = \|{:(-3, 2), (-1, 2):}\| =-4, M_(32) = \|{:(1, 2), (4, 2):}\| = -6, M_(33) = \|{:(1, -3), (4, -1):}\| =11` So, the cofactors of the corresponding elements of `Delta` are `C_(11) = (-1)^(1+1) * M_(11) = M_(11) = -12, C_(12) = (-1)^(1+2) * M_(12) = -M_(12) = -2,` `C_(13) = (-1)^(1+3) * M_(13) = M_(13) = 23, C_(21) = (-1)^(2+1) * M_(21) = -M_(21) = 16,` `C_(22) = (-1)^(2+2) * M_(22) = M_(22) = -4, C_(23) = (-1)^(2+3) * M_(23) = -M_(23) = -14,` `C_(31) = (-1)^(3+1) * M_(31) = M_(31) = -4, C_(32) = (-1)^(3+2) * M_(32) = -M_(32) = 6,` `C_(33) = (-1)^(3+3) * M_(33) = M_(33) = 11`.

Discussion

447.	Evaluate : \(\begin{vmatrix}cos\,15° & \ sin\,15° \\[0.3em]sin\,75° & cos\,75° \\[0.3em]\end{vmatrix}\)
Answer» Expanding the determinant, we get cos 15°. cos 75° – sin 15°. sin 75° = cos (15° + 75°) = cos 90° = 0 [Note: cos (A + B) = cos A. cos B – sin A. sin B]

447.

Evaluate : \(\begin{vmatrix}cos\,15° & \ sin\,15° \\[0.3em]sin\,75° & cos\,75° \\[0.3em]\end{vmatrix}\)

Answer»

Expanding the determinant, we get

cos 15°. cos 75° – sin 15°. sin 75°

= cos (15° + 75°)

= cos 90° = 0

[Note: cos (A + B) = cos A. cos B – sin A. sin B]

Discussion

448.	Find the minor of the element of second row and third column (a23) in the following determinant :\(\begin{vmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 &5 &-7\end{vmatrix}\)
Answer» We have, \(\begin{vmatrix} 2 & -3 & 5 \\[0.3em] 6 & 0 & 4 \\[0.3em] 1 &5 &-7 \end{vmatrix}\) Minor of an element a₂₃ = M₂₃ = \(\begin{vmatrix} 2 & -3 \\[0.3em] 1 & 5 \\[0.3em] \end{vmatrix}\) = 10 + 3 = 13

448.

Find the minor of the element of second row and third column (a23) in the following determinant :\(\begin{vmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 &5 &-7\end{vmatrix}\)

Answer»

We have,

\(\begin{vmatrix} 2 & -3 & 5 \\[0.3em] 6 & 0 & 4 \\[0.3em] 1 &5 &-7 \end{vmatrix}\)

Minor of an element a₂₃ = M₂₃

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

= 10 + 3 = 13

Discussion

449.	The vertices of `/_ ` ABC are A(-2, 4), B (2, -6) and C(5, 4). The area of `/_ ` ABC isA. 17.5 sq unitsB. 35 sq unitsC. 32 sq unitsD. 28 sq units
Answer» Correct Answer - B `Delta = (1)/(2)\|[-2, 4, 1], [2, -6, 1], [5, 4, 1]\| = (1)/(2)\|[-2, 4, 1], [4, -10, 0], [7, 0, 0]\| = 35` sq. units

Discussion

450.	Show that if `x_(1),x_(2),x_(3) ne 0` `\|{:(x_(1) +a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}\|` `=x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1))+(a_(2)b_(2))/(x_(2))+(a_(3)b_(3))/(x_(3)))`
Answer» The given determinant can be written as the sum of two determinants `\|{:(x_(1),,a_(1)b_(2),,a_(1)b_(3)),(0,,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(0,,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}\|+\|{:(a_(1)b_(1),,a_(1)b_(2),,a_(1)b_(3)),(a_(2)b_(1),,x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(1),,a_(3)b_(2),,x_(3)+a_(3)b_(3)):}\|` Expanding the first determinant along `C_(1)` we get `x_(1) \|{:(x_(2)+a_(2)b_(2),,a_(2)b_(3)),(a_(3)b_(2),,x_(3)+a_(3)b_(3)):}\|` `=x_(1)[(x_(2)+a_(2)b_(2))(x_(3)+a_(3)b_(3))-a_(3)b_(2)a_(2)b_(3)]` `=x_(1)(x_(2)x_(3)+x_(3)a_(2)b_(2)+x_(2)a_(3)b_(3)+a_(2)b_(2)a_(3)b_(3)-a_(3)b_(2)b_(3)]` `=x_(1)x_(2)x_(3)+x_(1)x_(3)a_(2)b_(2)+x_(1)x_(2)a_(3)b_(3)` In the second determinant taking `b_(1)` common from `C_(1)` and then applying `C_(1) to C_(2) -b_(2)C_(1)" and " C_(3) to -b_(3)dC_(3)` we obtain `b_(1) \|{:(a_(1),,0,,0),(a_(2),,x_(2),,0),(a_(3),,0,,x_(3)):}\|=a_(2)b_(1)x_(2)x_(3)` Therefore the given determinant is `x_(1)x_(2)x_(3)+x_(1)x_(3)a_(2)b_(2)+x_(1)x_(2)a_(3)b_(3)+a_(1)b_(1)x_(2)x_(3)` `=x_(1)x_(2)x_(3) (1+(a_(1)b_(1))/(x_(1)) +(a_(2)b_(2))/(x_(2)) +(a_(3)b_(3))/(x_(3)))`

Discussion

Explore topic-wise InterviewSolutions in .

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Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}8 &2 & 7 \\[0.3em]12 & 3 & 5 \\[0.3em]16 &4 & 3\end{vmatrix}\)

Evaluate : \(\begin{vmatrix}cos\,15° & \ sin\,15° \\[0.3em]sin\,75° & cos\,75° \\[0.3em]\end{vmatrix}\)

Find the minor of the element of second row and third column (a23) in the following determinant :\(\begin{vmatrix}2 & -3 & 5 \\[0.3em]6 & 0 & 4 \\[0.3em]1 &5 &-7\end{vmatrix}\)