NCERT Solutions & Concepts: Class 12 Determinants Miscellaneous Exercise

Why the Miscellaneous Exercise is Critical

In CBSE Class 12 board exams, questions from the Determinants Miscellaneous Exercise are highly favored because they test multiple conceptual layers simultaneously. Instead of applying basic mechanical formulas, you are expected to analyze mathematical relationships, construct logical proofs using determinant properties, and solve complex system equations using the Matrix Inversion Method. Understanding the interplay between a matrix and its adjoint, such as the fundamental identity $A \cdot \text{adj}(A) = |A|I$, is key to simplifying tedious calculations. Approaching these multi-step challenges requires algebraic precision, systematic sign-tracking during expansions, and continuous verification of your operations.

A Systematic Strategy for Solving Complex Determinant Problems

1. Step 1: Simplify Using Row/Column Operations — Before expanding a high-order determinant, apply row or column operations to create maximum zeros in any single row or column. Look for patterns to factor out common terms first.
2. Step 2: Calculate Cofactors and Adjoint Methodically — Compute minors and cofactors carefully. Remember to apply the alternating sign factor $(-1)^{i+j}$ for each element. Transpose the cofactor matrix to obtain the Adjoint.
3. Step 3: Analyze Consistency in Systems of Equations — For a system $AX = B$, calculate the determinant $|A|$. If $|A| \neq 0$, the system has a unique solution given by $X = A^{-1}B$. If $|A| = 0$, verify consistency by checking $(\text{adj } A)B$.
4. Step 4: Back-Substitute to Verify — Always check your final answers by substituting the calculated values of $x, y, z$ back into the original algebraic equations to ensure absolute accuracy.

Fully Solved Miscellaneous Problems

• Example 1: Prove that the determinant of $A = \begin{vmatrix} x & \sin\theta & \cos\theta \\ -\sin\theta & -x & 1 \\ \cos\theta & 1 & x \end{vmatrix}$ is independent of $\theta$. Step 1: Expand the determinant along the first row ($R_1$): $\Delta = x(-x^2 - 1) - \sin\theta(-x\sin\theta - \cos\theta) + \cos\theta(-\sin\theta + x\cos\theta)$ Step 2: Distribute the terms algebraically: $\Delta = -x^3 - x + x\sin^2\theta + \sin\theta\cos\theta - \sin\theta\cos\theta + x\cos^2\theta$ Step 3: Simplify by canceling opposing terms and grouping: $\Delta = -x^3 - x + x(\sin^2\theta + \cos^2\theta)$ Step 4: Use the trigonometric identity $\sin^2\theta + \cos^2\theta = 1$: $\Delta = -x^3 - x + x(1) = -x^3$ Conclusion: Since the final value $-x^3$ does not contain any $\theta$ terms, the determinant is independent of $\theta$.
• Example 2: Solve the system of linear equations using the matrix method: $2x + 3y + 3z = 5$, $x - 2y + z = -4$, $3x - y - 2z = 3$. Step 1: Write the system in matrix form $AX = B$ where: $A = \begin{bmatrix} 2 & 3 & 3 \\ 1 & -2 & 1 \\ 3 & -1 & -2 \end{bmatrix}$, $X = \begin{bmatrix} x \\ y \\ z \end{bmatrix}$, and $B = \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix}$ Step 2: Calculate the determinant $|A|$: $|A| = 2(4 + 1) - 3(-2 - 3) + 3(-1 + 6) = 2(5) - 3(-5) + 3(5) = 10 + 15 + 15 = 40$ Since $|A| \neq 0$, the inverse $A^{-1}$ exists. Step 3: Compute the cofactor matrix of $A$: $C_{11} = 5, C_{12} = 5, C_{13} = 5$ $C_{21} = 3, C_{22} = -13, C_{23} = 11$ $C_{31} = 9, C_{32} = 1, C_{33} = -7$ Step 4: Find the adjoint of $A$ (transpose of cofactors): $\text{adj}(A) = \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix}$ Step 5: Calculate $X = A^{-1}B = \frac{1}{40} \text{adj}(A)B$: $X = \frac{1}{40} \begin{bmatrix} 5 & 3 & 9 \\ 5 & -13 & 1 \\ 5 & 11 & -7 \end{bmatrix} \begin{bmatrix} 5 \\ -4 \\ 3 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 25 - 12 + 27 \\ 25 + 52 + 3 \\ 25 - 44 - 21 \end{bmatrix} = \frac{1}{40} \begin{bmatrix} 40 \\ 80 \\ -40 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ -1 \end{bmatrix}$ Final Answer: $x = 1, y = 2, z = -1$.

Pro-Tips to Avoid Traps on Exam Day

1. Beware the Alternating Sign Grid: The most common point loss on 6-mark matrix inversion questions stems from calculation slip-ups when applying $(-1)^{i+j}$ signs to cofactor elements. Always write down the signs explicitly: $\begin{bmatrix} + & - & + \\ - & + & - \\ + & - & + \end{bmatrix}$.

1. Verify Through Quick Substitution: Once you obtain values for $x, y, z$, plug them back into any of your three original equations. If both sides match, you have successfully locked in your marks.

1. Property Limitations: Remember that operations like $R_i \to R_i + k R_j$ do not change the determinant value, but multiplying a whole row by a constant scale factor $k$ changes the determinant to $k\Delta$. Always keep track of factors extracted outside of the determinant.

Practice Questions with Solutions

• Q: Evaluate the determinant of $A = \begin{vmatrix} 1 & x & y \\ 1 & x+y & y \\ 1 & x & x+y \end{vmatrix}$. A: Step 1: Apply row operations to create zeros in the first column. Perform $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$. Step 2: The transformed determinant becomes: $\Delta = \begin{vmatrix} 1 & x & y \\ 0 & y & 0 \\ 0 & 0 & x \end{vmatrix}$ Step 3: Since this is an upper triangular matrix, expand along the first column: $\Delta = 1 \cdot (y \cdot x - 0 \cdot 0) = xy$ Final answer: $xy$
• Q: Show that the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ satisfies the equation $A^2 - 4A + I = O$, and find $A^{-1}$ using this equation. A: Step 1: Compute $A^2$: $A^2 = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 4+3 & 6+6 \\ 2+2 & 3+4 \end{bmatrix} = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix}$ Step 2: Substitute $A^2$ into the polynomial expression: $A^2 - 4A + I = \begin{bmatrix} 7 & 12 \\ 4 & 7 \end{bmatrix} - \begin{bmatrix} 8 & 12 \\ 4 & 8 \end{bmatrix} + \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7-8+1 & 12-12+0 \\ 4-4+0 & 7-8+1 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$. Step 3: To find $A^{-1}$, multiply the equation by $A^{-1}$: $A^{-1}(A^2 - 4A + I) = A^{-1}O \implies A - 4I + A^{-1} = O$ $A^{-1} = 4I - A = \begin{bmatrix} 4 & 0 \\ 0 & 4 \end{bmatrix} - \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$ Final answer: $A^{-1} = \begin{bmatrix} 2 & -3 \\ -1 & 2 \end{bmatrix}$
• Q: Solve using matrix inverse rules: if $A = \begin{bmatrix} 1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$, find $X$ where $AX = B$. A: Step 1: Find $|A|$: $|A| = 1(8 - 6) - (-1)(0 + 9) + 2(0 - 6) = 1(2) + 1(9) + 2(-6) = 2 + 9 - 12 = -1$ Step 2: Calculate cofactors of $A$: $C_{11} = 2, C_{12} = -9, C_{13} = -6$ $C_{21} = 0, C_{22} = -2, C_{23} = -1$ $C_{31} = -1, C_{32} = 3, C_{33} = 2$ Step 3: Write the adjoint matrix of $A$: $\text{adj}(A) = \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix}$ Step 4: Compute inverse $A^{-1} = \frac{1}{|A|} \text{adj}(A)$: $A^{-1} = \frac{1}{-1} \begin{bmatrix} 2 & 0 & -1 \\ -9 & -2 & 3 \\ -6 & -1 & 2 \end{bmatrix} = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix}$ Step 5: Compute $X = A^{-1}B$: $X = \begin{bmatrix} -2 & 0 & 1 \\ 9 & 2 & -3 \\ 6 & 1 & -2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix} = \begin{bmatrix} -2(1) + 0(1) + 1(2) \\ 9(1) + 2(1) - 3(2) \\ 6(1) + 1(1) - 2(2) \end{bmatrix} = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}$ Final answer: $X = \begin{bmatrix} 0 \\ 5 \\ 3 \end{bmatrix}$
• Q: If $a, b, c$ are real numbers and the determinant $\Delta = \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 0$, prove that either $a+b+c = 0$ or $a=b=c$. A: Step 1: Apply the column operation $C_1 \to C_1 + C_2 + C_3$: $\Delta = \begin{vmatrix} 2(a+b+c) & c+a & a+b \\ 2(a+b+c) & a+b & b+c \\ 2(a+b+c) & b+c & c+a \end{vmatrix}$ Step 2: Factor out $2(a+b+c)$ from the first column: $\Delta = 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 1 & a+b & b+c \\ 1 & b+c & c+a \end{vmatrix}$ Step 3: Apply row operations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$: $\Delta = 2(a+b+c) \begin{vmatrix} 1 & c+a & a+b \\ 0 & b-c & c-a \\ 0 & b-a & c-b \end{vmatrix}$ Step 4: Expand along the first column: $\Delta = 2(a+b+c)[(b-c)(c-b) - (c-a)(b-a)]$ Simplifying the brackets yields: $-(a+b+c)(a^2+b^2+c^2-ab-bc-ca) = 0$ Step 5: This equation can be rewritten as: $-\frac{1}{2}(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] = 0$ Therefore, either $(a+b+c) = 0$ or $(a-b)^2 + (b-c)^2 + (c-a)^2 = 0$ (which forces $a=b=c$). Final answer: Proved.

Frequently Asked Questions