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Q6:
Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$
Solution :
Step 1: Setting up the Determinant Matrix
Let the given determinant be $\Delta$. We are tasked to evaluate:
$\Delta = \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$
Step 2: Applying Row Operations for Simplification
To simplify the determinant, we utilize the property that the value of a determinant remains unchanged under the operation $R_i \rightarrow R_i - R_j$ [Property of Determinants: Invariance under Elementary Row Operations].
Perform the following operations:
- $R_2 \rightarrow R_2 - R_1$
- $R_3 \rightarrow R_3 - R_1$
Applying $R_2 \rightarrow R_2 - R_1$:
$R_2 = [1 - 1, (x+y) - x, y - y] = [0, y, 0]$
Applying $R_3 \rightarrow R_3 - R_1$:
$R_3 = [1 - 1, x - x, (x+y) - y] = [0, 0, x]$
The updated determinant is:
$\Delta = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$
Step 3: Expansion along the First Column
Since the first column ($C_1$) contains two zeros, expanding along $C_1$ is the most efficient method [Laplace Expansion Theorem].
$\Delta = 1 \cdot \begin{vmatrix} y & 0 \\ 0 & x \end{vmatrix} - 0 \cdot \begin{vmatrix} x & y \\ 0 & x \end{vmatrix} + 0 \cdot \begin{vmatrix} x & y \\ y & 0 \end{vmatrix}$
Step 4: Evaluating the 2x2 Determinant
For a $2 \times 2$ matrix $\begin{vmatrix} a & b \\ c & d \end{vmatrix}$, the determinant is given by $ad - bc$.
$\Delta = 1 \cdot (y \cdot x - 0 \cdot 0)$
$\Delta = xy$
Visual Representation of Matrix Structure
Final Solution: $xy$
More Questions from Class 12 Mathematics Determinants EXERCISE Miscellaneous
- Q1: Prove that the determinant $\begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$.
- Q2: Evaluate $\begin{vmatrix} \cos\alpha\cos\beta & \cos\alpha\sin\beta & -\sin\alpha \\ -\sin\beta & \cos\beta & 0 \\ \sin\alpha\cos\beta & \sin\alpha\sin\beta & \cos\alpha \end{vmatrix}$.
- Q3: If $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ -15 & 6 & -5 \\ 5 & -2 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & -2 \\ -1 & 3 & 0 \\ 0 & -2 & 1 \end{bmatrix}$, find $(AB)^{-1}$.
- Q4(i): Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (i) $[adj A]^{-1} = adj (A^{-1})$
- Q4(ii): Let $A = \begin{bmatrix} 1 & -2 & 1 \\ -2 & 3 & 1 \\ 1 & 1 & 5 \end{bmatrix}$. Verify that (ii) $(A^{-1})^{-1} = A$
- Q5: Evaluate $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$
- Q7: Solve the system of equations $\frac{2}{x} + \frac{3}{y} + \frac{10}{z} = 4$, $\frac{4}{x} - \frac{6}{y} + \frac{5}{z} = 1$, $\frac{6}{x} + \frac{9}{y} - \frac{20}{z} = 2$
- Q8: If $x, y, z$ are nonzero real numbers, then the inverse of matrix $A = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ is
- Q9: Let $A = \begin{bmatrix} 1 & \sin\theta & 1 \\ -\sin\theta & 1 & \sin\theta \\ -1 & -\sin\theta & 1 \end{bmatrix}$, where $0 \leq \theta \leq 2\pi$. Then
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