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NCERT Exemplar · Class 12 Mathematics Determinants

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EXERCISE 4.3 1–10 (part 1 of 6)

  1. Using the properties of determinants in Exercises $\displaystyle 1$ to $\displaystyle 6$, evaluate:

    Exercise 1

    ∣x2−x+1x−1x+1x+1∣\displaystyle \left|\begin{array}{cc}x^2-x+1 & x-1 \\ x+1 & x+1\end{array}\right|

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    NCERT’s answer
    \(\displaystyle x^3-x^2+2\)
    Take \(\displaystyle (x+1)\) common from \(\displaystyle R_2\): \[D = (x+1)\begin{vmatrix} x^2-x+1 & x-1 \\ 1 & 1 \end{vmatrix} \] \[D = (x+1)\left[(x^2-x+1) - (x-1)\right] \] \[D = (x+1)(x^2-2x+2) \] \[D = x^3 - x^2 + 2 \]Answer: \(\displaystyle x^3 - x^2 + 2\)

  2. Exercise 2

    ∣a+xyzxa+yzxya+z∣\displaystyle \left|\begin{array}{ccc}a+x & y & z \\ x & a+y & z \\ x & y & a+z\end{array}\right|

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    NCERT’s answer
    \(\displaystyle a^2(a+x+y+z)\)
    \[D = \begin{vmatrix} a+x & y & z \\ x & a+y & z \\ x & y & a+z \end{vmatrix} \] \[C_1 \to C_1 + C_2 + C_3: \quad D = \begin{vmatrix} a+x+y+z & y & z \\ a+x+y+z & a+y & z \\ a+x+y+z & y & a+z \end{vmatrix} \] \[D = (a+x+y+z)\begin{vmatrix} 1 & y & z \\ 1 & a+y & z \\ 1 & y & a+z \end{vmatrix} \] \[R_2 \to R_2 - R_1,\ R_3 \to R_3 - R_1: \quad D = (a+x+y+z)\begin{vmatrix} 1 & y & z \\ 0 & a & 0 \\ 0 & 0 & a \end{vmatrix} \] \[D = a^2(a+x+y+z) \]Answer: \(\displaystyle a^2(a+x+y+z)\)

  3. Exercise 3

    ∣0xy2xz2x2y0yz2x2zzy20∣\displaystyle \left|\begin{array}{ccc}0 & x y^2 & x z^2 \\ x^2 y & 0 & y z^2 \\ x^2 z & z y^2 & 0\end{array}\right|

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    NCERT’s answer
    \(\displaystyle 2 x^3 y^3 z^3\)
    Take \(\displaystyle x, y, z\) out of \(\displaystyle R_1, R_2, R_3\), then \(\displaystyle x^2, y^2, z^2\) out of \(\displaystyle C_1, C_2, C_3\): \[D = xyz\begin{vmatrix} 0 & y^2 & z^2 \\ x^2 & 0 & z^2 \\ x^2 & y^2 & 0 \end{vmatrix} \] \[D = xyz \cdot x^2y^2z^2\begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} \] \[\begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = 0(0-1) - 1(0-1) + 1(1-0) = 2 \] \[D = 2x^3y^3z^3 \]Answer: \(\displaystyle 2x^3y^3z^3\)

  4. Exercise 4

    ∣3x−x+y−x+zx−y3yz−yx−zy−z3z∣\displaystyle \left|\begin{array}{ccc}3 x & -x+y & -x+z \\ x-y & 3 y & z-y \\ x-z & y-z & 3 z\end{array}\right|

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    NCERT’s answer
    \(\displaystyle 3(x+y+z)(x y+y z+z x)\)
    \[C_1 \to C_1 + C_2 + C_3: \quad D = \begin{vmatrix} x+y+z & -x+y & -x+z \\ x+y+z & 3y & z-y \\ x+y+z & y-z & 3z \end{vmatrix} \] \[D = (x+y+z)\begin{vmatrix} 1 & y-x & z-x \\ 1 & 3y & z-y \\ 1 & y-z & 3z \end{vmatrix} \] \[R_2 \to R_2 - R_1,\ R_3 \to R_3 - R_1: \quad D = (x+y+z)\begin{vmatrix} 1 & y-x & z-x \\ 0 & x+2y & x-y \\ 0 & x-z & x+2z \end{vmatrix} \] \[D = (x+y+z)\left[(x+2y)(x+2z) - (x-y)(x-z)\right] \] \[D = (x+y+z)(3xy + 3yz + 3zx) \]Answer: \(\displaystyle 3(x+y+z)(xy+yz+zx)\)

  5. Exercise 5

    ∣x+4xxxx+4xxxx+4∣\displaystyle \left|\begin{array}{ccc}x+4 & x & x \\ x & x+4 & x \\ x & x & x+4\end{array}\right|

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    NCERT’s answer
    \(\displaystyle 16(3 x+4)\)
    \[C_1 \to C_1 + C_2 + C_3: \quad D = \begin{vmatrix} 3x+4 & x & x \\ 3x+4 & x+4 & x \\ 3x+4 & x & x+4 \end{vmatrix} \] \[D = (3x+4)\begin{vmatrix} 1 & x & x \\ 1 & x+4 & x \\ 1 & x & x+4 \end{vmatrix} \] \[R_2 \to R_2 - R_1,\ R_3 \to R_3 - R_1: \quad D = (3x+4)\begin{vmatrix} 1 & x & x \\ 0 & 4 & 0 \\ 0 & 0 & 4 \end{vmatrix} \] \[D = 16(3x+4) \]Answer: \(\displaystyle 16(3x+4) = 48x + 64\)

  6. Exercise 6

    ∣a−b−c2a2a2bb−c−a2b2c2cc−a−b∣\displaystyle \left|\begin{array}{ccc}a-b-c & 2 a & 2 a \\ 2 b & b-c-a & 2 b \\ 2 c & 2 c & c-a-b\end{array}\right|

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    NCERT’s answer
    \(\displaystyle (a+b+c)^3\)
    \[R_1 \to R_1 + R_2 + R_3: \quad D = \begin{vmatrix} a+b+c & a+b+c & a+b+c \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \] \[D = (a+b+c)\begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \] \[C_2 \to C_2 - C_1,\ C_3 \to C_3 - C_1: \quad D = (a+b+c)\begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} \] \[D = (a+b+c)\cdot(a+b+c)^2 \]Answer: \(\displaystyle (a+b+c)^3\)
  7. Using the proprties of determinants in Exercises $\displaystyle 7$ to $\displaystyle 9$, prove that:

    Exercise 7

    ∣y2z2yzy+zz2x2zxz+xx2y2xyx+y∣=0\displaystyle \left|\begin{array}{lll}y^2 z^2 & y z & y+z \\ z^2 x^2 & z x & z+x \\ x^2 y^2 & x y & x+y\end{array}\right|=0

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    For \(\displaystyle xyz \neq 0\), multiply \(\displaystyle R_1, R_2, R_3\) by \(\displaystyle x, y, z\) and divide by \(\displaystyle xyz\): \[D = \frac{1}{xyz}\begin{vmatrix} xy^2z^2 & xyz & xy+xz \\ x^2yz^2 & xyz & yz+xy \\ x^2y^2z & xyz & zx+yz \end{vmatrix} \] Take \(\displaystyle xyz\) out of \(\displaystyle C_1\) and \(\displaystyle C_2\): \[D = xyz\begin{vmatrix} yz & 1 & xy+xz \\ zx & 1 & yz+xy \\ xy & 1 & zx+yz \end{vmatrix} \] \[C_3 \to C_3 + C_1: \quad D = xyz\begin{vmatrix} yz & 1 & xy+yz+zx \\ zx & 1 & xy+yz+zx \\ xy & 1 & xy+yz+zx \end{vmatrix} \] \[D = xyz\,(xy+yz+zx)\begin{vmatrix} yz & 1 & 1 \\ zx & 1 & 1 \\ xy & 1 & 1 \end{vmatrix} \] \[C_2 = C_3 \ \Rightarrow\ D = 0 \] Both sides are polynomials in \(\displaystyle x, y, z\), so the identity holds for all values.Answer: \(\displaystyle D = 0\)

  8. Exercise 8

    ∣y+zzyzz+xxyxx+y∣=4xyz\displaystyle \left|\begin{array}{ccc}y+z & z & y \\ z & z+x & x \\ y & x & x+y\end{array}\right|=4 x y z

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    \[C_1 \to C_1 - C_2 - C_3: \quad D = \begin{vmatrix} 0 & z & y \\ -2x & z+x & x \\ -2x & x & x+y \end{vmatrix} \] \[R_3 \to R_3 - R_2: \quad D = \begin{vmatrix} 0 & z & y \\ -2x & z+x & x \\ 0 & -z & y \end{vmatrix} \] Expand along \(\displaystyle C_1\): \[D = (-2x)(-1)^{2+1}\begin{vmatrix} z & y \\ -z & y \end{vmatrix} \] \[D = 2x\,(zy + zy) \] \[D = 4xyz \]Answer: \(\displaystyle D = 4xyz\)

  9. Exercise 9

    ∣a2+2a2a+112a+1a+21331∣=(a−1)3\displaystyle \left|\begin{array}{ccc}a^2+2 a & 2 a+1 & 1 \\ 2 a+1 & a+2 & 1 \\ 3 & 3 & 1\end{array}\right|=(a-1)^3

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    Apply \(\displaystyle R_1\to R_1-R_2\) and \(\displaystyle R_2\to R_2-R_3\): \[\Delta=\begin{vmatrix} a^2-1 & a-1 & 0\\ 2a-2 & a-1 & 0\\ 3 & 3 & 1\end{vmatrix} \] Take \(\displaystyle (a-1)\) out of \(\displaystyle R_1\) and of \(\displaystyle R_2\): \[\Delta=(a-1)^2\begin{vmatrix} a+1 & 1 & 0\\ 2 & 1 & 0\\ 3 & 3 & 1\end{vmatrix} \] Expand along \(\displaystyle C_3\): \[\Delta=(a-1)^2\,[(a+1)\cdot 1-1\cdot 2] \] \[\Delta=(a-1)^2(a-1)=(a-1)^3 \] Answer: \(\displaystyle \Delta=(a-1)^3\)

  10. Exercise 10

    If A+B+C=0\displaystyle \mathrm{A}+\mathrm{B}+\mathrm{C}=0, then prove that ∣1cos⁡Ccos⁡Bcos⁡C1cos⁡Acos⁡Bcos⁡A1∣=0\displaystyle \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|=0

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    Expand along \(\displaystyle R_1\): \[\Delta=(1-\cos^2A)-\cos C(\cos C-\cos A\cos B)+\cos B(\cos A\cos C-\cos B) \] \[\Delta=1-\cos^2A-\cos^2B-\left(\cos^2C-2\cos A\cos B\cos C\right) \] Since \(\displaystyle C=-(A+B)\): \[\cos C=\cos(A+B) \] \[\cos C-2\cos A\cos B=\cos(A+B)-2\cos A\cos B=-\cos(A-B) \] \[\cos^2C-2\cos A\cos B\cos C=-\cos(A+B)\cos(A-B)=-(\cos^2A-\sin^2B) \] Substitute: \[\Delta=1-\cos^2A-\cos^2B+\cos^2A-\sin^2B \] \[\Delta=1-\cos^2B-\sin^2B=0 \] Answer: \(\displaystyle \Delta=0\)