643 + Interview Questions in DETERMINANTS IN GENERAL AWARENESS Page 7 InterviewSolution

301.	Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1 & 5 \\[0.3em] 79 & 7 &9\\[0.3em] 29 & 5 & 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if \|A\| = 0. Also, A is called non singular if \|A\| ≠ 0.
Answer» We know that C₁⇒ C₁-C₂, would not change anything for the determinant. Applying the same in above determinant, we get \(\begin{bmatrix} 40 &1 & 5 \\[0.3em] 72 & 7 &9\\[0.3em] 24 & 5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁=8 × C₃ Applying above equation we get, \(\begin{bmatrix} 0 &1 & 5 \\[0.3em] 0 & 7 &9\\[0.3em] 0 & 5 & 3 \end{bmatrix}\) We know that if a row or column of a determinant is 0. Then it is singular determinant.

301.

Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1 & 5 \\[0.3em] 79 & 7 &9\\[0.3em] 29 & 5 & 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if |A| = 0. Also, A is called non singular if |A| ≠ 0.

Answer»

We know that C₁⇒ C₁-C₂, would not change anything for the determinant.

Applying the same in above determinant, we get

\(\begin{bmatrix} 40 &1 & 5 \\[0.3em] 72 & 7 &9\\[0.3em] 24 & 5 & 3 \end{bmatrix}\) Now it can clearly be seen that C₁=8 × C₃

Applying above equation we get,

\(\begin{bmatrix} 0 &1 & 5 \\[0.3em] 0 & 7 &9\\[0.3em] 0 & 5 & 3 \end{bmatrix}\)

We know that if a row or column of a determinant is 0. Then it is singular determinant.

Discussion

302.	Let A be a square matric of order 3 × 3. Write the value of \|2A\|, where \|A\| = 4.
Answer» ∵ \|2A\| = 2ⁿ \|A\|, where n is order of matrix A. Here, \|A\| = 4 and n = 3 \|2A\| = 2³ × 4 = 32

302.

Let A be a square matric of order 3 × 3. Write the value of |2A|, where |A| = 4.

Answer»

∵ |2A| = 2ⁿ |A|, where n is order of matrix A.

Here,

|A| = 4 and n = 3

|2A| = 2³ × 4 = 32

Discussion

303.	Find the area of the triangle whose vertices are: `(-2,-3),(3,2),(-1,-8)`
Answer» Here, `(x_(1) =-2, y_(1) =-3), (x_(2) =3, y_(2) =2) " and " (x_(3) =-1, y_(3) =-8)` `therefore (1)/(2) \|[x_(1), y_(1), 1],[x_(2), y_(2), 1],[x_(3), y_(3), 1]\| =(1)/(2) * \|[-2, -3, 1],[3, 2, 1],[-1, -8, 1]\|` `=(1)/(2) * \|[-2, -3, 1],[5, 5, 0],[1, -5, 0]\| [R_(2) to (R_(2) -R_(1)) " and" R_(3) to (R_(3) -R_(1))]` ` =(1)/(2) 1 \|[5, 5], [1, -5]\| = (1)/(2) xx (-25 -5) = -15` Hence, ar `(triangleABC) =\|-15\| =15` square units.

Discussion

304.	Show that points`A (a , b + c)`, `B (b , c + a)`, `C (c , a + b)`are collinear.
Answer» Points A, B, C are collinear `hArr ar(triangleABC) =0` Now, `ar(triangleABC) = (1)/(2) * \|[a, a+b+c, 1],[b, a+b+c, 1],[c, a+b+c, 1]\| [C_(2) to (C_(2) + C_(1))]` ` = (1)/(2)(a+b+c)*\|[a, 1, 1],[b, 1, 1],[c, 1, 1]\|` `=(1)/(2)(a+b+c) xx 0 =0 [because C_(2) " and"C_(3) "are identical"]` Hence, the given points are collinear.

Discussion

305.	Find the value of k in order that the points (5, 5), (k, 1) and (10, 7) are collinear.
Answer» The given points are collinear `hArr \|[5, 5, 1],[k, 1, 1],[10, 7, 1]\| =0 ` `hArr \|[5, 5, 1],[k-5, -4, 0],[5, 2, 0]\| =0 [R_(2) to R_(2) -R_(1) " and "R_(3) to R_(3) -R_(1)] ` `hArr 1. [2(k-5) +20] =0` `hArr 2k+10=0 hArr k =-5` Hence, k =-5

Discussion

306.	Solve by matrix method : `2x+y=5` `x-y = 1`
Answer» Write the given equations in matrix form `[{:(2,1),(1,-1):}][{:(x),(y):}]=[{:(5),(1):}]` `rArr" " AX=B` `"Where "A=[{:(2,1),(1,-1):}],X=[{:(x),(y):}]" and B"=[{:(5),(1):}]` `\|A\|=[{:(2,1),(1,-1):}]=-2-1 =-3 ne0` `therefore" A is invertible",` Cofactors of the elements of matrix A. `c_(11)=(-1)^(1+1)(-1)=-1` `x_(12)=(-1)^(1+2)1=-1` `c_(21)=(-1)^(2+1)1=-1` `c_(22)=(-1)^(2+2)2=2` `therefore" adj. A"=[{:(c_(11),c_(12)),(c_(21),c_(22)):}]` `[{:(-1,-1),(-1,2):}]=[{:(-1,-1),(-1,2):}]` `"and "A^(-1)=1/(\|A\|)" adj. A"=-1/3[{:(-1,-1),(-1,2):}]` Now AX=B `rArr" "AX=B` `rArr[{:(x),(y):}]=-1/3[{:(-1,-1),(-1,2):}]=-1/3[{:(-5,-1),(-5,+2):}]=[{:(2),(1):}]` `therefore" "x=2, y=1.`

Discussion

307.	If `A=[{:(3,-3,4),(2,-3,4),(0,-1,1):}]`, then show that `A^(3)=A^(-1)`.
Answer» `A^(2)=A.A` `=[{:(3,-3,4),(2,-3,4),(0,-1,1):}][{:(3,-3,4),(2,-3,4),(0,-1,1):}]` `=[{:(9-6+0,-9+9-4,12-12+4),(6-6+0,-6+9-4,8-12+4),(0-2+0,0+3-1,0-4+1):}]` `=[{:(3,-4,4),(0,-1,0),(-2,2,-3):}]` `"and "A^(4)=A^(2).A^(2)` `=[{:(3,-4,4),(0,-1,0),(-2,2,-2):}][{:(3,-4,4),(0,-1,0),(-2,2,-3):}]` `[{:(9+0-8,-12+4+8,12+0-12),(0+0+0,0+1+0,0+0+),(-6+0+6,8-2-6,-8+0+9):}]` `=[{:(1,0,0),(0,1,0),(0,0,1):}]=I` `therefore" "A^(4)-I` `=[{:(3,-3,4),(2,-3,4),(0,0,1):}]=3\|{:(-3,4),(-1,1):}\|-2\|{:(-3,4),(-1,1):}\|+0` `("Expanding along" C_(1))` ` =3(-3+4)-2(-3+4)=3-2=1ne0` A is invertible. `rArr A^(-1)"exists".` `A^(4)=I` `rArr" "A^(-1)A^(4)=A^(-1)I` `rArr" "A^(3)=A^(-1)`.

Discussion

308.	Find a matrix B of order `2xx2` such that : `[{:(1,1),(-2,3):}]B=[{:(3,4),(-1,2):}]`
Answer» `LetA=[{:(1,1),(-2,3):}]" and "C=[{:(3,4),(-1,2):}]` `therefore" "\|A\|=[{:(1,1),(-2,3):}]=3-(-2)=5ne0` `therefore" A is invertible".` Cofactors of martrix A `c_(11)=(-1)^(1+1).3=3` `c_(12)=(-1)^(1+2).(-2)=2` `c_(21)=(-1)^(2+1).1=-1` `c_(22)=(-1)^(2+2).1=1` `therefore" adj.A"=[{:(c_(11),c_(12)),(c_(21),c_(22)):}]=[{:(3,2),(-1,1):}]=[{:(3,-1),(2,1):}]` `"and "A^(-1)=1/(\|A\|)"adj.A"=1/5[{:(3,-1),(2,1):}]` From eq. (1) AB = C `rArr" " A^(-1) (AB) = A^(-1) C` `rArr" " (A^(-1)A)B=1/5[{:(3,-1),(2,1):}][{:(3,4),(-1,2):}]` (From associative law) `rArr" "IB=1/5[{:(9+1,12-2),(6-1,8+2):}](becauseA^(-1)A=I)` `rArr" "B=[{:(2,2),(1,2):}]`

Discussion

309.	If `A=[{:(1,-1),(2,3):}]`, shown that `A^(2)-44+5I=o`. Hence Find `A^(-1)`.
Answer» `A=[{:(1,-1),(2,3):}]` `rArr" "A^(2)=A.A=[{:(1,-1),(2,3):}][{:(1,-1),(2,3):}]` `[{:(1-2,-1-3),(2+6,-2+9):}]=[{:(-1,-4),(8,7):}]` `"Now, L.H.S."=A^(2)-4A+5I` `[{:(-1,-4),(8,7):}]-4[{:(-1,-1),(2,3):}]+5[{:(1,0),(0,1):}]` `[{:(-1,-4),(8,7):}]-[{:(4,-4),(8,12):}]+[{:(5,0),(0,):}]` `[{:(0,0),(0,0):}]=o=R.H.S.` `"Now "\|A\|=\|{:(1,-1),(2,3):}\|=3-(-2)=5ne0` therefore, `A^(-1)` exists. `therefore" "A^(2)-4A+5I=0 ` `rArr A^(-1)(A^(2)-4A+5I)=A^(1).O` `rArr" "A-4I+5A^(-1)=4I-A` `=[{:(1,0),(0,1):}]-[{:(1,-1),(2,3):}]=[{:(3,1),(-2,1):}]` `rArr=A^(-1)=1/5[{:(3,1),(-2,1):}].`

Discussion

310.	If `A=[{:(3,2,1),(0,-1,-2),(-3,4,2):}]`, find `A^(-1)`
Answer» `\|A\|=[{:(3,2,1),(0,-1,-2),(-3,4,2):}]` `=3[{:(-1,-2),(4,2):}]-2\|{:(0,-2),(-3,2):}\|+1\|{:(0,-1),(-3,4):}\|` `=3(-2+8)-2(0-6)+1(0-3) =18+12-3=27ne0` `therefore" A is invertible matrix".` Cofactore of matrix A `C_(11)=(-1)^(1+1)\|{:(-1,-2),(4,2):}\|=(-2+8)=6` `C_(12)=(-1)^(1+2)\|{:(0,-2),(-3,2):}\|=-(0-6)=6` `C_(13)=(-1)^(1+3)\|{:(0,-1),(-3,4):}\|=0-3=-3` `C_(21)=(-1)^(2+1)\|{:(2,1),(4,2):}\|=-(4-4)=0` `C_(22)=(-1)^(2+2)\|{:(3,1),(-3,2):}\|=6+3=6` `C_(23)=(-1)^(2+3)\|{:(3,2),(-3,4):}\|=-(12+6)=-18` `C_(31)=(-1)^(3+1)\|{:(2,1),(-1,-2):}\|=-4+1=-3` `C_(32)=(-1)^(3+2)\|{:(3,2),(0,-2):}\|=-(-0)=6` `C_(33)=(-1)^(3+3)\|{:(3,2),(0,-1):}\|=-3-0=-3` `therefore" adj.A"=[{:(c_(11),c_(12),c_(13)),(c_(21),c_(22),c_(23)),(c_(31),c_(32),c_(33)):}]` `=[{:(6,6,-3),(0,9,-18),(-3,6,-3):}]=[{:(6,0,-3),(6,9,6),(-3,-18,-3):}]` `"and "A^(-1)=1/(\|A\|)."adj.A"=1/27[{:(6,0,-3),(6,9,6),(-3,-18,-3):}]` `=3/27[{:(2,0,-1),(2,3,2),(-1,-6,-1):}]=1/9[{:(2,0,-1),(2,3,2),(-1,-6,-1):}]`

Discussion

311.	If `A=\|{:(3,5),(2,4):}\|`, find `A^(-1)`.
Answer» `\|A\|=\|{:(3,5),(2,4):}\|` =12-10 `=2ne0` `rArr" "A is invertible.` `"Now"" "C_(11)=(-1)^(1+1).4=4` `C_(12)=(-1)^(1+2).42=-2` `C_(21)=(-1)^(2+1).5=-5` `C_(22)=(-1)^(2+3).3=3` ` therefore"adj.A="[{:(C_(11),C_(12)),(C_(21),C_(22)):}]=[{:(4,-2),(-5,3):}]=[{:(4,-5),(-2,3):}]` `therefore"adj.A^=1/(\|A\|)."adj.A=1/2[{:(4,-5),(-2,3):}]`

Discussion

312.	Find the equation of a line passing through the points (3,5) and (-2, 1).
Answer» Equation of a line passing throught the points (3,5) and (-2, 1) is : `=1/2\|{:(x,y,1),(3,5,1),(-2,1,1):}\|=0` `rArr x(5-1) -y(3+2)+1(3+10)=0 ` `("Expanding along "R_(1))` `rArr" "4x-5y+13=0.`

Discussion

313.	Prove that the points (0,3), (4,6) and (-8, -3) are collinear.
Answer» `Delta=1/2\|{:(0,3,2),(4,6,1),(-8,-3,1):}\|` `=1/2\|{:(0,3,2),(4,6,1),(-8,-3,1):}\|" "(C_(2)toC_(2)-3C_(3))` `=1/2\|{:(4,3),(-8,-6):}\|" "("Expanding along "R_(1))` `=1/2(-24+24)=0` `therefore" The given points are collinear"`.

Discussion

314.	which of the following is`//`are true for `Delta= \|{:(a^(2),,1,,a+c),(0,,b^(2)+1,,b+c),(0,,b+c,,c^(2)+1):}\|` ?A. `Delta ge 0` for real values of a,b,cB. `Delta le 0` for real values of a,b,cC. `Delta =\|{:(bc-1,,0,,0),(1,,ac,,-a),(-b,,-a,,ab):}\|`D. `Delta =0` if bc =1 where a,b,c are non-zero
Answer» Correct Answer - A::C::D `\|{:(a^(2),,1,,a+c),(0,,b^(2)+1,,b+c),(0,,b+c,,c^(1)+1):}\|=\|{:(a,,0,,1),(0,,b,,1),(0,,1,,c):}\|\|{:(a,,0,,1),(0,,b,,1),(0,,1,,c):}\|` `=\|{:(a,,0,,1),(0,,b,,1),(0,,1,,c):}\|^(2)` `=(abc -a)^(2)= a^(2) (bc-1)^(2)` `=\|{:(bc-1,,0,,0),(1,,ac,,-a),(-b,,-a,,ab):}\|` {Determinant formed by cofactors)

Discussion

315.	Prove that`\|a b+c a^2b c+a b^2c a+b c^2\|=-(a+b+c)xx(a-b)(b-c)(c-a)dot`
Answer» We have `Delta = \|{:(a,,b+c,,a^(2)),(b,,c+a,,b^(2)),(c,,a+b,,c^(2)):}\|` Applying `C_(2) to C_(2) +C_(1)` and then taking `(a+b+c)` common from `C_(2)` we get ` Delta =(a+b+c)xx \|{:(a,,1,,a^(2)),(b,,1,,b^(2)),(c,,1,,c^(2)):}\|` `=- (a+b+c) xx \|{:(1,,a,,a^(2)),(1,,b,,b^(2)),(1,,c,,c^(2)):}\| " (Applying " C_(1) to C_(2)")"` `=- (a+b+c)(a-b)(b-c)(c-a)`

Discussion

316.	Find the area of triangle whose vertices are (2,7), (1,1) and (10,8).
Answer» Area of tringle `=1/2\|{:(2,7,1),(1,1,1),(10,8,1):}\|` `=1/2\|{:(1,6,0),(-9,-7,0),(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,

Discussion

317.	`" if " \|{:(yz-x^(2),,zx-y^(2),,xy-z^(2)),(xz-y^(2),,xy-z^(2),,yz-x^(2)),(xy-z^(2),,yz-x^(2),,zx-y^(2)):}\|=\|{:(r^(2),,u^(2),,u^(2)),(u^(2),,r^(2),,u^(2)),(u^(2),,u^(2),,r^(2)):}\|` thenA. `r^(2)=x+y+z`B. `r^(2) =x^(2) =y^(2) +z^(2)`C. `u^(2) =yz+zx+xy`D. `u^(2) =xyz`
Answer» Correct Answer - B::C In the left - hand determinant each element is the cofactor of the elements of the determinant `\|{:(x,,y,,z),(y,,z,,x),(z,,x,,y):}\|=Delta " (says) "` hence `Delta^(2) = \|{:(x,,y,,z),(y,,z,,x),(z,,x,,y):}\|\|{:(x,,y,,z),(y,,z,,x),(z,,x,,y):}\|` `=\|{:(x^(2)+y^(2)+z^(2),,xy+yz+zx,,xz+yx+zy),(Sigmaxy,,Sigmax^(2),,Sigmaxy),(Sigmaxy,,Sigmaxy,,Sigmax^(2)):}\|` `\|{:(r^(2),,u^(2),,u^(2)),(u^(2),,r^(2),,u^(2)),(u^(2),,u^(2),,r^(2)):}\|" "underset(xy+yz+zx=u^(2))("Where "x^(2)+y^(2)+z^(2)=r^(2))`

Discussion

318.	Prove that `\|{:(x^(2),,x^(2)-(y-z)^(2),,yz),(y^(2),,y^(2)-(z-x)^(2),,zx),(z^(2),,z^(2)-(x-y)^(2),,xy):}\|` `=(x-y) (y-z) (z-x)(x+y+z) (x^(2)+y^(2)+z^(2))`
Answer» we have `Delta =\|{:(x^(2),,x^(2)-(y-z)^(2),,yz),(y^(2),,y^(2)-(z-x)^(2),,zx),(z^(2),,z^(2)-(x-y)^(2),,xy):}\|` Applying `C_(2) to C_(2) -2 C_(1) -2C_(3)` we get `Delta =\|{:(x^(2),,-(x^(2)+y^(2)+z^(2)),,yz),(y^(2),,-(x^(2)+y^(2)+z^(2)),,zx),(z^(2),,-(x^(2)+y^(2)+z^(2)),,xy):}\|` `=- (x^(2) +Y^(2) +z^(2)) \|{:(x^(2),,1,,yz),(y^(2),,1,,zx),(z^(2),,1,,xy):}\|` Multiplying `R_(1),R_(2) " and " R_(3) by x, y` and z, respectively we get `Delta =- ((x^(2)+y^(2)+z^(2)))/(xyz) \|{:(x^(3),,x,,yz),(y^(3),,y,,zx),(z^(3),,z,,xy):}\|` `=- (x^(2) +y^(2) +z^(2)) \|{:(x^(3),,x,,1),(y^(3),,y,,1),(z^(3),,z,,1):}\| "( Taking xyz common from " c_(3)")"` `=(x^(2)+y^(2)+z^(2)) \|{:(1,,x,,x^(3)),(1,,y,,y^(3)),(1,,z,,z^(3)):}\| "(Applying " C_(1) hArr C_(3)")"` ` =(x-y) (y-z) (z-x) (x+y+z) (x^(2) +y^(2)+z^(2))`

Discussion

319.	if `\|{:(3,1,-4),(3,2,5),(1,-1,3):}\|=49`, then evaluate `\|{:(6,3,-8/3),(6,6,10/3),(2,-3,2):}\|`.
Answer» `\|{:(3,1,-4),(3,2,5),(1,-1,3):}\|=49` `\|{:(6,3,-8/3),(6,6,10/3),(2,-3,2):}\|=2xx3xx2/3\|{:(3,1,4),(3,2,5),(1,-1,3):}\|`

Discussion

320.	Prove that `\|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}\|=0`
Answer» `L.H.S.=\|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}\|` `=\|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}\|(C_(3)toC_(3)+C_(2))` `=(x+y+z)\|{:(1,x+alpha,1),(1,t+beta,1),(1,z+gamma,1):}\|` `=0 " "(therefore C_(1)" and "C_(3)" are same")` `=R.H.S." "Hence Proved`.

Discussion

321.	Solve the equation `\|{:(x+2,1,-3),(1,x-3,-2),(-3,-2,1):}\|=0.`
Answer» `\|{:(x+2,1,-3),(1,x-3,-2),(-3,-2,1):}\|=0.` `rArr" "\|{:(x-2,1,-3),(-(x-2),x-3,-2),(0,-2,1):}\|=0.` `rArr" (x-2) "\|{:(1,1,-3),(-1,x-3,-2),(0,-2,1):}\|=0.` `rArr" (x-2) "\|{:(1,1,-3),(-1,x-2,-5),(0,-2,1):}\|=0 ("From "R_(2)toR_(2)+R_(1))` `rArr(x-2).1\|{:(x-2,-5),(-2,1):}\|=0 ("Expanding with "C_(1))` `rArr (x-2)(x-2-10)=0` `rArr (x-2)(x-12)=0` ` rArr x=2 or x=12`.

Discussion

322.	Prove that : `\|{:(1,b,c),(b,c,a),(c,a,b):}\|=3 a b c-a^(3)-b^(3)-c^(3)`
Answer» `\|{:(1,b,c),(b,c,a),(c,a,b):}\|` `\|{:(1+b+c,b,c),(b+c+a,c,a),(c+a+b,a,b):}\|(C_(1)toC_(1)+C_(2)+C_(3))` `=(a+b+c)\|{:(1,b,c),(1,c,a),(1,a,b):}\|` `=(a+b+c)\|{1\|{:(c,a),(a,b):}\|-1\|{:(b,c),(a,b):}\|+1\|{:(b,c),(c,a):}\|}` `("Expanding with "C_(1))` `(a+b+c)(bc-a^(2)-b^(2)+ac+ab-c^(2))` `=(a+b+C)(ab+bc+ca-a^(2)-b^(2)-c^(2))` `=3a b c -a^(3)-c^(3)= R.H.S.`

Discussion

323.	Prove that `\|{:(1,b,c),(a^(2),b^(2),c^(2)),(b+c,c+a,a+b):}\|=(a-b)(b-c)(a+b+c)`
Answer» `\|{:(1,b,c),(a^(2),b^(2),c^(2)),(b+c,c+a,a+b):}\|` `\|{:(1+b+c,a+b+c,a+b+c),(a^(2),b^(2),c^(2)),(b+c,c+a,a+b):}\|(R_(1)toR_(1)+R_(3))` `=(a+b+c)\|{:(1,1,1),(a^(2),b^(2),c^(2)),(b+c,c+a,a+b):}\|` `=(a+b+c)\|{:(0,0,1),(a^(2)-b^(2),b^(2)-c^(2),c^(2)),(b-c,c-b,a+b):}\|` `(C_(1)toC_(1)-C_(2))` `(C_(2)toC_(2)-C_(3))` `=(a+b+c)(a-b)(b-c)\|{:(0,0,1),(a+b,b+c,c^(2)),(-1,-1,a+b):}\|` `=(a+b+c)(a-b)(b-c).1\|underset(-1)(a+b)" "underset(-1)(b+c)\|` `(" Expanding with "R_(1))` `=(a+b+c)(a-b)(b-c){-(a+b)+(b+c)} =(a+b+c)(a-b)(b-c)(c-a)=R.H.S.`

Discussion

324.	Without, prove that : `\|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}\|=2\|{:(a,b,c),(p,q,r),(x,y,z):}\|`
Answer» `\|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}\|` `=\|{:(2(a+b+c),b+c,c+b),(2(p+q+r),q+r,r+p),(2(x+y+z),y+z,z+x):}\|` `(C_(1)toC_(1)+C_(2)+C_(3))` `\|{:(a+b+c,b+c,c+a),(p+q+r,q+r,r+p),(x+y+z,y+z,z+x):}\|` `=2\|{:(1,b+c,c+a),(p,q+r,r+p),(x,y+z,z+x):}\|` `(C_(1)toC_(1)-C_(2))` `=2\|{:(1,b+c,c),(p,q+r,r),(x,y+z,z):}\|` `(C_(3)toC_(3)-C_(1))` `=2\|{:(1,b,c),(p,q,r),(x,y,z):}\|(C_(2)toC_(2)-C_(3))` =R.H.S. Hence Proved.

Discussion

325.	Prove that : `=2\|{:(1,1,1),(x,y,rz),(x^(2),y^(2),z^(2)):}\|=(x-y)(y-z)(z-x)`
Answer» `L.H.S.=2\|{:(1,1,1),(x,y,rz),(x^(2),y^(2),z^(2)):}\|` `=\|{:(0,0,1),(x-y,y-z,z),(x^(2)-y^(2),y^(2)-z^(2),z^(2)):}\|{:(((C_(1)to,C_(1)-,C_(2)))),(((C_(2)to,C_(2)-,C_(3)))):}` `(x-y)(y-z)\|{:(0,0,1),(1,1,z),(x+y,y+z,z^(2)):}\|` `(x-y)(y-z)1\|underset(x+y)(1)" "underset(y+z)(1)\|` `("Expending with "R_(1))` ` =(x-y){(y-z)-(x+y))}` `=(x-y) (y-z) (z-x) = R.H.S.`

Discussion

326.	Using the property of determinants and without expanding, prove that:`\|[0,a,-b],[-a,0,-c],[ b, c,0]\|=0`
Answer» `D= \|[0,a,-b],[-a,0,-c],[b,c,0]\|` We know, if we interchange rows and columns of a determinant, value of the determinant does not change. So, if we interchange rows and columns of given determinant, then, `D = \|[0,-a,b],[a,0,c],[-b,-c,0]\| = (-1)^3\|[0,a,-b],[-a,0,-c],[b,c,0]\| = -D` `:. D = -D` `=>2D = 0` `=> D = 0` Therefore, value of the given determinant is `0`.

Discussion

327.	if a=cos 0 + in sin 0,=cos 20 -I sin 2 0, c= cos 0 + I sin 3 0 and if `\|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\| =0 ` thenA. `=0 2 k pi, k in Z`B. ` 0= (2k +1) pi k in Z`C. `0= (4k +1) pi k in Z`D. none of these
Answer» Correct Answer - A `Delta =\|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}\|` `=-(a+b+c)(a^(2)+b^(2)+c^(2)-ab-bc-ca)` `=(1)/(2) (a+b+c)[(a-b)^(2)+(b-c)^(2)+(c-a)^(2)]=0` `rArr a+b+c=0" or " a=b=c ` `" if " a+b+c =0 , ` we have `cos0 + cos 2 0 + cos 3 0 =0` and s sin 0 -sin 2 0 + sin 3 0 =0 `" or " cos 2 0 (2 cos 0 + 1) =0 " and " sin 2 0 (1 -2 cos 0 ) =0 (1)` which is not possible as cos 2 0 = 0 given sin 2 0 `ne 0` cos 0 `ne 1//2` And cos 0 = `1//2` gives sin 2 0 `ne 0 cos 0 ne 1//2` . Therefore , Eq. (1) does not hold simultaneously .Therefore , `a+b+c ne 0` `rArr a=b=c` `:. e^(i0) =e^(-2i0) =e^(3i0)` which is satisfied only by by `e^(i 0) =1, i.e., cos 0 =1 sin 0 =0` `so 0 = 2k pi , k in Z`

Discussion

328.	Without, prove that : `\|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}\|=2\|{:(a,b,c),(p,q,r),(x,y,z):}\|`
Answer» `\|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}\|` `=\|{:(2(a+b+c),b+c,c+b),(2(p+q+r),q+r,r+p),(2(x+y+z),y+z,z+x):}\|` `(C_(1)toC_(1)+C_(2)+C_(3))` `\|{:(a+b+c,b+c,c+a),(p+q+r,q+r,r+p),(x+y+z,y+z,z+x):}\|` `=2\|{:(1,b+c,c+a),(p,q+r,r+p),(x,y+z,z+x):}\|` `(C_(1)toC_(1)-C_(2))` `=2\|{:(1,b+c,c),(p,q+r,r),(x,y+z,z):}\|` `(C_(3)toC_(3)-C_(1))` `=2\|{:(1,b,c),(p,q,r),(x,y,z):}\|(C_(2)toC_(2)-C_(3))` =R.H.S. Hence Proved.

Discussion

329.	[ 28.If f(x)=\|[0,x-a,x-bx+a,0,x-cx+b,x+c,0]\| then,[ 1) f(a)=0, 2) f(b)=0]]A. `f(a)=0`B. `f(b)=0`C. `f(0)=0`D. `f(1)=0`
Answer» Correct Answer - C We have `f(x)=\|(0,x-a,x-b),(x+a,0,x-c),(x+b,x+c,0)\|` `implies f(a)=\|(0,0,a-b),(2a,0,a-c),(a+b,a+c,0)\|` `=[(a-b)}2a.(a+c)}]!=0` `:. f(b)=\|(0,b-a,0),(b+a,0,b-c),(2b,b+c,0)\|` `=-(b-a)[2b(b-c)]` `=-2b(b-a)(b-c)!=0` `:.f(0)=\|(0,-a,-b),(a,0,-c),(b,c,0)\|` `=a(bc)-b(ac)` `=abc-abc=0`

Discussion

330.	Can students Score high marks if they practice Determinants Class 12 Mock Test
Answer» Start practicing for Determinants class 12 mock test and become proficient in this topic before you appear in the CBSE class 12 exams. Our teachers have provided the class 12 Determinants Online Mock Test series and based on MCQ questions with answers in the below links for CBSE Class 12 Determinants Online Mock Test. These tests are very helpful in preparations for all major exams, so make a habit of attempting these CBSE class 12 mock tests and experience the improvement in your skills. You should practice these Determinants Class 12 Online Tests having objective-based questions and compare your answers with the solutions provided by us. Students can attempt the mock tests for CBSE Class 12 as many number of times as they want and also download a passing certificate after they have cleared the online test. The free Determinants online mock tests Class 12 should be used by students to check their understanding of their concepts of Class 12 Determinants. Sarthaks eConnect brings here the biggest collection of Class 12 question based on online tests for CBSE Class 12 Determinants 2021 exams to help you prepare properly and get highest rank in exams. Click here to start practice: - Class 12 Determinants Mock Test

330.

Can students Score high marks if they practice Determinants Class 12 Mock Test

Answer»

Start practicing for Determinants class 12 mock test and become proficient in this topic before you appear in the CBSE class 12 exams. Our teachers have provided the class 12 Determinants Online Mock Test series and based on MCQ questions with answers in the below links for CBSE Class 12 Determinants Online Mock Test. These tests are very helpful in preparations for all major exams, so make a habit of attempting these CBSE class 12 mock tests and experience the improvement in your skills.

You should practice these Determinants Class 12 Online Tests having objective-based questions and compare your answers with the solutions provided by us. Students can attempt the mock tests for CBSE Class 12 as many number of times as they want and also download a passing certificate after they have cleared the online test.

The free Determinants online mock tests Class 12 should be used by students to check their understanding of their concepts of Class 12 Determinants. Sarthaks eConnect brings here the biggest collection of Class 12 question based on online tests for CBSE Class 12 Determinants 2021 exams to help you prepare properly and get highest rank in exams.

Click here to start practice: - Class 12 Determinants Mock Test

Discussion

331.	If `A=[[2,a,-3] , [0,2,5] , [1,1,3]]` then,find the value of a for which `A^(-1)` exists.A. `a=2`B. `a!=2`C. `a!=-2`D. none of these
Answer» Correct Answer - D We have `A=\|(2, lamda, -3),(0,2,5),(1,1,3)\|` Expanding along `R_(1)` `\|A\|=2(6-5)-lamda(-5)-3(-2)=2+5lamda+6` We know that `A^(-1)` exists if `A` is non singular matrix i.e. `\|A\|!=0` `:.2+5lamda+6!=0` `implies 5lamda!=-8` `:. lamda!=(-8)/5` So, `A^(-1)` exists if and only if `lamda!=(-8)/5`.

Discussion

332.	If A is a singular matrix, then write the value of \|A\|.
Answer» Since a singular matrix is a matrix whose determinant is 0, Therefore the determinant of A is 0.

Discussion

333.	Find the value of the `"determinant" \|{:(sqrt13+sqrt3,2sqrt5,sqrt5),(sqrt26+sqrt15,5,sqrt10),(sqrt65+3,sqrt15,5):}\|`
Answer» `\|{:(sqrt13+sqrt3,2sqrt5,sqrt5),(sqrt26+sqrt15,5,sqrt10),(sqrt65+3,sqrt15,5):}\|` `\|{:(sqrt13,2sqrt5,sqrt5),(sqrt26+5,5,sqrt10),(sqrt65+3,sqrt15,5):}\|+\|{:(2sqrt3,2sqrt5,sqrt5),(sqrt15,5, sqrt10),(3,sqrt15,5):}\|` `sqrt13xxsqrt15xxsqrt5\|{:(1,2,1),(sqrt2,sqrt5,sqrt2),(sqrt5,sqrt3,sqrt5):}\|+sqrt3xxsqrt5xxsqrt5\|{:(2,2,1),(sqrt5,sqrt5,sqrt2),(sqrt3,sqrt3,sqrt5):}\|` =0+0 =0.

Discussion

334.	If a determinant of order `3xx3` is formed by using the numbers 1 or -1 then minimum value of determinant is :A. -2B. -4C. 0D. -8
Answer» Correct Answer - B `\|{:(1-2 sin^(2)x,,sin^(2)x,,1-8 sin^(2)x(1-sin^(2)x)),(sin^(2)x,,1-sin^(2)x,,1-sin^(2)x),(1-8sin^(2)x(1-sin^(2)x),,1-sin^(2)x,,1-sin^(2)x):}\|` The required constant terms is `\|{:(1,,0,,1),(0,,1,,1),(1,,1,,1):}\|+\|{:(1,,0,,0),(0,,1,,1),(1,,1,,0):}\|=1(10 -1)=-1`

Discussion

335.	If `\|[x^n, x^(n+2),x^(n+3)],[y^n, y^(n+2),y^(n+3)],[z^n, z^(n+2),z^(n+3)]\|=(x-y)(y-z)(z-x)(1/x+1/y+1/z),`then `n`equalsA. 1B. -1C. 2D. -2
Answer» Correct Answer - B The degree or each terms of the determinant is `n + (n+2)+ (n+3) =3n+5` and the degree of each terms of the expression on R.H.S. is 2. `:. 3n+ 5=2 " or " n=-1`

Discussion

336.	witout expanding at any stage Prove that `\|{:(0,,a,,-b),(-a,,0,,-c),(b,,c,,0):}\| =0`
Answer» `\|{:(0,,a,,-b),(-a,,0,,-c),(b,,c,,0):}\|=\|{:(0,,-a,,b),(a,,0,,c),(-b,,-c,,0):}\|` (Taking transpose) ` =(-1)^(3)\|{:(0,,a,,-b),(-a,,0,,-c),(b,,c,,0):}\|` (Taking (-1) common from each row) `:. Delta=-Delta` `rArr 2Delta =0` `rArr Delta=0`

Discussion

337.	for `x,x,z gt 0` Prove that `\|{:(1,,log_(x)y,,log_(x)z),(log_(y)x,,1,,log_(y)z),(log_(z) x,,log_(z)y,,1):}\| =0`
Answer» We have \|{:(1,,(log y)/(log x),,(log y)/(log x)),((log x)/(logy),,1,,(log z)/(log y)),((log x)/(log y),,(log y)/(log z),,1):}\|` (Changing base of logarithm) ltbrtgt Taking `(1)/(log x), (1)/(log y) " and " (1)/(log z) " common from " R_(1) ,R_(2) " and " R_(3)f " respectively we get "` `Delta= (1)/(" log x log y log z ") \|{:(log x,,log y,,log z),(log x,,log y,,log z),(log x,,log y,,log z):}\|` `:. Delta =0 " ""(As "R_(1) ,R_(2) " and "R_(3) " are identical "`

Discussion

338.	Prove that ` \|{:(ax,,by,,cz),(x^(2),,y^(2),,z^(2)),(1,,1,,1):}\|=\|{:(a,,c,,c),(x,,y,,z),(yz,,xz,,xy):}\|`
Answer» ` \|{:(ax,,by,,cz),(x^(2),,y^(2),,z^(2)),(1,,1,,1):}\|` Taking x,y,z common from`C_(1)C_(2) " and " C_(3)` respeectively we get ` \|{:(a,,c,,c),(x,,y,,z),((1)/(x),,(1)/(y),,(1)/(z)):}\|=\|{:(a,,b,,c),(x,,y,,z),(yz,,xz,,xy):}\|` (Taking xzy to `R_(3)`)

Discussion

339.	Using the property of determinants and without expanding, prove that:`\|[b+c, q+r, y+z],[ c+a, r+p, z+x],[ a+b, p+q, x+y]\|=2\|[a, p, x],[ b, q ,y],[ c, r, z]\|`
Answer» `L.H.S. = \|[b+c,q+r,y+z],[c+A,r+p,z+x],[a+b,p+q,x+y]\|` Applying `R_1->R_1+R_2+R_3` `= \|[2(a+b+c),2(p+q+r),2(x+y+z)],[c+a,r+p,z+x],[a+b,p+q,x+y]\|` `= 2\|[a+b+c,p+q+r,x+y+z],[c+a,r+p,z+x],[a+b,p+q,x+y]\|` Applying `R_2->R_2-R_1` and `R_3->R_3-R_1` `=2\|[a+b+c,p+q+r,x+y+z],[-b,-q,-y],[-c,-r,-z]\|` Applying `R_1->R_1+R_2+R_3` `=2\|[a,p,x],[-b,-q,-y],[-c,-r,-z]\|` `=2(-1)(-1)\|[a,p,x],[b,q,y],[c,r,z]\|` `=2\|[a,p,x],[b,q,y],[c,r,z]\|=R.H.S.`

Discussion

340.	Evaluate the following: ` \|[a,h,g],[h,b,f],[g,f,c]\| `
Answer» Expanding by 1st column, we get `Delta = a \|{:(b, f), (f, c):}\|-h \|{:(h, g), (f, c):}\| +g * \|{:(h, g), (b, f):}\|` ` = a(bc-f^(2)) -h * (ch-fg) + g * (fh -bg)` ` = abc-af^(2)-ch^(2) + fgh + fgh -bg^(2)` ` =(abc +2fgh -af^(2)-bg^(2)-ch^(2)).`

Discussion

341.	If the determinant `\|(cos2x,sin^2 x,cos 4x),(sin^2 x,cos 2x,cos^2 x),(cos 4x,cos^2 x,cos 2x)\|` is expanded in powers of `sin x`, then the constant term isA. 1B. 0C. -1D. 2
Answer» Correct Answer - C `\|{:(1-2 sin^(2)x,,sin^(2)x,,1-8 sin^(2)x(1-sin^(2)x)),(sin^(2)x,,1-sin^(2)x,,1-sin^(2)x),(1-8sin^(2)x(1-sin^(2)x),,1-sin^(2)x,,1-sin^(2)x):}\|` The required constant terms is `\|{:(1,,0,,1),(0,,1,,1),(1,,1,,1):}\|+\|{:(1,,0,,0),(0,,1,,1),(1,,1,,0):}\|=1(10 -1)=-1`

Discussion

342.	consider the fourth -degree polynomial equation `\|{:(a_(1)+b_(1)x,,a_(1)x^(2)+b_(1),,c_(1)),(a_(2)+b_(2)x^(2),,a_(2)x^(2)+b_(2),,c_(2)),(a_(3)+b_(3)x^(2),,a_(3)x^(2)+b_(3),,c_(3)):}\|=0` Without expanding the determinant find all the roots of the equation.
Answer» `\|{:(a_(1)+b_(1)x,,a_(1)x^(2)+b_(1),,c_(1)),(a_(2)+b_(2)x^(2),,a_(2)x^(2)+b_(2),,c_(2)),(a_(3)+b_(3)x^(2),,a_(3)x^(2)+b_(3),,c_(3)):}\|=0 ("As " C_(1) " and " C_(2) " are indentical")` So `x= +-1` are roots of the given equation . From Sarrus rule we observe that the degree of equation is 4.

Discussion

343.	Evaluate `\|{:(2, 7, 65), (3, 8, 75), (5, 9, 86):}\|`
Answer» Given determinant `=\|{:(2, 7, 65), (3, 8, 75), (5, 9, 86):}\|` `=\|{:(2, 7, 2), (3, 8, 3), (5, 9, 5):}\| " "["applying" C_(3) to C_(3) - 9C_(2)]` `=0 " " [because C_(1) " and " C_(3) " are identical"].`

Discussion

344.	Using the property of determinants and without expanding, prove that:`\|[2, 7, 65],[ 3 ,8 ,75 ],[4, 9,86]\|=0`
Answer» `L.H.S. = \|[2,7,65],[3,8,75],[5,9,86]\|` Applying `C_3->C_3-C_1` `= \|[2,7,63],[3,8,72],[5,9,81]\|` `=9\|[2,7,7],[3,8,8],[5,9,9]\|` Applying `C_3->C_3-C_2` `=9\|[2,7,0],[3,8,0],[5,9,0]\|` 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`. `:.` Value of above determinant is `0`. `:. 9\|[2,7,0],[3,8,0],[5,9,0]\| = 9**0 = 0 = R.H.S.`

Discussion

345.	`" if " Delta_(r) = \|{:(r,,612,,915),(101r^(2),,2r,,3r),(r,,(1)/(r),,(1)/(r^(2))):}\|` then the value of `lim_( n to oo) .(1)/(n^(3)) (Sigma_(r=1)^(n) Delta_(r) " is " "____"`
Answer» Correct Answer - 2 `Detla_(r )= \|{:(r,,612,,915),(101r^(2),,2r,,3r),(r,,(1)/(r),,(1)/(r^(2))):}\|` (Taking r common from `C_(1) " and " R_(2)`) `= \|{:(1,,612,,915),(101,,2,,3),(r^(2),,r,,1):}\|` ` :. .underset(h to oo)("lim") .(1)/(n^(3)) .overset(n)underset(r=1)(Sigma) \|{:(1,,612,,915),(101,,2,,3),(r^(2),,r,,1):}\|` `= underset( h to oo)("lim") .(1)/(n^(3)) . \|{:(1,,612,,915),(101,,2,,3),((n(n+1)(2n+1))/(6),,(n(n+1))/(2),,n):}\|` ` = underset(h to oo) ("lim") \|{:(1,,612,,915),(101,,2,,3),(((n+1)(2n+1))/(6n^(2)),,(n+1)/(2n^(2)),,(1)/(n^(2))):}\|` `= \|{:(1,,612,,915),(101,,2,,3),((1)/(3),,0,,0):}\|` `=(1)/(3) (3 xx 612 -2xx 915)` `=2`

Discussion

346.	`" Let " Delta_(r)=\|{:(r-1,,n,,6),((r-1)^(2),,2n^(2),,4n-2),((r-1)^(2),,3n^(3),,3n^(2)-3n):}\|. " Show that " Sigma_(r=1)^(n) Delta_(r)` is constant.
Answer» Since `c_(1)` has variable terms and `c_(2) " and " c_(3)` have constant terms summation is taken to `C_(1)` Therefore, `overset(n)underset(r=1)(Sigma) \|{:(overset(n)underset(1)(Sigma)(r-1),,n,,6),(overset(n)underset(1)(Sigma)(r-1)^(2),,2n^(2),,4n-2),(overset(n)underset(1)(Sigma)(r-1)^(3),,3n^(3),,3n^(2)-3n):}\|` `\|{:((1)/(2)(n-1)n,,n,,6),((1)/(6) (n-1)(2n-1),,2n^(2),,4n-2),((1)/(4) (n-1)^(2)n^(2),,3n^(3),,3n^(2)-3n):}\|` Taking `(1)/(12) n(n-1)` common from `C_(1) " and " n` common from `C_(2)` we get ` Sigma Delta_(r)=(1)/(12)n^(2)(n-1) xx\|{:(6,,1,,6),(2(2n-1),,2n,,2(2n-1)),(3n(n-1),,3n^(2),,3n(n-1)):}\|` `=0 " Which is constant " [:. C_(1) " and " C_(3) " are identical "]`

Discussion

347.	`D_(r) = \|(2^(r -1),2.3^(r-1),4.5^(r -1)),(alpha,beta,gamma),(2^(n) -1,3^(n) -1,5^(n) -1)\|`. Then, the value of `Sigma_(r=1)^(n) D_(r)` isA. `alpha beta gamma`B. `2^(n) alpha + 2^(n) beta + 4^(n) gamma`C. `2 alpha + 3 beta + 4 gamma`D. none of these
Answer» Correct Answer - D

Discussion

348.	State whether the matrix \(\begin{bmatrix}2 & 3 \\6 & 4 \\\end{bmatrix}\) is singular or non-singular.
Answer» Let A = \(\begin{bmatrix} 2 & 3 \\ 6 & 4 \\ \end{bmatrix}\) Then \|A\| = \(\begin{bmatrix} 2 & 3 \\ 6 & 4 \\ \end{bmatrix}\) = 2 × 4 - 3 × 6 = -10 (Expanding along R₁) Since \|A\| ≠ 0, therefore A is a non-singular matrix.

348.

State whether the matrix \(\begin{bmatrix}2 & 3 \\6 & 4 \\\end{bmatrix}\) is singular or non-singular.

Answer»

Let A = \(\begin{bmatrix} 2 & 3 \\ 6 & 4 \\ \end{bmatrix}\)

Then |A| = \(\begin{bmatrix} 2 & 3 \\ 6 & 4 \\ \end{bmatrix}\)

= 2 × 4 - 3 × 6

= -10 (Expanding along R₁)

Since |A| ≠ 0, therefore A is a non-singular matrix.

Discussion

349.	` " if " Delta_(r) =\|{:(2^(r-1),,2xx3^(r-1),,4xx5^(r-1)),(alpha,,beta,,gamma),(2^(n)-1,,3^(n)-1,,5^(n)-1):}\|` then find the value of `Sigma_(r=1)^(n) Delta r`.
Answer» `overset(n)underset(r=1)(Sigma)2^(r-1)=1+2+2^(2)+.......+2^(n+1)=1xx((2^(n)-1))/(2-1)=2^(n)-1` `overset(n)underset(r=1)(Sigma)2xx3^(r-1)=2(1+3+3^(2)+...+3^(n-1))=(2(3^(n)-1))/(3-1)=3^(n)-1` `overset(n)underset(r=1)(Sigma)4xx5^(r-1)=4(1+5+5^(2)+....+5^(n)-1)=(4(5^(n)-1))/(5-1)=5^(n)-1` `:. overset(n)underset(r=1)(Sigma)Delta_(r ) =\|{:(2^(n-1),,3^(n-1),,5^(n-1)),(alpha,,beta,,gamma),(2^(n-1),,3^(n-1),,5^(n-1)):}\|=0`

Discussion

350.	Show that each of the following systems of linear equations has infinite number of solutions and solve : x + 2y = 5 3x + 6y = 15
Answer» Given : - Two equation x + 2y = 5 and 3x + 6y = 15 Tip : - We know that For a system of 2 simultaneous linear equation with 2 unknowns (i) If D ≠ 0, then the given system of equations is consistent and has a unique solution given by x = \(\frac{D_1}{D}\), y = \(\frac{D_2}{D}\) (ii) If D = 0 and D₁ = D₂ = 0, then the system is consistent and has infinitely many solution. (iii) If D = 0 and one of D₁ and D₂ is non – zero, then the system is inconsistent. Now, We have, x + 2y = 5 3x + 6y = 15 Lets find D ⇒ D = \(\begin{vmatrix} 1 & 2 \\[0.3em] 3 & 6 \\[0.3em] \end{vmatrix}\) ⇒ D = – 6 – 6 ⇒ D = 0 Again, D₁ by replacing 1^st column by B Here, B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\) ⇒ D₁ = \(\begin{vmatrix} 5 & 2 \\[0.3em] 15 & 6 \\[0.3em] \end{vmatrix}\) ⇒ D₁ = 30 – 30 ⇒ D₁ = 0 And, D₂ by replacing 2^ndcolumn by B Here, B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\) ⇒ D₂ = \(\begin{vmatrix} 1& 5 \\[0.3em] 3 & 15 \\[0.3em] \end{vmatrix}\) ⇒ D₂ = 15 – 15 ⇒ D₂ = 0 So, here we can see that D = D₁ = D₂ = 0 Thus, The system is consistent with infinitely many solutions. Let y = k then, ⇒ x + 2y = 5 ⇒ x = 5 – 2k By changing value of k you may get infinite solutions.

350.

Show that each of the following systems of linear equations has infinite number of solutions and solve : x + 2y = 5 3x + 6y = 15

Answer»

Given : -

Two equation

x + 2y = 5 and 3x + 6y = 15

Tip : - We know that

For a system of 2 simultaneous linear equation with 2 unknowns

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

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

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

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

Now,

We have,

x + 2y = 5

3x + 6y = 15

Lets find D

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

⇒ D = – 6 – 6

⇒ D = 0

Again,

D₁ by replacing 1^st column by B

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\)

⇒ D₁ = \(\begin{vmatrix} 5 & 2 \\[0.3em] 15 & 6 \\[0.3em] \end{vmatrix}\)

⇒ D₁ = 30 – 30

⇒ D₁ = 0

And,

D₂ by replacing 2^ndcolumn by B

Here,

B = \(\begin{vmatrix} 5 \\[0.3em] 15\\[0.3em] \end{vmatrix}\)

⇒ D₂ = \(\begin{vmatrix} 1& 5 \\[0.3em] 3 & 15 \\[0.3em] \end{vmatrix}\)

⇒ D₂ = 15 – 15

⇒ D₂ = 0

So, here we can see that

D = D₁ = D₂ = 0

Thus,

The system is consistent with infinitely many solutions.

Let

y = k

then,

⇒ x + 2y = 5

⇒ x = 5 – 2k

By changing value of k you may get infinite solutions.

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Without expanding the determinant, prove that \(\begin{vmatrix} 41 &amp;1 &amp; 5 \\[0.3em] 79 &amp; 7 &amp;9\\[0.3em] 29 &amp; 5 &amp; 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if |A| = 0. Also, A is called non singular if |A| ≠ 0.

Let A be a square matric of order 3 × 3. Write the value of |2A|, where |A| = 4.

Find the area of the triangle whose vertices are: `(-2,-3),(3,2),(-1,-8)`

Show that points`A (a , b + c)`, `B (b , c + a)`, `C (c , a + b)`are collinear.

Find the value of k in order that the points (5, 5), (k, 1) and (10, 7) are collinear.

Solve by matrix method : `2x+y=5` `x-y = 1`

If `A=[{:(3,-3,4),(2,-3,4),(0,-1,1):}]`, then show that `A^(3)=A^(-1)`.

Find a matrix B of order `2xx2` such that : `[{:(1,1),(-2,3):}]B=[{:(3,4),(-1,2):}]`

If `A=[{:(1,-1),(2,3):}]`, shown that `A^(2)-44+5I=o`. Hence Find `A^(-1)`.

If `A=[{:(3,2,1),(0,-1,-2),(-3,4,2):}]`, find `A^(-1)`

If `A=|{:(3,5),(2,4):}|`, find `A^(-1)`.

Find the equation of a line passing through the points (3,5) and (-2, 1).

Prove that the points (0,3), (4,6) and (-8, -3) are collinear.

which of the following is`//`are true for `Delta= |{:(a^(2),,1,,a+c),(0,,b^(2)+1,,b+c),(0,,b+c,,c^(2)+1):}|` ?A. `Delta ge 0` for real values of a,b,cB. `Delta le 0` for real values of a,b,cC. `Delta =|{:(bc-1,,0,,0),(1,,ac,,-a),(-b,,-a,,ab):}|`D. `Delta =0` if bc =1 where a,b,c are non-zero

Prove that`|a b+c a^2b c+a b^2c a+b c^2|=-(a+b+c)xx(a-b)(b-c)(c-a)dot`

Find the area of triangle whose vertices are (2,7), (1,1) and (10,8).

`" if " |{:(yz-x^(2),,zx-y^(2),,xy-z^(2)),(xz-y^(2),,xy-z^(2),,yz-x^(2)),(xy-z^(2),,yz-x^(2),,zx-y^(2)):}|=|{:(r^(2),,u^(2),,u^(2)),(u^(2),,r^(2),,u^(2)),(u^(2),,u^(2),,r^(2)):}|` thenA. `r^(2)=x+y+z`B. `r^(2) =x^(2) =y^(2) +z^(2)`C. `u^(2) =yz+zx+xy`D. `u^(2) =xyz`

Prove that `|{:(x^(2),,x^(2)-(y-z)^(2),,yz),(y^(2),,y^(2)-(z-x)^(2),,zx),(z^(2),,z^(2)-(x-y)^(2),,xy):}|` `=(x-y) (y-z) (z-x)(x+y+z) (x^(2)+y^(2)+z^(2))`

if `|{:(3,1,-4),(3,2,5),(1,-1,3):}|=49`, then evaluate `|{:(6,3,-8/3),(6,6,10/3),(2,-3,2):}|`.

Prove that `|{:(1,x+alpha,y+z-alpha),(1,t+beta,+x-beta),(1,z+gamma,x+y-gamma):}|=0`

Solve the equation `|{:(x+2,1,-3),(1,x-3,-2),(-3,-2,1):}|=0.`

Prove that : `|{:(1,b,c),(b,c,a),(c,a,b):}|=3 a b c-a^(3)-b^(3)-c^(3)`

Prove that `|{:(1,b,c),(a^(2),b^(2),c^(2)),(b+c,c+a,a+b):}|=(a-b)(b-c)(a+b+c)`

Without, prove that : `|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}|=2|{:(a,b,c),(p,q,r),(x,y,z):}|`

Prove that : `=2|{:(1,1,1),(x,y,rz),(x^(2),y^(2),z^(2)):}|=(x-y)(y-z)(z-x)`

Using the property of determinants and without expanding, prove that:`|[0,a,-b],[-a,0,-c],[ b, c,0]|=0`

if a=cos 0 + in sin 0,=cos 20 -I sin 2 0, c= cos 0 + I sin 3 0 and if `|{:(a,,b,,c),(b,,c,,a),(c,,a,,b):}| =0 ` thenA. `=0 2 k pi, k in Z`B. ` 0= (2k +1) pi k in Z`C. `0= (4k +1) pi k in Z`D. none of these

Without, prove that : `|{:(1+b,b+c,c+a),(p+q,q+r,r+p),(x+y,y+z,z+x),(x+y,y+z,z+x):}|=2|{:(a,b,c),(p,q,r),(x,y,z):}|`

[ 28.If f(x)=|[0,x-a,x-bx+a,0,x-cx+b,x+c,0]| then,[ 1) f(a)=0, 2) f(b)=0]]A. `f(a)=0`B. `f(b)=0`C. `f(0)=0`D. `f(1)=0`

Can students Score high marks if they practice Determinants Class 12 Mock Test

If `A=[[2,a,-3] , [0,2,5] , [1,1,3]]` then,find the value of a for which `A^(-1)` exists.A. `a=2`B. `a!=2`C. `a!=-2`D. none of these

If A is a singular matrix, then write the value of |A|.

Find the value of the `"determinant" |{:(sqrt13+sqrt3,2sqrt5,sqrt5),(sqrt26+sqrt15,5,sqrt10),(sqrt65+3,sqrt15,5):}|`

If a determinant of order `3xx3` is formed by using the numbers 1 or -1 then minimum value of determinant is :A. -2B. -4C. 0D. -8

If `|[x^n, x^(n+2),x^(n+3)],[y^n, y^(n+2),y^(n+3)],[z^n, z^(n+2),z^(n+3)]|=(x-y)(y-z)(z-x)(1/x+1/y+1/z),`then `n`equalsA. 1B. -1C. 2D. -2

witout expanding at any stage Prove that `|{:(0,,a,,-b),(-a,,0,,-c),(b,,c,,0):}| =0`

for `x,x,z gt 0` Prove that `|{:(1,,log_(x)y,,log_(x)z),(log_(y)x,,1,,log_(y)z),(log_(z) x,,log_(z)y,,1):}| =0`

Prove that ` |{:(ax,,by,,cz),(x^(2),,y^(2),,z^(2)),(1,,1,,1):}|=|{:(a,,c,,c),(x,,y,,z),(yz,,xz,,xy):}|`

Using the property of determinants and without expanding, prove that:`|[b+c, q+r, y+z],[ c+a, r+p, z+x],[ a+b, p+q, x+y]|=2|[a, p, x],[ b, q ,y],[ c, r, z]|`

Evaluate the following: ` |[a,h,g],[h,b,f],[g,f,c]| `

If the determinant `|(cos2x,sin^2 x,cos 4x),(sin^2 x,cos 2x,cos^2 x),(cos 4x,cos^2 x,cos 2x)|` is expanded in powers of `sin x`, then the constant term isA. 1B. 0C. -1D. 2

consider the fourth -degree polynomial equation `|{:(a_(1)+b_(1)x,,a_(1)x^(2)+b_(1),,c_(1)),(a_(2)+b_(2)x^(2),,a_(2)x^(2)+b_(2),,c_(2)),(a_(3)+b_(3)x^(2),,a_(3)x^(2)+b_(3),,c_(3)):}|=0` Without expanding the determinant find all the roots of the equation.

Evaluate `|{:(2, 7, 65), (3, 8, 75), (5, 9, 86):}|`

Using the property of determinants and without expanding, prove that:`|[2, 7, 65],[ 3 ,8 ,75 ],[4, 9,86]|=0`

`" if " Delta_(r) = |{:(r,,612,,915),(101r^(2),,2r,,3r),(r,,(1)/(r),,(1)/(r^(2))):}|` then the value of `lim_( n to oo) .(1)/(n^(3)) (Sigma_(r=1)^(n) Delta_(r) " is " "____"`

`" Let " Delta_(r)=|{:(r-1,,n,,6),((r-1)^(2),,2n^(2),,4n-2),((r-1)^(2),,3n^(3),,3n^(2)-3n):}|. " Show that " Sigma_(r=1)^(n) Delta_(r)` is constant.

`D_(r) = |(2^(r -1),2.3^(r-1),4.5^(r -1)),(alpha,beta,gamma),(2^(n) -1,3^(n) -1,5^(n) -1)|`. Then, the value of `Sigma_(r=1)^(n) D_(r)` isA. `alpha beta gamma`B. `2^(n) alpha + 2^(n) beta + 4^(n) gamma`C. `2 alpha + 3 beta + 4 gamma`D. none of these

State whether the matrix \(\begin{bmatrix}2 &amp; 3 \\6 &amp; 4 \\\end{bmatrix}\) is singular or non-singular.

` " if " Delta_(r) =|{:(2^(r-1),,2xx3^(r-1),,4xx5^(r-1)),(alpha,,beta,,gamma),(2^(n)-1,,3^(n)-1,,5^(n)-1):}|` then find the value of `Sigma_(r=1)^(n) Delta r`.

Show that each of the following systems of linear equations has infinite number of solutions and solve : x + 2y = 5 3x + 6y = 15

Without expanding the determinant, prove that \(\begin{vmatrix} 41 &1 & 5 \\[0.3em] 79 & 7 &9\\[0.3em] 29 & 5 & 3 \end{vmatrix}\) = 0SINGULAR MATRIX A square matrix A is said to be singular if |A| = 0. Also, A is called non singular if |A| ≠ 0.

State whether the matrix \(\begin{bmatrix}2 & 3 \\6 & 4 \\\end{bmatrix}\) is singular or non-singular.