Solved Examples

Example 1: Constructing a Matrix

Question: Construct a 3x2 matrix where $a_{ij} = i - j$.

Explanation:

1. Understand the Order: A 3x2 matrix has 3 rows and 2 columns. The general form is $$\begin{bmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \\ a_{31} & a_{32} \end{bmatrix}$$
2. Use the Formula: We find the value of each element using the rule $a_{ij} = i - j$ (where 'i' is the row number and 'j' is the column number).
   • $a_{11}$ (i=1, j=1): $1-1=0$
   • $a_{12}$ (i=1, j=2): $1-2=-1$
   • $a_{21}$ (i=2, j=1): $2-1=1$
   • $a_{22}$ (i=2, j=2): $2-2=0$
   • $a_{31}$ (i=3, j=1): $3-1=2$
   • $a_{32}$ (i=3, j=2): $3-2=1$
3. Construct the Matrix: Place the calculated values into their corresponding positions.

Answer: $A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \\ 2 & 1 \end{bmatrix}$.

Example 2: Equality of Matrices

Question: Find x, y, a, b if $\begin{bmatrix} x-y & a+b \\ a-b & x+y \end{bmatrix} = \begin{bmatrix} 2 & 5 \\ 1 & 10 \end{bmatrix}$.

Explanation: By the definition of matrix equality, the corresponding elements must be equal. This gives us two independent systems of linear equations.

System 1 (for x and y):

1. $x-y=2$
2. $x+y=10$

Adding the two equations: $$(x-y)+(x+y) = 2+10$$ $$\implies 2x=12 \implies x=6$$ Substituting $x=6$ into the second equation: $$6+y=10 \implies y=4$$

System 2 (for a and b):

1. $a+b=5$
2. $a-b=1$

Adding the two equations: $$(a+b)+(a-b) = 5+1$$ $$\implies 2a=6 \implies a=3$$ Substituting $a=3$ into the first equation: $$3+b=5 \implies b=2$$

Answer: $x=6, y=4, a=3, b=2$.

Example 3: Possible Matrix Orders

Question: If a matrix has 13 elements, what are the possible orders?

Explanation:

1. Understand Order: The order of a matrix is (number of rows, $m$) $\times$ (number of columns, $n$). The total number of elements is $m \times n$.

2. Find Factor Pairs: We need to find pairs of natural numbers ($m, n$) whose product is 13.

3. Analyze 13: The number 13 is a prime number. Its only natural number factors are 1 and 13.

4. List Possibilities: The only possible pairs are (1, 13) and (13, 1).

Answer: The possible orders are 1 x 13 (a row matrix) and 13 x 1 (a column matrix).

Example 4: Matrix Operations

Question: If $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$, find $3A - B$.

Explanation:

1. Scalar Multiplication: First, calculate the matrix $3A$ by multiplying every element of A by 3. $$3A = 3\begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$$ $$= \begin{bmatrix} 3(1) & 3(2) \\ 3(3) & 3(4) \end{bmatrix} = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix}$$
2. Matrix Subtraction: Now, subtract the corresponding elements of B from 3A. $$3A - B = \begin{bmatrix} 3 & 6 \\ 9 & 12 \end{bmatrix} - \begin{bmatrix} 5 & 6 \\ 7 & 8 \end{bmatrix}$$ $$= \begin{bmatrix} 3-5 & 6-6 \\ 9-7 & 12-8 \end{bmatrix}$$
3. Calculate Result: $$3A - B = \begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$$

Answer: $\begin{bmatrix} -2 & 0 \\ 2 & 4 \end{bmatrix}$.

Example 5: Matrix Polynomial

Question: Find $A^2 - 5A + 6I$ if $A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$.

Explanation: We need to compute each term of the polynomial $f(A)=A^2-5A+6I$, where $I$ is the 3x3 identity matrix.

1. Calculate $A^2$: $$A^2 = A \cdot A = \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix} \begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} 4+0+1 & 0+0-1 & 2+0+0 \\ 4+2+3 & 0+1-3 & 2+3+0 \\ 2-2+0 & 0-1+0 & 1-3+0 \end{bmatrix}$$ $$= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix}$$
2. Calculate $5A$: $$5A = 5\begin{bmatrix} 2 & 0 & 1 \\ 2 & 1 & 3 \\ 1 & -1 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix}$$
3. Calculate $6I$: $$6I = 6\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$$
4. Combine the matrices: $$A^2 - 5A + 6I$$ $$= \begin{bmatrix} 5 & -1 & 2 \\ 9 & -2 & 5 \\ 0 & -1 & -2 \end{bmatrix} - \begin{bmatrix} 10 & 0 & 5 \\ 10 & 5 & 15 \\ 5 & -5 & 0 \end{bmatrix} + \begin{bmatrix} 6 & 0 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 6 \end{bmatrix}$$ $$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0 \\ 9-10+0 & -2-5+6 & 5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}$$ $$= \begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$$

Answer: $\begin{bmatrix} 1 & -1 & -3 \\ -1 & -1 & -10 \\ -5 & 4 & 4 \end{bmatrix}$.

Example 6: Non-Commutativity of Multiplication

Question: If $A = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$ and $B = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}$, find AB and BA.

Explanation: 1. Calculate AB: $$AB = \begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} (1)(2)+(-1)(1) & (1)(1)+(-1)(0) \\ (2)(2)+3(1) & (2)(1)+3(0) \end{bmatrix}$$ $$= \begin{bmatrix} 2-1 & 1+0 \\ 4+3 & 2+0 \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 7 & 2 \end{bmatrix}$$

2. Calculate BA: $$BA = \begin{bmatrix} 2 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 2 & 3 \end{bmatrix}$$ $$= \begin{bmatrix} (2)(1)+1(2) & (2)(-1)+1(3) \\ (1)(1)+0(2) & (1)(-1)+0(3) \end{bmatrix}$$ $$= \begin{bmatrix} 2+2 & -2+3 \\ 1+0 & -1+0 \end{bmatrix} = \begin{bmatrix} 4 & 1 \\ 1 & -1 \end{bmatrix}$$

3. Conclusion: We observe that $AB \ne BA$, which illustrates that matrix multiplication is not commutative.

Answer: $AB = \begin{bmatrix} 1 & 1 \\ 7 & 2 \end{bmatrix}$ and $BA = \begin{bmatrix} 4 & 1 \\ 1 & -1 \end{bmatrix}$.

Example 7: Solving a Matrix Equation

Question: Solve for x: $\begin{bmatrix} x & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -2 & 0 \end{bmatrix} = O$.

Explanation:

1. Set up the multiplication: The product of a (1x2) matrix and a (2x2) matrix will be a (1x2) matrix.

   The zero matrix $O$ on the right side must also be of order (1x2), so $$O = \begin{bmatrix} 0 & 0 \end{bmatrix}$$

2. Perform Multiplication: $$\begin{bmatrix} (x)(1)+(1)(-2) & (x)(0)+(1)(0) \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$ $$\begin{bmatrix} x-2 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \end{bmatrix}$$

3. Equate Elements and Solve: By equating the corresponding elements, we get:

   • $x-2 = 0 \implies x=2$
   • $0 = 0$ (This is consistent)

Answer: x=2.

Example 8: Powers of a Matrix

Question: If $A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$, find $A^{100}$.

Explanation: We will find the first few powers of A to identify a pattern.

1. Calculate $A^2$: $$A^2 = A \cdot A = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$$ We can rewrite this as $$A^2 = 2\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} = 2A$$
2. Calculate $A^3$: $$A^3 = A^2 \cdot A = (2A) \cdot A$$ $$= 2(A^2) = 2(2A) = 4A = 2^2A$$
3. Identify the Pattern: We see a pattern: $$A^n = 2^{n-1}A$$
4. Apply the Pattern for n=100: $$A^{100} = 2^{100-1}A = 2^{99}A$$

Answer: $A^{100} = 2^{99}A = 2^{99}\begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix}$.

Example 9: Binomial Expansion with Nilpotent Matrix

Question: If $A = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, find $(aI+bA)^3$.

Explanation:

1. Analyze Matrix A: Let's find the powers of A. $A$ is a nilpotent matrix. $$A^2 = \begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} = O$$ Since $A^2$ is the zero matrix, all higher powers ($A^3, A^4, ...$) will also be the zero matrix.

2. Expand the Binomial: Since $I$ (identity) and $A$ commute ($IA=AI=A$), we can use the binomial expansion $$(X+Y)^3 = X^3+3X^2Y+3XY^2+Y^3$$ $$(aI+bA)^3 = (aI)^3 + 3(aI)^2(bA) + 3(aI)(bA)^2 + (bA)^3$$

3. Simplify: $$= a^3I^3 + 3a^2b(I^2A) + 3ab^2(IA^2) + b^3A^3$$

   • $I^3 = I$ and $I^2A = A$.
   • $A^2 = O$ (from step 1), so $IA^2 = O$.
   • $A^3 = O$ (from step 1).

   The expression simplifies to: $$a^3I + 3a^2bA + O + O$$

Answer: $$a^3I + 3a^2bA = \begin{bmatrix} a^3 & 0 \\ 0 & a^3 \end{bmatrix} + \begin{bmatrix} 0 & 3a^2b \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} a^3 & 3a^2b \\ 0 & a^3 \end{bmatrix}$$

Example 10: Matrix Multiplication Word Problem

Question: A trust fund has Rs 30,000 to invest in two types of bonds. The first bond pays 5% interest per year, and the second pays 7%. Using matrix multiplication, determine how to divide Rs 30,000 among the two types of bonds if the trust must obtain an annual total interest of Rs 1,800.

Explanation:

1. Define Variables: Let the investment in the first bond be Rs $x$. Then the investment in the second bond is Rs $(30000-x)$.

2. Set up Matrices:

   • Investment Matrix (A): A 1x2 row matrix representing the amount in each bond. $$A = \begin{bmatrix} x & 30000-x \end{bmatrix}$$
   • Interest Rate Matrix (R): A 2x1 column matrix representing the interest rate for each bond. $$R = \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix}$$
   • Total Interest Matrix (T): The desired total interest, as a 1x1 matrix. $$T = \begin{bmatrix} 1800 \end{bmatrix}$$

3. Form the Matrix Equation: The total interest is the product of the investment matrix and the rate matrix: $A \cdot R = T$. $$\begin{bmatrix} x & 30000-x \end{bmatrix} \begin{bmatrix} 0.05 \\ 0.07 \end{bmatrix} = \begin{bmatrix} 1800 \end{bmatrix}$$

4. Solve the Equation: $$(x)(0.05) + (30000-x)(0.07) = 1800$$ $$0.05x + 2100 - 0.07x = 1800$$ $$-0.02x = 1800 - 2100$$ $$-0.02x = -300$$ $$x = \frac{-300}{-0.02} = 15000$$

5. Find Both Investments:

   • Investment in first bond (at 5%): $x = 15,000$.

   • Investment in second bond (at 7%): $30000-x = 15,000$.

Answer: The investment is Rs 15,000 in each bond.

Example 11: Verifying $(AB)^T = B^TA^T$

Question: Verify the reversal law $(AB)^T = B^TA^T$ for $A = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B = \begin{bmatrix} 4 & 1 \\ 2 & 6 \end{bmatrix}$.

Explanation: 1. Calculate the Left-Hand Side (LHS):

• First, find $AB$: $$AB = \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix}\begin{bmatrix} 4 & 1 \\ 2 & 6 \end{bmatrix}$$ $$= \begin{bmatrix} 1(4)+2(2) & 1(1)+2(6) \\ 3(4)+4(2) & 3(1)+4(6) \end{bmatrix}$$ $$= \begin{bmatrix} 8 & 13 \\ 20 & 27 \end{bmatrix}$$
• Now, find the transpose: $$(AB)^T = \begin{bmatrix} 8 & 20 \\ 13 & 27 \end{bmatrix}$$

2. Calculate the Right-Hand Side (RHS):

• First, find $A^T$ and $B^T$:

  $A^T = \begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$ and $B^T = \begin{bmatrix} 4 & 2 \\ 1 & 6 \end{bmatrix}$.

• Now, find the product $B^TA^T$: $$B^TA^T = \begin{bmatrix} 4 & 2 \\ 1 & 6 \end{bmatrix}\begin{bmatrix} 1 & 3 \\ 2 & 4 \end{bmatrix}$$ $$= \begin{bmatrix} 4(1)+2(2) & 4(3)+2(4) \\ 1(1)+6(2) & 1(3)+6(4) \end{bmatrix}$$ $$= \begin{bmatrix} 8 & 20 \\ 13 & 27 \end{bmatrix}$$

3. Conclusion: Since LHS = RHS, the property is verified.

Answer: Verified, as both sides equal $\begin{bmatrix} 8 & 20 \\ 13 & 27 \end{bmatrix}$.

Example 12: Properties of Skew-Symmetric Matrix

Question: If the matrix $A = \begin{bmatrix} 0 & a & 3 \\ 2 & b & -1 \\ c & 1 & 0 \end{bmatrix}$ is skew-symmetric, find a, b, c.

Explanation: A skew-symmetric matrix must satisfy two conditions:

1. All diagonal elements are zero: $a_{ii} = 0$.

2. Non-diagonal elements are negative opposites: $a_{ij} = -a_{ji}$.

• Condition 1: We check the diagonal. $a_{22} = b$. Therefore, we must have b=0.
• Condition 2: We check the non-diagonal elements.
  • $a_{12} = -a_{21} \implies a = -(2) \implies$ a=-2.
  • $a_{13} = -a_{31} \implies 3 = -(c) \implies$ c=-3.
  • (We can check the last pair: $a_{23} = -a_{32} \implies -1 = -(1)$, which is true.)

Answer: a=-2, b=0, c=-3.

Example 13: Sum of Symmetric and Skew-Symmetric

Question: Express $A = \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix}$ as the sum of a symmetric (P) and a skew-symmetric (Q) matrix.

Explanation:

We use the formulas $$P = \frac{1}{2}(A+A^T)$$ and $$Q = \frac{1}{2}(A-A^T)$$

1. Find $A^T$: $$A^T = \begin{bmatrix} 3 & 1 \\ 5 & -1 \end{bmatrix}$$
2. Find P (Symmetric Part): $$A+A^T = \begin{bmatrix} 3+3 & 5+1 \\ 1+5 & -1-1 \end{bmatrix}$$ $$= \begin{bmatrix} 6 & 6 \\ 6 & -2 \end{bmatrix}$$ $$P = \frac{1}{2}(A+A^T) = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix}$$
3. Find Q (Skew-Symmetric Part): $$A-A^T = \begin{bmatrix} 3-3 & 5-1 \\ 1-5 & -1-(-1) \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 4 \\ -4 & 0 \end{bmatrix}$$ $$Q = \frac{1}{2}(A-A^T) = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$
4. Verification: $$P+Q = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$$ $$= \begin{bmatrix} 3 & 5 \\ 1 & -1 \end{bmatrix} = A$$

Answer: $A = \begin{bmatrix} 3 & 3 \\ 3 & -1 \end{bmatrix} + \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$.

Example 14: Property of Symmetric Matrices

Question: If A and B are symmetric matrices of the same order, show that AB+BA is a symmetric matrix.

Explanation:

1. Given: A and B are symmetric, so $A^T=A$ and $B^T=B$.

2. Goal: We need to show that the matrix $P = AB+BA$ is symmetric, which means we must prove $P^T=P$.

3. Take the Transpose: $P^T = (AB+BA)^T$.

4. Apply Transpose Properties: Property of sum: $$P^T = (AB)^T + (BA)^T$$ Reversal law of multiplication: $$P^T = (B^TA^T) + (A^TB^T)$$

5. Substitute the 'Given': Since $A^T=A$ and $B^T=B$, we get: $$P^T = (BA) + (AB)$$

6. Conclusion: $$P^T = BA+AB = AB+BA = P$$ Since $P^T=P$, the matrix $AB+BA$ is symmetric.

Answer: Proven.

Example 15: Property of $A^TA$

Question: Show that for any square matrix A, $A^TA$ is symmetric.

Explanation:

1. Define the Matrix: Let $P = A^TA$.

2. Goal: We need to show that $P$ is symmetric, i.e., $P^T=P$.

3. Take the Transpose: $P^T = (A^TA)^T$.

4. Apply Reversal Law: $$(XY)^T = Y^TX^T$$ Here, $X=A^T$ and $Y=A$. $$P^T = (A)^T (A^T)^T$$

5. Simplify: Using the property $(A^T)^T = A$, we get $P^T = A^T A$.

6. Conclusion: Since $P = A^TA$ and $P^T = A^TA$, we have $P^T=P$. Therefore, $A^TA$ is always symmetric.

Answer: Proven.

Example 16: Property of $A^2$

Question: If A is a skew-symmetric matrix, what is $A^2$?

Explanation: Let's test the nature of $A^2$ by checking its transpose.

1. Given: A is skew-symmetric, so $A^T = -A$.

2. Goal: We want to find the property of $P = A^2$. Let's find $P^T$.

3. Take the Transpose: $P^T = (A^2)^T = (A \cdot A)^T$.

4. Apply Reversal Law: $P^T = A^T A^T$.

5. Substitute the 'Given': Replace $A^T$ with $-A$. $P^T = (-A)(-A) = A^2$.

6. Conclusion: Since $P^T = A^2 = P$, the matrix $A^2$ is symmetric.

Answer: $A^2$ is a symmetric matrix.

Example 17: Finding 2x2 Inverse (Shortcut)

Question: Find the inverse of $A = \begin{bmatrix} 1 & 3 \\ 2 & 7 \end{bmatrix}$ using the shortcut.

Explanation:

1. Recall the 2x2 Inverse Formula: For a matrix $M = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$, the inverse is $$M^{-1} = \frac{1}{ad-bc}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$$
2. Calculate the Determinant: $$ad-bc = (1)(7) - (3)(2) = 7-6 = 1$$
3. Apply the Formula: $$A^{-1} = \frac{1}{1}\begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$$

Answer: $A^{-1} = \begin{bmatrix} 7 & -3 \\ -2 & 1 \end{bmatrix}$.

Example 18: Non-Invertible Matrix

Question: Find the inverse of $A = \begin{bmatrix} 2 & 1 \\ 4 & 2 \end{bmatrix}$, if it exists.

Explanation: A matrix is invertible if and only if its determinant is non-zero.

1. Calculate the Determinant: The determinant is $ad-bc$. $$\det(A) = (2)(2) - (1)(4) = 4 - 4 = 0$$
2. Conclusion: Since the determinant is zero, the matrix is singular.

Answer: The inverse does not exist.

Example 19: Finding Inverse with Elementary Operations

Question: Find the inverse of $A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix}$ using elementary row operations.

Explanation:

1. Set up: Start with the augmented matrix $[A|I]$. Our goal is to turn this into $[I|A^{-1}]$.

   $[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 2 & -1 & 0 & 1 \end{array}]$.

2. Create 0 in (2,1): Apply $R_2 \to R_2 - 2R_1$. $$[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & -5 & -2 & 1 \end{array}]$$

3. Create 1 in (2,2): Apply $R_2 \to -\frac{1}{5}R_2$. $$[\begin{array}{cc|cc} 1 & 2 & 1 & 0 \\ 0 & 1 & 2/5 & -1/5 \end{array}]$$

4. Create 0 in (1,2): Apply $R_1 \to R_1 - 2R_2$. $$R_1 = [1 \quad 2 \quad 1 \quad 0] - 2[0 \quad 1 \quad 2/5 \quad -1/5] = [1 \quad 0 \quad 1/5 \quad 2/5]$$ The matrix becomes $[\begin{array}{cc|cc} 1 & 0 & 1/5 & 2/5 \\ 0 & 1 & 2/5 & -1/5 \end{array}]$.

5. Conclusion: The LHS is now $I$, so the RHS is $A^{-1}$.

Answer: $A^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}$.

Example 20: Finding Inverse from a Matrix Equation

Question: If $A^2-A+I=O$, find the inverse of A.

Explanation:

1. Goal: To find $A^{-1}$, we must rearrange the equation into the form $A \cdot B = I$ or $B \cdot A = I$, where $B$ will be the inverse.

2. Rearrange the Equation: Start with $A^2-A+I=O$. Isolate the identity matrix $I$. $$I = A - A^2$$

3. Factor out A: Factor $A$ out of the right-hand side. We can pre-factor or post-factor.

   • Post-factor: $I = A(I - A)$.

   • Pre-factor: $I = (I - A)A$.

4. Conclusion: Both factorizations match the definition of an inverse $AB=BA=I$, where $$B = I-A$$

Answer: $A^{-1} = I-A$.

Example 21: Property of Invertible Transpose

Question: If A is an invertible matrix of order 2, find $(A^T)^{-1}$.

Explanation: This question asks for the inverse of the transpose of A. We use a standard property.

1. Start with the Inverse Definition: $AA^{-1} = I$.

2. Take the Transpose of both sides: $(AA^{-1})^T = I^T$.

3. Apply Reversal Law: $(A^{-1})^T (A^T) = I$ (since $I^T=I$).

4. Analyze the Result: This equation shows that the matrix $(A^{-1})^T$ is the left-inverse of the matrix $A^T$. For square matrices, the left-inverse is the inverse.

   Therefore, $(A^T)^{-1} = (A^{-1})^T$.

Answer: $(A^T)^{-1} = (A^{-1})^T$. The inverse of the transpose is the transpose of the inverse.

Example 22: Solving a Matrix Equation for X

Question: If A and B are invertible matrices, solve the equation $AXB=C$ for X.

Explanation: We need to isolate X. We must be careful to multiply in the correct order, as matrix multiplication is not commutative.

1. Given Equation: $AXB=C$.

2. Pre-multiply by $A^{-1}$: To remove A from the left, we pre-multiply (multiply on the left) both sides by $A^{-1}$.

   $A^{-1}(AXB) = A^{-1}C$

   $(A^{-1}A)XB = A^{-1}C$

   $IXB = A^{-1}C \implies XB=A^{-1}C$.

3. Post-multiply by $B^{-1}$: To remove B from the right, we post-multiply (multiply on the right) both sides by $B^{-1}$.

   $(XB)B^{-1} = (A^{-1}C)B^{-1}$

   $X(BB^{-1}) = A^{-1}CB^{-1}$

   $XI = A^{-1}CB^{-1}$.

Answer: $X=A^{-1}CB^{-1}$.

Example 23: Determinant Property

Question: If A is a square matrix such that $A^2=I$, what are the possible values for the determinant of A?

Explanation: This question uses properties of determinants (covered in the next chapter).

1. Start with the Equation: $A^2=I$.

2. Take the Determinant of both sides: $|A^2| = |I|$.

3. Apply Determinant Properties:

   • $|A^2| = |A|^2$.
   • $|I| = 1$.

4. Solve: The equation becomes $|A|^2=1$. Taking the square root gives $|A| = \pm 1$.

Answer: The possible values for the determinant are 1 and -1.

Example 24: Determinant Property

Question: If A is a 3x3 skew-symmetric matrix, what is the value of its determinant?

Explanation: This uses a property from the next chapter.

1. Property of Determinants: We know that $|A^T| = |A|$.

2. Property of Skew-Symmetric: We know that $A^T = -A$.

3. Combine: $|A| = |A^T| = |-A|$.

4. Property of Scalar Multiplication: For an $n \times n$ matrix, $|kA| = k^n|A|$. Here $n=3$ and $k=-1$.

   $|-A| = (-1)^3|A| = -|A|$.

5. Solve: Our equation becomes $|A| = -|A|$. This implies $2|A|=0$, so $|A|=0$.

Answer: The determinant of any odd-order skew-symmetric matrix is 0.

Example 25: Proving Symmetry

Question: Show that if A is a symmetric matrix, then $B^TAB$ is also symmetric.

Explanation:

1. Given: A is symmetric, so $A^T = A$.

2. Goal: We need to prove that the matrix $P = B^TAB$ is symmetric, which means we must show $P^T = P$.

3. Take the Transpose: $P^T = (B^TAB)^T$. We use the reversal law $(XY)^T = Y^TX^T$ twice.

   $$P^T = (B)^T (B^T A)^T = B^T ( A^T (B^T)^T )$$

4. Simplify: Since $(B^T)^T = B$, this becomes $P^T = B^T A^T B$.

5. Use the 'Given': Since A is symmetric, $A^T = A$. Substituting this in:

   $P^T = B^T A B$.

6. Conclusion: We have shown $P^T = P$. Therefore, $B^TAB$ is symmetric.

Answer: Proven.

Example 26: Contradictory Matrix Equation

Question: If $A = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$ and $A^2=B$, find $\alpha$.

Explanation:

1. Calculate $A^2$: $$A^2 = \begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} \alpha & 0 \\ 1 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} (\alpha)(\alpha)+0(1) & (\alpha)(0)+0(1) \\ (1)(\alpha)+1(1) & (1)(0)+1(1) \end{bmatrix}$$ $$= \begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1 \end{bmatrix}$$

2. Set up the Equation: We are given $A^2=B$. $$\begin{bmatrix} \alpha^2 & 0 \\ \alpha+1 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 5 & 1 \end{bmatrix}$$

3. Equate Elements and Solve:

   • From $a_{11}$: $\alpha^2 = 1 \implies \alpha = \pm 1$.

   • From $a_{21}$: $\alpha+1 = 5 \implies \alpha = 4$.

4. Conclusion: The value of $\alpha$ must be 4 to satisfy the second equation, but this does not satisfy the first equation ($4^2 \ne 1$). This is a contradiction.

Answer: There is no such value of $\alpha$ that satisfies the equation.

Example 27: Matrix Polynomial

Question: Let $f(x)=x^2-5x+6$. Find $f(A)$ if $A=\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$.

Explanation: To find $f(A)$, we replace $x$ with $A$ and the constant 6 with $6I$, where $I$ is the identity matrix. $$f(A)=A^2-5A+6I$$

1. Calculate $A^2$: $$A^2 = \begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix}$$ $$= \begin{bmatrix} 4+0 & 0+0 \\ 2+1 & 0+1 \end{bmatrix}$$ $$= \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix}$$

2. Calculate $5A$: $$5A = 5\begin{bmatrix} 2 & 0 \\ 1 & 1 \end{bmatrix} = \begin{bmatrix} 10 & 0 \\ 5 & 5 \end{bmatrix}$$

3. Calculate $6I$: $$6I = 6\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$

4. Combine: $$f(A) = \begin{bmatrix} 4 & 0 \\ 3 & 1 \end{bmatrix} - \begin{bmatrix} 10 & 0 \\ 5 & 5 \end{bmatrix} + \begin{bmatrix} 6 & 0 \\ 0 & 6 \end{bmatrix}$$ $$= \begin{bmatrix} 4-10+6 & 0-0+0 \\ 3-5+0 & 1-5+6 \end{bmatrix}$$ $$= \begin{bmatrix} 0 & 0 \\ -2 & 2 \end{bmatrix}$$

Answer: $f(A) = \begin{bmatrix} 0 & 0 \\ -2 & 2 \end{bmatrix}$.