Matrices and Determinants questions.

Steps to Solve

1. Question 1: Understanding Matrix Types

The question asks to identify the correct statement relating scalar, identity, and diagonal matrices.

• A scalar matrix is a diagonal matrix where all diagonal elements are equal.
• An identity matrix is a scalar matrix where all diagonal elements are 1.

2. Question 2: Identifying Skew-Symmetric Matrices

The question defines a skew-symmetric matrix as $A = (a_{ij})$ where $a_{ij} = -a_{ji}$ for all $i, j$.

3. Question 3: Properties of Triangular Matrices

This question checks the understanding of triangular matrices. A triangular matrix $A = (a_{ij})$ has the property that either $a_{ij} = 0$ for all $i > j$ (upper triangular) or $a_{ij} = 0$ for all $i < j$ (lower triangular). It must be a square matrix. The elements in the principal diagonal may or may not be zero.

4. Question 14: Solving for Matrix B

Given the equations $2A + 3B = \begin{pmatrix} 1 & 3 \ 2 & -1 \ 4 & -2 \end{pmatrix}$ and $A + 2B = \begin{pmatrix} 5 & 0 \ 3 & 1 \ 6 & 2 \end{pmatrix}$, we need to solve for $B$. Multiply the second equation by 2 to get $2A + 4B = \begin{pmatrix} 10 & 0 \ 6 & 2 \ 12 & 4 \end{pmatrix}$.

Subtract the first equation from the modified second equation: $(2A + 4B) - (2A + 3B) = B = \begin{pmatrix} 10 & 0 \ 6 & 2 \ 12 & 4 \end{pmatrix} - \begin{pmatrix} 1 & 3 \ 2 & -1 \ 4 & -2 \end{pmatrix} = \begin{pmatrix} 9 & -3 \ 4 & 3 \ 8 & 6 \end{pmatrix}$

However, this does not match any of the answer options. Let's recheck the equations given in the problem. We have $2A + 3B = \begin{pmatrix} 2 & -1 & 4 \ -1 & 3 & 2 \ 4 & 3 & 5 \end{pmatrix}$ and $A + 2B = \begin{pmatrix} 5 & 0 & 3 \ 1 & 6 & 2 \ 1 & 6 & 2 \end{pmatrix}$. Multiplying the second equation by 2 gives $2A + 4B = \begin{pmatrix} 10 & 0 & 6 \ 2 & 12 & 4 \ 2 & 12 & 4 \end{pmatrix}$. Subtracting the first equation gives $B = \begin{pmatrix} 10 & 0 & 6 \ 2 & 12 & 4 \ 2 & 12 & 4 \end{pmatrix} - \begin{pmatrix} 2 & -1 & 4 \ -1 & 3 & 2 \ 4 & 3 & 5 \end{pmatrix} = \begin{pmatrix} 8 & 1 & 2 \ 3 & 9 & 2 \ -2 & 9 & -1 \end{pmatrix}$. This still does not match any answer given.

Multiply the first equation by 2 to get $4A + 6B = \begin{pmatrix} 4 & -2 & 8 \ -2 & 6 & 4 \ 8 & 6 & 10 \end{pmatrix}$. Multiply the second equation by 3 to get $3A + 6B = \begin{pmatrix} 15 & 0 & 9 \ 3 & 18 & 6 \ 3 & 18 & 6 \end{pmatrix}$. $(4A + 6B) - (3A + 6B) = A = \begin{pmatrix} 4 & -2 & 8 \ -2 & 6 & 4 \ 8 & 6 & 10 \end{pmatrix} - \begin{pmatrix} 15 & 0 & 9 \ 3 & 18 & 6 \ 3 & 18 & 6 \end{pmatrix} = \begin{pmatrix} -11 & -2 & -1 \ -5 & -12 & -2 \ 5 & -12 & 4 \end{pmatrix}$. Then $A + 2B = \begin{pmatrix} 5 & 0 & 3 \ 1 & 6 & 2 \ 1 & 6 & 2 \end{pmatrix}$ so $2B = \begin{pmatrix} 5 & 0 & 3 \ 1 & 6 & 2 \ 1 & 6 & 2 \end{pmatrix} - A = \begin{pmatrix} 5 & 0 & 3 \ 1 & 6 & 2 \ 1 & 6 & 2 \end{pmatrix} - \begin{pmatrix} -11 & -2 & -1 \ -5 & -12 & -2 \ 5 & -12 & 4 \end{pmatrix} = \begin{pmatrix} 16 & 2 & 4 \ 6 & 18 & 4 \ -4 & 18 & -2 \end{pmatrix}$. $B = \begin{pmatrix} 8 & 1 & 2 \ 3 & 9 & 2 \ -2 & 9 & -1 \end{pmatrix}$. Comparing again shows no answers match. There may be an error in the provided answer choices. By observation, the first row of B $\begin{pmatrix} 8 & 1 & 2 \end{pmatrix}$ matches (a), (b), and (c). The second row of B $\begin{pmatrix} 3 & 9 & 2 \end{pmatrix}$, is closest to $\begin{pmatrix} -1 & 10 & 1 \end{pmatrix}$ in option a) and c). This is probably a typo.

5. Question 15: Matrix Operations and Simplification

Given $A = \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix}$, we need to find $A^+ + A^- - A^2$. Assuming $A^+$ means transpose $A^T$, and $A^-$ means inverse $A^{-1}$. So, $A^T = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix}$.

$|A| = (2)(-2) - (-1)(3) = -4 + 3 = -1$.

$A^{-1} = \frac{1}{|A|} \begin{pmatrix} -2 & 1 \ -3 & 2 \end{pmatrix} = \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix}$ which equals A

$A^2 = \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix}\begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} = \begin{pmatrix} 4-3 & -2+2 \ 6-6 & -3+4 \end{pmatrix} = \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = I$. Therefore, $A^T + A^{-1} - A^2 = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix} + \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 2 \ 2 & -4 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \ 2 & -5 \end{pmatrix}$. The answer is none of these.

If $A^+$ means adjoint $adj(A)$, $adj(A) = \begin{pmatrix} -2 & 1 \ -3 & 2 \end{pmatrix}$

If $A^+$ means transpose $A^T$, and $A^-$ means $-A$. So, $A^T = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix}$. $A^2 = I$. Therefore, $A^T - A - A^2 = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix} - \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 0 & 4 \ -4 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & 4 \ -4 & -1 \end{pmatrix}$. The answer is none of these.

If $A^+$ means transpose $A^T$, and $A^-$ is just A. So, $A^T = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix}$. $A^2 = I$. $A^T + A - A^2 = \begin{pmatrix} 2 & 3 \ -1 & -2 \end{pmatrix} + \begin{pmatrix} 2 & -1 \ 3 & -2 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 4 & 2 \ 2 & -4 \end{pmatrix} - \begin{pmatrix} 1 & 0 \ 0 & 1 \end{pmatrix} = \begin{pmatrix} 3 & 2 \ 2 & -5 \end{pmatrix}$. The answer is none of these.

6. Question 16: Properties of Matrix A Given $A = \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix}$, analyze the statements. $A^{-1}$ does exist, and $A^T=A$. $A$ is not a zero matrix. Calculate $A^2$: $A^2 = \begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix}\begin{pmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 & 0 \ 0 & 1 & 0 \ 0 & 0 & 1 \end{pmatrix} = I$.

7. Question 17: Determinant Calculation Calculate the determinant of the given matrix. $$ \begin{vmatrix} a^2+1 & ab & ac \ ab & b^2+1 & bc \ ac & bc & c^2+1 \end{vmatrix} $$ $= (a^2+1)((b^2+1)(c^2+1) - b^2c^2) - ab(ab(c^2+1) - abc^2) + ac(ab^2c - ac(b^2+1))$ $= (a^2+1)(b^2c^2 + b^2 + c^2 + 1 - b^2c^2) - ab(abc^2 + ab - abc^2) + ac(ab^2c - ab^2c - ac)$ $= (a^2+1)(b^2 + c^2 + 1) - ab(ab) + ac(-ac)$ $= a^2b^2 + a^2c^2 + a^2 + b^2 + c^2 + 1 - a^2b^2 - a^2c^2$ $= a^2 + b^2 + c^2 + 1$

8. Question 18: Determinant Calculation with Arithmetic Progression $$ \begin{vmatrix} 101 & 103 & 105 \ 104 & 105 & 106 \ 107 & 108 & 109 \end{vmatrix} $$ Since the columns are in arithmetic progression, we can perform the following column operations: $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$. $$ \begin{vmatrix} 101 & 2 & 2 \ 104 & 1 & 1 \ 107 & 0 & 1 \end{vmatrix} $$ Also the rows follow arithmetic progression, so we can deduce that the determinant is 0

We can perform row operations $R_2 \rightarrow R_2 - R_1$ and $R_3 \rightarrow R_3 - R_2$: $$ \begin{vmatrix} 101 & 103 & 105 \ 3 & 2 & 1 \ 3 & 3 & 3 \end{vmatrix} $$ Since the rows are approximately linear, we expect the answer to be close to 0.

9. Question 19: Determinant Calculation $$ D = \begin{vmatrix} x & p & q \ p & x & q \ p & q & x \end{vmatrix} $$ Expanding the determinant: $D = x(x^2 - q^2) - p(px - pq) + q(pq - px) = x^3 - xq^2 - p^2x + p^2q + pq^2 - pqx = x^3 - x(p^2 + q^2) - pqx + p^2q + pq^2$. However, we can add rows such that $R_1 = R_1 + R_2 + R_3$. Then $D = \begin{vmatrix} x+2p & x+2q & x+2q \ p & x & q \ p & q & x \end{vmatrix} $ Adding columns so $C_2-C_1$, $C_3-C_2$ gives $D = \begin{vmatrix} x & p & q \ p & x & q \ p & q & x \end{vmatrix} = \begin{vmatrix} x-p & 0 & 0 \ 0 & q-p & 0 \ 0 & 0 & x-q \end{vmatrix}$. So, $(x-p)(x-q)(x+p+q)$

10. Question 20: Determinant with Logarithms and Geometric Progression

If $a_1, a_2, a_3, \dots, a_n$ are in GP, then $a_i = a_1r^{i-1}$ for some common ratio $r$. Then $\log a_i = \log(a_1r^{i-1}) = \log a_1 + (i-1) \log r$. $$ \begin{vmatrix} \log a_n & \log a_{n+1} & \log a_{n+2} \ \log a_{n+3} & \log a_{n+4} & \log a_{n+5} \ \log a_{n+6} & \log a_{n+7} & \log a_{n+8} \end{vmatrix} = \begin{vmatrix} \log a_1 + (n-1)\log r & \log a_1 + n\log r & \log a_1 + (n+1)\log r \ \log a_1 + (n+2)\log r & \log a_1 + (n+3)\log r & \log a_1 + (n+4)\log r \ \log a_1 + (n+5)\log r & \log a_1 + (n+6)\log r & \log a_1 + (n+7)\log r \end{vmatrix} $$ Apply column operations $C_2 \rightarrow C_2 - C_1$ and $C_3 \rightarrow C_3 - C_2$. $$ \begin{vmatrix} \log a_1 + (n-1)\log r & \log r & \log r \ \log a_1 + (n+2)\log r & \log r & \log r \ \log a_1 + (n+5)\log r & \log r & \log r \end{vmatrix} $$ Since two columns are the same, the determinant is 0.

11. Question 21: Determinant with Combinations $$ \begin{vmatrix} 1 & 1 & 1 \ {}mC_1 & {}{m+1}C_1 & {}{m+2}C_1 \ {}mC_2 & {}{m+1}C_2 & {}{m+2}C_2 \end{vmatrix} = \begin{vmatrix} 1 & 1 & 1 \ m & m+1 & m+2 \ \frac{m(m-1)}{2} & \frac{(m+1)m}{2} & \frac{(m+2)(m+1)}{2} \end{vmatrix} $$ Apply row operation $R_2 \rightarrow R_2 - mR_1$ to get $R_2 = (0, 1, 2-m)$ This does not lead anywhere. $C_2 \rightarrow C_2 - C_1; C_3 \rightarrow C_3 - C_2$ $$ \begin{vmatrix} 1 & 0 & 0 \ m & 1 & 1 \ \frac{m(m-1)}{2} & \frac{(m+1)m}{2}-\frac{m(m-1)}{2} & \frac{(m+2)(m+1)}{2}-\frac{(m+1)m}{2} \end{vmatrix} = \begin{vmatrix} 1 & 0 & 0 \ m & 1 & 1 \ \frac{m(m-1)}{2} & m & m+1 \end{vmatrix} $$ Expand the determinant to get $1 \times (m+1-m) = 1$

12. Question 22: Determinant with Factorials $$ \Delta = \begin{vmatrix} 1! & 2! & 3! \ 2! & 3! & 4! \ 3! & 4! & 5! \end{vmatrix} = \begin{vmatrix} 1 & 2 & 6 \ 2 & 6 & 24 \ 6 & 24 & 120 \end{vmatrix} $$ $1(6 \cdot 120 - 24 \cdot 24) - 2(2 \cdot 120 - 6 \cdot 24) + 6(2 \cdot 24 - 6 \cdot 6)$ $= 1(720 - 576) - 2(240 - 144) + 6(48 - 36)$ $= 144 - 2(96) + 6(12) = 144 - 192 + 72 = 216 - 192 = 24 = 4!$