Skip to content
Question

Q.(a) Evaluate : ∣111xyzx2y2z2∣\begin{vmatrix}1 & 1 & 1\\ x & y & z\\ x^2 & y^2 & z^2\end{vmatrix}

(OR)
(b) Find the inverse of the matrix [12−2−1300−21]\begin{bmatrix}1 & 2 & -2\\ -1 & 3 & 0\\ 0 & -2 & 1\end{bmatrix}.
CBSECBSE Class XII Board 2024Subjective· 3mImportance★★★★★
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

  1. The determinant equals (x−y)(y−z)(z−x)(x-y)(y-z)(z-x);
  2. since det⁡A=1\det A=1, the inverse equals the adjoint [326112225]\begin{bmatrix}3&2&6\\1&1&2\\2&2&5\end{bmatrix}.

  1. Expansion / column operations on a 3×33\times3 determinant.
  2. A−1=1det⁡A adj⁡AA^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}A, where adj⁡A\operatorname{adj}A is the transpose of the cofactor matrix.

Part (a): Evaluate the determinant

  1. Apply C1→C1−C2C_1\to C_1-C_2 and C2→C2−C3C_2\to C_2-C_3 to the determinant D=∣111xyzx2y2z2∣D=\begin{vmatrix}1&1&1\\x&y&z\\x^2&y^2&z^2\end{vmatrix}: D=∣001x−yy−zzx2−y2y2−z2z2∣.D=\begin{vmatrix}0&0&1\\x-y&y-z&z\\x^2-y^2&y^2-z^2&z^2\end{vmatrix}.
  2. Factor (x−y)(x-y) from C1C_1 and (y−z)(y-z) from C2C_2 (using x2−y2=(x−y)(x+y)x^2-y^2=(x-y)(x+y) etc.): D=(x−y)(y−z)∣00111zx+yy+zz2∣.D=(x-y)(y-z)\begin{vmatrix}0&0&1\\1&1&z\\x+y&y+z&z^2\end{vmatrix}.
  3. Expand along the first row (only the (1,3)(1,3) entry is non-zero): D=(x−y)(y−z)[(1)(y+z)−(1)(x+y)]=(x−y)(y−z)(z−x).D=(x-y)(y-z)\big[(1)(y+z)-(1)(x+y)\big]=(x-y)(y-z)(z-x).
  4. Check with x=0,y=1,z=2x=0,y=1,z=2: original =2=2 and (0−1)(1−2)(2−0)=(−1)(−1)(2)=2.(0-1)(1-2)(2-0)=(-1)(-1)(2)=2. ✓

Part (b): Inverse of A=[12−2−1300−21]A=\begin{bmatrix}1&2&-2\\-1&3&0\\0&-2&1\end{bmatrix} …

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.