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101.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}49 & 1 & 6 \\[0.3em]39 & 7 & 4 \\[0.3em]26 &2 & 3\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}49 & 1 & 6 \\[0.3em]39 & 7 & 4 \\[0.3em]26 &2 & 3\end{vmatrix}\)

Applying, 

C1→C1 – 8C3

⇒ Δ = \(\begin{vmatrix}1 & 1 & 6 \\[0.3em]7 & 7 & 4 \\[0.3em]2 &2 & 3\end{vmatrix}\)

As, 

C1 = C2 hence, the determinant is zero.
Discussion
102.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix} 1/a& a^2 & bc \\[0.3em] 1/b& b^2 &ac \\[0.3em] 1/c &c^2& ab\end{vmatrix}\) 

Multiplying R1, R2 and R3 with a, b and c respectively we get,

⇒ Δ = \(\begin{vmatrix} 1& a^3 & abc \\[0.3em] 1& b^3 &abc \\[0.3em] 1 &c^3& abc\end{vmatrix}\) 

Taking, a b c common from C3 gives,

⇒ Δ = \(\begin{vmatrix} 1& a^3 & 1 \\[0.3em] 1& b^3 &1 \\[0.3em] 1 &c^3& 1\end{vmatrix}\) 

As, 

C1 = C3 hence value of determinant is zero.
Discussion
103.

Without expanding, show that the value of each of the following determinants is zero:\(\begin{vmatrix}1 &a &a^2-bc \\[0.3em]1 & b & b^2-ac \\[0.3em]1 &c & c^2-ab\end{vmatrix}\) 

Answer»

Let Δ = \(\begin{vmatrix}1 &a &a^2-bc \\[0.3em]1 & b & b^2-ac \\[0.3em]1 &c & c^2-ab\end{vmatrix}\) 

⇒ Δ = \(\begin{vmatrix}1 &a &a^2 \\[0.3em]1 & b & b^2\\[0.3em]1 &c & c^2\end{vmatrix}\) - \(\begin{vmatrix}1 &a &bc \\[0.3em]1 & b & ac \\[0.3em]1 &c & ab\end{vmatrix}\) 

Applying R2→R2 – R1 and R3 → R3 – R1, we get,

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

Taking (b – a) and (c – a) common from R2 and R3 respectively,

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

= [(b – a)(c – a)][(c + a) – (b + a) – ( – b + c)] 

= [(b – a)(c – a)][c + a + b – a – b – c] 

= [(b – a)(c – a)][0] = 0
Discussion
104.

Without expanding, show that the value of each of the following determinants is zero :\(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\)

Answer»

Let Δ = \(\begin{vmatrix}a+b & 2a+b & 3a+b \\[0.3em]2a+b & 3a+b & 4a+b \\[0.3em]4a+b & 5a+b& 6a+b\end{vmatrix}\)

Applying C3 → C3 – C2, we get,

 Δ = \(\begin{vmatrix}a+b & 2a+b & a \\[0.3em]2a+b & 3a+b & a \\[0.3em]4a+b & 5a+b& a\end{vmatrix}\)

Applying C2→C2 – C1 gives,

  Δ = \(\begin{vmatrix}a+b & a & a \\[0.3em]2a+b & a & a \\[0.3em]4a+b & a& a\end{vmatrix}\)

As, 

C2 = C3, 

So the value of the determinant is zero.
Discussion
105.

if `f (x) = |{:(x,,a,,a),(a,,x,,a),(a,,a,,x):}|=0` thenA. `f(x) =0" and " f(x) =0` has one common rootB. `f(x) =0" and " f(x) =0` has one common rootC. sum of roots of f(x) =0 is -3aD. none of these

Answer» Correct Answer - B
Applying `C_(1) to C_(1)+C_(2) +C_(3)` we get
`Delta = |{:(x+2a,,a,,a),(x+2a,,x,,a),(x+2a,,a,,x):}|=(x +2a) |{:(1,,a,,a),(1,,x,,a),(1,,a,,x):}|`
Applying `R_(1) to R_(2) -R_(2) " and "R_(2) to R_(2) - R_(3)` we get
`Delta =(x+2a) |{:(0,,a-x,,0),(0,,x-a,,a-x),(1,,a,,x):}|=(x-a)^(2) (x+2a)`
Discussion
106.

the value of the determinant`|{:(.^(n)C_(r-1),,.^(n)C_(r),,(r+1)^(n+2)C_(r+1)),(.^(n)C_(r),,.^(n)C_(r+1),,(r+2)^(n+2)C_(r+2)),(.^(n)C_(r+1),,.^(n)C_(r+2),,(r+3)^(n+2)C_(r+3)):}|` isA. `n^(2)+n-1)`B. `0`C. `.^(n+3)C_(r+3)`D. `.^(n)C_(r-1)+^(n)C_(r)+^(n)C_(r+1)`

Answer» Correct Answer - B
`Delta = |{:(.^(n)C_(r-1),,.^(n)C_(r),,(r+1)^(n+2)C_(r+1)),(.^(n)C_(r),,.^(n)C_(r+1),,(n+2)^(n+1)C_(r+1)),(.^(n)C_(r+1),,.^(n)C_(r+2),,(r+3)^(n+2)C_(r+3)):}|`
Applying `C_(1) to C_(1)+C_(2) " and using " .^(n)C_(r)=(n)/(r )^(n-1) C_(r+1) " in " C_(3)` we get
` Delta =|{:(.^(n+1)C_(r),,.^(n)C_(r),,(n+2)^(n+1)C_(r)),(.^(n+1)C_(r+1),,.^(n)C_(r+1),,(n+2)^(n+1)C_(r+1)),(.^(n+1)C_(r+2),,.^(n)C_(r+2),,(n+2)^(n+1)C_(r+2)):}|`
`=(n+2) |{:(.^(n+1)C_(r),,.^(n)C_(r),,.^(n+1)C_(r)),(.^(n+1)C_(r+1),,.^(n)C_(r+1),,.^(n+1)C_(r+1)),(.^(n+1)C_(r+2),,.^(n)C_(r+2),,.^(n+1)C_(r+2)):}|`
( as `C_(1) " and " C_(3)`are identical )
Discussion
107.

The greatest value of n for which the determinant `Delta = |(1,1,1),(.^(n)C_(1),.^(n+3)C_(1),.^(n+6)C_(1)),(.^(n)C_(2),.^(n+3)C_(2),.^(n+6)C_(2))|` is divisible by `3^(n)`, isA. 7B. 5C. 3D. 1

Answer» Correct Answer - C
We have,
`Delta = (1)/(2) |(1,1,1),(n,n +3,n +6),(n(n-1),(n+2) (n+3),(n+5) (n +6))|`
`rArr Delta = (1)/(2) |(1,0,0),(n,3,6),(n(n-1),6n + 6,12n + 30)|`
`rArr Delta = 9 |(1,0,0),(n,1,1),(n(n-1),2(n +1),2n +5)| = 27`
Clearly, `Delta = 27` is divisible by 3, `3^(2) and 3^(3)`. Hence, n = 3
Discussion
108.

The value of the determinant `|(1,1,1),(.^(m)C_(1),.^(m +1)C_(1),.^(m+2)C_(1)),(.^(m)C_(2),.^(m +1)C_(2),.^(m+2)C_(2))|` is equal toA. 1B. -1C. 0D. none of these

Answer» Correct Answer - A
`|{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+2)C_(1)),(.^(m)C_(2),,.^(m+1)C_(2),,.^(m+2)C_(2)):}|`
`=|{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+1)C_(0)+.^(m+1)C_(1)),(.^(m)C_(1),,.^(m+1)C_(2),,.^(m+1)C_(1)+.^(m+1)C_(2)):}|
|{:(1,,1,,1),(.^(m)C_(1),,.^(m+1)C_(1),,.^(m+1)C_(0)),(.^(m)C_(2),,,^(m+1)C_(2),,.^(m+1)C_(1)):}|" ""[Applying " C_(3) to C_(3) -C_(2)"]"`
`=|{:(1,,1,,1),(.^(m)C_(1),,.^(m)C_(0).^(m)C_(1),,.^(m+1)C_(0)),(.^(m)C_(2),,.^(m)C_(1)+.^(m)C_(2),,.^(m+1)C_(1)):}|`
`=|{:(1,,0,,0),(.^(m)C_(1),,.^(m)C_(0),,.^(m+1)C_(0)),(.^(m)C_(2),,.^(m)C_(1),,.^(m+1)C_(1)):}|" ""[Applying " C_(2) to C_(2) -C_(1)" ]"`
`=.^(m)C_(0).^(m+1)C_(1)-^(m+1)C_(0).^(m)C_(1)`
`=m +1-m`
`=1`
Discussion
109.

for the equation `|{:(1,,x,,x^(2)),(x^(2),,1,,x),(x,,x^(2),,6):}|=0`A. There are exactly two distinct rootsB. there is one pair of equation real rootsC. There are three pairs of equal rootsD. Modulus of each root is 2

Answer» Correct Answer - C
`Delta = (1+ x+x^(2)) |{:(1,,1,,1),(x^(2),,1,,x),(x,,x^(2),,1):}| =(1-x^(3))^(2)`
Therefore `Delta =0` has roots `1,1 ,omega, omega, omega^(2),omega^(2)`
Discussion
110.

If `|x^n x^(n+2)x^(2n)1x^a a x^(n+5)x^(a+6)x^(2n+5)|=0,AAx in R ,w h e r en in N ,`then value of `a`is`n`b. `n-1`c. `n+1`d. none of theseA. nB. n-1C. n+1D. none of these

Answer» Correct Answer - C
Taking `x^(5)` common from last row we bet
`x^(5) |{:(x^(n),,x^(n+2),,x^(2n)),(1,,x^(a),,a),(x^(n),,x^(a+1),,x^(2n)):}|=0 ,AA x in R`
`rArr a +1 = n+2 " or " a = n+1`
(as it will make first and third row is identical )
Discussion
111.

`" if " |{:(x,,3,,6),(3,,6,,x),(6,,x,,3):}|= |{:(2,,x,,7),(x,,7,,2),(7,,2,,x):}|=|{:(4,,5,,x),(5,,x,,4),(x,,4,,5):}|=0 ` then x is equal to

Answer» Correct Answer - B
In each determinant applying `R_(1) to R_(1)+R_(2)+R_(3)` and then taking out `(x+9)` common we get
Discussion
112.

if f(x) =`|{:(mx,,mx-p,,mx-p),(n,,n+p,,n-p),(mx+2n,,mx+2n+p,,mx+2n-p):}|` then y=f(x) representsA. a straight line parallet to x-axisB. a striaght line parallel to y-axisC. parabolaD. a striaght line with negative slope

Answer» Correct Answer - B
`R_(3) to R_(3)- 2R_(2)` hence two indentical rows `rArr f(x) =` constant
Discussion
113.

If `|(1,1,1),(a,b,c),(a^(3),b^(3),c^(3))|= (a -b) (b -c) (c -a) (a + b+c)` where a, b, c are all different, then the determinant `|(1,1,1),((x-a)^(2),(x -b)^(2),(x -c)^(2)),((x -b) (x -c),(x -c) (x -a),(x -a) (x -b))|` vanishes whenA. `a+b+c=0`B. `x=(1)/(3) (a+b+c)`C. `x=(1)/(2) (a+b+c)`D. `x=a+b+c`

Answer» Correct Answer - B
we have
`|{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}|=(a-b)(b-c)(c-) (a+b+c)`
`" Also " |{:(1,,1,,1),(a,,b,,c),(a^(3),,b^(3),,c^(3)):}|`
`=abc |{:((1)/(a),,(1)/(b),,(1)/(c)),(1,,1,,1),(a^(2),,b^(2),,c^(2)):}|" ""(taking a,b,c common from "R_(1),R_(2),R_(3)")"`
`=|{:(bc,,ac,,ab),(1,,1,,1),(a^(2),,b^(2),,c^(2)):}|` (Multiplying `R_(1)` by abc)
`=|{:(1,,1,,1),(a^(2),,b^(2),,c^(2)),(bc,,ac,,ab):}|`
`" Then " D = |{:(1,,1,,1),((x-a)^(2),,(x-b)^(2),,(x-c)^(2)),((x-b)(x-c),,(x-c)(x-a),,(x-a)(x-b)):}|`
Now given that a,b and C are all different So D=0 when
`x=(1)/(3) (a+b+c)`
Discussion
114.

If `a_1, a_2, a_3,.......` are in G.P. then the value of determinant `|(log(a_n), log(a_(n+1)), log(a_(n+2))), (log(a_(n+3)), log(a_(n+4)), log(a_(n+5))), (log(a_(n+6)), log(a_(n+7)), log(a_(n+8)))|` equals(A) 0 (B) 1 (C) 2 (D) 3

Answer» Correct Answer - A
Discussion
115.

The value of the determinant `|k a k^2+a^2 1k b k^2+b^2 1k c k^2+c^2 1|`is`k(a+b)(b+c)(c+a)``k a b c(a^2+b^(f2)+c^2)``k(a-b)(b-c)(c-a)``k(a+b-c)(b+c-a)(c+a-b)`A. `k(a+b)(b+c)(c+a)`B. `k abc (a^(2)+b^(2)+c^(2))`C. `k(a-b)(b-c)(c-a)`D. `k(a+b-c)(b+c-a)(c+a-b)`

Answer» Correct Answer - C
we have
`|{:(ka,,k^(2)+a^(2),,1),(kb,,k^(2)+b^(2),,1),(kc,,k^(2)+c^(2),,1):}|=|{:(ka,,k^(2),,1),(kb,,k^(2),,1),(kc,,k^(2),,1):}|+|{:(ka,,a^(2),,1),(kb,,b^(2),,1),(kc,,c^(2),,1):}|`
`=0+k |{:(a,,a^(2),,1),(b,,b^(2),,1),(c,,c^(2),,1):}|`
`=k(a-b) (b-c) (c-a)`
Discussion
116.

If `a_1, a_2, a_3,.......` are in G.P. then the value of determinant `|(log(a_n), log(a_(n+1)), log(a_(n+2))), (log(a_(n+3)), log(a_(n+4)), log(a_(n+5))), (log(a_(n+6)), log(a_(n+7)), log(a_(n+8)))|` equalsA. 1B. 0C. 2aD. a

Answer» Correct Answer - A
we have
`(a_(n+1))^(2) =a_(n)a_(n)+2`
`rArr 2 log a_(n+1) =log a_(n)+ log a_(n+2)`
Similarly
`2 log a_(n+4) =log a_(n+3)+ log a_(n+2)`
`2log a_(n+7)= log a_(n+6) + log a+(n+8)`
Substituting these values in second column of determinant we get
`Delta =(1)/(2) |{:(log a_(n),,loga_(n)+loga_(n+2),,log a_(n+2)),(log a_(n+3),,log a_(n+3)+log a_(n+5),,log a_(n+5)),(log a_(n+6),,log_(n+6)+loga_(n+8),,log a_(n+8)):}|`
`=(1)/(2) (0) =0" ""[Using " C_(2) to C_(2) -C_(1)-C_(3)"]"`
Discussion
117.

Which of the following is not the root of the equation `|x-6-1 2-3xx-3-3 2xx+2|=0?``2`b. `0`c. `1`d. `-3`A. 2B. 0C. 1D. -3

Answer» Correct Answer - B
Operation `R_(1) to R_(1)-R_(2)` gives
`Delta =|{:(x-2,,3(x-2),,-(x-2)),(2,,-3x,,x-3),(-3,,2x,,x+2):}|`
`=(x-2)|{:(1,,3,,-1),(2,,-3x,,x-3),(-3,,2x,,x+2):}|`
`=(x-2) |{:(1,,3,,-1),(0,,-3(x+2),,x-1),(0,,2x+9,,x-1):}|" " underset(R_(3) to R_(3) +3R_(1)]([R_(2) to R_(2) -2R_(1)]]`
`=(x-2){-(3x+6)(x-1)-(x-1)(2x+9)}`
`=-(x-2)(x-1) (5x+15)`
Therefore `Delta =0 " gives " x=2 ,1,=3`
Discussion
118.

Value of `|x+y z z x y+z x y y z+x|,w h e r ex ,y ,z`are nonzero real number, is equal to`x y z`b. `2x y z`c. `3x y z`d. `4x y z`A. xyzB. 2xyzC. 3xyzD. 4xyz

Answer» Correct Answer - D
Applying `R_(1) to R_(1) -(R_(2) +R_(3))` we get
`D = |{:(0,,-2y,,-2x),(x,,y+z,,x),(y,,y,,z+x):}| `
`=2|{:(0,,-y,,-x),(x,,y+z,,x),(y,,y,,z+x):}|`
`=2 |{:(0,,-y,,-x),(x,,z,,0),(y,,0,,z):}|" " underset(R_(3) to R_(3) +R_(1))((R_(2) to R_(2)+R_(1) " and "))`
`=4xyz`
Discussion
119.

If the system of equations `x-k y-z=0, k x-y-z=0,x+y-z=0`has a nonzero solution, then the possible value of `k`are`-1,2`b. `1,2`c. `0,1`d. `-1,1`A. `-1, 2`B. `1, 2`C. 0, 1D. `-1, 1`

Answer» Correct Answer - D
The given system of equations has non-zero i.e., non -trivial solution
`:. |(1,-k,-1),(k,-1,-1),(1,1,-1)|= 0`
`rArr|(1,-k,-1),(k -1,-1 +k,0),(0,1 +k,0)| = 0 " " [("Applying " R_(2) rarr R_(2) - R_(1)),(and R_(3) rarrR_(3) - R_(1))]`
`rArr - (k^(2) -1) = 0 rArr k = +- 1`
Discussion
120.

For which value of k, det\(\begin{vmatrix}k&2\\[0.3em]4&-3\end{vmatrix}\) will be zero ?

Answer»

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

⇒ - 3k - 8 = 0 

⇒ - 3k = 8 

⇒ k = - 8/3
Discussion
121.

The system of equations `alphax+y+z=alpha-1, x+alphay+z=alpha-1, x+y+alphaz=alpha-1` has no solution if alpha is (A) 1 (B) not -2 (C) either -2 or 1 (D) -2A. 1B. not `-2`C. either `-2 or 1`D. `-2`

Answer» Correct Answer - D
For `alpha = 1`, the system reduces to a homogeneous system which is always consistent. So, `alpha != 1`.
For `alpha != 1`, we have
`D = |(alpha,1,1),(1,alpha,1),(1,1,alpha)| = |(alpha + 2 ,alpha + 2,alpha + 2),(1,alpha,1),(1,1,alpha)| ["Applying" R_(1) rarr R_(1) + R_(2) + R_(3)]`
`rArr D = (alpha + 2) |(1,1,1),(1,alpha,1),(1,1,alpha)| = (alpha + 2) |(1,0,0),(1,alpha -1,0),(1,0,alpha -1)| ["Applying " C_(2) rarr C_(2) - C_(1), C_(3) rarr C_(3) - C_(1)]`
`rArr D = (alpha +2) (alpha -1)^(2)`
and `D_(1) = |(alpha - 1,1,1),(alpha -1,alpha,1),(alpha -1,1,alpha)| = (alpha -1) |(1,1,1),(1,alpha,1),(1,1,alpha)|`
`rArr D_(1) = (alpha -1) |(1,0,0),(1,alpha -1,0),(1,0,alpha -1)| [("Applying " C_(2) rarr C_(2) - C_(1)),(C_(3) rarr C_(3) - C_(1))]`
`rArr D_(1) = (alpha -1)^(3)`
Clearly, `D = 0 " for " alpha = -2 " but " D_(1) != 0`
So, the system is inconsistent for `alpha = -2`
Discussion
122.

`|[a,a^2,a^3-1],[b,b^2,b^3-1],[c,c^2,c^3-1]|=0 ` prove that `abc= I `

Answer» `|[a,a^2,a^3-1],[b,b^2,b^3-1],[c,c^2,c^3-1]| = 0`
`=>|[a,a^2,a^3],[b,b^2,b^3],[c,c^2,c^3]|-|[a,a^2,1],[b,b^2,1],[c,c^2,1]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-|[a,a^2,1],[b,b^2,1],[c,c^2,1]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-(-1)^2|[1,a,a^2],[1,b,b^2],[1,c,c^2]| = 0`
`=>abc|[1,a,a^2],[1,b,b^2],[1,c,c^2]|-|[1,a,a^2],[1,b,b^2],[1,c,c^2]| = 0`
`=>|[1,a,a^2],[1,b,b^2],[1,c,c^2]|[abc-I] = 0`
As, `|[1,a,a^2],[1,b,b^2],[1,c,c^2]|` can not be `0`.
`:. abc - I = 0`
`=>abc = I.`
Discussion
123.

Show that `|[1,alpha,alpha^2],[1,beta,beta^2],[1,gamma,gamma^2]|=(alpha-beta)(beta-gamma)(gamma-alpha)`

Answer» `L.H.S. = |[1,alpha,alpha^2],[1,beta,beta^2],[1,gamma,gamma^2]|`
Applying `R_1->R_1-R_3` and `R_2->R_2-R_3`
`= |[0,alpha-gamma,alpha^2-gamma^2],[0,beta-gamma,beta^2-gamma^2],[1,gamma,gamma^2]|`
`=(alpha-gamma)(beta-gamma)|[0,1,alpha+gamma],[0,1,beta+gamma],[1,gamma,gamma^2]|`
`=(alpha-gamma)(beta-gamma)[beta+gamma - alpha-gamma]`
`= (alpha-gamma)(beta-gamma)(beta-alpha)`
`=(-1)(-1)(alpha-beta)(beta-gamma)(gamma-alpha)`
`=(alpha-beta)(beta-gamma)(gamma-alpha) = R.H.S.`
Discussion
124.

If A = \(\begin{bmatrix}3& 4 \\[0.3em]1 & 2 \\[0.3em]\end{bmatrix}\) find the value of 3|A|.

Answer»

Find the determinant of A and then multiply it by 3 

|A|=2 

3|A|=3 × 2 

=6
Discussion
125.

If \(\begin{bmatrix}2{\text{x}} & {\text{x}} +3 \\[0.3em]2({\text{x}} +1) & {\text{x}} +1\end{bmatrix}\) = \(\begin{bmatrix}1 & 5 \\[0.3em]3 & 3\end{bmatrix}\), Write the value of x.

Answer»

Simply by equating both sides we can get the value of x. 

2x2+2x - 2(x2+4x+3)= -12 

⇒ -6x -6 = -12 

⇒ -6x = -6 

⇒ x = 1
Discussion
126.

If \(\begin{vmatrix}2x& 5 \\[0.3em]8 & x\end{vmatrix}\) = \(\begin{vmatrix}6& -2 \\[0.3em]7 & 3\end{vmatrix}\), Write the value of x.

Answer»

this question is having the same logic as above. 

2x2-40 = 18+14 

⇒ 2x2 = 72 

⇒ x2 = 36 

⇒ x = \(\pm\)6
Discussion
127.

If \(\begin{bmatrix}3x & 7 & \\[0.3em]-2& 4 \\[0.3em]\end{bmatrix}\) = \(\begin{bmatrix}8 & 7 \\[0.3em]6& 4 \\[0.3em]\end{bmatrix}\), Write the  value of x.

Answer»

Here the determinant is compared so we need to take determinant both sides then find x. 

12x +14 = 32 - 42 

⇒ 12x = -10-14 

⇒ 12x = -24 

⇒ x = -2
Discussion
128.

Find the value of x, if \(\begin{vmatrix} 2& 3 \\[0.3em] 4& 5 \end{vmatrix}\)= \(\begin{vmatrix} x& 3 \\[0.3em] 2x& 5 \end{vmatrix}\)

Answer»

\(\begin{vmatrix} 2& 3 \\[0.3em] 4& 5 \end{vmatrix}\)= \(\begin{vmatrix} x& 3 \\[0.3em] 2x& 5 \end{vmatrix}\) 

⇒ 2 × 5 – 4 × 3 = x × 5 – 2x × 3 

⇒ 10 – 12 = 5x – 6x 

⇒ –x = –2 

⇒ x = 2
Discussion
129.

If A = \(\begin{bmatrix}1 & 0 & 1 \\[0.3em]0&1 & 2 \\[0.3em]0 & 0 & 4\end{bmatrix}\), then show that |3A| = 27|A|.

Answer»

|A| = \(\begin{vmatrix} 1 & 0 & 1 \\[0.3em] 0&1 & 2 \\[0.3em] 0 & 0 & 4 \end{vmatrix}\) 

Expanding along the first row,

|A| = 1\(\begin{vmatrix} 1 & 2 \\[0.3em] 0&4 \\[0.3em] \end{vmatrix}\) - 0\(\begin{vmatrix} 0 & 2 \\[0.3em] 0&4 \\[0.3em] \end{vmatrix}\) + 1\(\begin{vmatrix} 0 & 1 \\[0.3em] 0&0 \\[0.3em] \end{vmatrix}\)

= 1(1×4 – 2×0) – 0(0×4 – 0×2) + 1(0×0 – 0×1) 

= 1(4 – 0) + 0 + 1(0 + 0) 

= 1×4 

= 4 

Now,

|3A| = \(\begin{vmatrix} 3 & 0 & 3 \\[0.3em] 0&3 & 6 \\[0.3em] 0 & 0 & 12 \end{vmatrix}\) 

Expanding along the first row,

 |3A| = 3\(\begin{vmatrix} 3 & 6 \\[0.3em] 0&12 \\[0.3em] \end{vmatrix}\) - 0\(\begin{vmatrix} 0 & 6 \\[0.3em] 0&12 \\[0.3em] \end{vmatrix}\) + 3\(\begin{vmatrix} 0 & 3 \\[0.3em] 0&0 \\[0.3em] \end{vmatrix}\)

= 3(3×12 – 6×0) – 0(0×12 – 0×6) + 3(0×0 – 0×3) 

= 3(36 – 0) + 0 + 3(0 + 0) 

= 3×36 = 108 

= 27 × 4 

= 27 |A| 

Hence, 

|3A|= 27 |A|
Discussion
130.

Find the value of x, if \(\begin{vmatrix}2& 4 \\[0.3em]5& 1\end{vmatrix}\)= \(\begin{vmatrix}2x& 4 \\[0.3em]6& x\end{vmatrix}\)

Answer»

\(\begin{vmatrix}2& 4 \\[0.3em]5& 1\end{vmatrix}\)= \(\begin{vmatrix}2x& 4 \\[0.3em]6& x\end{vmatrix}\) 

⇒ 2 × 1 – 4 × 5 = 2x × x – 4 × 6 

⇒ 2 – 20 = 2x2 – 24 

⇒ 2x2 = –18 +24 

⇒ 2x2 = 6 

⇒ x2 = 3 

⇒ x = ±√ 3
Discussion
131.

For what value of x matrix A is singular?A = \(\begin{bmatrix}x-1 & 1 & 1 \\[0.3em]1 & x-1 & 1 \\[0.3em]1 &1 & x-1\end{bmatrix}\)

Answer»

|A| = \(\begin{vmatrix}x-1 & 1 & 1 \\[0.3em]1 & x-1 & 1 \\[0.3em]1 &1 & x-1\end{vmatrix}\) 

Expanding along the first row,

|A| = (x - 1)\(\begin{vmatrix}x-1 & 1 \\[0.3em]1 & x-1 \\[0.3em]\end{vmatrix}\) - 1\(\begin{vmatrix}1 & 1 \\[0.3em]1 & x-1 \\[0.3em]\end{vmatrix}\) + \(\begin{vmatrix}1 & x-1 \\[0.3em]1 & 1 \\[0.3em]\end{vmatrix}\) 

= (x–1) ((x–1) (x–1)– 1×1) – 1((x–1) – 1×1) + 1(1×1 – 1×(x–1)) 

= (x–1) (x2 – 2x + 1 – 1) – 1(x–1 – 1) + 1(1 – x+1) 

= x(x–1) (x– 2) – 1(x–2) – (x–2) 

= (x– 2) {x(x–1) – 1 – 1} 

= (x– 2) (x2 – x – 2)

For singular |A| = 0,

(x– 2) (x2 – x – 2) = 0 

(x– 2) (x2 – 2x + x – 2) = 0 

(x–2)(x–2)(x+1) = 0 

∴ x = –1 or 2 

Also, 

|A| = 28 

⇒ 7x2 + 3x – 6 = 28

⇒ 7x2 + 3x – 34 = 0 

⇒ 7x2 + 17x – 14x – 34 = 0 

⇒ x(7x+ 17) – 2(7x +17) = 0 

⇒ (x–2)(7x +17) = 0
Discussion
132.

For what value of x matrix A is singular?A = \(\begin{bmatrix}1+x & 7 \\[0.3em]3-x & 8\end{bmatrix}\)

Answer»

|A| = 0

\(\begin{bmatrix}1+x & 7 \\[0.3em]3-x & 8\end{bmatrix}\)= 0

⇒ (1 + x) × 8 – 7 × (3 – x) = 0 

⇒ 8 + 8x – 21 + 7x = 0 

⇒ 15x – 13 = 0

⇒ x = \(\frac{13}{15}\)
Discussion
133.

Find the integral value of x, if \(\begin{vmatrix}x^2 & x & 1 \\[0.3em]0& 2 & 1 \\[0.3em]3 &1 &4\end{vmatrix}\) = 28

Answer»

|A| = \(\begin{vmatrix}x^2 & x & 1 \\[0.3em]0& 2 & 1 \\[0.3em]3 &1 &4\end{vmatrix}\)

Expanding along the first row,

|A| = x2\(\begin{vmatrix}2 & 1 \\[0.3em]1&4 \\[0.3em]\end{vmatrix}\) -   x\(\begin{vmatrix}0 & 1 \\[0.3em]3&4 \\[0.3em]\end{vmatrix}\) + 1\(\begin{vmatrix}0 & 2 \\[0.3em]3&1 \\[0.3em]\end{vmatrix}\)

= x2(2×4 – 1×1) – x(0×4 – 1×3) + 1(0×1 – 2×3) 

= x2(8 – 1) – x(0 – 3) + 1(0 – 6) 

= 7x2 + 3x –6 

Also, 

|A| = 28 

⇒ 7x2 + 3x – 6 =28 

⇒ 7x2 + 3x – 34 = 0 

⇒ 7x2 + 17x – 14x – 34 = 0 

⇒ x(7x+ 17) – 2(7x +17) = 0 

⇒ (x – 2)(7x +17) = 0

x = 2, \(-\frac{17}{7}\)

Integer value of x is 2.
Discussion
134.

Find the value of x, if\(\begin{vmatrix} x+1& x-1 \\[0.3em] x-3& x+2 \end{vmatrix}\) = \(\begin{vmatrix} 4& -1 \\[0.3em]1& 3 \end{vmatrix}\)

Answer»

Discussion
135.

Find the value of x, if\(\begin{vmatrix} 3x& 7 \\[0.3em] 2& 4 \end{vmatrix}\) = 10

Answer»

 \(\begin{vmatrix} 3x& 7 \\[0.3em] 2& 4 \end{vmatrix}\) = 10

⇒ 3x × 4 – 7 × 2 = 10

⇒ 12x – 14 = 10 

⇒ 12x= 10 +14 

⇒ 12x = 24 

⇒ x = 2
Discussion
136.

Evaluate : Δ = \(\begin{vmatrix}0& sin\,\alpha& -cos\,\alpha \\[0.3em]-sin\,\alpha& 0 & sin\,\beta \\[0.3em]cos\,\alpha & -sin\,\beta &0\end{vmatrix}\)

Answer»

Δ = \(\begin{vmatrix}0& sin\,\alpha& -cos\,\alpha \\[0.3em]-sin\,\alpha& 0 & sin\,\beta \\[0.3em]cos\,\alpha & -sin\,\beta &0\end{vmatrix}\)

Expanding along the first row,

|A| = 0\(\begin{vmatrix}0& sin\,\beta \\[0.3em]-sin\,\beta& 0 \\[0.3em]\end{vmatrix}\) - sin α \(\begin{vmatrix}-sin\,\alpha& sin\,\beta \\[0.3em]cos\,\alpha& 0 \\[0.3em]\end{vmatrix}\) - cos α \(\begin{vmatrix}-sin\,\alpha& 0 \\[0.3em]cos\,\alpha& -sin\,\beta \\[0.3em]\end{vmatrix}\)

⇒ |A| = 0(0 – sinβ(–sinβ) ) –sinα(–sinα× 0 – sinβ cosα ) – cosα((–sinα)(–sinβ) – 0× cosα ) 

|A| = 0 + sinα sinβ cosα – cosα sinα sinβ 

|A| = 0
Discussion
137.

If `A=[2-1 1-1 2-1 1-1 2]`. Verify that `A^3-6A^2+9A-4I=O`and hence find `A^(-1)`.

Answer» `A=[{:(2,-1,1),(-1,2,-1),(1,-1,2):}],`
`therefore" "A^(2)=A.A=[{:(2,-1,1),(-1,2,-1),(1,-1,2):}][{:(2,-1,1),(-1,2,-1),(1,-1,2):}]`
`[{:(4+1+1+,-2-2-1,2+1+2),(-2-2-1,1+4+1,-1-2-2),(2+1+2,-1-2-2,1+1+4):}]=[{:(6,-5,5),(-5,6,-5),(5,-5,6):}]`
`A^(3)=A^(2).A=[{:(6,-5,5),(-5,6,-5),(5,-5,6):}][{:(2,-1,1),(-1,2,-1),(1,-1,2):}]`
`[{:(12+5+5,-6-10-5,6+5+10),(-10-6-5,5+12+5,-5-6-10),(10+5+6,-5-10-6,5+5+12):}]`
`[{:(22,-21,21),(-21,22,21),(21,-21,22):}]`
`"Now L.H.S.=" A^(3)-6A^(2)+9A-4I`
`[{:(22,-21,21),(-21,22,21),(21,-21,22):}]-6[{:(6,-5,5),(-5,6,-5),(5,-5,6):}]`
`+9[{:(2,-1,1),(-1,2,-1),(1,-1,2):}]-4[{:(1,0,0),(0,1,0),(0,0,1):}]`
`=[{:(22,-21,21),(-21,22,21),(21,-21,22):}]+[{:(-36,30,-30),(30,-30,30),(-30,30,-30):}]`
`+[{:(18,-9,9),(-9,18,-9),(9,9-,18):}]+[{:(-4,0,0),(0,-4,0),(0,0,-4):}]`
`=[{:(0,0,0),(0,0,0),(0,0,0):}]=0=R.H.S.`
`"Now |A|="[{:(2,-1,1),(-1,2,-1),(1,-1,2):}]`
=2(4-1)-(-1)(-2+1)+1(1-2)
`=6-1-1=ne0`
`therefore A^(-1)`exists.
We have proved that `A^(3)-6A^(2)+9A-4I=0`
`rArr A^(-1)(A^(3)-6A^(2)+9A-4I)=A^(-1)0`
`rArr" "A^(2)-6A+9I-4A^(-1)=0`
`rArr" "4A^(-1)=A^(-1)A^(2)-6A+9I`
`[{:(6,-5,5),(-5,6,-5),(5,-5,6):}]-6[{:(2,-1,1),(-1,2,-1),(1,-1,2):}]+9[{:(1,0,0),(0,1,0),(0,0,1):}]`
`[{:(6,-5,5),(-5,6,-5),(5,-5,6):}]+[{:(-12,6,-6),(6,-12,6),(-6,6,-12):}]+[{:(9,0,0),(0,9,0),(0,0,9):}]`
`A^(-1)=1/4[{:(3,1,-1),(1,3,1),(-1,1,3):}]`
`rArr" "A^(-1)=1/4[{:(3,1,-1),(1,3,1),(-1,1,3):}]`
Discussion
138.

Find the value of x, if\(\begin{vmatrix} 3& x \\[0.3em] x& 1 \end{vmatrix}\)= \(\begin{vmatrix} 3& 2 \\[0.3em] 4& 1 \end{vmatrix}\)

Answer»

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

⇒ 3 × 1 – x × x = 3 × 1 – 4 × 2 

⇒ 3 – x2 = 3 – 8 

⇒ – x2 = –5 –3 

⇒ –x2 = –8 

⇒ x = ±2√ 2
Discussion
139.

If `a !=b`, then the system of equation `ax + by + bz = 0` `bx + ay + bz = 0` and `bx + by + az = 0` will have a non-trivial solution, ifA. `a + b = 0`B. `a + 2b = 0`C. `2a + b = 0`D. `a + 4b = 0`

Answer» Correct Answer - B
Discussion
140.

`|[x+lambda, 2x, 2x], [2x, x+lambda, 2x], [2x, 2x, x+lambda]| =(5x+ lambda)(lambda-x)^(2)`

Answer» Apply `R_(1) to R_(1) + R_(2) +R_(3) " and take" (5x + lambda) "common from "R_(1)`
Discussion
141.

`|[x+4,2x,2x] , [2x,x+4,2x] , [2x,2x,x+4]|=(5x+4)(x-4)^2`

Answer» Apply `R_(1) to R_(1) + R_(2) + R_(3) "and take (5x +4) common from"R_(1)`
Discussion
142.

`|[x, a, a], [a, x, a], [a, a, x]| =(x +2)(x-a)^(2)`

Answer» Apply `C_(1) to C_(1) + C_(2) + C_(3) " and take (x+2a) common from"C_(1)`
Discussion
143.

`|[1, 1, 1], [a, b, c], [bc, ca, ab]| = (a-b)(b-c)(c-a)`

Answer» Apply `C_(1) to C_(1) -C_(3) " and "C_(2) to C_(2) -C_(3)`
Discussion
144.

`|[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» Apply `R_(2) to R_(2) -R_(1) "and "R_(3) to R_(3) - 3R_(1)`
Discussion
145.

Using properties of determinants, prove that`|a+x y z x a+y z x y a+z|=a^2(a+x+y+z)`

Answer» Apply `C_(1) to C_(1) + C_(2) + C_(3) "and take (a+x+y+z) common from "C_(1)`
Discussion
146.

Show that`|1 1+p1+p+q2 3+2p1+3p+2q3 6+3p 106 p+3q|=1.`

Answer» `" Let " Delta =|{:(1,,1+p,,1+p+q),(2,,3+2p,,1+3p+2q),(3,,6+3p,,1+6p+3q):}|`
Applying `C_(2) to C_(2) -pC " and C_(3) to C_(3) -qC_(1)` we get
`Delta= |{:(1,,1,,1+p),(2,,3,,1+3),(3,,6,,1+6p):}|`
`=|{:(1,,1,,1),(2,,3,,1),(3,,6,,1):}|" ""[Applying "C_(3)to C_(3) -pC_(3)"]"`
`=|{:(0,,0,,1),(1,,2,,1),(2,,5,,1):}|" ""[Applying "C_(1)to C_(1)-C_(3),C_(2)to C_(2)-"]"`
`=5-4=1" ""[Expanding ]"`
Discussion
147.

Show that: `|3a-a+b-a+c-b+a3b-b+c-c+a-c+b3c|=3(a+b+c)(a b+b c+c a)dot`

Answer» `" Let " Delta =|{:(3a,,-a+b,,-a+c),(-b+a,,3b,,-b+c),(-c+a,,-c+b,,3c):}|`
`=(a+b+c) |{:(1,,-a+b,,-a+c),(1,,3b,,-b+c),(1,,-c+b,,3c):}|`
`=(a+b+c) |{:(1,,-a+b,,-a+c),(0,,2b+a,,-b+a),(0,,-c+a,,2c+a):}|`
`"[Applyint " R_(2) toR_(2) -R_(1) ,R_(3) to R_(3) -R_(1)]`
`=(a+b+c) |{:(2b+a,,-b+a),(-c+a,,2c+a):}|` [Expanding along `C_(1)]`
`=(a+b+c){(2b+a)(2c+a)-(-b+a)(-c+a)}`
`=(a+b+c){(4bc+2ab+2ca+a^(2))`
`-(bc-ab-ac+a^(2))`
`=(a+b+c)(3bc+3ab+3ca)`
`=3(a+b+c)(ab+bc+ca)`
Discussion
148.

Show that `|{:(a-x,,c,,b),(c ,,b-x,,a),( b,, a,,c-x):}|=0` where `a+b+c ne0`

Answer» Correct Answer - `x=a+b+c,ne sqrt((1)/(2)(a-b)^(2)+(b-c)(2)+(c-a)^(2))`
Applying `C_(1) to C_(1) +C_(2)+C_(3)` we get
`Delta=|{:(a+b+c-x,,c,,b),(a+b+c-x,,b-x,,a),(a+b+c-x,,a,,c-x):}|`
`=(a+b+c-x) |{:(1,,c,,b),(1,,b-x,,a),(1,,a,,c-x):}|`
Applying `R_(1) to R_(1)-R_(2),R_(2)toR_(2)-R_(3),` we get
`Delta =(a+b+c-x) |{:(0,,c-b+x,,b-a),(0,,b-a-x,,a-c+x),(1,,a,,c-x):}|`
`=(a+b+c -x){(x^(2)+x(a-b)+c-b)(a-c)`
`+(x+a-b)(b-a)`
`=(a+b+c-x){(x^(2)+x(a-b)+(c-b)(a-c)`
`+x(b-a)-(a-b)^(2)}`
`=(a+b+c-x)(x^(2)+ac-c^(2)-ab+bc-a^(2)-b^(2)+2ab)`
`=(a+b+c-x)(x^(2)-a^(2)-b^(2)-c^(2)+ab+bc+ca)`
`=(a+b+c-x)(x^(2)-(1)/(2){(a-b)^(2)+(b-c)^(2)+(c-a)^(2))})`
Since `Delta =0` we have
`x=a +b +c, +- sqrt((1)/(2)((a-b)^(2)+(b-c)^(2)+(c-a)^(2)))`
Discussion
149.

`a!=p , b!=q,c!=r` and `|(p,b,c),(a,q,c),(a,b,r)|=0` the value of `p/(p-a)+q/(q-b)+r/(r-c)=`A. 3B. 2C. 1D. 0

Answer» Correct Answer - B
We have,
`|(p,b,c),(p + q,q +b,2c),(a,b,r)|= 0`
`rArr |(p,b,c),(a -p,q -b,0),(a -p,0,r -c)| = 0 [("Applying " R_(2) rarr R_(2) - 2R_(1)),(R_(3) rarr R_(3) - R_(1))]`
`rArr p (q -b) (r -c) - (a - p) ( r-c) - (a -p) c (q -b) = 0`
`rArr p(q -b)(r -c) + b (p-a) (r -c) + c (p -a) (q -b) = 0`
`rArr (p)/(p -a) + (b)/(q -b) + (c)/(r -c) = 0 " " [("Dividing throughout by"),((p-q) (q -b) (r -c))]`
`rArr (p)/(p-q) + ((b)/(q -b) + 1) + ((c)/(r -c) + 1) = 2`
`rArr (p)/( p-q) + (q)/(q -b) + (r)/(r -c) = 2`
Discussion
150.

`{:("If",a = 1 + 2 + 4 + ..."to n terms"),(,b = 1 + 3 + 9 + ..."to n terms"),(,c =1 + 5 + 25+ ..."to n terms"):}` then `|(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|=`A. `30^(n)`B. `10^(n)`C. 0D. `2^(n) + 3^(n) + 5^(n)`

Answer» Correct Answer - C
We have,
`{:(a = 1 + 2 + 4,..."to n terms",= 2^(n) -1),(b = 1+ 3 + 9 +,..."to n terms",= (1)/(2) (3^(n) -1)),(c =1 + 5 + 25 +,..."to n terms",= (1)/(4) (5^(n) -1)):}`
`:. |(a,2b,4c),(2,2,2),(2^(n),3^(n),5^(n))|`
`= |(2^(n) -1,3^(n) -1,5^(n) -1),(2,2,2),(2^(n),3^(n),5^(n))|`
`= 2 |(2^(n) -1,3^(n) -1,5^(n) -1),(1,1,1),(1,1,1)|` Applying `R_(3) rarr R_(3) - R_(1)` and taking 2 common `R_(2)
`=2 xx 0 = 0`
Discussion