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NCERT Exemplar · Class 12 Mathematics Determinants

58 questions · 58 still being checked

EXERCISE 4.3 21–30 (part 3 of 6)

  1. Exercise 21

    If a+b+c≠0\displaystyle a+b+c \neq 0 and ∣abcbcacab∣=0\displaystyle \left|\begin{array}{lll}a & b & c \\ b & c & a \\ c & a & b\end{array}\right|=0, then prove that a=b=c\displaystyle a=b=c.

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    Let \(\displaystyle \Delta\) be the determinant. Apply \(\displaystyle R_1\to R_1+R_2+R_3\), then \(\displaystyle C_2\to C_2-C_1,\ C_3\to C_3-C_1\). \[\Delta=(a+b+c)\begin{vmatrix}1&1&1\\b&c&a\\c&a&b\end{vmatrix}=(a+b+c)\begin{vmatrix}1&0&0\\b&c-b&a-b\\c&a-c&b-c\end{vmatrix} \] \[\Delta=(a+b+c)\left[(c-b)(b-c)-(a-b)(a-c)\right] \] \[\Delta=-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) \] \[\Delta=-\tfrac12(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \] Since \(\displaystyle \Delta=0\) and \(\displaystyle a+b+c\neq0\): \[(a-b)^2+(b-c)^2+(c-a)^2=0 \] Squares of real numbers are non-negative, so each is zero. \[a-b=0,\quad b-c=0,\quad c-a=0 \] Answer: \(\displaystyle a=b=c\).

  2. Exercise 22

    Prove that ∣bc−a2ca−b2ab−c2ca−b2ab−c2bc−a2ab−c2bc−a2ca−b2∣\displaystyle \left|\begin{array}{lll}b c-a^2 & c a-b^2 & a b-c^2 \\ c a-b^2 & a b-c^2 & b c-a^2 \\ a b-c^2 & b c-a^2 & c a-b^2\end{array}\right| is divisible by a+b+c\displaystyle a+b+c and find the quotient.

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    Put \(\displaystyle p=bc-a^2,\ q=ca-b^2,\ r=ab-c^2\) and \(\displaystyle S=a^2+b^2+c^2-ab-bc-ca\). The determinant is \(\displaystyle D\). \[D=\begin{vmatrix}p&q&r\\q&r&p\\r&p&q\end{vmatrix} \] Apply \(\displaystyle R_1\to R_1+R_2+R_3\), then \(\displaystyle C_2\to C_2-C_1,\ C_3\to C_3-C_1\). \[D=(p+q+r)\begin{vmatrix}1&0&0\\q&r-q&p-q\\r&p-r&q-r\end{vmatrix} \] \[D=(p+q+r)\left[(r-q)(q-r)-(p-q)(p-r)\right]=-(p+q+r)(p^2+q^2+r^2-pq-qr-rp) \] Now express in \(\displaystyle a,b,c\). \[p+q+r=-S \] \[p-q=(b-a)(a+b+c),\quad q-r=(c-b)(a+b+c),\quad r-p=(a-c)(a+b+c) \] \[p^2+q^2+r^2-pq-qr-rp=\tfrac12\left[(p-q)^2+(q-r)^2+(r-p)^2\right]=(a+b+c)^2S \] \[D=-(-S)(a+b+c)^2S=(a+b+c)^2S^2 \] \[\frac{D}{a+b+c}=(a+b+c)\,S^2 \] So \(\displaystyle D\) is divisible by \(\displaystyle a+b+c\).Answer: quotient \(\displaystyle =(a+b+c)(a^2+b^2+c^2-ab-bc-ca)^2\).

  3. Exercise 23

    If x+y+z=0\displaystyle x+y+z=0, prove that ∣xaybzcyczaxbzbxcya∣=xyz∣abccabbca∣\displaystyle \left|\begin{array}{ccc}x a & y b & z c \\ y c & z a & x b \\ z b & x c & y a\end{array}\right|=x y z\left|\begin{array}{ccc}a & b & c \\ c & a & b \\ b & c & a\end{array}\right|

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    Expand the left side along the first row. \[\text{LHS}=xa\,(yz\,a^2-x^2bc)-yb\,(y^2ac-xz\,b^2)+zc\,(xy\,c^2-z^2ab) \] \[\text{LHS}=xyz\,(a^3+b^3+c^3)-abc\,(x^3+y^3+z^3) \] Use \(\displaystyle x+y+z=0\). \[x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)=0 \] \[x^3+y^3+z^3=3xyz \] \[\text{LHS}=xyz\,(a^3+b^3+c^3)-3xyz\,abc=xyz\,(a^3+b^3+c^3-3abc) \] Expand the determinant on the right. \[\begin{vmatrix}a&b&c\\c&a&b\\b&c&a\end{vmatrix}=a(a^2-bc)-b(ca-b^2)+c(c^2-ab)=a^3+b^3+c^3-3abc \] \[\text{LHS}=xyz\begin{vmatrix}a&b&c\\c&a&b\\b&c&a\end{vmatrix}=\text{RHS} \] Answer: the identity holds when \(\displaystyle x+y+z=0\).
  4. Choose the correct answer from given four options in each of the Exercises from $\displaystyle 24$ to 37.

    Exercise 24

    If ∣2x58x∣=∣6−273∣\displaystyle \left|\begin{array}{cc}2 x & 5 \\ 8 & x\end{array}\right|=\left|\begin{array}{cc}6 & -2 \\ 7 & 3\end{array}\right|, then value of x\displaystyle x is
    (A)
    3\displaystyle 3 (B) ±3\displaystyle \pm 3
    (C)
    ±6\displaystyle \pm 6

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    NCERT’s answer
    C
    (C) \(\displaystyle \pm 6\)Expand both sides. \[2x^2-40=18+14 \] \[x^2=36 \] \[x=\pm 6 \]

  5. Exercise 25

    The value of determinant ∣a−bb+cab−ac+abc−aa+bc∣\displaystyle \left|\begin{array}{lll}a-b & b+c & a \\ b-a & c+a & b \\ c-a & a+b & c\end{array}\right|
    (A)
    a3+b3+c3\displaystyle a^3+b^3+c^3
    (B)
    3bc\displaystyle 3 b c
    (C)
    a3+b3+c3−3abc\displaystyle a^3+b^3+c^3-3 a b c
    (D)
    none of these

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    (D) none of these\(\displaystyle C_2 \to C_2 + C_3\) makes every entry of \(\displaystyle C_2\) equal to \(\displaystyle a+b+c\); then \(\displaystyle R_2 \to R_2 - R_1,\; R_3 \to R_3 - R_1\).\[D = (a+b+c)\begin{vmatrix} a-b & 1 & a \\ b-a & 1 & b \\ c-a & 1 & c \end{vmatrix} \]\[D = (a+b+c)\begin{vmatrix} a-b & 1 & a \\ 2(b-a) & 0 & b-a \\ c+b-2a & 0 & c-a \end{vmatrix} \]Expand along \(\displaystyle C_2\):\[D = -(a+b+c)\left[\,2(b-a)(c-a) - (b-a)(c+b-2a)\,\right] \]\[D = -(a+b+c)(b-a)(c-b) = (a+b+c)(a-b)(c-b) \]At \(\displaystyle a=1,\ b=2,\ c=3\): \(\displaystyle D=-6\), while (A), (B), (C) give \(\displaystyle 36,\ 18,\ 18\).

  6. Exercise 26

    The area of a triangle with vertices (−3,0),(3,0)\displaystyle (-3,0),(3,0) and (0,k)\displaystyle (0, k) is 9\displaystyle 9 sq. units. The value of k\displaystyle k will be
    (A)
    9\displaystyle 9 (B) 3\displaystyle 3 (C) −9\displaystyle -9 (D) 6\displaystyle 6

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    NCERT’s answer
    B
    (B) \(\displaystyle 3\)Let \(\displaystyle A(-3,0),\ B(3,0),\ C(0,k)\).NCERT_Solution_Class12_Maths_Exemplar_Ch4_Ex4-3_Q26\[\Delta = \frac{1}{2}\begin{vmatrix} -3 & 0 & 1 \\ 3 & 0 & 1 \\ 0 & k & 1 \end{vmatrix} \]Expand along \(\displaystyle C_2\):\[\Delta = \frac{1}{2}\left|\,-k\,(-3-3)\,\right| = \frac{1}{2}\,|6k| = 3|k| \]\[3|k| = 9 \;\Rightarrow\; |k| = 3 \]\(\displaystyle k=\pm 3\); only \(\displaystyle 3\) is among the options.

  7. Exercise 27

    The determinant ∣b2−abb−cbc−acab−a2a−bb2−abbc−acc−aab−a2∣\displaystyle \left|\begin{array}{lll}b^2-a b & b-c & b c-a c \\ a b-a^2 & a-b & b^2-a b \\ b c-a c & c-a & a b-a^2\end{array}\right| equals
    (A)
    abc(b−c)(c−a)(a−b)\displaystyle a b c(b-c)(c-a)(a-b)
    (B)
    (b−c)(c−a)(a−b)\displaystyle (b-c)(c-a)(a-b)
    (C)
    (a+b+c)(b−c)(c−a)(a−b)\displaystyle (a+b+c)(b-c)(c-a)(a-b)
    (D)
    None of these

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    NCERT’s answer
    D
    (D) None of theseTake \(\displaystyle (b-a)\) out of \(\displaystyle C_1\) and out of \(\displaystyle C_3\):\[D = (b-a)^2\begin{vmatrix} b & b-c & c \\ a & a-b & b \\ c & c-a & a \end{vmatrix} \]\(\displaystyle C_2 \to C_2 - C_1 + C_3\):\[\big(b-c-b+c,\; a-b-a+b,\; c-a-c+a\big) = (0,\,0,\,0) \]\[D = 0 \](A), (B), (C) are not zero in general; at \(\displaystyle a=1,\ b=2,\ c=3\) they give \(\displaystyle 12,\ 2,\ 12\).

  8. Exercise 28

    The number of distinct real roots of ∣sin⁡xcos⁡xcos⁡xcos⁡xsin⁡xcos⁡xcos⁡xcos⁡xsin⁡x∣=0\displaystyle \left|\begin{array}{ccc}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0 in the interval −π4≤x≤π4\displaystyle -\frac{\pi}{4} \leq x \leq \frac{\pi}{4} is
    (A)
    0\displaystyle 0 (B) 2\displaystyle 2 (C) 1\displaystyle 1 (D) 3\displaystyle 3

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    NCERT’s answer
    C
    (C) \(\displaystyle 1\)\(\displaystyle C_1 \to C_1 + C_2 + C_3\) gives the factor \(\displaystyle \sin x + 2\cos x\); then \(\displaystyle R_2 \to R_2 - R_1,\; R_3 \to R_3 - R_1\).\[\Delta = (\sin x + 2\cos x)\begin{vmatrix} 1 & \cos x & \cos x \\ 1 & \sin x & \cos x \\ 1 & \cos x & \sin x \end{vmatrix} \]\[\Delta = (\sin x + 2\cos x)\begin{vmatrix} 1 & \cos x & \cos x \\ 0 & \sin x - \cos x & 0 \\ 0 & 0 & \sin x - \cos x \end{vmatrix} \]\[\Delta = (\sin x + 2\cos x)(\sin x - \cos x)^2 = 0 \]\[\tan x = -2 \quad\text{or}\quad \tan x = 1 \]On \(\displaystyle \left[-\frac{\pi}{4}, \frac{\pi}{4}\right]\), \(\displaystyle |\tan x| \le 1\), so \(\displaystyle \tan x=-2\) is rejected; only \(\displaystyle x=\frac{\pi}{4}\) remains.

  9. Exercise 29

    If A,B\displaystyle \mathrm{A}, \mathrm{B} and C are angles of a triangle, then the determinant ∣−1cos⁡Ccos⁡Bcos⁡C−1cos⁡Acos⁡Bcos⁡A−1∣ is equal to \left|\begin{array}{ccc} -1 & \cos \mathrm{C} & \cos \mathrm{B} \\ \cos \mathrm{C} & -1 & \cos \mathrm{A} \\ \cos \mathrm{B} & \cos \mathrm{A} & -1 \end{array}\right| \text { is equal to }
    (A)
    0\displaystyle 0 (B) −1\displaystyle -1 (C) 1\displaystyle 1 (D) None of these

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\)Expand along \(\displaystyle R_1\):\[\Delta = -(1-\cos^2 A) - \cos C(-\cos C - \cos A\cos B) + \cos B(\cos A\cos C + \cos B) \]\[\Delta = \cos^2 A + \cos^2 B + \cos^2 C + 2\cos A\cos B\cos C - 1 \]Since \(\displaystyle C = \pi - (A+B)\):\[\cos C = \sin A\sin B - \cos A\cos B \]\[\cos^2 C + 2\cos A\cos B\cos C = \cos C(\cos C + 2\cos A\cos B) = \sin^2 A\sin^2 B - \cos^2 A\cos^2 B \]\[\cos^2 A + \cos^2 B + \sin^2 A\sin^2 B - \cos^2 A\cos^2 B = \cos^2 A\sin^2 B + \sin^2 A\sin^2 B + \cos^2 B = 1 \]\[\Delta = 1 - 1 = 0 \]

  10. Exercise 30

    Let f(t)=∣cos⁡tt12sin⁡tt2tsin⁡ttt∣\displaystyle f(t)=\left|\begin{array}{ccc}\cos t & t & 1 \\ 2 \sin t & t & 2 t \\ \sin t & t & t\end{array}\right|, then lim⁡t→0f(t)t2\displaystyle \lim _{t \rightarrow 0} \frac{f(t)}{t^2} is equal to
    (A)
    0\displaystyle 0 (B) - 1\displaystyle 1 (C) 2\displaystyle 2 (D) 3\displaystyle 3

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    NCERT’s answer
    A
    (A) \(\displaystyle 0\)Expand along \(\displaystyle R_1\):\[f(t) = \cos t\,(t^2 - 2t^2) - t\,(2t\sin t - 2t\sin t) + 1\cdot(2t\sin t - t\sin t) \]\[f(t) = t\sin t - t^2\cos t \]\[\frac{f(t)}{t^2} = \frac{\sin t}{t} - \cos t \]\[\lim_{t\to 0}\frac{f(t)}{t^2} = 1 - 1 = 0 \]