NCERT Class 12 Maths Chapter 3: Matrices Miscellaneous Exercise

Decoding the Advanced Mathematical Themes

The Miscellaneous Exercise of Chapter 3 (Matrices) is structured around three foundational pillars that test your theoretical depth. First is Mathematical Induction on Matrices. Often, you are given a matrix $A$ and asked to prove a general formula for $A^n$ for all $n \in \mathbb{N}$. Standard algebraic induction rules apply, but you must strictly respect matrix multiplication rules (specifically, that matrix multiplication is non-commutative, so $AB \neq BA$ in general). The second pillar is Symmetric and Skew-Symmetric Matrices. You will explore advanced proofs demonstrating how matrices like $B'AB$ inherit properties from $A$. Lastly, the exercise covers Matrix Equations, where you must solve for an unknown matrix $X$ from equations like $AX = B$ or $XA = B$. Because division is not defined for matrices, you must master pre-multiplication and post-multiplication techniques, or set up a system of linear equations using general matrix elements.

Step-by-Step Framework for Mathematical Induction in Matrices

1. Define the Proposition P(n) — Express the given matrix statement clearly as $P(n)$ for all $n \in \mathbb{N}$.
2. Verify the Base Case P(1) — Substitute $n = 1$ into both the LHS and RHS of the formula to prove the statement holds true.
3. Formulate the Induction Hypothesis P(k) — Assume that the statement $P(k)$ is true for some positive integer $k$. Write down this assumed matrix equation.
4. Prove the Inductive Step P(k+1) — Show that $P(k+1)$ is true using the hypothesis. Express $A^{k+1}$ as $A^k \cdot A^1$. Substitute the assumed value of $A^k$ and use row-by-column matrix multiplication to simplify the expression to match the $P(k+1)$ structure.

Worked Example: Proof by Induction

• Example 1: Let $A = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$. Prove by induction that $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$ for all $n \in \mathbb{N}$. Step 1: Let $P(n)$ be the statement: $A^n = \begin{bmatrix} 1+2n & -4n \\ n & 1-2n \end{bmatrix}$. Step 2: Base case ($n=1$): LHS = $A^1 = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ RHS = \begin{bmatrix} 1+2(1) & -4(1) \\ 1 & 1-2(1) \end{bmatrix} = \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$. Since LHS = RHS, $P(1)$ is true. Step 3: Assume $P(k)$ is true for some $k \in \mathbb{N}$: $A^k = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix}$. Step 4: Prove $P(k+1)$ is true, i.e., $A^{k+1} = \begin{bmatrix} 1+2(k+1) & -4(k+1) \\ k+1 & 1-2(k+1) \end{bmatrix} = \begin{bmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{bmatrix}$. Using laws of exponents, write $A^{k+1} = A^k \cdot A$. $A^{k+1} = \begin{bmatrix} 1+2k & -4k \\ k & 1-2k \end{bmatrix} \begin{bmatrix} 3 & -4 \\ 1 & -1 \end{bmatrix}$ Multiply row by column: $= \begin{bmatrix} (1+2k)(3) + (-4k)(1) & (1+2k)(-4) + (-4k)(-1) \\ (k)(3) + (1-2k)(1) & (k)(-4) + (1-2k)(-1) \end{bmatrix}$ $= \begin{bmatrix} 3+6k-4k & -4-8k+4k \\ 3k+1-2k & -4k-1+2k \end{bmatrix} = \begin{bmatrix} 2k+3 & -4k-4 \\ k+1 & -2k-1 \end{bmatrix}$. Hence, $P(k+1)$ is true. By the Principle of Mathematical Induction, the statement holds for all $n \in \mathbb{N}$.

CBSE Board Exam Tips & Pitfalls

1. Never Assume Commutativity: In matrix multiplication, $AB$ is generally NOT equal to $BA$. While verifying proofs, do not simplify $(A+B)^2$ directly as $A^2 + 2AB + B^2$ unless you are explicitly given that $AB = BA$. Instead, write $(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2$.
2. The Transpose of a Product: Remember the reversal law: $(AB)' = B'A'$. Many students incorrectly write $(AB)' = A'B'$. Applying this incorrectly during proofs of symmetric or skew-symmetric matrices will lead to a complete loss of marks.
3. Clear Inductive Logic: When writing induction proofs, do not skip writing down the statement of the inductive hypothesis ($P(k)$) and concluding with the standard induction sentence. Evaluators look for these precise step markings.

Practice Questions with Solutions

• Q: If $A$ and $B$ are symmetric matrices of the same order, prove that $AB - BA$ is a skew-symmetric matrix. A: Step 1: Identify given conditions. Since $A$ and $B$ are symmetric matrices, we have $A' = A$ and $B' = B$. Step 2: Let $X = AB - BA$. To show $X$ is skew-symmetric, we must prove $X' = -X$. Step 3: Take the transpose of both sides: $X' = (AB - BA)'$. Using the property $(P - Q)' = P' - Q'$, we get: $X' = (AB)' - (BA)'$. Step 4: Apply the reversal law of transpose $(CD)' = D'C'$: $X' = B'A' - A'B'$. Step 5: Substitute the given symmetric values $A' = A$ and $B' = B$: $X' = BA - AB$. Step 6: Factor out a negative sign: $X' = -(AB - BA) = -X$. Final answer: Since $X' = -X$, the matrix $AB - BA$ is verified to be a skew-symmetric matrix.
• Q: If $A$ is a square matrix such that $A^2 = A$, then find the value of $(I + A)^3 - 7A$. A: Step 1: Expand $(I + A)^3$ using binomial expansion (allowed here because the identity matrix $I$ commutes with any matrix $A$, i.e., $IA = AI = A$): $(I+A)^3 = I^3 + 3I^2A + 3IA^2 + A^3$. Step 2: Simplify the powers of $I$ and $A$. We know $I^3 = I$, $I^2 = I$, and $IA^2 = A^2$. So, $(I + A)^3 = I + 3A + 3A^2 + A^3$. Step 3: Simplify $A^3$ and $A^2$ using the given condition $A^2 = A$: $A^3 = A^2 \cdot A = A \cdot A = A^2 = A$. Step 4: Substitute $A^2 = A$ and $A^3 = A$ back into the expanded equation: $(I + A)^3 = I + 3A + 3(A) + A = I + 7A$. Step 5: Calculate $(I + A)^3 - 7A$: $(I + A)^3 - 7A = (I + 7A) - 7A = I$. Final answer: The simplified value is the Identity matrix $I$.
• Q: Let $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$. Show that $(aI + bA)^n = a^n I + n a^{n-1} b A$, where $I$ is the identity matrix of order 2 and $n \in \mathbb{N}$. A: Step 1: Let the proposition be $P(n): (aI + bA)^n = a^n I + n a^{n-1} b A$. Step 2: Test for $n=1$: LHS = $(aI + bA)^1 = aI + bA$. RHS = $a^1 I + 1 a^{1-1} b A = aI + a^0 b A = aI + b A$. Since LHS = RHS, $P(1)$ is true. Step 3: Assume $P(k)$ is true for some $k \in \mathbb{N}$: $(aI + bA)^k = a^k I + k a^{k-1} b A$. Step 4: Prove $P(k+1)$ is true, i.e., $(aI + bA)^{k+1} = a^{k+1} I + (k+1) a^{k} b A$. $(aI + bA)^{k+1} = (aI + bA)^k \cdot (aI + bA)^1$. Substitute the induction hypothesis: $= (a^k I + k a^{k-1} b A) (aI + bA)$ $= a^k I(aI) + a^k I(bA) + (k a^{k-1} b A)(aI) + (k a^{k-1} b A)(bA)$ $= a^{k+1} I^2 + a^k b A + k a^k b A + k a^{k-1} b^2 A^2$. Step 5: Note that $I^2 = I$. Calculate $A^2$: $A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$ (zero matrix). So, the term $k a^{k-1} b^2 A^2 = O$. Step 6: Substitute these values back into the expression: $= a^{k+1} I + (a^k b + k a^k b) A + O$ $= a^{k+1} I + (k+1) a^k b A$. This is exactly equal to the RHS of $P(k+1)$. Therefore, by PMI, the statement is true for all $n \in \mathbb{N}$. Final answer: Proved by Principle of Mathematical Induction.
• Q: Find the matrix $X$ such that $X \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$. A: Step 1: Determine the order of matrix $X$. Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix}$ (order $2 \times 3$) and $B = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$ (order $2 \times 3$). Since $X A = B$, the order of $X$ must be $2 \times 2$ for the product to be defined. Step 2: Let $X = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$. Step 3: Write the matrix equation: $\begin{bmatrix} a & b \\ c & d \end{bmatrix} \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$. Multiply LHS: $\begin{bmatrix} a+4b & 2a+5b & 3a+6b \\ c+4d & 2c+5d & 3c+6d \end{bmatrix} = \begin{bmatrix} -7 & -8 & -9 \\ 2 & 4 & 6 \end{bmatrix}$. Step 4: Set up systems of equations by equating corresponding elements. For the first row: 1) $a + 4b = -7$ 2) $2a + 5b = -8$ Multiply (1) by 2: $2a + 8b = -14$. Subtract (2) from this: $3b = -6 \implies b = -2$. Substitute $b = -2$ in (1): $a - 8 = -7 \implies a = 1$. Check third term: $3(1) + 6(-2) = -9$ (consistent). Step 5: For the second row: 3) $c + 4d = 2$ 4) $2c + 5d = 4$ Multiply (3) by 2: $2c + 8d = 4$. Subtract (4) from this: $3d = 0 \implies d = 0$. Substitute $d = 0$ in (3): $c + 0 = 2 \implies c = 2$. Check third term: $3(2) + 6(0) = 6$ (consistent). Final answer: $X = \begin{bmatrix} 1 & -2 \\ 2 & 0 \end{bmatrix}$.

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