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NCERT Exemplar · Class 12 Mathematics Determinants

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EXERCISE 4.3 11–20 (part 2 of 6)

  1. Exercise 11

    If the co-ordinates of the vertices of an equilateral triangle with sides of length ‘a\displaystyle a’ are (x1,y1),(x2,y2),(x3,y3)\displaystyle \left(x_1, y_1\right),\left(x_2, y_2\right),\left(x_3, y_3\right), then ∣x1y11x2y21x3y31∣2=3a44\displaystyle \left|\begin{array}{lll}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{array}\right|^2=\frac{3 a^4}{4}.

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    NCERT_Solution_Class12_Maths_Exemplar_Ch4_Ex4-3_Q11 Let \(\displaystyle \Delta\) be the determinant and take \(\displaystyle A(x_1,y_1)\), \(\displaystyle B(x_2,y_2)\), \(\displaystyle C(x_3,y_3)\). \[\text{Area}(\triangle ABC)=\frac12\left|\Delta\right| \] \[\text{Area}(\triangle ABC)=\frac{\sqrt3}{4}a^2 \quad \text{(equilateral, side } a\text{)} \] \[\frac12|\Delta|=\frac{\sqrt3}{4}a^2 \] \[|\Delta|=\frac{\sqrt3}{2}a^2 \] \[\Delta^2=\frac{3}{4}a^4 \] Answer: \(\displaystyle \Delta^2=\dfrac{3a^4}{4}\)

  2. Exercise 12

    Find the value of θ\displaystyle \theta satisfying [11sin⁡3θ−43cos⁡2θ7−7−2]=0\displaystyle \left[\begin{array}{ccc}1 & 1 & \sin 3 \theta \\ -4 & 3 & \cos 2 \theta \\ 7 & -7 & -2\end{array}\right]=0.

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    NCERT’s answer
    \(\displaystyle \theta=n \pi\) or \(\displaystyle n \pi+(-1)^n\left(\frac{\pi}{6}\right)\)
    Expand along \(\displaystyle R_1\): \[\Delta=(-6+7\cos2\theta)-(8-7\cos2\theta)+\sin3\theta\,(28-21) \] \[\Delta=14\cos2\theta+7\sin3\theta-14=0 \] \[2\cos2\theta+\sin3\theta-2=0 \] With \(\displaystyle \cos2\theta=1-2\sin^2\theta\) and \(\displaystyle \sin3\theta=3\sin\theta-4\sin^3\theta\): \[-4\sin^2\theta+3\sin\theta-4\sin^3\theta=0 \] \[\sin\theta\,(2\sin\theta-1)(2\sin\theta+3)=0 \] \(\displaystyle \sin\theta=-\tfrac32\) is impossible, so \[\sin\theta=0 \;\Rightarrow\; \theta=n\pi \] \[\sin\theta=\tfrac12 \;\Rightarrow\; \theta=n\pi+(-1)^n\frac{\pi}{6} \] Answer: \(\displaystyle \theta=n\pi\) or \(\displaystyle \theta=n\pi+(-1)^n\dfrac{\pi}{6}\), \(\displaystyle n\in\mathbb Z\)

  3. Exercise 13

    If [4−x4+x4+x4+x4−x4+x4+x4+x4−x]=0\displaystyle \left[\begin{array}{lll}4-x & 4+x & 4+x \\ 4+x & 4-x & 4+x \\ 4+x & 4+x & 4-x\end{array}\right]=0, then find values of x\displaystyle x.

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    NCERT’s answer
    \(\displaystyle x=0,-12\)
    Apply \(\displaystyle C_1\to C_1+C_2+C_3\) (every row sums to \(\displaystyle 12+x\)): \[\Delta=(12+x)\begin{vmatrix}1&4+x&4+x\\1&4-x&4+x\\1&4+x&4-x\end{vmatrix} \] Apply \(\displaystyle R_2\to R_2-R_1\), \(\displaystyle R_3\to R_3-R_1\): \[\Delta=(12+x)\begin{vmatrix}1&4+x&4+x\\0&-2x&0\\0&0&-2x\end{vmatrix} \] \[\Delta=(12+x)(-2x)(-2x)=4x^2(12+x)=0 \] Answer: \(\displaystyle x=0\) or \(\displaystyle x=-12\)

  4. Exercise 14

    If a1,a2,a3,…,ar\displaystyle a_1, a_2, a_3, \ldots, a_r are in G.P., then prove that the determinant ∣ar+1ar+5ar+9ar+7ar+11ar+15ar+11ar+17ar+21∣\displaystyle \left|\begin{array}{ccc}a_{r+1} & a_{r+5} & a_{r+9} \\ a_{r+7} & a_{r+11} & a_{r+15} \\ a_{r+11} & a_{r+17} & a_{r+21}\end{array}\right| is independent of r\displaystyle r.

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    Let the common ratio be \(\displaystyle k\), so \(\displaystyle a_n=a_1k^{\,n-1}\). \[\Delta=\begin{vmatrix} a_1k^{r} & a_1k^{r+4} & a_1k^{r+8}\\ a_1k^{r+6} & a_1k^{r+10} & a_1k^{r+14}\\ a_1k^{r+10} & a_1k^{r+16} & a_1k^{r+20}\end{vmatrix} \] Compare the first two rows: \[R_2=k^6R_1 \] Two proportional rows give \[\Delta=0 \] Answer: \(\displaystyle \Delta=0\) for every \(\displaystyle r\), so it is independent of \(\displaystyle r\).

  5. Exercise 15

    Show that the points (a+5,a−4),(a−2,a+3)\displaystyle (a+5, a-4),(a-2, a+3) and (a,a)\displaystyle (a, a) do not lie on a straight line for any value of a\displaystyle a.

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    NCERT_Solution_Class12_Maths_Exemplar_Ch4_Ex4-3_Q15 Three points are collinear only if the triangle they form has zero area. \[\Delta=\begin{vmatrix}a+5&a-4&1\\a-2&a+3&1\\a&a&1\end{vmatrix} \] Apply \(\displaystyle R_1\to R_1-R_2\), \(\displaystyle R_2\to R_2-R_3\): \[\Delta=\begin{vmatrix}7&-7&0\\-2&3&0\\a&a&1\end{vmatrix} \] \[\Delta=7\cdot3-(-7)(-2)=7 \] \[\text{Area}=\frac12|\Delta|=\frac72\neq0 \] Answer: The area is \(\displaystyle \tfrac72\) whatever \(\displaystyle a\) is, so the points are never collinear.

  6. Exercise 16

    Show that the ΔABC\displaystyle \Delta \mathrm{ABC} is an isosceles triangle if the determinant Δ=[1111+cos⁡A1+cos⁡B1+cos⁡Ccos⁡2A+cos⁡Acos⁡2B+cos⁡Bcos⁡2C+cos⁡C]=0.\Delta=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 1+\cos \mathrm{A} & 1+\cos \mathrm{B} & 1+\cos \mathrm{C} \\ \cos ^2 \mathrm{A}+\cos \mathrm{A} & \cos ^2 \mathrm{B}+\cos \mathrm{B} & \cos ^2 \mathrm{C}+\cos \mathrm{C} \end{array}\right]=0 .

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    Put \(\displaystyle x=\cos A\), \(\displaystyle y=\cos B\), \(\displaystyle z=\cos C\). Apply \(\displaystyle R_2\to R_2-R_1\), then \(\displaystyle R_3\to R_3-R_2\): \[\Delta=\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix} \] Apply \(\displaystyle C_1\to C_1-C_2\), \(\displaystyle C_2\to C_2-C_3\): \[\Delta=\begin{vmatrix}0&0&1\\x-y&y-z&z\\x^2-y^2&y^2-z^2&z^2\end{vmatrix} \] \[\Delta=(x-y)(y-z)\begin{vmatrix}0&0&1\\1&1&z\\x+y&y+z&z^2\end{vmatrix} \] \[\Delta=(x-y)(y-z)(z-x)=0 \] \[\cos A=\cos B \ \text{ or } \ \cos B=\cos C \ \text{ or } \ \cos C=\cos A \] Cosine is one-one on \(\displaystyle (0,\pi)\), so two angles are equal, say \[B=C \;\Rightarrow\; AB=AC \quad \text{(sides opposite equal angles)} \] NCERT_Solution_Class12_Maths_Exemplar_Ch4_Ex4-3_Q16 Answer: Two angles of \(\displaystyle \triangle ABC\) are equal, so it is isosceles.

  7. Exercise 17

    Find A−1\displaystyle A^{-1} if A=[011101110]\displaystyle A=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right] and show that A−1=A2−3I2\displaystyle A^{-1}=\frac{A^2-3 I}{2}.

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    \[|A| = 0(0-1)-1(0-1)+1(1-0) = 2 \neq 0 \] Cofactors: diagonal ones are \(\displaystyle -1\), all others are \(\displaystyle 1\). \[\operatorname{adj}A=\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix} \] \[A^{-1}=\frac{\operatorname{adj}A}{|A|}=\frac12\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix} \] Now square \(\displaystyle A\). \[A^2=\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}\begin{bmatrix}0&1&1\\1&0&1\\1&1&0\end{bmatrix}=\begin{bmatrix}2&1&1\\1&2&1\\1&1&2\end{bmatrix} \] \[A^2-3I=\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}=2A^{-1} \] \[A^{-1}=\frac{A^2-3I}{2} \] Answer: \(\displaystyle A^{-1}=\dfrac12\begin{bmatrix}-1&1&1\\1&-1&1\\1&1&-1\end{bmatrix}\), and it equals \(\displaystyle \dfrac{A^2-3I}{2}\).

  8. Exercise 18

    If A=[120−2−1−20−11]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right], find A−1\displaystyle \mathrm{A}^{-1}. Using A−1\displaystyle A^{-1}, solve the system of linear equations x−2y=10,2x−y−z=8,−2y+z=7\displaystyle x-2 y=10,2 x-y-z=8,-2 y+z=7.

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    NCERT’s answer
    \(\displaystyle x=0, y=-5, z=-3\)
    \[|A| = 1(-1-2)-2(-2-0)+0 = 1 \] Cofactors: \[A_{11}=-3,\ A_{12}=2,\ A_{13}=2,\ A_{21}=-2,\ A_{22}=1,\ A_{23}=1,\ A_{31}=-4,\ A_{32}=2,\ A_{33}=3 \] \[A^{-1}=\frac{\operatorname{adj}A}{1}=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix} \] The coefficient matrix of the system is \(\displaystyle A^{T}\). \[\begin{bmatrix}1&-2&0\\2&-1&-1\\0&-2&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}10\\8\\7\end{bmatrix} \] \[X=(A^{T})^{-1}B=(A^{-1})^{T}B=\begin{bmatrix}-3&2&2\\-2&1&1\\-4&2&3\end{bmatrix}\begin{bmatrix}10\\8\\7\end{bmatrix} \] \[X=\begin{bmatrix}-30+16+14\\-20+8+7\\-40+16+21\end{bmatrix}=\begin{bmatrix}0\\-5\\-3\end{bmatrix} \] Answer: \(\displaystyle A^{-1}=\begin{bmatrix}-3&-2&-4\\2&1&2\\2&1&3\end{bmatrix}\); \(\displaystyle x=0,\ y=-5,\ z=-3\).

  9. Exercise 19

    Using matrix method, solve the system of equations 3x+2y−2z=3,x+2y+3z=6,2x−y+z=2\displaystyle 3 x+2 y-2 z=3, x+2 y+3 z=6,2 x-y+z=2.

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    NCERT’s answer
    \(\displaystyle x=1, y=1, z=1\)
    \[\begin{bmatrix}3&2&-2\\1&2&3\\2&-1&1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\6\\2\end{bmatrix},\qquad AX=B \] \[|A|=3(2+3)-2(1-6)-2(-1-4)=15+10+10=35 \] Cofactors: \[A_{11}=5,\ A_{12}=5,\ A_{13}=-5,\ A_{21}=0,\ A_{22}=7,\ A_{23}=7,\ A_{31}=10,\ A_{32}=-11,\ A_{33}=4 \] \[\operatorname{adj}A=\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix} \] \[X=A^{-1}B=\frac1{35}\begin{bmatrix}5&0&10\\5&7&-11\\-5&7&4\end{bmatrix}\begin{bmatrix}3\\6\\2\end{bmatrix} \] \[X=\frac1{35}\begin{bmatrix}15+0+20\\15+42-22\\-15+42+8\end{bmatrix}=\frac1{35}\begin{bmatrix}35\\35\\35\end{bmatrix} \] Answer: \(\displaystyle x=1,\ y=1,\ z=1\).

  10. Exercise 20

    Given A=[22−4−42−42−15],B=[1−10234012]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}2 & 2 & -4 \\ -4 & 2 & -4 \\ 2 & -1 & 5\end{array}\right], \mathrm{B}=\left[\begin{array}{ccc}1 & -1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2\end{array}\right], find BA and use this to solve the system of equations y+2z=7,x−y=3,2x+3y+4z=17\displaystyle y+2 z=7, x-y=3,2 x+3 y+4 z=17.

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    NCERT’s answer
    \(\displaystyle x=2, y=-1, z=4\)
    \[BA=\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix} \] \[BA=\begin{bmatrix}2+4+0&2-2+0&-4+4+0\\4-12+8&4+6-4&-8-12+20\\0-4+4&0+2-2&0-4+10\end{bmatrix}=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}=6I \] \[B^{-1}=\frac{A}{6} \] Write the system in the order \(\displaystyle x-y=3,\ 2x+3y+4z=17,\ y+2z=7\). \[\begin{bmatrix}1&-1&0\\2&3&4\\0&1&2\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}3\\17\\7\end{bmatrix},\qquad BX=C \] \[X=B^{-1}C=\frac16\begin{bmatrix}2&2&-4\\-4&2&-4\\2&-1&5\end{bmatrix}\begin{bmatrix}3\\17\\7\end{bmatrix} \] \[X=\frac16\begin{bmatrix}6+34-28\\-12+34-28\\6-17+35\end{bmatrix}=\frac16\begin{bmatrix}12\\-6\\24\end{bmatrix} \] Answer: \(\displaystyle BA=6I\); \(\displaystyle x=2,\ y=-1,\ z=4\).