RBSE Solutions Class 12 Maths Chapter 3 Matrix More Questions

Question 1. If A = \( \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \), then find A.
Answer: We are given the matrix A. To find A, we usually need to find a transformed version like \( A^2 \) or \( A^{-1} \). In this context, "find A" likely implies finding \( A^2 \) or some characteristic of A, as shown in the solution steps. Let's calculate \( A^2 \):
\( A^2 = A \cdot A \)
\( = \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 1 \\ 1 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} (1 \times 1) + (1 \times 1) & (1 \times 1) + (1 \times 1) \\ (1 \times 1) + (1 \times 1) & (1 \times 1) + (1 \times 1) \end{bmatrix} \)
\( = \begin{bmatrix} 1+1 & 1+1 \\ 1+1 & 1+1 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix} \)
We can also write this as \( 2A \). This shows how multiplying A by itself results in a scaled version of A. So, \( A^2 = 2A \).
In simple words: When you multiply matrix A by itself, you get a new matrix where every number is doubled compared to the original A. This means \( A^2 \) is the same as \( 2 \times A \).

🎯 Exam Tip: When a question asks to "find A" but provides A, it often implies calculating a power of A like \( A^2 \) or demonstrating a property related to A, as shown in the provided solution.

Question 2. If A = \( \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} \), then find (A - 2I) (A - 3I).
Answer: First, we need to calculate \( (A - 2I) \) and \( (A - 3I) \), where I is the identity matrix \( \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Now, let's find \( A - 2I \):
\( A - 2I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 2 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 2 & 0 \\ 0 & 2 \end{bmatrix} \)
\( = \begin{bmatrix} 4-2 & 2-0 \\ -1-0 & 1-2 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \)
Next, we find \( A - 3I \):
\( A - 3I = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - 3 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 2 \\ -1 & 1 \end{bmatrix} - \begin{bmatrix} 3 & 0 \\ 0 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 4-3 & 2-0 \\ -1-0 & 1-3 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \)
Finally, we multiply the two resulting matrices \( (A - 2I) \) and \( (A - 3I) \):
\( (A - 2I) (A - 3I) = \begin{bmatrix} 2 & 2 \\ -1 & -1 \end{bmatrix} \begin{bmatrix} 1 & 2 \\ -1 & -2 \end{bmatrix} \)
\( = \begin{bmatrix} (2 \times 1) + (2 \times (-1)) & (2 \times 2) + (2 \times (-2)) \\ ((-1) \times 1) + ((-1) \times (-1)) & ((-1) \times 2) + ((-1) \times (-2)) \end{bmatrix} \)
\( = \begin{bmatrix} 2-2 & 4-4 \\ -1+1 & -2+2 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
This result is a null matrix, where all elements are zero. This calculation is a basic example of finding eigenvalues if you consider \( (A - \lambda I) \).
In simple words: First, subtract a scaled identity matrix (2I) from A, and then subtract another scaled identity matrix (3I) from A. After you have these two new matrices, multiply them together. The final answer is a matrix with all zeros.

🎯 Exam Tip: Remember to perform matrix scalar multiplication (e.g., 2I, 3I) before subtraction. Also, be careful with signs during matrix multiplication, especially when dealing with negative numbers.

Question 4. If A = \( \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \) and B = \( \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \) where \( i = \sqrt{-1} \), then find BA.
Answer: We need to find the product of matrices B and A, in that specific order. Remember that matrix multiplication is not commutative, so BA is generally different from AB.
\( BA = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \)
To multiply, we take rows of the first matrix and columns of the second:
\( = \begin{bmatrix} (0 \times (-i)) + (i \times 0) & (0 \times 0) + (i \times i) \\ (i \times (-i)) + (0 \times 0) & (i \times 0) + (0 \times i) \end{bmatrix} \)
\( = \begin{bmatrix} 0+0 & 0+i^2 \\ -i^2+0 & 0+0 \end{bmatrix} \)
Since \( i = \sqrt{-1} \), we know that \( i^2 = -1 \). Let's substitute this value:
\( = \begin{bmatrix} 0 & -1 \\ -(-1) & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \)
This resulting matrix is a rotation matrix when multiplied by a vector, highlighting the unique properties of complex numbers in linear algebra.
In simple words: To find BA, multiply the rows of matrix B by the columns of matrix A. Remember that \( i \) squared (\( i^2 \)) is equal to minus one (-1). This will help you get the correct final matrix.

🎯 Exam Tip: Always pay attention to the order of matrix multiplication (BA vs. AB) as the result can be different. Correctly substitute \( i^2 = -1 \) for accurate calculations.

Question 5. RBSE Solutions for Class 12 Maths Chapter 3 Matrix Miscellaneous Exercise, then find matrices A and B.
Answer: This question appears to be incomplete as stated, likely referencing a previous part or a set of equations to find matrices A and B. However, the provided solution calculates A and B based on an assumption of two simultaneous matrix equations, typically \( A+B \) and \( A-B \) or similar. Let's follow the steps shown, assuming two equations were given in context (i) and (ii), which are not explicitly stated in the question text. The solution calculates \( 2A \) by adding (i) and (ii), and \( 2B \) by subtracting (ii) from (i).
Assuming we have (i) \( A+B = \begin{bmatrix} 3 & 5 & -7 \\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix} \) and (ii) \( A-B = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \).
Adding (i) and (ii) to find 2A:
\( 2A = \begin{bmatrix} 3 & 5 & -7 \\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix} + \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 3+1 & 5+1 & -7+1 \\ -1+1 & 1+1 & 4+0 \\ 11+1 & 8+0 & 0+0 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 6 & -8 \\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{bmatrix} \)
To find A, we divide 2A by 2 (or multiply by \( \frac{1}{2} \)):
\( A = \frac{1}{2} \begin{bmatrix} 4 & 6 & -8 \\ 0 & 2 & 4 \\ 12 & 8 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 3 & -4 \\ 0 & 1 & 2 \\ 6 & 4 & 0 \end{bmatrix} \)
Subtracting (ii) from (i) to find 2B:
\( 2B = \begin{bmatrix} 3 & 5 & -7 \\ -1 & 1 & 4 \\ 11 & 8 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 0 \\ 1 & 0 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 3-1 & 5-1 & -7-1 \\ -1-1 & 1-1 & 4-0 \\ 11-1 & 8-0 & 0-0 \end{bmatrix} \)
\( = \begin{bmatrix} 2 & 4 & -8 \\ -2 & 0 & 4 \\ 10 & 8 & 0 \end{bmatrix} \)
To find B, we divide 2B by 2 (or multiply by \( \frac{1}{2} \)):
\( B = \frac{1}{2} \begin{bmatrix} 2 & 4 & -8 \\ -2 & 0 & 4 \\ 10 & 8 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 2 & -3 \\ -1 & 0 & 2 \\ 5 & 4 & 0 \end{bmatrix} \)
So, the matrices A and B are determined. This method is similar to solving simultaneous equations for variables, but applied to matrices.
In simple words: Imagine you have two matrix equations where A and B are unknown. Add the two equations to find a matrix that is twice A. Then, divide by two to get A. Subtract the second equation from the first to find a matrix that is twice B, and then divide by two to get B.

🎯 Exam Tip: When given sum and difference of two matrices, adding them gives twice the first matrix, and subtracting gives twice the second matrix. Always verify that the dimensions of the matrices are compatible for addition and subtraction.

Question 7. Order of matrix A is 3 x 4 and B is a matrix, such that AᵀB and ABᵀ defined, then write the order of B.
Answer: Let the order of matrix B be \( m \times n \).
We are given that the order of matrix A is \( 3 \times 4 \).
First, let's find the order of Aᵀ (A transpose). If A is \( 3 \times 4 \), then Aᵀ is \( 4 \times 3 \).
The product AᵀB is defined. For two matrices to be multiplied, the number of columns of the first matrix must be equal to the number of rows of the second matrix.
Order of Aᵀ is \( 4 \times 3 \). Order of B is \( m \times n \).
For AᵀB to be defined, the number of columns of Aᵀ (which is 3) must equal the number of rows of B (which is m).
\( \implies m = 3 \)
Next, the product ABᵀ is defined.
Order of A is \( 3 \times 4 \). If B is \( m \times n \), then Bᵀ is \( n \times m \).
For ABᵀ to be defined, the number of columns of A (which is 4) must equal the number of rows of Bᵀ (which is n).
\( \implies n = 4 \)
Therefore, the order of matrix B is \( m \times n = 3 \times 4 \). This ensures all multiplications are possible according to matrix algebra rules.
In simple words: When you multiply two matrices, the number of columns in the first one must match the number of rows in the second. By using this rule for both AᵀB and ABᵀ, we can figure out the unknown number of rows and columns for matrix B.

🎯 Exam Tip: Remember that if matrix A is of order \( p \times q \), then its transpose Aᵀ is of order \( q \times p \). For a matrix product XY to be defined, the number of columns in X must equal the number of rows in Y.

Question 8. If A = \( \begin{bmatrix} -2 & -1 & 1 \\ -1 & 7 & 4 \\ 1 & -x & -3 \end{bmatrix} \) is a symmetric matrix, then determine x.
Answer: A matrix A is symmetric if it is equal to its transpose, i.e., \( A = A^T \). This means that the element in the \( i \)-th row and \( j \)-th column (denoted as \( a_{ij} \)) must be equal to the element in the \( j \)-th row and \( i \)-th column (denoted as \( a_{ji} \)).
Given matrix A is:
\( A = \begin{bmatrix} -2 & -1 & 1 \\ -1 & 7 & 4 \\ 1 & -x & -3 \end{bmatrix} \)
For A to be symmetric, we must have \( a_{ij} = a_{ji} \). Let's compare the off-diagonal elements:
\( a_{12} = a_{21} \implies -1 = -1 \) (This is consistent)
\( a_{13} = a_{31} \implies 1 = 1 \) (This is consistent)
\( a_{23} = a_{32} \implies 4 = -x \)
From the last comparison, \( 4 = -x \).
To find x, we simply multiply both sides by -1.
\( \implies x = -4 \)
Thus, for the matrix to be symmetric, x must be -4. This property is crucial in many areas of mathematics and physics.
In simple words: A symmetric matrix is like a mirror image across its main diagonal. This means the number at (row 2, column 3) must be the same as the number at (row 3, column 2). By setting these equal, we can find the value of x.

🎯 Exam Tip: Remember the definition of a symmetric matrix: \( A = A^T \), which implies \( a_{ij} = a_{ji} \) for all i and j. Focus on comparing the off-diagonal elements to find unknown variables.

Question 9. Construct a matrix of order 3 x 3, B = [bij], whose elements are bij= (i) (j).
Answer: We need to construct a 3x3 matrix B, where each element \( b_{ij} \) is found by multiplying its row number (i) by its column number (j).
The general form of a 3x3 matrix is:
\( B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \\ b_{31} & b_{32} & b_{33} \end{bmatrix} \)
Now, let's calculate each element using the rule \( b_{ij} = i \times j \):
\( b_{11} = 1 \times 1 = 1 \)
\( b_{12} = 1 \times 2 = 2 \)
\( b_{13} = 1 \times 3 = 3 \)
\( b_{21} = 2 \times 1 = 2 \)
\( b_{22} = 2 \times 2 = 4 \)
\( b_{23} = 2 \times 3 = 6 \)
\( b_{31} = 3 \times 1 = 3 \)
\( b_{32} = 3 \times 2 = 6 \)
\( b_{33} = 3 \times 3 = 9 \)
Substituting these values into the matrix B, we get:
\( B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{bmatrix} \)
This matrix is also symmetric, which is a common characteristic when elements are defined by a product of their indices.
In simple words: For each spot in the 3x3 grid, multiply its row number by its column number. Put that answer in the spot. For example, the top-left spot is row 1, column 1, so it's \( 1 \times 1 = 1 \).

🎯 Exam Tip: Clearly understand what 'i' and 'j' represent (row and column index respectively) when defining matrix elements. Be systematic in calculating each element to avoid errors.

Question 10. If A = \( \begin{bmatrix} 2 & 3 & -4 \\ -1 & 2 & 3 \end{bmatrix} \) and B = \( \begin{bmatrix} -1 & 2 \\ 3 & 4 \\ -5 & -6 \end{bmatrix} \) then find A+Bᵀ.
Answer: To find \( A+B^T \), we first need to calculate the transpose of matrix B, denoted as \( B^T \). The transpose of a matrix is found by swapping its rows and columns.
Given matrix B is of order 3x2:
\( B = \begin{bmatrix} -1 & 2 \\ 3 & 4 \\ -5 & -6 \end{bmatrix} \)
Its transpose \( B^T \) will be of order 2x3:
\( B^T = \begin{bmatrix} -1 & 3 & -5 \\ 2 & 4 & -6 \end{bmatrix} \)
Now, we can add matrix A and matrix \( B^T \). For matrix addition, both matrices must have the same order. Matrix A is \( 2 \times 3 \) and \( B^T \) is also \( 2 \times 3 \), so addition is possible.
\( A + B^T = \begin{bmatrix} 2 & 3 & -4 \\ -1 & 2 & 3 \end{bmatrix} + \begin{bmatrix} -1 & 3 & -5 \\ 2 & 4 & -6 \end{bmatrix} \)
We add the corresponding elements of the two matrices:
\( = \begin{bmatrix} 2+(-1) & 3+3 & -4+(-5) \\ -1+2 & 2+4 & 3+(-6) \end{bmatrix} \)
\( = \begin{bmatrix} 2-1 & 6 & -4-5 \\ 1 & 6 & 3-6 \end{bmatrix} \)
\( = \begin{bmatrix} 1 & 6 & -9 \\ 1 & 6 & -3 \end{bmatrix} \)
The resulting matrix is \( \begin{bmatrix} 1 & 6 & -9 \\ 1 & 6 & -3 \end{bmatrix} \). This calculation demonstrates basic matrix operations.
In simple words: First, flip matrix B so its rows become columns and columns become rows. This is called the transpose of B. Then, add this new matrix to matrix A by adding the numbers that are in the same position in both matrices.

🎯 Exam Tip: Always calculate the transpose correctly before performing addition or subtraction. Ensure that both matrices have the same dimensions (number of rows and columns) before attempting to add or subtract them.

Question 11. Express matrix A as the sum of symmetric and skew-symmetric matrices, where A = \( \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} \).
Answer: Any square matrix A can be uniquely expressed as the sum of a symmetric matrix P and a skew-symmetric matrix Q, where \( P = \frac{1}{2}(A + A^T) \) and \( Q = \frac{1}{2}(A - A^T) \).
First, let's find the transpose of A, \( A^T \):
\( A = \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} \implies A^T = \begin{bmatrix} 6 & 5 \\ 2 & 4 \end{bmatrix} \)
Now, let's calculate \( A + A^T \):
\( A + A^T = \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} + \begin{bmatrix} 6 & 5 \\ 2 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 6+6 & 2+5 \\ 5+2 & 4+4 \end{bmatrix} \)
\( = \begin{bmatrix} 12 & 7 \\ 7 & 8 \end{bmatrix} \)
So, the symmetric part P is:
\( P = \frac{1}{2}(A + A^T) = \frac{1}{2} \begin{bmatrix} 12 & 7 \\ 7 & 8 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & \frac{7}{2} \\ \frac{7}{2} & 4 \end{bmatrix} \)
Next, let's calculate \( A - A^T \):
\( A - A^T = \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} - \begin{bmatrix} 6 & 5 \\ 2 & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 6-6 & 2-5 \\ 5-2 & 4-4 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} \)
So, the skew-symmetric part Q is:
\( Q = \frac{1}{2}(A - A^T) = \frac{1}{2} \begin{bmatrix} 0 & -3 \\ 3 & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & -\frac{3}{2} \\ \frac{3}{2} & 0 \end{bmatrix} \)
Finally, we express A as the sum of P and Q:
\( A = P + Q \)
\( = \begin{bmatrix} 6 & \frac{7}{2} \\ \frac{7}{2} & 4 \end{bmatrix} + \begin{bmatrix} 0 & -\frac{3}{2} \\ \frac{3}{2} & 0 \end{bmatrix} \)
\( = \begin{bmatrix} 6+0 & \frac{7}{2} - \frac{3}{2} \\ \frac{7}{2} + \frac{3}{2} & 4+0 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & \frac{4}{2} \\ \frac{10}{2} & 4 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 2 \\ 5 & 4 \end{bmatrix} \)
This confirms that A is indeed the sum of a symmetric and a skew-symmetric matrix. This decomposition is a fundamental concept in linear algebra.
In simple words: To split a matrix A into a symmetric part and a skew-symmetric part, first find its transpose (Aᵀ). Then, add A and Aᵀ and divide by two to get the symmetric part. Next, subtract Aᵀ from A and divide by two to get the skew-symmetric part. Adding these two new matrices will give you back the original matrix A.

🎯 Exam Tip: A symmetric matrix has \( P = P^T \), and a skew-symmetric matrix has \( Q = -Q^T \). Always verify these properties for your calculated P and Q matrices.

Question 12. If A = \( \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \), then prove that :
(i) \( (A^T)^T = A \)
(ii) \( A + A^T \) is a symmetric matrix.
(iii) \( A - A^T \) is a skew-symmetric matrix.
(iv) \( AA^T \) and \( A^T A \) are symmetric matrix.
Answer: We will prove each part one by one using the given matrix A.
Given matrix: \( A = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \)

(i) Prove \( (A^T)^T = A \):
First, find the transpose of A, \( A^T \). Swap rows and columns:
\( A^T = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
Now, find the transpose of \( A^T \), denoted as \( (A^T)^T \). Swap rows and columns of \( A^T \):
\( (A^T)^T = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \)
We can see that \( (A^T)^T \) is indeed equal to A.
\( \implies (A^T)^T = A \)
This property shows that transposing a matrix twice brings you back to the original matrix.

(ii) Prove \( A + A^T \) is a symmetric matrix:
A matrix X is symmetric if \( X = X^T \). Let \( X = A + A^T \). We need to show that \( (A + A^T)^T = A + A^T \).
First, calculate \( A + A^T \):
\( A + A^T = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} + \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2+2 & 1+(-1) & 1+0 \\ -1+1 & 0+0 & 2+1 \\ 0+1 & 1+2 & 3+3 \end{bmatrix} \)
\( = \begin{bmatrix} 4 & 0 & 1 \\ 0 & 0 & 3 \\ 1 & 3 & 6 \end{bmatrix} \)
Now, find the transpose of \( (A + A^T) \):
\( (A + A^T)^T = \begin{bmatrix} 4 & 0 & 1 \\ 0 & 0 & 3 \\ 1 & 3 & 6 \end{bmatrix}^T = \begin{bmatrix} 4 & 0 & 1 \\ 0 & 0 & 3 \\ 1 & 3 & 6 \end{bmatrix} \)
Since \( (A + A^T)^T = A + A^T \), the matrix \( A + A^T \) is symmetric. This is a general property: the sum of a matrix and its transpose is always symmetric.

(iii) Prove \( A - A^T \) is a skew-symmetric matrix:
A matrix Y is skew-symmetric if \( Y = -Y^T \). Let \( Y = A - A^T \). We need to show that \( (A - A^T)^T = -(A - A^T) \).
First, calculate \( A - A^T \):
\( A - A^T = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} - \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} 2-2 & 1-(-1) & 1-0 \\ -1-1 & 0-0 & 2-1 \\ 0-1 & 1-2 & 3-3 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 2 & 1 \\ -2 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix} \)
Now, find the transpose of \( (A - A^T) \):
\( (A - A^T)^T = \begin{bmatrix} 0 & 2 & 1 \\ -2 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix}^T = \begin{bmatrix} 0 & -2 & -1 \\ 2 & 0 & -1 \\ 1 & 1 & 0 \end{bmatrix} \)
Now, let's find \( -(A - A^T) \):
\( -(A - A^T) = -\begin{bmatrix} 0 & 2 & 1 \\ -2 & 0 & 1 \\ -1 & -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 & -1 \\ 2 & 0 & -1 \\ 1 & 1 & 0 \end{bmatrix} \)
Since \( (A - A^T)^T = -(A - A^T) \), the matrix \( A - A^T \) is skew-symmetric. The diagonal elements of a skew-symmetric matrix are always zero.

(iv) Prove \( AA^T \) and \( A^T A \) are symmetric matrices:
A matrix Z is symmetric if \( Z = Z^T \). We need to show \( (AA^T)^T = AA^T \) and \( (A^T A)^T = A^T A \).
First, calculate \( AA^T \):
\( AA^T = \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(2)+(1)(1)+(1)(1) & (2)(-1)+(1)(0)+(1)(2) & (2)(0)+(1)(1)+(1)(3) \\ (-1)(2)+(0)(1)+(2)(1) & (-1)(-1)+(0)(0)+(2)(2) & (-1)(0)+(0)(1)+(2)(3) \\ (0)(2)+(1)(1)+(3)(1) & (0)(-1)+(1)(0)+(3)(2) & (0)(0)+(1)(1)+(3)(3) \end{bmatrix} \)
\( = \begin{bmatrix} 4+1+1 & -2+0+2 & 0+1+3 \\ -2+0+2 & 1+0+4 & 0+0+6 \\ 0+1+3 & 0+0+6 & 0+1+9 \end{bmatrix} \)
\( = \begin{bmatrix} 6 & 0 & 4 \\ 0 & 5 & 6 \\ 4 & 6 & 10 \end{bmatrix} \)
Now, find the transpose of \( AA^T \):
\( (AA^T)^T = \begin{bmatrix} 6 & 0 & 4 \\ 0 & 5 & 6 \\ 4 & 6 & 10 \end{bmatrix}^T = \begin{bmatrix} 6 & 0 & 4 \\ 0 & 5 & 6 \\ 4 & 6 & 10 \end{bmatrix} \)
Since \( (AA^T)^T = AA^T \), the matrix \( AA^T \) is symmetric. This is a general property for any matrix A.
Next, calculate \( A^T A \):
\( A^T A = \begin{bmatrix} 2 & -1 & 0 \\ 1 & 0 & 1 \\ 1 & 2 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 & 1 \\ -1 & 0 & 2 \\ 0 & 1 & 3 \end{bmatrix} \)
\( = \begin{bmatrix} (2)(2)+(-1)(-1)+(0)(0) & (2)(1)+(-1)(0)+(0)(1) & (2)(1)+(-1)(2)+(0)(3) \\ (1)(2)+(0)(-1)+(1)(0) & (1)(1)+(0)(0)+(1)(1) & (1)(1)+(0)(2)+(1)(3) \\ (1)(2)+(2)(-1)+(3)(0) & (1)(1)+(2)(0)+(3)(1) & (1)(1)+(2)(2)+(3)(3) \end{bmatrix} \)
\( = \begin{bmatrix} 4+1+0 & 2+0+0 & 2-2+0 \\ 2+0+0 & 1+0+1 & 1+0+3 \\ 2-2+0 & 1+0+3 & 1+4+9 \end{bmatrix} \)
\( = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 2 & 4 \\ 0 & 4 & 14 \end{bmatrix} \)
Now, find the transpose of \( A^T A \):
\( (A^T A)^T = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 2 & 4 \\ 0 & 4 & 14 \end{bmatrix}^T = \begin{bmatrix} 5 & 2 & 0 \\ 2 & 2 & 4 \\ 0 & 4 & 14 \end{bmatrix} \)
Since \( (A^T A)^T = A^T A \), the matrix \( A^T A \) is symmetric. Both \( AA^T \) and \( A^T A \) are always symmetric, which is a useful property in various mathematical applications.
In simple words: This question asks us to prove several basic rules about matrix transposes and types of matrices. (i) Flipping a matrix twice brings it back to the start. (ii) Adding a matrix and its flip always makes a symmetric matrix (where it's the same forwards and backwards). (iii) Subtracting a matrix and its flip always makes a skew-symmetric matrix (where it's the negative of itself when flipped). (iv) Multiplying a matrix by its flip (in either order) also always results in a symmetric matrix.

🎯 Exam Tip: Remember the definitions: \( X \) is symmetric if \( X = X^T \), and \( Y \) is skew-symmetric if \( Y = -Y^T \). Use properties like \( (X+Y)^T = X^T+Y^T \) and \( (XY)^T = Y^T X^T \) for proofs.

Question 13. If A = \( \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} \); B = \( \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} \), and 3A - 2B + C is a null matrix, then determine matrix 'C'.
Answer: We are given two matrices A and B, and a matrix equation \( 3A - 2B + C = O \), where O is a null matrix (a matrix with all elements as zero). We need to find matrix C.
From the given equation, we can rearrange it to solve for C:
\( C = O - 3A + 2B \)
\( C = 2B - 3A \) (Since subtracting a null matrix doesn't change anything)
First, let's calculate \( 3A \):
\( 3A = 3 \begin{bmatrix} 4 & 2 \\ 1 & 3 \end{bmatrix} = \begin{bmatrix} 3 \times 4 & 3 \times 2 \\ 3 \times 1 & 3 \times 3 \end{bmatrix} = \begin{bmatrix} 12 & 6 \\ 3 & 9 \end{bmatrix} \)
Next, let's calculate \( 2B \):
\( 2B = 2 \begin{bmatrix} -2 & 1 \\ 3 & 2 \end{bmatrix} = \begin{bmatrix} 2 \times (-2) & 2 \times 1 \\ 2 \times 3 & 2 \times 2 \end{bmatrix} = \begin{bmatrix} -4 & 2 \\ 6 & 4 \end{bmatrix} \)
Now, substitute these into the equation for C:
\( C = 2B - 3A \)
\( C = \begin{bmatrix} -4 & 2 \\ 6 & 4 \end{bmatrix} - \begin{bmatrix} 12 & 6 \\ 3 & 9 \end{bmatrix} \)
Subtract the corresponding elements:
\( C = \begin{bmatrix} -4-12 & 2-6 \\ 6-3 & 4-9 \end{bmatrix} \)
\( C = \begin{bmatrix} -16 & -4 \\ 3 & -5 \end{bmatrix} \)
So, the matrix C is \( \begin{bmatrix} -16 & -4 \\ 3 & -5 \end{bmatrix} \). This calculation involves scalar multiplication and matrix subtraction, common matrix operations.
In simple words: We are given an equation with matrices A, B, and C. First, multiply matrix A by 3 and matrix B by 2. Then, to find C, subtract the scaled A matrix from the scaled B matrix.

🎯 Exam Tip: When dealing with matrix equations like \( pA + qB + rC = O \), remember that you can rearrange them just like algebraic equations to solve for an unknown matrix. Always perform scalar multiplication before addition or subtraction.

Question 14. Construct a matrix B = [bij] of the order 2 x 3, whose elements are bij = (i +2j)²/2
Answer: We need to construct a 2x3 matrix B, where each element \( b_{ij} \) is determined by the formula \( b_{ij} = \frac{(i + 2j)^2}{2} \). Here, 'i' represents the row number (1 or 2) and 'j' represents the column number (1, 2, or 3).
The general form of a 2x3 matrix is:
\( B = \begin{bmatrix} b_{11} & b_{12} & b_{13} \\ b_{21} & b_{22} & b_{23} \end{bmatrix} \)
Let's calculate each element:
For \( b_{11} \): \( i=1, j=1 \)
\( b_{11} = \frac{(1 + 2 \times 1)^2}{2} = \frac{(1+2)^2}{2} = \frac{3^2}{2} = \frac{9}{2} \)
For \( b_{12} \): \( i=1, j=2 \)
\( b_{12} = \frac{(1 + 2 \times 2)^2}{2} = \frac{(1+4)^2}{2} = \frac{5^2}{2} = \frac{25}{2} \)
For \( b_{13} \): \( i=1, j=3 \)
\( b_{13} = \frac{(1 + 2 \times 3)^2}{2} = \frac{(1+6)^2}{2} = \frac{7^2}{2} = \frac{49}{2} \)
For \( b_{21} \): \( i=2, j=1 \)
\( b_{21} = \frac{(2 + 2 \times 1)^2}{2} = \frac{(2+2)^2}{2} = \frac{4^2}{2} = \frac{16}{2} = 8 \)
For \( b_{22} \): \( i=2, j=2 \)
\( b_{22} = \frac{(2 + 2 \times 2)^2}{2} = \frac{(2+4)^2}{2} = \frac{6^2}{2} = \frac{36}{2} = 18 \)
For \( b_{23} \): \( i=2, j=3 \)
\( b_{23} = \frac{(2 + 2 \times 3)^2}{2} = \frac{(2+6)^2}{2} = \frac{8^2}{2} = \frac{64}{2} = 32 \)
Now, we assemble these elements into the matrix B:
\( B = \begin{bmatrix} \frac{9}{2} & \frac{25}{2} & \frac{49}{2} \\ 8 & 18 & 32 \end{bmatrix} \)
This matrix construction demonstrates how specific rules can define elements of a matrix. This type of element definition is often used in coding or generating specific matrix types.
In simple words: To build this matrix, for each spot, take its row number (i) and add it to two times its column number (2j). Square that result, and then divide it by two. Place this final number in the correct spot in the 2x3 matrix.

🎯 Exam Tip: Pay close attention to the order of operations in the formula: multiplication before addition, then squaring, then division. Be careful with calculations for each element.

Question 15. If A = \( \begin{bmatrix} -1 & 0 & 1 \\ 2 & 3 & -2 \\ 1 & 0 & 0 \end{bmatrix} \); B = \( \begin{bmatrix} 0 & -1 & 0 \\ 0 & 5 & 7 \end{bmatrix} \) and C = \( \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \), then find the element of 1st row of ABC.
Answer: To find the element of the 1st row of ABC, we first need to multiply matrices A and B, and then multiply the result by C. Alternatively, we can first multiply B and C, then multiply A by the result. The order of operations for matrix multiplication is associative, so \( (AB)C = A(BC) \). Let's calculate AB first, then multiply by C.
Matrix A is \( 3 \times 3 \), Matrix B is \( 2 \times 3 \). We notice that AB is not defined because the number of columns in A (3) does not match the number of rows in B (2). This means there might be an error in the question or the provided matrices as AB cannot be multiplied. Let's assume the question implicitly expects to find a product like \( A \cdot B^T \cdot C \) or if B and C can be multiplied first. If we consider \( B \cdot C \): Matrix B is \( 2 \times 3 \) and Matrix C is \( 3 \times 1 \). This product IS defined.
Let's calculate \( BC \):
\( BC = \begin{bmatrix} 0 & -1 & 0 \\ 0 & 5 & 7 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \)
\( = \begin{bmatrix} (0 \times (-1)) + (-1 \times 0) + (0 \times 1) \\ (0 \times (-1)) + (5 \times 0) + (7 \times 1) \end{bmatrix} \)
\( = \begin{bmatrix} 0+0+0 \\ 0+0+7 \end{bmatrix} \)
\( = \begin{bmatrix} 0 \\ 7 \end{bmatrix} \)
Now, we need to calculate \( A(BC) \). Matrix A is \( 3 \times 3 \) and \( BC \) is \( 2 \times 1 \). This product \( A(BC) \) is also NOT defined because the number of columns in A (3) does not match the number of rows in \( BC \) (2).

There appears to be a mismatch in the dimensions of the matrices provided for a standard matrix product ABC. Let's re-examine the provided solution steps in the original document. The solution provided in the source proceeds as follows, which is effectively computing \( (A \cdot \text{some_matrix_X}) \cdot \text{C} \) where 'some_matrix_X' is not B, or the B matrix is assumed to have different dimensions or form a different expression than stated. The source solution directly multiplies A with some other matrix and then that result with C. The numbers in the calculation steps (e.g., \( -1 \times 0 + 0 \times 2 + 1 \times 7 \)) do not directly correspond to \( A \cdot B \) or \( A \cdot (B \cdot C) \) with the given B. However, I must follow the provided solution for numerical calculations if possible, even if the question setup seems to have an issue. The OCR suggests the following intermediate multiplication step to find the first row of \( AB' \) (where B' is some matrix) followed by multiplying by C:
The source shows an intermediate step of `[-1x0+0x2+1x7]` leading to `[0]` and `[2x0+3x2-2x7]` leading to `[-8]`. This implies a product of the first row of A with some column, which then becomes part of a larger matrix. Let's trace the actual calculation shown in the source, as if it is finding the first row of \( (A \cdot \text{IntermediateMatrix}) \cdot C \). The next step in the source shows: \( \begin{bmatrix} -1 & 5 & 7 \\ 2 & -13 & -8 \end{bmatrix} \begin{bmatrix} -1 \\ 0 \\ 1 \end{bmatrix} \)
This intermediate matrix \( \begin{bmatrix} -1 & 5 & 7 \\ 2 & -13 & -8 \end{bmatrix} \) is derived from some preceding operations. Assuming this intermediate matrix is the result of AB (despite dimensional issues if B is the given one), let's proceed with this multiplication with C.
To find the element of the 1st row of the final product, we multiply the 1st row of \( \begin{bmatrix} -1 & 5 & 7 \\ 2 & -13 & -8 \end{bmatrix} \) by the column of C:
\( \text{First row element} = (-1 \times (-1)) + (5 \times 0) + (7 \times 1) \)
\( = 1 + 0 + 7 \)
\( = 8 \)
This calculation is what the source arrives at for the first element. This implies the intermediate matrix was intended to be \( (AB) \), and its multiplication by C results in the value 8 in the first row. Despite the stated matrix dimensions not allowing \( AB \) directly, following the numerical steps of the solution leads to this result. Matrix multiplication is fundamental to many linear transformations and data manipulations.
In simple words: The problem asks for one number from the first row of the matrix created by multiplying A, B, and C. Since the given matrices do not directly allow \( AB \), we use the intermediate result shown in the steps. We take the first row of that intermediate matrix and multiply it by the column of C, adding up the products to get the final answer.

🎯 Exam Tip: Always double-check matrix dimensions before attempting multiplication. For \( A_{m \times n} \cdot B_{p \times q} \), n must equal p. If there's a discrepancy, carefully re-read the question or assume an intermediate step is provided.

Question 16. If A = \( \begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} \), then find AAᵀ.
Answer: We need to find the product of matrix A and its transpose, \( A^T \).
First, find the transpose of A, \( A^T \). Swap the rows and columns of A:
\( A^T = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \)
Now, multiply A by \( A^T \):
\( AA^T = \begin{bmatrix} \cos \theta & \sin \theta \\ - \sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \)
Perform the matrix multiplication:
\( = \begin{bmatrix} (\cos \theta)(\cos \theta) + (\sin \theta)(\sin \theta) & (\cos \theta)(-\sin \theta) + (\sin \theta)(\cos \theta) \\ (-\sin \theta)(\cos \theta) + (\cos \theta)(\sin \theta) & (-\sin \theta)(-\sin \theta) + (\cos \theta)(\cos \theta) \end{bmatrix} \)
\( = \begin{bmatrix} \cos^2 \theta + \sin^2 \theta & -\cos \theta \sin \theta + \sin \theta \cos \theta \\ -\sin \theta \cos \theta + \sin \theta \cos \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix} \)
Using the trigonometric identity \( \sin^2 \theta + \cos^2 \theta = 1 \):
\( = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
This is the identity matrix \( I_2 \). This result shows that A is an orthogonal matrix, a type of matrix important in rotations and transformations.
In simple words: To find AAᵀ, first flip matrix A to get Aᵀ. Then, multiply A by Aᵀ. When you do this, using the rule that \( \cos^2 \theta + \sin^2 \theta = 1 \), all the numbers outside the main diagonal become zero, and the numbers on the diagonal become one. This gives you an identity matrix.

🎯 Exam Tip: Recognize trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) as they are frequently used in matrix problems involving trigonometric functions. This result also identifies A as an orthogonal matrix.

Question 18. If B = \( \begin{bmatrix} a & b \\ c & d \end{bmatrix} \), then prove \( B^2 - (a+d)B = (bc - ad)I_2 \), where \( I_2 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \).
Answer: Not available in the provided page range (solution starts on page 15).

🎯 Exam Tip: For proofs involving matrices and scalars, always work methodically, calculating each term (like \( B^2 \), \( (a+d)B \)) separately before combining them. Remember that \( I_2 \) is the \( 2 \times 2 \) identity matrix.

Question 19. If \( A = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \) and \( B = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \), then find the matrix form of the following \( (aA + bB) (aA - bB) \).
Answer: First, we will find \( aA \) and \( bB \):
\( aA = a \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} \)
\( bB = b \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} \)

Next, we calculate the sum and difference of these matrices:
\( aA + bB = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} + \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \)
\( aA - bB = \begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} - \begin{bmatrix} 0 & b \\ -b & 0 \end{bmatrix} = \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)

Now, we multiply these two resulting matrices:
\( (aA + bB)(aA - bB) = \begin{bmatrix} a & b \\ -b & a \end{bmatrix} \begin{bmatrix} a & -b \\ b & a \end{bmatrix} \)
\( = \begin{bmatrix} a \cdot a + b \cdot b & a \cdot (-b) + b \cdot a \\ (-b) \cdot a + a \cdot b & (-b) \cdot (-b) + a \cdot a \end{bmatrix} \)
\( = \begin{bmatrix} a^2 + b^2 & -ab + ab \\ -ab + ab & b^2 + a^2 \end{bmatrix} \)
\( = \begin{bmatrix} a^2 + b^2 & 0 \\ 0 & a^2 + b^2 \end{bmatrix} \)
This matrix can be factored as a scalar multiplied by the identity matrix:
\( = (a^2 + b^2) \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} \)
So, \( (aA + bB)(aA - bB) = (a^2 + b^2) I_2 \). This resembles a common algebraic identity.
In simple words: We first found `aA` and `bB`. Then we added and subtracted them. Finally, we multiplied the two new matrices. The result is a simple matrix that is `(a^2 + b^2)` times the identity matrix.

🎯 Exam Tip: Remember that matrix multiplication is generally not commutative, so \( AB \neq BA \). However, in special cases like this one, properties similar to algebraic identities can emerge.

Question 20. If \( A = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \) and \( B = \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} \), then prove that \( (A - B)^2 \neq A^2 - 2AB + B^2 \).
Answer: We need to calculate both sides of the inequality separately.

Left Hand Side (LHS): \( (A - B)^2 \)
First, find \( A - B \):
\( A - B = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} - \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2-1 & 1-4 \\ -1-(-1) & 2-1 \end{bmatrix} = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} \)

Now, square the result:
\( (A - B)^2 = \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} 1 & -3 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + (-3) \cdot 0 & 1 \cdot (-3) + (-3) \cdot 1 \\ 0 \cdot 1 + 1 \cdot 0 & 0 \cdot (-3) + 1 \cdot 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1 + 0 & -3 - 3 \\ 0 + 0 & 0 + 1 \end{bmatrix} = \begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \) ... (i)

Right Hand Side (RHS): \( A^2 - 2AB + B^2 \)
First, find \( A^2 \):
\( A^2 = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} = \begin{bmatrix} 2 \cdot 2 + 1 \cdot (-1) & 2 \cdot 1 + 1 \cdot 2 \\ (-1) \cdot 2 + 2 \cdot (-1) & (-1) \cdot 1 + 2 \cdot 2 \end{bmatrix} \)
\( = \begin{bmatrix} 4-1 & 2+2 \\ -2-2 & -1+4 \end{bmatrix} = \begin{bmatrix} 3 & 4 \\ -4 & 3 \end{bmatrix} \)

Next, find \( AB \):
\( AB = \begin{bmatrix} 2 & 1 \\ -1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 + 1 \cdot (-1) & 2 \cdot 4 + 1 \cdot 1 \\ (-1) \cdot 1 + 2 \cdot (-1) & (-1) \cdot 4 + 2 \cdot 1 \end{bmatrix} \)
\( = \begin{bmatrix} 2-1 & 8+1 \\ -1-2 & -4+2 \end{bmatrix} = \begin{bmatrix} 1 & 9 \\ -3 & -2 \end{bmatrix} \)

Then, find \( 2AB \):
\( 2AB = 2 \begin{bmatrix} 1 & 9 \\ -3 & -2 \end{bmatrix} = \begin{bmatrix} 2 & 18 \\ -6 & -4 \end{bmatrix} \)

Now, find \( B^2 \):
\( B^2 = \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & 4 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} 1 \cdot 1 + 4 \cdot (-1) & 1 \cdot 4 + 4 \cdot 1 \\ (-1) \cdot 1 + 1 \cdot (-1) & (-1) \cdot 4 + 1 \cdot 1 \end{bmatrix} \)
\( = \begin{bmatrix} 1-4 & 4+4 \\ -1-1 & -4+1 \end{bmatrix} = \begin{bmatrix} -3 & 8 \\ -2 & -3 \end{bmatrix} \)

Finally, calculate \( A^2 - 2AB + B^2 \):
\( A^2 - 2AB + B^2 = \begin{bmatrix} 3 & 4 \\ -4 & 3 \end{bmatrix} - \begin{bmatrix} 2 & 18 \\ -6 & -4 \end{bmatrix} + \begin{bmatrix} -3 & 8 \\ -2 & -3 \end{bmatrix} \)
\( = \begin{bmatrix} 3-2+(-3) & 4-18+8 \\ -4-(-6)+(-2) & 3-(-4)+(-3) \end{bmatrix} \)
\( = \begin{bmatrix} 3-2-3 & 4-18+8 \\ -4+6-2 & 3+4-3 \end{bmatrix} = \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix} \) ... (ii)

From (i) and (ii), we can clearly see that \( \begin{bmatrix} 1 & -6 \\ 0 & 1 \end{bmatrix} \neq \begin{bmatrix} -2 & -6 \\ 0 & 4 \end{bmatrix} \).
Thus, \( (A - B)^2 \neq A^2 - 2AB + B^2 \) is proven.
In simple words: We calculated `(A-B)` first, then squared it. Then we calculated `A` squared, `B` squared, and `2AB` separately. When we put them all together, the two final matrices were different. This shows that the basic algebra rule for `(x-y)^2` does not work for matrices in the same way.

🎯 Exam Tip: This problem highlights a key difference between matrix algebra and scalar algebra: matrix multiplication is generally not commutative (\( AB \neq BA \)). This means common algebraic identities involving products, like \((A-B)^2 = A^2 - 2AB + B^2\), do not hold true for matrices unless \( AB = BA \).

Question 22. If \( i = \sqrt{-1} \), then prove that :
(i) \( A^2 = B^2 = C^2 = -I_2 \)
(ii) \( AB = - BA = -C \)
where \( A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \), \( B = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \), \( C = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \).
Answer: We will prove each part separately using the given matrices.

Part (i): Prove \( A^2 = B^2 = C^2 = -I_2 \)
For matrix \( A \):
\( A^2 = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 \cdot 0 + (-1) \cdot 1 & 0 \cdot (-1) + (-1) \cdot 0 \\ 1 \cdot 0 + 0 \cdot 1 & 1 \cdot (-1) + 0 \cdot 0 \end{bmatrix} \)
\( = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = -I_2 \)

For matrix \( B \):
\( B^2 = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 0 \cdot 0 + i \cdot i & 0 \cdot i + i \cdot 0 \\ i \cdot 0 + 0 \cdot i & i \cdot i + 0 \cdot 0 \end{bmatrix} \)
\( = \begin{bmatrix} i^2 & 0 \\ 0 & i^2 \end{bmatrix} \)
Since \( i = \sqrt{-1} \), then \( i^2 = -1 \):
\( = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I_2 \)

For matrix \( C \):
\( C^2 = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} = \begin{bmatrix} i \cdot i + 0 \cdot 0 & i \cdot 0 + 0 \cdot (-i) \\ 0 \cdot i + (-i) \cdot 0 & 0 \cdot 0 + (-i) \cdot (-i) \end{bmatrix} \)
\( = \begin{bmatrix} i^2 & 0 \\ 0 & i^2 \end{bmatrix} \)
Since \( i^2 = -1 \):
\( = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I_2 \)
Thus, \( A^2 = B^2 = C^2 = -I_2 \) is proven.

Part (ii): Prove \( AB = - BA = -C \)
First, calculate \( AB \):
\( AB = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} = \begin{bmatrix} 0 \cdot 0 + (-1) \cdot i & 0 \cdot i + (-1) \cdot 0 \\ 1 \cdot 0 + 0 \cdot i & 1 \cdot i + 0 \cdot 0 \end{bmatrix} \)
\( = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \)
We know that \( C = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \). So, \( -C = \begin{bmatrix} -i & 0 \\ 0 & i \end{bmatrix} \).
Therefore, \( AB = -C \).

Next, calculate \( BA \):
\( BA = \begin{bmatrix} 0 & i \\ i & 0 \end{bmatrix} \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 0 \cdot 0 + i \cdot 1 & 0 \cdot (-1) + i \cdot 0 \\ i \cdot 0 + 0 \cdot 1 & i \cdot (-1) + 0 \cdot 0 \end{bmatrix} \)
\( = \begin{bmatrix} i & 0 \\ 0 & -i \end{bmatrix} \)
This matrix is equal to \( C \). So, \( BA = C \).
Therefore, \( -BA = -C \).

Combining these results, we have \( AB = -C \) and \( -BA = -C \).
Thus, \( AB = -BA = -C \) is proven.
In simple words: For Part (i), we multiplied each matrix by itself and found that all of them resulted in the negative of the identity matrix. For Part (ii), we multiplied `A` by `B` and `B` by `A`. We found that `AB` was equal to `(-C)` and `BA` was equal to `C`, which means `(-BA)` is also `(-C)`. This proves that `AB = -BA = -C`.

🎯 Exam Tip: Complex matrices like these are often used in advanced topics like quantum mechanics (related to Pauli matrices). When proving matrix identities, calculate each side of the equation carefully and explicitly, demonstrating step-by-step equality or inequality.

Question 23. If \( A = \begin{bmatrix} 3 & 1 \\ -1 & 2 \end{bmatrix} \) and \( f(A) = A^2 – 5A + 7I \), then find \( f(A) \).
Answer: We need to substitute the matrix A into the function \( f(A) \).

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

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

Since \( A \) is a \( 2 \times 2 \) matrix, \( I \) refers to the \( 2 \times 2 \) identity matrix. So, calculate \( 7I \):
\( 7I = 7 \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)

Now, substitute these into the expression for \( f(A) \):
\( f(A) = A^2 - 5A + 7I \)
\( = \begin{bmatrix} 8 & 5 \\ -5 & 3 \end{bmatrix} - \begin{bmatrix} 15 & 5 \\ -5 & 10 \end{bmatrix} + \begin{bmatrix} 7 & 0 \\ 0 & 7 \end{bmatrix} \)
\( = \begin{bmatrix} 8-15+7 & 5-5+0 \\ -5-(-5)+0 & 3-10+7 \end{bmatrix} \)
\( = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
So, \( f(A) \) is the null matrix of order 2.
In simple words: We first found `A` multiplied by itself, then `A` multiplied by 5, and the identity matrix multiplied by 7. We then put all these results together in the formula for `f(A)`. After doing the addition and subtraction, the final matrix was all zeros.

🎯 Exam Tip: Problems like this demonstrate the Cayley-Hamilton theorem, which states that every square matrix satisfies its own characteristic equation. If \( f(A) \) results in a null matrix, it implies that the polynomial is the characteristic polynomial of A.

Question 24. Prove that \( \begin{bmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{bmatrix} \begin{bmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \) when \( \alpha - \beta = (2m - 1)\frac{\pi}{2}, m \in N \).
Answer: Let the two given matrices be \( M_1 \) and \( M_2 \). We need to multiply them.

\( M_1 M_2 = \begin{bmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{bmatrix} \begin{bmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{bmatrix} \)

Multiplying the matrices, we get the elements of the resulting matrix:
(1,1) element: \( \cos^2 \alpha \cos^2 \beta + \cos \alpha \sin \alpha \cos \beta \sin \beta \)
\( = \cos \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \)

(1,2) element: \( \cos^2 \alpha \cos \beta \sin \beta + \cos \alpha \sin \alpha \sin^2 \beta \)
\( = \cos \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \)

(2,1) element: \( \cos \alpha \sin \alpha \cos^2 \beta + \sin^2 \alpha \cos \beta \sin \beta \)
\( = \sin \alpha \cos \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \)

(2,2) element: \( \cos \alpha \sin \alpha \cos \beta \sin \beta + \sin^2 \alpha \sin^2 \beta \)
\( = \sin \alpha \sin \beta (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \)

Notice that all elements have a common factor: \( (\cos \alpha \cos \beta + \sin \alpha \sin \beta) \).
Using the trigonometric identity \( \cos(A - B) = \cos A \cos B + \sin A \sin B \), this common factor is equal to \( \cos(\alpha - \beta) \).

So, the product of the matrices is:
\( M_1 M_2 = \cos(\alpha - \beta) \begin{bmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{bmatrix} \)

Now, we apply the given condition: \( \alpha - \beta = (2m - 1)\frac{\pi}{2} \), where \( m \in N \).
This means that \( \alpha - \beta \) is an odd multiple of \( \frac{\pi}{2} \) (e.g., \( \frac{\pi}{2}, \frac{3\pi}{2}, \frac{5\pi}{2}, \ldots \)).
For any odd multiple of \( \frac{\pi}{2} \), the cosine value is 0.
So, \( \cos(\alpha - \beta) = \cos\left((2m - 1)\frac{\pi}{2}\right) = 0 \).

Substituting this value back into the matrix product:
\( M_1 M_2 = 0 \cdot \begin{bmatrix} \cos \alpha \cos \beta & \cos \alpha \sin \beta \\ \sin \alpha \cos \beta & \sin \alpha \sin \beta \end{bmatrix} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \)
This proves that the product of the two matrices is a null matrix under the given condition.
In simple words: We multiplied the two matrices together. All the terms in the new matrix had a common part, which simplified to `cos(alpha - beta)`. Since we were told that `alpha - beta` is always an odd multiple of `pi/2`, `cos(alpha - beta)` becomes zero. Because of this, the entire resulting matrix also becomes zero.

🎯 Exam Tip: This problem beautifully combines matrix multiplication with trigonometric identities. Recognizing the `cos(A-B)` identity is crucial. Pay close attention to conditions like `(2m-1)π/2`, as they often simplify expressions to zero or one.