Samacheer Kalvi Class 10 Maths Solutions Chapter 3 Algebra Exercise 3.18

 

Question 1. Find the order of the product matrix AB if

Answer:
For the product of two matrices AB to exist, the number of columns in matrix A must be equal to the number of rows in matrix B. If A is an \( m \times n \) matrix and B is an \( n \times p \) matrix, then the product AB is an \( m \times p \) matrix. We will find the order of AB for each case given.
(i) Order of A \( = 3 \times 3 \), Order of B \( = 3 \times 3 \). Number of columns in A (3) = Number of rows in B (3). So, the order of AB \( = 3 \times 3 \).
(ii) Order of A \( = 4 \times 3 \), Order of B \( = 3 \times 2 \). Number of columns in A (3) = Number of rows in B (3). So, the order of AB \( = 4 \times 2 \).
(iii) Order of A \( = 4 \times 2 \), Order of B \( = 2 \times 2 \). Number of columns in A (2) = Number of rows in B (2). So, the order of AB \( = 4 \times 2 \).
(iv) Order of A \( = 4 \times 5 \), Order of B \( = 5 \times 1 \). Number of columns in A (5) = Number of rows in B (5). So, the order of AB \( = 4 \times 1 \).
(v) Order of A \( = 1 \times 1 \), Order of B \( = 1 \times 3 \). Number of columns in A (1) = Number of rows in B (1). So, the order of AB \( = 1 \times 3 \).
In simple words: To multiply two matrices, the number of columns in the first matrix must match the number of rows in the second. The new matrix will have the rows of the first and the columns of the second.

🎯 Exam Tip: Always double-check that the inner dimensions (columns of the first matrix and rows of the second) match before calculating the order of the product matrix. If they don't match, the product is undefined.

 

Question 2. A has 'a' rows and 'a + 3 ' columns. B has '6' rows and '17 – b' columns, and if both products AB and BA exist, find a, b?
Answer:
Given that A has \( a \) rows and \( (a+3) \) columns. So, the order of A is \( a \times (a+3) \).
Given that B has \( 6 \) rows and \( (17-b) \) columns. So, the order of B is \( 6 \times (17-b) \).

For the product AB to exist, the number of columns in A must be equal to the number of rows in B.
\( a+3 = 6 \)
\( \implies a = 6 - 3 \)
\( \implies a = 3 \) ............ (1)

For the product BA to exist, the number of columns in B must be equal to the number of rows in A.
\( 17-b = a \)
Now substitute the value of \( a \) from (1) into this equation:
\( 17-b = 3 \)
\( \implies b = 17 - 3 \)
\( \implies b = 14 \)
So, the values are \( a = 3 \) and \( b = 14 \).
In simple words: For two matrices to be multiplied in both ways (AB and BA), the number of rows of one must equal the columns of the other, and vice versa. This helps us find the unknown numbers 'a' and 'b'.

🎯 Exam Tip: Remember the rule for matrix multiplication: the number of columns of the first matrix must equal the number of rows of the second. Apply this rule for both AB and BA separately to form equations.

 

Question 3. A has 'a' rows and 'a + 3 ' columns. B has rows and 'b' columns, and if both products AB and BA exist, find a,b?
Answer:
Given that A has \( a \) rows and \( (a+3) \) columns. So, the order of A is \( a \times (a+3) \).
Given that B has \( 6 \) rows and \( b \) columns (assuming the "rows" meant 6 rows from Question 2, as the problem is similar). So, the order of B is \( 6 \times b \).

For the product AB to exist, the number of columns in A must be equal to the number of rows in B.
\( a+3 = 6 \)
\( \implies a = 6 - 3 \)
\( \implies a = 3 \) ............ (1)

For the product BA to exist, the number of columns in B must be equal to the number of rows in A.
\( b = a \)
Now substitute the value of \( a \) from (1) into this equation:
\( b = 3 \)
So, the values are \( a = 3 \) and \( b = 3 \).
In simple words: When two matrices can be multiplied in both orders, their dimensions must follow specific rules. We use these rules to find the values of 'a' and 'b'.

🎯 Exam Tip: Always make sure to write down the order of both matrices clearly before applying the conditions for existence of product AB and BA. This helps in setting up the correct equations.

 

Question 4. If \( A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix}, B = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix} \) find AB, BA and check if AB = BA?
Answer:
Given matrices are \( A = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix} \).

First, calculate the product AB:
\( AB = \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix} \times \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} (2 \times 1) + (5 \times 2) & (2 \times -3) + (5 \times 5) \\ (4 \times 1) + (3 \times 2) & (4 \times -3) + (3 \times 5) \end{pmatrix} \)
\( = \begin{pmatrix} 2 + 10 & -6 + 25 \\ 4 + 6 & -12 + 15 \end{pmatrix} \)
\( = \begin{pmatrix} 12 & 19 \\ 10 & 3 \end{pmatrix} \)

Next, calculate the product BA:
\( BA = \begin{pmatrix} 1 & -3 \\ 2 & 5 \end{pmatrix} \times \begin{pmatrix} 2 & 5 \\ 4 & 3 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 2) + (-3 \times 4) & (1 \times 5) + (-3 \times 3) \\ (2 \times 2) + (5 \times 4) & (2 \times 5) + (5 \times 3) \end{pmatrix} \)
\( = \begin{pmatrix} 2 - 12 & 5 - 9 \\ 4 + 20 & 10 + 15 \end{pmatrix} \)
\( = \begin{pmatrix} -10 & -4 \\ 24 & 25 \end{pmatrix} \)

Comparing AB and BA:
\( AB = \begin{pmatrix} 12 & 19 \\ 10 & 3 \end{pmatrix} \) and \( BA = \begin{pmatrix} -10 & -4 \\ 24 & 25 \end{pmatrix} \)
Since the elements are not the same, \( AB \ne BA \). This shows that matrix multiplication is generally not commutative.
In simple words: We multiply matrix A by B, and then B by A. We find that the results are different. This means that for matrices, the order of multiplication usually matters.

🎯 Exam Tip: Remember that matrix multiplication is generally not commutative, meaning \( AB \ne BA \). Always perform both calculations separately to verify for specific matrices.

 

Question 5. Given that \( A = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix}, B = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{pmatrix}, C = \begin{pmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{pmatrix} \) verify that A(B + C) = AB + AC
Answer:
Given matrices are \( A = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix} \), \( B = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{pmatrix} \), and \( C = \begin{pmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{pmatrix} \).

First, calculate the left-hand side: \( A(B + C) \).
Calculate \( B + C \):
\( B + C = \begin{pmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{pmatrix} + \begin{pmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{pmatrix} \)
\( = \begin{pmatrix} 1+1 & -1+3 & 2+2 \\ 3-4 & 5+1 & 2+3 \end{pmatrix} \)
\( = \begin{pmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{pmatrix} \)

Now, calculate \( A(B+C) \):
\( A(B+C) = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix} \times \begin{pmatrix} 2 & 2 & 4 \\ -1 & 6 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 2) + (3 \times -1) & (1 \times 2) + (3 \times 6) & (1 \times 4) + (3 \times 5) \\ (5 \times 2) + (-1 \times -1) & (5 \times 2) + (-1 \times 6) & (5 \times 4) + (-1 \times 5) \end{pmatrix} \)
\( = \begin{pmatrix} 2 - 3 & 2 + 18 & 4 + 15 \\ 10 + 1 & 10 - 6 & 20 - 5 \end{pmatrix} \)
\( = \begin{pmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{pmatrix} \) ............ (1)

Next, calculate the right-hand side: \( AB + AC \).
Calculate \( AB \):
\( AB = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix} \times \begin{pmatrix} 1 & -1 & 2 \\ 3 & 5 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 1) + (3 \times 3) & (1 \times -1) + (3 \times 5) & (1 \times 2) + (3 \times 2) \\ (5 \times 1) + (-1 \times 3) & (5 \times -1) + (-1 \times 5) & (5 \times 2) + (-1 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 1 + 9 & -1 + 15 & 2 + 6 \\ 5 - 3 & -5 - 5 & 10 - 2 \end{pmatrix} \)
\( = \begin{pmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{pmatrix} \)

Calculate \( AC \):
\( AC = \begin{pmatrix} 1 & 3 \\ 5 & -1 \end{pmatrix} \times \begin{pmatrix} 1 & 3 & 2 \\ -4 & 1 & 3 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 1) + (3 \times -4) & (1 \times 3) + (3 \times 1) & (1 \times 2) + (3 \times 3) \\ (5 \times 1) + (-1 \times -4) & (5 \times 3) + (-1 \times 1) & (5 \times 2) + (-1 \times 3) \end{pmatrix} \)
\( = \begin{pmatrix} 1 - 12 & 3 + 3 & 2 + 9 \\ 5 + 4 & 15 - 1 & 10 - 3 \end{pmatrix} \)
\( = \begin{pmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{pmatrix} \)

Now, calculate \( AB + AC \):
\( AB + AC = \begin{pmatrix} 10 & 14 & 8 \\ 2 & -10 & 8 \end{pmatrix} + \begin{pmatrix} -11 & 6 & 11 \\ 9 & 14 & 7 \end{pmatrix} \)
\( = \begin{pmatrix} 10-11 & 14+6 & 8+11 \\ 2+9 & -10+14 & 8+7 \end{pmatrix} \)
\( = \begin{pmatrix} -1 & 20 & 19 \\ 11 & 4 & 15 \end{pmatrix} \) ............ (2)

From (1) and (2), we can see that \( A(B + C) = AB + AC \). This verifies the distributive property of matrix multiplication over matrix addition.
In simple words: We showed that multiplying a matrix A by the sum of two other matrices (B and C) gives the same result as multiplying A by B and A by C separately, and then adding those results. This proves a useful rule for working with matrices.

🎯 Exam Tip: When verifying matrix properties, always calculate each side of the equation separately and show all intermediate steps clearly. A single error in calculation can lead to a wrong conclusion. Distributive property is often tested.

 

Question 6. Show that the matrices \( A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix} \) satisfy commutative property AB = BA
Answer:
Given matrices are \( A = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix} \).

First, calculate the product AB:
\( AB = \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 1) + (2 \times -3) & (1 \times -2) + (2 \times 1) \\ (3 \times 1) + (1 \times -3) & (3 \times -2) + (1 \times 1) \end{pmatrix} \)
\( = \begin{pmatrix} 1 - 6 & -2 + 2 \\ 3 - 3 & -6 + 1 \end{pmatrix} \)
\( = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix} \) ............ (1)

Next, calculate the product BA:
\( BA = \begin{pmatrix} 1 & -2 \\ -3 & 1 \end{pmatrix} \times \begin{pmatrix} 1 & 2 \\ 3 & 1 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 1) + (-2 \times 3) & (1 \times 2) + (-2 \times 1) \\ (-3 \times 1) + (1 \times 3) & (-3 \times 2) + (1 \times 1) \end{pmatrix} \)
\( = \begin{pmatrix} 1 - 6 & 2 - 2 \\ -3 + 3 & -6 + 1 \end{pmatrix} \)
\( = \begin{pmatrix} -5 & 0 \\ 0 & -5 \end{pmatrix} \) ............ (2)

From (1) and (2), we observe that \( AB = BA \).
Therefore, the given matrices A and B satisfy the commutative property for multiplication.
In simple words: We multiplied matrix A by B and then B by A. Since both calculations gave the exact same result, these two specific matrices follow a rule called the commutative property, which means their multiplication order doesn't change the answer.

🎯 Exam Tip: While matrix multiplication is generally not commutative, specific pairs of matrices might satisfy this property. Always perform the calculations for both AB and BA to prove it. These types of questions test your calculation accuracy.

 

Question 7. Show that (i) A(BC) = (AB)C (ii) (A-B)C = AC – BC (iii) (A-B)\(^T\) = A\(^T\) – B\(^T\)
Answer:
Given matrices are \( A = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} \), \( B = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \), and \( C = \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \).

(i) To show \( A(BC) = (AB)C \):
First, calculate \( BC \):
\( BC = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (4 \times 2) + (0 \times 1) & (4 \times 0) + (0 \times 2) \\ (1 \times 2) + (5 \times 1) & (1 \times 0) + (5 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 8 + 0 & 0 + 0 \\ 2 + 5 & 0 + 10 \end{pmatrix} \)
\( = \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix} \)

Next, calculate \( A(BC) \):
\( A(BC) = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} \times \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 8) + (2 \times 7) & (1 \times 0) + (2 \times 10) \\ (1 \times 8) + (3 \times 7) & (1 \times 0) + (3 \times 10) \end{pmatrix} \)
\( = \begin{pmatrix} 8 + 14 & 0 + 20 \\ 8 + 21 & 0 + 30 \end{pmatrix} \)
\( = \begin{pmatrix} 22 & 20 \\ 29 & 30 \end{pmatrix} \) ............ (1)

Now, calculate \( AB \):
\( AB = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} \times \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 4) + (2 \times 1) & (1 \times 0) + (2 \times 5) \\ (1 \times 4) + (3 \times 1) & (1 \times 0) + (3 \times 5) \end{pmatrix} \)
\( = \begin{pmatrix} 4 + 2 & 0 + 10 \\ 4 + 3 & 0 + 15 \end{pmatrix} \)
\( = \begin{pmatrix} 6 & 10 \\ 7 & 15 \end{pmatrix} \)

Next, calculate \( (AB)C \):
\( (AB)C = \begin{pmatrix} 6 & 10 \\ 7 & 15 \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (6 \times 2) + (10 \times 1) & (6 \times 0) + (10 \times 2) \\ (7 \times 2) + (15 \times 1) & (7 \times 0) + (15 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 12 + 10 & 0 + 20 \\ 14 + 15 & 0 + 30 \end{pmatrix} \)
\( = \begin{pmatrix} 22 & 20 \\ 29 & 30 \end{pmatrix} \) ............ (2)

From (1) and (2), we have \( A(BC) = (AB)C \). This proves the associative property of matrix multiplication. The grouping of matrices does not change the final product.

(ii) To show \( (A-B)C = AC – BC \):
First, calculate \( A-B \):
\( A - B = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} - \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} 1-4 & 2-0 \\ 1-1 & 3-5 \end{pmatrix} \)
\( = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix} \)

Next, calculate \( (A-B)C \):
\( (A-B)C = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (-3 \times 2) + (2 \times 1) & (-3 \times 0) + (2 \times 2) \\ (0 \times 2) + (-2 \times 1) & (0 \times 0) + (-2 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} -6 + 2 & 0 + 4 \\ 0 - 2 & 0 - 4 \end{pmatrix} \)
\( = \begin{pmatrix} -4 & 4 \\ -2 & -4 \end{pmatrix} \) ............ (3)

Now, calculate \( AC \):
\( AC = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 2) + (2 \times 1) & (1 \times 0) + (2 \times 2) \\ (1 \times 2) + (3 \times 1) & (1 \times 0) + (3 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 2 + 2 & 0 + 4 \\ 2 + 3 & 0 + 6 \end{pmatrix} \)
\( = \begin{pmatrix} 4 & 4 \\ 5 & 6 \end{pmatrix} \)

Next, calculate \( BC \):
\( BC = \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \times \begin{pmatrix} 2 & 0 \\ 1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (4 \times 2) + (0 \times 1) & (4 \times 0) + (0 \times 2) \\ (1 \times 2) + (5 \times 1) & (1 \times 0) + (5 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 8 + 0 & 0 + 0 \\ 2 + 5 & 0 + 10 \end{pmatrix} \)
\( = \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix} \)

Now, calculate \( AC - BC \):
\( AC - BC = \begin{pmatrix} 4 & 4 \\ 5 & 6 \end{pmatrix} - \begin{pmatrix} 8 & 0 \\ 7 & 10 \end{pmatrix} \)
\( = \begin{pmatrix} 4-8 & 4-0 \\ 5-7 & 6-10 \end{pmatrix} \)
\( = \begin{pmatrix} -4 & 4 \\ -2 & -4 \end{pmatrix} \) ............ (4)

From (3) and (4), we have \( (A-B)C = AC - BC \). This shows that matrix multiplication distributes over matrix subtraction from the right.

(iii) To show \( (A-B)^T = A^T – B^T \):
First, calculate \( A-B \):
\( A - B = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} - \begin{pmatrix} 4 & 0 \\ 1 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} 1-4 & 2-0 \\ 1-1 & 3-5 \end{pmatrix} \)
\( = \begin{pmatrix} -3 & 2 \\ 0 & -2 \end{pmatrix} \)

Next, find the transpose of \( (A-B) \):
\( (A-B)^T = \begin{pmatrix} -3 & 0 \\ 2 & -2 \end{pmatrix} \) ............ (5)

Now, find the transpose of A, \( A^T \):
\( A^T = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} \)

Find the transpose of B, \( B^T \):
\( B^T = \begin{pmatrix} 4 & 1 \\ 0 & 5 \end{pmatrix} \)

Next, calculate \( A^T - B^T \):
\( A^T - B^T = \begin{pmatrix} 1 & 1 \\ 2 & 3 \end{pmatrix} - \begin{pmatrix} 4 & 1 \\ 0 & 5 \end{pmatrix} \)
\( = \begin{pmatrix} 1-4 & 1-1 \\ 2-0 & 3-5 \end{pmatrix} \)
\( = \begin{pmatrix} -3 & 0 \\ 2 & -2 \end{pmatrix} \) ............ (6)

From (5) and (6), we have \( (A-B)^T = A^T - B^T \). This proves that the transpose of a difference of matrices is the difference of their transposes.
In simple words: We showed three important rules for matrices. First, how you group matrices for multiplication doesn't change the answer. Second, matrix multiplication spreads out over subtraction, just like regular numbers. Third, if you subtract two matrices and then flip them (transpose), it's the same as flipping each matrix first and then subtracting them.

🎯 Exam Tip: Matrix operations like multiplication and subtraction have several important properties. Carefully apply the rules for each operation, especially for transposes, ensuring rows become columns and vice-versa, to avoid calculation errors.

 

Question 8. If \( A = \begin{pmatrix} \cos \theta & 0 \\ 0 & \cos \theta \end{pmatrix}, B = \begin{pmatrix} \sin \theta & 0 \\ 0 & \sin \theta \end{pmatrix} \) then show that \( A^2 + B^2 = I \).
Answer:
Given matrices are \( A = \begin{pmatrix} \cos \theta & 0 \\ 0 & \cos \theta \end{pmatrix} \) and \( B = \begin{pmatrix} \sin \theta & 0 \\ 0 & \sin \theta \end{pmatrix} \).

First, calculate \( A^2 = A \times A \):
\( A^2 = \begin{pmatrix} \cos \theta & 0 \\ 0 & \cos \theta \end{pmatrix} \times \begin{pmatrix} \cos \theta & 0 \\ 0 & \cos \theta \end{pmatrix} \)
\( = \begin{pmatrix} (\cos \theta \times \cos \theta) + (0 \times 0) & (\cos \theta \times 0) + (0 \times \cos \theta) \\ (0 \times \cos \theta) + (\cos \theta \times 0) & (0 \times 0) + (\cos \theta \times \cos \theta) \end{pmatrix} \)
\( = \begin{pmatrix} \cos^2 \theta + 0 & 0 + 0 \\ 0 + 0 & 0 + \cos^2 \theta \end{pmatrix} \)
\( = \begin{pmatrix} \cos^2 \theta & 0 \\ 0 & \cos^2 \theta \end{pmatrix} \)

Next, calculate \( B^2 = B \times B \):
\( B^2 = \begin{pmatrix} \sin \theta & 0 \\ 0 & \sin \theta \end{pmatrix} \times \begin{pmatrix} \sin \theta & 0 \\ 0 & \sin \theta \end{pmatrix} \)
\( = \begin{pmatrix} (\sin \theta \times \sin \theta) + (0 \times 0) & (\sin \theta \times 0) + (0 \times \sin \theta) \\ (0 \times \sin \theta) + (\sin \theta \times 0) & (0 \times 0) + (\sin \theta \times \sin \theta) \end{pmatrix} \)
\( = \begin{pmatrix} \sin^2 \theta + 0 & 0 + 0 \\ 0 + 0 & 0 + \sin^2 \theta \end{pmatrix} \)
\( = \begin{pmatrix} \sin^2 \theta & 0 \\ 0 & \sin^2 \theta \end{pmatrix} \)

Now, calculate \( A^2 + B^2 \):
\( A^2 + B^2 = \begin{pmatrix} \cos^2 \theta & 0 \\ 0 & \cos^2 \theta \end{pmatrix} + \begin{pmatrix} \sin^2 \theta & 0 \\ 0 & \sin^2 \theta \end{pmatrix} \)
\( = \begin{pmatrix} \cos^2 \theta + \sin^2 \theta & 0 + 0 \\ 0 + 0 & \cos^2 \theta + \sin^2 \theta \end{pmatrix} \)

Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
This is the identity matrix I. Thus, \( A^2 + B^2 = I \).
In simple words: We squared matrix A and matrix B separately. Then we added their squared results. Because of a basic trigonometry rule, the answer turned out to be the identity matrix, which is like the number '1' for matrices.

🎯 Exam Tip: This problem combines matrix multiplication with trigonometric identities. Remember that \( \cos^2 \theta + \sin^2 \theta = 1 \) is a key identity often used in such problems.

 

Question 9. If \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \) prove that \( AA^T = I \).
Answer:
Given matrix is \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \).

First, find the transpose of A, \( A^T \):
\( A^T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \)

Now, calculate the product \( AA^T \):
\( AA^T = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \times \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \)
\( = \begin{pmatrix} (\cos \theta \times \cos \theta) + (\sin \theta \times \sin \theta) & (\cos \theta \times -\sin \theta) + (\sin \theta \times \cos \theta) \\ (-\sin \theta \times \cos \theta) + (\cos \theta \times \sin \theta) & (-\sin \theta \times -\sin \theta) + (\cos \theta \times \cos \theta) \end{pmatrix} \)
\( = \begin{pmatrix} \cos^2 \theta + \sin^2 \theta & -\sin \theta \cos \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \sin^2 \theta + \cos^2 \theta \end{pmatrix} \)

Using the trigonometric identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\( = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
This is the identity matrix I.
Therefore, \( AA^T = I \).
In simple words: We took a special matrix A and flipped its rows and columns to get its transpose, A\(^T\). Then, we multiplied A by A\(^T\). The final answer was the identity matrix, which is a special matrix with ones on the main diagonal and zeros everywhere else.

🎯 Exam Tip: This question demonstrates that for an orthogonal matrix (where \( AA^T = I \)), its transpose is also its inverse. Remember the transpose rule (rows become columns) and the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \).

 

Question 10. Verify that \( A^2 = I \) when \( A = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix} \).
Answer:
Given matrix is \( A = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix} \).

We need to calculate \( A^2 = A \times A \):
\( A^2 = \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix} \times \begin{pmatrix} 5 & -4 \\ 6 & -5 \end{pmatrix} \)
\( = \begin{pmatrix} (5 \times 5) + (-4 \times 6) & (5 \times -4) + (-4 \times -5) \\ (6 \times 5) + (-5 \times 6) & (6 \times -4) + (-5 \times -5) \end{pmatrix} \)
\( = \begin{pmatrix} 25 - 24 & -20 + 20 \\ 30 - 30 & -24 + 25 \end{pmatrix} \)
\( = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
This is the identity matrix I.
Thus, \( A^2 = I \). This verifies the property for the given matrix A.
In simple words: We took the given matrix A and multiplied it by itself to find A squared. The calculation resulted in the identity matrix, which has ones on the main line and zeros everywhere else. This proves the statement.

🎯 Exam Tip: When verifying a property like \( A^2 = I \), perform the matrix multiplication carefully, element by element. Ensure each step is accurate to arrive at the correct identity matrix.

 

Question 11. If \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) and \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \) show that \( A^2 – (a + d)A = (bc – ad)I_2 \).
Answer:
Given matrix is \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) and identity matrix \( I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

First, calculate \( A^2 = A \times A \):
\( A^2 = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \times \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)
\( = \begin{pmatrix} (a \times a) + (b \times c) & (a \times b) + (b \times d) \\ (c \times a) + (d \times c) & (c \times b) + (d \times d) \end{pmatrix} \)
\( = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} \)

Next, calculate \( (a+d)A \):
\( (a+d)A = (a+d) \begin{pmatrix} a & b \\ c & d \end{pmatrix} \)
\( = \begin{pmatrix} (a+d)a & (a+d)b \\ (a+d)c & (a+d)d \end{pmatrix} \)
\( = \begin{pmatrix} a^2 + ad & ab + bd \\ ac + dc & ad + d^2 \end{pmatrix} \)

Now, calculate the left-hand side: \( A^2 - (a+d)A \):
\( A^2 - (a+d)A = \begin{pmatrix} a^2 + bc & ab + bd \\ ac + dc & bc + d^2 \end{pmatrix} - \begin{pmatrix} a^2 + ad & ab + bd \\ ac + dc & ad + d^2 \end{pmatrix} \)
\( = \begin{pmatrix} (a^2+bc)-(a^2+ad) & (ab+bd)-(ab+bd) \\ (ac+dc)-(ac+dc) & (bc+d^2)-(ad+d^2) \end{pmatrix} \)
\( = \begin{pmatrix} a^2+bc-a^2-ad & 0 \\ 0 & bc+d^2-ad-d^2 \end{pmatrix} \)
\( = \begin{pmatrix} bc - ad & 0 \\ 0 & bc - ad \end{pmatrix} \)

Finally, calculate the right-hand side: \( (bc-ad)I_2 \):
\( (bc-ad)I_2 = (bc-ad) \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
\( = \begin{pmatrix} bc-ad & 0 \\ 0 & bc-ad \end{pmatrix} \)

Since the left-hand side equals the right-hand side, the identity is proven: \( A^2 – (a + d)A = (bc – ad)I_2 \). This is a form of the Cayley-Hamilton theorem for 2x2 matrices.
In simple words: We proved a mathematical rule for a 2x2 matrix. When you square a matrix and subtract a special multiple of the original matrix, the answer is always a multiple of the identity matrix. The number it's multiplied by is found by a simple calculation from the matrix's parts.

🎯 Exam Tip: This problem involves the Cayley-Hamilton theorem. When working with variables in matrices, carefully perform each step of multiplication and subtraction, ensuring each element is calculated correctly. The determinant of matrix A is \( ad-bc \).

 

Question 12. If \( A = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix}, B = \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix} \) verify that \( (AB)^T = B^T A^T \).
Answer:
Given matrices are \( A = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix} \) and \( B = \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix} \).

First, calculate the left-hand side: \( (AB)^T \).
Calculate \( AB \):
\( AB = \begin{pmatrix} 5 & 2 & 9 \\ 1 & 2 & 8 \end{pmatrix} \times \begin{pmatrix} 1 & 7 \\ 1 & 2 \\ 5 & -1 \end{pmatrix} \)
\( = \begin{pmatrix} (5 \times 1)+(2 \times 1)+(9 \times 5) & (5 \times 7)+(2 \times 2)+(9 \times -1) \\ (1 \times 1)+(2 \times 1)+(8 \times 5) & (1 \times 7)+(2 \times 2)+(8 \times -1) \end{pmatrix} \)
\( = \begin{pmatrix} 5+2+45 & 35+4-9 \\ 1+2+40 & 7+4-8 \end{pmatrix} \)
\( = \begin{pmatrix} 52 & 30 \\ 43 & 3 \end{pmatrix} \)

Now, find the transpose of AB:
\( (AB)^T = \begin{pmatrix} 52 & 43 \\ 30 & 3 \end{pmatrix} \) ............ (1)

Next, calculate the right-hand side: \( B^T A^T \).
Find the transpose of A, \( A^T \):
\( A^T = \begin{pmatrix} 5 & 1 \\ 2 & 2 \\ 9 & 8 \end{pmatrix} \)

Find the transpose of B, \( B^T \):
\( B^T = \begin{pmatrix} 1 & 1 & 5 \\ 7 & 2 & -1 \end{pmatrix} \)

Now, calculate the product \( B^T A^T \):
\( B^T A^T = \begin{pmatrix} 1 & 1 & 5 \\ 7 & 2 & -1 \end{pmatrix} \times \begin{pmatrix} 5 & 1 \\ 2 & 2 \\ 9 & 8 \end{pmatrix} \)
\( = \begin{pmatrix} (1 \times 5)+(1 \times 2)+(5 \times 9) & (1 \times 1)+(1 \times 2)+(5 \times 8) \\ (7 \times 5)+(2 \times 2)+(-1 \times 9) & (7 \times 1)+(2 \times 2)+(-1 \times 8) \end{pmatrix} \)
\( = \begin{pmatrix} 5+2+45 & 1+2+40 \\ 35+4-9 & 7+4-8 \end{pmatrix} \)
\( = \begin{pmatrix} 52 & 43 \\ 30 & 3 \end{pmatrix} \) ............ (2)

From (1) and (2), we can see that \( (AB)^T = B^T A^T \). This verifies the property of the transpose of a product of matrices.
In simple words: We showed that if you multiply two matrices (A and B) and then flip the result, it's the same as flipping each matrix first and then multiplying them in the reverse order. This is a key rule for working with matrix transposes.

🎯 Exam Tip: Remember the order reversal property for the transpose of a product: \( (AB)^T = B^T A^T \), not \( A^T B^T \). Careful calculation of each matrix element is crucial for verifying such identities.

 

Question 13. If \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \) show that \( A^2 – 5A + 7I_2 = 0 \).
Answer:
Given matrix is \( A = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \). The identity matrix \( I_2 = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \).

First, calculate \( A^2 = A \times A \):
\( A^2 = \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \times \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} (3 \times 3) + (1 \times -1) & (3 \times 1) + (1 \times 2) \\ (-1 \times 3) + (2 \times -1) & (-1 \times 1) + (2 \times 2) \end{pmatrix} \)
\( = \begin{pmatrix} 9 - 1 & 3 + 2 \\ -3 - 2 & -1 + 4 \end{pmatrix} \)
\( = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} \)

Next, calculate \( 5A \):
\( 5A = 5 \begin{pmatrix} 3 & 1 \\ -1 & 2 \end{pmatrix} \)
\( = \begin{pmatrix} 5 \times 3 & 5 \times 1 \\ 5 \times -1 & 5 \times 2 \end{pmatrix} \)
\( = \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} \)

Next, calculate \( 7I_2 \):
\( 7I_2 = 7 \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \)
\( = \begin{pmatrix} 7 \times 1 & 7 \times 0 \\ 7 \times 0 & 7 \times 1 \end{pmatrix} \)
\( = \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \)

Now, substitute these into the expression \( A^2 – 5A + 7I_2 \):
\( A^2 – 5A + 7I_2 = \begin{pmatrix} 8 & 5 \\ -5 & 3 \end{pmatrix} - \begin{pmatrix} 15 & 5 \\ -5 & 10 \end{pmatrix} + \begin{pmatrix} 7 & 0 \\ 0 & 7 \end{pmatrix} \)
\( = \begin{pmatrix} 8 - 15 + 7 & 5 - 5 + 0 \\ -5 - (-5) + 0 & 3 - 10 + 7 \end{pmatrix} \)
\( = \begin{pmatrix} 8 - 15 + 7 & 5 - 5 \\ -5 + 5 & 3 - 10 + 7 \end{pmatrix} \)
\( = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \)
This is the zero matrix.
Therefore, \( A^2 – 5A + 7I_2 = 0 \). This demonstrates the Cayley-Hamilton theorem for the given matrix A.
In simple words: We took a matrix A, calculated A squared, subtracted 5 times A, and added 7 times the identity matrix. After all the calculations, the result was a matrix full of zeros. This shows a special property where the matrix satisfies its own characteristic equation.

🎯 Exam Tip: When evaluating a matrix polynomial like \( A^2 - 5A + 7I \), perform each term (matrix multiplication, scalar multiplication) separately and accurately. Then, combine the matrices using addition and subtraction. Don't forget to include the identity matrix for constant terms.