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NCERT Exemplar · Class 12 Mathematics Determinants

58 questions · 58 still being checked

EXERCISE 4.3 51–58 (part 6 of 6)

  1. State True or False for the statements of the following Exercises:

    Exercise 51

    If A and B are matrices of order 3\displaystyle 3 and ∣A∣=5,∣ B∣=3\displaystyle |\mathrm{A}|=5,|\mathrm{~B}|=3, then ∣3AB∣=27×5×3=405\displaystyle |3 \mathrm{AB}|=27 \times 5 \times 3=405.

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    NCERT’s answer
    True
    True — for a matrix of order \(\displaystyle 3\), \(\displaystyle |k\mathrm{M}|=k^{3}|\mathrm{M}|\). \[|3\mathrm{AB}|=3^{3}\,|\mathrm{AB}| \] \[=27\,|\mathrm{A}|\,|\mathrm{B}| \] \[=27\times 5\times 3=405 \]

  2. Exercise 52

    If the value of a third order determinant is 12\displaystyle 12, then the value of the determinant formed by replacing each element by its co-factor will be 144.

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    NCERT’s answer
    True
    True — the cofactor matrix is \(\displaystyle \mathrm{C}=(\operatorname{adj}\mathrm{A})^{T}\). \[|\mathrm{C}|=|\operatorname{adj}\mathrm{A}| \] \[|\operatorname{adj}\mathrm{A}|=|\mathrm{A}|^{n-1}=12^{3-1}=144 \]

  3. Exercise 53

    ∣x+1x+2x+ax+2x+3x+bx+3x+4x+c∣=0\displaystyle \left|\begin{array}{lll}x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c\end{array}\right|=0, where a,b,c\displaystyle a, b, c are in A.P.

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    NCERT’s answer
    True
    True — apply \(\displaystyle R_3\to R_3-R_2\), then \(\displaystyle R_2\to R_2-R_1\): \[\Delta=\begin{vmatrix} x+1 & x+2 & x+a\\ 1 & 1 & b-a\\ 1 & 1 & c-b\end{vmatrix} \] \[a,b,c \text{ in A.P.}\ \Rightarrow\ 2b=a+c\ \Rightarrow\ b-a=c-b \] \[R_2=R_3\ \Rightarrow\ \Delta=0 \]

  4. Exercise 54

    ∣adj⁡.A∣=∣A∣2\displaystyle |\operatorname{adj} . \mathrm{A}|=|\mathrm{A}|^2, where A is a square matrix of order two.

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    NCERT’s answer
    False
    False — for order \(\displaystyle n=2\), \(\displaystyle |\operatorname{adj}\mathrm{A}|=|\mathrm{A}|^{n-1}=|\mathrm{A}|\). Counter-example: \[\mathrm{A}=\begin{bmatrix}2&0\\0&2\end{bmatrix},\qquad |\mathrm{A}|=4,\qquad \operatorname{adj}\mathrm{A}=\begin{bmatrix}2&0\\0&2\end{bmatrix} \] \[|\operatorname{adj}\mathrm{A}|=4\neq 16=|\mathrm{A}|^{2} \]

  5. Exercise 55

    The determinant ∣sin⁡Acos⁡Asin⁡A+cos⁡Bsin⁡Bcos⁡Asin⁡B+cos⁡Bsin⁡Ccos⁡Asin⁡C+cos⁡B∣\displaystyle \left|\begin{array}{lll}\sin \mathrm{A} & \cos \mathrm{A} & \sin \mathrm{A}+\cos \mathrm{B} \\ \sin \mathrm{B} & \cos \mathrm{A} & \sin \mathrm{B}+\cos \mathrm{B} \\ \sin \mathrm{C} & \cos \mathrm{A} & \sin \mathrm{C}+\cos \mathrm{B}\end{array}\right| is equal to zero.

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    NCERT’s answer
    True
    True — apply \(\displaystyle C_3\to C_3-C_1\): \[\Delta=\begin{vmatrix}\sin A&\cos A&\cos B\\ \sin B&\cos A&\cos B\\ \sin C&\cos A&\cos B\end{vmatrix} \] \[\Delta=\cos A\cos B\begin{vmatrix}\sin A&1&1\\ \sin B&1&1\\ \sin C&1&1\end{vmatrix} \] \[C_2=C_3\ \Rightarrow\ \Delta=0 \]

  6. Exercise 56

    If the determinant ∣x+ap+ul+fy+bq+vm+gz+cr+wn+h∣\displaystyle \left|\begin{array}{lll}x+a & p+u & l+f \\ y+b & q+v & m+g \\ z+c & r+w & n+h\end{array}\right| splits into exactly K determinants of order 3\displaystyle 3, each element of which contains only one term, then the value of K is 8.

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    NCERT’s answer
    True
    True — each of the three columns is a sum of two columns, so it splits in two ways: \[2\times 2\times 2=2^{3}=8 \] Each piece has single-term entries, for example \[\begin{vmatrix}x&p&l\\ y&q&m\\ z&r&n\end{vmatrix},\qquad \begin{vmatrix}a&u&f\\ b&v&g\\ c&w&h\end{vmatrix},\ \ldots \]

  7. Exercise 57

    Let Δ=∣apxbqycrz∣=16\displaystyle \Delta=\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|=16, then Δ1=∣p+xa+xa+pq+yb+yb+qr+zc+zc+r∣=32\displaystyle \Delta_1=\left|\begin{array}{lll}p+x & a+x & a+p \\ q+y & b+y & b+q \\ r+z & c+z & c+r\end{array}\right|=32.

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    NCERT’s answer
    True
    True — column operations reduce \(\displaystyle \Delta_1\) to \(\displaystyle \Delta\).\[C_1 \to C_1 + C_2 + C_3 \]\[\Delta_1 = \begin{vmatrix} 2(a+p+x) & a+x & a+p \\ 2(b+q+y) & b+y & b+q \\ 2(c+r+z) & c+z & c+r \end{vmatrix} = 2\begin{vmatrix} a+p+x & a+x & a+p \\ b+q+y & b+y & b+q \\ c+r+z & c+z & c+r \end{vmatrix} \]\[C_2 \to C_2 - C_1, \quad C_3 \to C_3 - C_1 \]\[\Delta_1 = 2\begin{vmatrix} a+p+x & -p & -x \\ b+q+y & -q & -y \\ c+r+z & -r & -z \end{vmatrix} = 2\begin{vmatrix} a+p+x & p & x \\ b+q+y & q & y \\ c+r+z & r & z \end{vmatrix} \quad \text{(two sign changes)} \]\[C_1 \to C_1 - C_2 - C_3 \]\[\Delta_1 = 2\begin{vmatrix} a & p & x \\ b & q & y \\ c & r & z \end{vmatrix} = 2\Delta = 2(16) = 32 \]

  8. Exercise 58

    The maximum value of ∣1111(1+sin⁡θ)1111+cos⁡θ∣\displaystyle \left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & (1+\sin \theta) & 1 \\ 1 & 1 & 1+\cos \theta\end{array}\right| is 12\displaystyle \frac{1}{2}.

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

    NCERT’s answer
    True
    True — row operations make the determinant triangular.\[R_2 \to R_2 - R_1, \quad R_3 \to R_3 - R_1 \]\[\begin{vmatrix} 1 & 1 & 1 \\ 1 & 1+\sin\theta & 1 \\ 1 & 1 & 1+\cos\theta \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \\ 0 & \sin\theta & 0 \\ 0 & 0 & \cos\theta \end{vmatrix} = \sin\theta\cos\theta \]\[\sin\theta\cos\theta = \tfrac{1}{2}\sin 2\theta \]\[\sin 2\theta \le 1 \;\Rightarrow\; \tfrac{1}{2}\sin 2\theta \le \tfrac{1}{2} \]The value \(\displaystyle \tfrac12\) is reached at \(\displaystyle \theta = \tfrac{\pi}{4}\).\[\sin\tfrac{\pi}{4}\cos\tfrac{\pi}{4} = \tfrac{1}{\sqrt2}\cdot\tfrac{1}{\sqrt2} = \tfrac{1}{2} \]