SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Determinants

58 questions · 58 still being checked

EXERCISE 4.3 41–50 (part 5 of 6)

  1. Fill in the blanks

    Exercise 41

    If cos⁡2θ=0\displaystyle \cos 2 \theta=0, then ∣0cos⁡θsin⁡θcos⁡θsin⁡θ0sin⁡θ0cos⁡θ∣2=\displaystyle \left|\begin{array}{ccc}0 & \cos \theta & \sin \theta \\ \cos \theta & \sin \theta & 0 \\ \sin \theta & 0 & \cos \theta\end{array}\right|^2= ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{1}{2}\)
    \(\displaystyle \tfrac12\) (for \(\displaystyle \theta=\pi/4\)) \[\Delta = 0 - \cos\theta\,(\cos^2\theta - 0) + \sin\theta\,(0 - \sin^2\theta) \] \[\Delta = -(\cos^3\theta + \sin^3\theta) \] \[\cos 2\theta = 0 \Rightarrow 2\theta = \tfrac{\pi}{2} \Rightarrow \theta = \tfrac{\pi}{4} \] \[\sin\theta = \cos\theta = \tfrac{1}{\sqrt2} \Rightarrow \Delta = -\tfrac{2}{2\sqrt2} = -\tfrac{1}{\sqrt2} \] \[\Delta^2 = \tfrac12 \] The other branch \(\displaystyle \theta = \tfrac{3\pi}{4}\) makes \(\displaystyle \sin\theta + \cos\theta = 0\), so \(\displaystyle \Delta^2 = 0\).

  2. Exercise 42

    If A is a matrix of order 3×3\displaystyle 3 \times 3, then (A2)−1=\displaystyle \left(\mathrm{A}^2\right)^{-1}= ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \left(\mathrm{A}^{-1}\right)^2\)
    \(\displaystyle (\mathrm{A}^{-1})^2\) \[(A^2)^{-1} = (AA)^{-1} \] \[(AB)^{-1} = B^{-1}A^{-1} \Rightarrow (AA)^{-1} = A^{-1}A^{-1} \] \[(A^2)^{-1} = (A^{-1})^2 \]

  3. Exercise 43

    If A is a matrix of order 3×3\displaystyle 3 \times 3, then number of minors in determinant of A are ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    $\displaystyle 9$
    $\displaystyle 9$ \[M_{ij},\quad i = 1,2,3;\ j = 1,2,3 \] \[3 \times 3 = 9 \]

  4. Exercise 44

    The sum of the products of elements of any row with the co-factors of corresponding elements is equal to ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Value of the determinant
    the value of the determinant, \(\displaystyle |A|\) \[|A| = a_{i1}C_{i1} + a_{i2}C_{i2} + a_{i3}C_{i3} \quad \text{(any row } i\text{)} \] \[a_{i1}C_{j1} + a_{i2}C_{j2} + a_{i3}C_{j3} = 0 \quad (i \neq j) \]

  5. Exercise 45

    If x=−9\displaystyle x=-9 is a root of ∣x372x276x∣=0\displaystyle \left|\begin{array}{ccc}x & 3 & 7 \\ 2 & x & 2 \\ 7 & 6 & x\end{array}\right|=0, then other two roots are ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle x=2\) \(\displaystyle y=7\)
    \(\displaystyle 2\) and \(\displaystyle 7\) \[\Delta = x(x^2 - 12) - 3(2x - 14) + 7(12 - 7x) \] \[\Delta = x^3 - 67x + 126 \] \[x = -9 \Rightarrow (x+9) \text{ is a factor} \] \[x^3 - 67x + 126 = (x+9)(x^2 - 9x + 14) \] \[x^2 - 9x + 14 = (x-2)(x-7) \]

  6. Exercise 46

    ∣0xyzx−zy−x0y−zz−xz−y0∣=\displaystyle \left|\begin{array}{ccc}0 & x y z & x-z \\ y-x & 0 & y-z \\ z-x & z-y & 0\end{array}\right|= ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle (y-z)(z-x)(y-x+x y z)\)
    \(\displaystyle (y-z)(z-x)(xyz + y - x)\) \[\Delta = -xyz\,[\,0 - (y-z)(z-x)\,] + (x-z)\,[\,(y-x)(z-y) - 0\,] \] \[\Delta = xyz\,(y-z)(z-x) + (x-z)(y-x)(z-y) \] \[(x-z)(z-y) = (z-x)(y-z) \] \[\Delta = (y-z)(z-x)\,(xyz + y - x) \]

  7. Exercise 47

    If f(x)=∣(1+x)17(1+x)19(1+x)23(1+x)23(1+x)29(1+x)34(1+x)41(1+x)43(1+x)47∣=A+Bx+Cx2+…\displaystyle f(x)=\left|\begin{array}{lll}(1+x)^{17} & (1+x)^{19} & (1+x)^{23} \\ (1+x)^{23} & (1+x)^{29} & (1+x)^{34} \\ (1+x)^{41} & (1+x)^{43} & (1+x)^{47}\end{array}\right|=\mathrm{A}+\mathrm{B} x+\mathrm{C} x^2+\ldots, then A=\displaystyle \mathrm{A}= ____.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    Zero
    $\displaystyle 0$ \[f(0) = A \quad \text{(every other term contains } x\text{)} \] \[f(0) = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{vmatrix} \] \[R_1 = R_2 \Rightarrow f(0) = 0 \]
  8. State True or False for the statements of the following Exercises:

    Exercise 48

    (A3)−1=(A−1)3\displaystyle \left(\mathrm{A}^3\right)^{-1}=\left(\mathrm{A}^{-1}\right)^3, where A is a square matrix and ∣A∣≠0\displaystyle |\mathrm{A}| \neq 0.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    True
    True. \(\displaystyle |A| \neq 0\), so \(\displaystyle A^{-1}\) exists; apply the reversal law. \[(ABC)^{-1} = C^{-1}B^{-1}A^{-1} \] \[(A^3)^{-1} = (AAA)^{-1} = A^{-1}A^{-1}A^{-1} \] \[(A^3)^{-1} = (A^{-1})^3 \]

  9. Exercise 49

    (a A)−1=1a A−1\displaystyle (a \mathrm{~A})^{-1}=\frac{1}{a} \mathrm{~A}^{-1}, where a\displaystyle a is any real number and A is a square matrix.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    False
    False — the rule holds only for \(\displaystyle a\neq 0\): \[(a\mathrm{A})\left(\tfrac{1}{a}\mathrm{A}^{-1}\right)=\mathrm{I}\quad (a\neq 0) \] For \(\displaystyle a=0\): \[0\cdot\mathrm{A}=\mathrm{O},\qquad |\mathrm{O}|=0 \] so \(\displaystyle 0\cdot\mathrm{A}\) has no inverse, and \(\displaystyle \tfrac{1}{a}\) does not exist.

  10. Exercise 50

    ∣A−1∣≠∣A∣−1\displaystyle \left|\mathrm{A}^{-1}\right| \neq|\mathrm{A}|^{-1}, where A is non-singular matrix.

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    False
    False \[\mathrm{A}\mathrm{A}^{-1}=\mathrm{I} \] \[|\mathrm{A}|\,|\mathrm{A}^{-1}|=|\mathrm{I}|=1 \] \[|\mathrm{A}^{-1}|=\frac{1}{|\mathrm{A}|}=|\mathrm{A}|^{-1}\quad (|\mathrm{A}|\neq 0) \]