Define a null matrix and provide an example.

Step 1: Define a null matrix and provide an example. A null matrix (or zero matrix) is a matrix in which every element is zero. An example of a 2 × 3 null matrix is: 0 & 0 & 0 \\ 0 & 0 & 0 Step 2: Define a diagonal matrix and provide an example. A diagonal matrix is a square matrix in which all the elements outside the main diagonal are zero. An example of a 3 × 3 diagonal matrix is: 1 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 9 Step 3: State the condition for a skew-symmetric matrix. A square matrix A is classified as skew-symmetric if its transpose is equal to its negative, i.e., A^T = -A. Step 4: Solve Question 2.1. If A^T = -A, the matrix A is skew-symmetric. For a skew-symmetric matrix, the elements on the main diagonal must be zero. This is because for any diagonal element a_ii, the condition a_ii = -a_ii must hold, which implies 2a_ii = 0, so a_ii = 0. The correct option is C. The elements on the diagonal of the matrix are equal to C. 0 Step 5: Solve Question 2.2. We need to identify the false statement. A. If |A| = 0, then |adj(A)| = 0. This statement is true. If a matrix is singular (|A|=0), its adjoint is also singular (|adj(A)|=0). B. Adjoint of a diagonal matrix of order 3 × 3 is a diagonal matrix. This statement is true. The cofactors of off-diagonal elements of a diagonal matrix are zero, so its adjoint will also be diagonal. C. Product of two upper triangular matrices is an upper triangular matrix. This statement is true. This is a known property of triangular matrices. D. adj(AB) = adj(A)adj(B). This statement is false. The correct property is adj(AB) = adj(B)adj(A). The false statement is D. adj(AB) = adj(A)adj(B) Step 6: Solve Question 2.3. Given AB = 3 & 2 \\ 1 & 1 and B = 1 & 1 \\ 2 & 0 . We need to find A^-1. Let C = 3 & 2 \\ 1 & 1 . So AB = C. To find A^-1, we can multiply both sides by C^-1 on the right: A = C B^-1 Then A^-1 = (C B^-1)^-1 = (B^-1)^-1 C^-1 = B C^-1. First, calculate C^-1: The determinant of C is |C| = (3)(1) - (2)(1) = 3 - 2 = 1. The inverse of C is C^-1 = (1)/(|C|) 1 & -2 \\ -1 & 3 = (1)/(1) 1 & -2 \\ -1 & 3 = 1 & -2 \\ -1 & 3 . Now, calculate A^-1 = B C^-1: A^-1 = 1 & 1 \\ 2 & 0 1 & -2 \\ -1 & 3 A^-1 = (1)(1) + (1)(-1) & (1)(-2) + (1)(3) \\ (2)(1) + (0)(-1) & (2)(-2) + (0)(3) A^-1 = 1 - 1 & -2 + 3 \\ 2 + 0 & -4 + 0 A^-1 = 0 & 1 \\ 2 & -4 The correct option is C. A^-1 is C. 0 & 1 \\ 2 & -4 Step 7: Solve Question 2.4. Given AB' + BA' = 0 and A is skew-symmetric. If A is skew-symmetric, then A^T = -A. (Note: A' denotes the transpose, so A' = A^T). Substitute A^T = -A into the given equation: AB^T + B(-A) = 0 AB^T - BA = 0 AB^T = BA Now, we need to classify BA. Let's find the transpose of BA: (BA)^T = A^T B^T Since A^T = -A, substitute this into the expression: (BA)^T = (-A) B^T = -AB^T From the derived relation AB^T = BA, we can substitute AB^T with BA: (BA)^T = -BA This condition means that $BA