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Q5:
Evaluate $\begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$
Solution :
Step 1: Establishing the Initial Determinant
Let the given determinant be $\Delta$.
$$\Delta = \begin{vmatrix} x & y & x+y \\ y & x+y & x \\ x+y & x & y \end{vmatrix}$$
Step 2: Applying Property of Determinants for Simplification
We perform the column operation $C_1 \rightarrow C_1 + C_2 + C_3$ to simplify the first column [Per the property that the value of a determinant remains invariant under such elementary column transformations].
Summing the elements of each row for the first column:
- $C_{11}: x + y + (x+y) = 2(x+y)$
- $C_{21}: y + (x+y) + x = 2(x+y)$
- $C_{31}: (x+y) + x + y = 2(x+y)$
The determinant becomes:
$$\Delta = \begin{vmatrix} 2(x+y) & y & x+y \\ 2(x+y) & x+y & x \\ 2(x+y) & x & y \end{vmatrix}$$
Step 3: Factoring out the Common Constant
Extract $2(x+y)$ as a common factor from the first column [By the scalar multiplication property of determinants].
$$\Delta = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 1 & x+y & x \\ 1 & x & y \end{vmatrix}$$
Step 4: Row Reduction
Perform row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_1$ to introduce zeros in the first column.
$$\Delta = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 0 & (x+y)-y & x-(x+y) \\ 0 & x-y & y-(x+y) \end{vmatrix}$$
Simplifying the entries:
$$\Delta = 2(x+y) \begin{vmatrix} 1 & y & x+y \\ 0 & x & -y \\ 0 & x-y & -x-y \end{vmatrix}$$
Step 5: Expansion along the First Column
Expand along $C_1$ [By Laplace expansion]:
$$\Delta = 2(x+y) [1(x(-x-y) - (-y)(x-y)) - 0 + 0]$$
Expand the $2 \times 2$ determinant:
$$\Delta = 2(x+y) [-x^2 - xy + y(x-y)]$$
$$\Delta = 2(x+y) [-x^2 - xy + xy - y^2]$$
$$\Delta = 2(x+y) [-x^2 - y^2]$$
Step 6: Final Algebraic Consolidation
$$\Delta = -2(x+y)(x^2 + y^2)$$
Final Solution: $-2(x+y)(x^2 + y^2)$
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) $(A^{-1})^{-1} = A$
- Q6: Evaluate $\begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$
- 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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