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NCERT Exemplar · Class 12 Mathematics Determinants

58 questions · 58 still being checked

EXERCISE 4.3 31–40 (part 4 of 6)

  1. Choose the correct answer from given four options in each of the Exercises from $\displaystyle 24$ to 37.

    Exercise 31

    The maximum value of Δ=∣11111+sin⁡θ11+cos⁡θ11∣\displaystyle \Delta=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+\sin \theta & 1 \\ 1+\cos \theta & 1 & 1\end{array}\right| is (θ\displaystyle \theta is real number)
    (A)
    12\displaystyle \frac{1}{2}
    (B)
    32\displaystyle \frac{\sqrt{3}}{2}
    (C)
    2\displaystyle \sqrt{2}
    (D)
    234\displaystyle \frac{2 \sqrt{3}}{4}

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    NCERT’s answer
    A
    (A) \(\displaystyle \frac{1}{2}\)\(\displaystyle R_2 \to R_2 - R_1,\; R_3 \to R_3 - R_1\):\[\Delta = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ \cos\theta & 0 & 0 \end{vmatrix} \]Expand along \(\displaystyle R_2\):\[\Delta = \sin\theta\,(0 - \cos\theta) = -\sin\theta\cos\theta = -\frac{1}{2}\sin 2\theta \]\[\Delta_{\max} = \frac{1}{2} \quad (\sin 2\theta = -1) \]

  2. Exercise 32

    If f(x)=∣0x−ax−bx+a0x−cx+bx+c0∣\displaystyle f(x)=\left|\begin{array}{ccc}0 & x-a & x-b \\ x+a & 0 & x-c \\ x+b & x+c & 0\end{array}\right|, then
    (A)
    f(a)=0\displaystyle f(a)=0
    (B)
    f(b)=0\displaystyle f(b)=0
    (C)
    f(0)=0\displaystyle f(0)=0
    (D)
    f(1)=0\displaystyle f(1)=0

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    C
    (C) \(\displaystyle f(0)=0\)Expand along \(\displaystyle R_1\):\[f(x) = (x-a)(x-c)(x+b) + (x-b)(x+a)(x+c) \]\[f(0) = abc - abc = 0 \]\[f(a) = 2a(a-b)(a+c), \qquad f(b) = 2b(b-a)(b-c) \]At \(\displaystyle a=1,\ b=2,\ c=3\): \(\displaystyle f(a)=f(1)=-8\) and \(\displaystyle f(b)=-4\), so (A), (B), (D) fail.

  3. Exercise 33

    If A=[2λ−3025113]\displaystyle \mathrm{A}=\left[\begin{array}{ccc}2 & \lambda & -3 \\ 0 & 2 & 5 \\ 1 & 1 & 3\end{array}\right], then A−1\displaystyle \mathrm{A}^{-1} exists if
    (A)
    λ=2\displaystyle \lambda=2
    (B)
    λ≠2\displaystyle \lambda \neq 2
    (C)
    λ≠−2\displaystyle \lambda \neq-2
    (D)
    None of these

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    NCERT’s answer
    D
    (D) \(\displaystyle A^{-1}\) exists exactly when \(\displaystyle |A|\neq 0\); none of (A)–(C) states that condition. \[|A| = 2(6-5) - \lambda(0-5) + (-3)(0-2) \] \[|A| = 5\lambda + 8 \] \[|A|\neq 0 \iff \lambda \neq -\tfrac{8}{5} \]

  4. Exercise 34

    If A and B are invertible matrices, then which of the following is not correct?
    (A)
    adj⁡A=∣A∣.A−1\displaystyle \operatorname{adj} \mathrm{A}=|\mathrm{A}| . \mathrm{A}^{-1}
    (B)
    det⁡(A)−1=[det⁡(A)]−1\displaystyle \operatorname{det}(\mathrm{A})^{-1}=[\operatorname{det}(\mathrm{A})]^{-1}
    (C)
    (AB)−1=B−1 A−1\displaystyle (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}
    (D)
    (A+B)−1=B−1+A−1\displaystyle (\mathrm{A}+\mathrm{B})^{-1}=\mathrm{B}^{-1}+\mathrm{A}^{-1}

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    NCERT’s answer
    D
    (D) (A), (B), (C) are standard results; (D) fails. Take \(\displaystyle A=B=I\): \[(I+I)^{-1} = \tfrac{1}{2}I \] \[I^{-1}+I^{-1} = 2I \] \[\tfrac{1}{2}I \neq 2I \]

  5. Exercise 35

    If x,y,z\displaystyle x, y, z are all different from zero and ∣1+x1111+y1111+z∣=0\displaystyle \left|\begin{array}{ccc}1+x & 1 & 1 \\ 1 & 1+y & 1 \\ 1 & 1 & 1+z\end{array}\right|=0, then value of x−1+y−1+z−1\displaystyle x^{-1}+y^{-1}+z^{-1} is
    (A)
    xyz\displaystyle x y z
    (B)
    x−1y−1z−1\displaystyle x^{-1} y^{-1} z^{-1}
    (C)
    −x−y−z\displaystyle -x-y-z

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    NCERT’s answer
    D
    (D) Expand the determinant along the first row, then divide by \(\displaystyle xyz\). \[\Delta = (1+x)\left[(1+y)(1+z)-1\right] - \left[(1+z)-1\right] + \left[1-(1+y)\right] \] \[\Delta = (1+x)(y+z+yz) - z - y \] \[\Delta = xyz + xy + yz + zx = 0 \] \[1 + \frac{1}{z} + \frac{1}{x} + \frac{1}{y} = 0 \quad (\div\, xyz) \] \[x^{-1}+y^{-1}+z^{-1} = -1 \]

  6. Exercise 36

    The value of the determinant ∣xx+yx+2yx+2yxx+yx+yx+2yx∣\displaystyle \left|\begin{array}{ccc}x & x+y & x+2 y \\ x+2 y & x & x+y \\ x+y & x+2 y & x\end{array}\right| is
    (A)
    9x2(x+y)\displaystyle 9 x^2(x+y)
    (B)
    9y2(x+y)\displaystyle 9 y^2(x+y)
    (C)
    3y2(x+y)\displaystyle 3 y^2(x+y)
    (D)
    7x2(x+y)\displaystyle 7 x^2(x+y)

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    NCERT’s answer
    B
    (B) Add the columns into \(\displaystyle C_1\), then clear below the first entry. \[\Delta = 3(x+y)\left|\begin{array}{ccc}1 & x+y & x+2y \\ 1 & x & x+y \\ 1 & x+2y & x\end{array}\right| \quad (C_1\to C_1+C_2+C_3,\ \text{take out } 3(x+y)) \] \[\Delta = 3(x+y)\left|\begin{array}{ccc}1 & x+y & x+2y \\ 0 & -y & -y \\ 0 & y & -2y\end{array}\right| \quad (R_2\to R_2-R_1,\ R_3\to R_3-R_1) \] \[\Delta = 3(x+y)\left[(-y)(-2y)-(-y)(y)\right] \] \[\Delta = 3(x+y)\cdot 3y^2 = 9y^2(x+y) \]

  7. Exercise 37

    There are two values of a\displaystyle a which makes determinant, Δ=∣1−252a−1042a∣=86\displaystyle \Delta=\left|\begin{array}{ccc}1 & -2 & 5 \\ 2 & a & -1 \\ 0 & 4 & 2 a\end{array}\right|=86, then sum of these number is
    (A)
    4\displaystyle 4 (B) 5\displaystyle 5 (C) - 4\displaystyle 4 (D) 9\displaystyle 9

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    NCERT’s answer
    C
    (C) Expand along the first row, solve for \(\displaystyle a\), add the two roots. \[\Delta = 1(2a^2+4) - (-2)(4a) + 5(8) \] \[\Delta = 2a^2 + 8a + 44 = 86 \] \[a^2 + 4a - 21 = 0 \] \[(a+7)(a-3) = 0 \Rightarrow a = -7,\ 3 \] \[-7 + 3 = -4 \]
  8. Fill in the blanks

    Exercise 38

    If A is a matrix of order 3×3\displaystyle 3 \times 3, then ∣3 A∣=\displaystyle |3 \mathrm{~A}|= ____.

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    \(\displaystyle 27|\mathrm{A}|\)
    \(\displaystyle 27\,|A|\) Multiplying every one of the three rows by $\displaystyle 3$ pulls out a factor $\displaystyle 3$ from each row. \[|3A| = 3^3\,|A| = 27\,|A| \]

  9. Exercise 39

    If A is invertible matrix of order 3×3\displaystyle 3 \times 3, then ∣A−1∣\displaystyle \left|\mathrm{A}^{-1}\right| ____.

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    NCERT’s answer
    \(\displaystyle \frac{1}{|\mathrm{A}|}\)
    \(\displaystyle \dfrac{1}{|A|}\) Take the determinant of \(\displaystyle AA^{-1}=I\) and use the product rule. \[AA^{-1} = I \] \[|A|\,|A^{-1}| = |I| = 1 \] \[|A^{-1}| = \frac{1}{|A|} \quad (|A|\neq 0) \]

  10. Exercise 40

    If x,y,z∈R\displaystyle x, y, z \in \mathrm{R}, then the value of determinant ∣(2x+2−x)2(2x−2−x)21(3x+3−x)2(3x−3−x)21(4x+4−x)2(4x−4−x)21∣\displaystyle \left|\begin{array}{ccc}\left(2^x+2^{-x}\right)^2 & \left(2^x-2^{-x}\right)^2 & 1 \\ \left(3^x+3^{-x}\right)^2 & \left(3^x-3^{-x}\right)^2 & 1 \\ \left(4^x+4^{-x}\right)^2 & \left(4^x-4^{-x}\right)^2 & 1\end{array}\right| is equal to ____.

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    NCERT’s answer
    Zero
    $\displaystyle 0$ For \(\displaystyle k=2,3,4\), the first two entries of each row differ by a constant. \[(k^x+k^{-x})^2 - (k^x-k^{-x})^2 = 4\,k^x k^{-x} = 4 \] \[\Delta = \left|\begin{array}{ccc}4 & (2^x-2^{-x})^2 & 1 \\ 4 & (3^x-3^{-x})^2 & 1 \\ 4 & (4^x-4^{-x})^2 & 1\end{array}\right| \quad (C_1\to C_1-C_2) \] \[C_1 = 4\,C_3 \Rightarrow \Delta = 0 \]