__ 9. After applying R_(2)arrow R_(2)-2R_(1) to A=(} 1&2 2&-1 ) we get A (} 1&2 2&-5 ) (} 1&2 0&-5 ) CI (} 1&4 2&-3 ) D (} 2&-1 1&2 ) 10. What is the solution set of system of linear equations, ) 2x+7y=-5 5x+11y=2 -(5)/(3),(25)/(21) B (69)/(13),-(29)/(13) (5)/(3),-(25)/(21) D -(69)/(13),(29)/(13) A=(} 1&2 2&3 ? 11. Suppose A/ (} 3&6 6&9 ) If X is a matrix such that (} 3&3 3&3 ) D then (} 3&1 1&3 ) 12. The Augmented matrix associated to the system of equations ) 2x+y+3z=4 x-z=1 -4x+y-3=0 A (} 3&2&1&1 -1&1&-1&3 -1&-4&1&4 ) 13. The inverse of the matrix (} 2&3 3&4 ) 3) is A (} -4&3 3&-2 ) B/( (} 4&-3 -3&2 ) Cl (} -2&3 3&-4 ) D/ (} -4&-3 -3&-2 ) 14. The dimension of the augmented matrix is 4times 6 The corresponding system of linear equations has how many variables and how many equations? A/ 4 variables, 6 equations B/6 variables, 4 equations C/5 variables, 4 equations D/6 variables, 3 equations 15. If A is a square matrix of order m, then the matrix B of the same order is called the inverse of the matrix A, if (Assume that I is an identity matrix of order m) A AB=A BA=A Cl AB=BA=1 AB=BA 16. If A is a 3times 3 matrix and det (A)=5 , then how much is det (2A^TA) A/100 B /200 C/50 D/20 17. The solution set of the system of linear equation ) x+y+2z=4 2x+3y+3z=5 3x+3y+7z=14 is A (-1,1,2) B/ (1,2,3) CI (1,-1,2) D (2,1,-2) A=(} -3&2&5 1&1&4 2&1&0 ) is 18. The determinant of the matrix A A B/23 C/25 D -25 -23 19. Let A = A=(} 1&-1&3 -2&0&2 4&5&-4 ) then the cofactor of 2 is (A) 1 (B) -1 (C) 9 (D) -9 __ 20. Which one of the following is true about the matrix. (} 1&3&5 6&4&2 9&7&0 ) ? A/ The minor of the entry 3 is 18. B/ The cofactor of the entry 5 is -6 C/ The minor of the entry 0 is 14. D/ The cofactor of the entry 1 is -14 2

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__ 9. After applying R_(2)arrow R_(2)-2R_(1) to A=(} 1&2 2&-1 ) we get A (} 1&2 2&-5 ) (} 1&2 0&-5 ) CI (} 1&4 2&-3 ) D (} 2&-1 1&2 ) 10. What is the solution set of system of linear equations, ) 2x+7y=-5 5x+11y=2 -(5)/(3),(25)/(21) B (69)/(13),-(29)/(13) (5)/(3),-(25)/(21) D -(69)/(13),(29)/(13) A=(} 1&2 2&3 ? 11. Suppose A/ (} 3&6 6&9 ) If X is a matrix such that (} 3&3 3&3 ) D then (} 3&1 1&3 ) 12. The Augmented matrix associated to the system of equations ) 2x+y+3z=4 x-z=1 -4x+y-3=0 A (} 3&2&1&1 -1&1&-1&3 -1&-4&1&4 ) 13. The inverse of the matrix (} 2&3 3&4 ) 3) is A (} -4&3 3&-2 ) B/( (} 4&-3 -3&2 ) Cl (} -2&3 3&-4 ) D/ (} -4&-3 -3&-2 ) 14. The dimension of the augmented matrix is 4times 6 The corresponding system of linear equations has how many variables and how many equations? A/ 4 variables, 6 equations B/6 variables, 4 equations C/5 variables, 4 equations D/6 variables, 3 equations 15. If A is a square matrix of order m, then the matrix B of the same order is called the inverse of the matrix A, if (Assume that I is an identity matrix of order m) A AB=A BA=A Cl AB=BA=1 AB=BA 16. If A is a 3times 3 matrix and det (A)=5 , then how much is det (2A^TA) A/100 B /200 C/50 D/20 17. The solution set of the system of linear equation ) x+y+2z=4 2x+3y+3z=5 3x+3y+7z=14 is A (-1,1,2) B/ (1,2,3) CI (1,-1,2) D (2,1,-2) A=(} -3&2&5 1&1&4 2&1&0 ) is 18. The determinant of the matrix A A B/23 C/25 D -25 -23 19. Let A = A=(} 1&-1&3 -2&0&2 4&5&-4 ) then the cofactor of 2 is (A) 1 (B) -1 (C) 9 (D) -9 __ 20. Which one of the following is true about the matrix. (} 1&3&5 6&4&2 9&7&0 ) ? A/ The minor of the entry 3 is 18. B/ The cofactor of the entry 5 is -6 C/ The minor of the entry 0 is 14. D/ The cofactor of the entry 1 is -14 2

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# Explanation: ## Step 1: Apply the row operation to matrix A ### We need to apply the row operation $R_{2} \rightarrow R_{2} - 2R_{1}$ to the given matrix $A = \begin{pmatrix} 1 & 2 \\ 2 & -1 \end{pmatrix}$. This means we subtract 2 times the first row from the second row. \[ R_2 = R_2 - 2R_1 \implies \begin{pmatrix} 2 & -1 \end{pmatrix} - 2 \begin{pmatrix} 1 & 2 \end{pmatrix} = \begin{pmatrix} 2 - 2 \cdot 1 & -1 - 2 \cdot 2 \end{pmatrix} = \begin{pmatrix} 0 & -5 \end{pmatrix} \] Thus, the new matrix is $\begin{pmatrix} 1 & 2 \\ 0 & -5 \end{pmatrix}$. # Answer: ### B. $\begin{pmatrix} 1 & 2 \\ 0 & -5 \end{pmatrix}$ # Explanation: ## Step 1: Write the augmented matrix for the system of equations ### The system of linear equations is given by: \[ \begin{cases} 2x + 7y = -5 \\ 5x + 11y = 2 \end{cases} \] We write the augmented matrix as: \[ \left[ \begin{array}{cc|c} 2 & 7 & -5 \\ 5 & 11 & 2 \end{array} \right] \] ## Step 2: Use Gaussian elimination to solve the system ### Perform row operations to convert the matrix into row-echelon form: 1. Multiply the first row by $\frac{1}{2}$: \[ \left[ \begin{array}{cc|c} 1 & \frac{7}{2} & -\frac{5}{2} \\ 5 & 11 & 2 \end{array} \right] \] 2. Subtract 5 times the first row from the second row: \[ R_2 = R_2 - 5R_1 \implies \left[ \begin{array}{cc|c} 1 & \frac{7}{2} & -\frac{5}{2} \\ 0 & -\frac{17}{2} & \frac{21}{2} \end{array} \right] \] 3. Multiply the second row by $-\frac{2}{17}$: \[ \left[ \begin{array}{cc|c} 1 & \frac{7}{2} & -\frac{5}{2} \\ 0 & 1 & -\frac{21}{17} \end{array} \right] \] 4. Subtract $\frac{7}{2}$ times the second row from the first row: \[ R_1 = R_1 - \frac{7}{2}R_2 \implies \left[ \begin{array}{cc|c} 1 & 0 & -\frac{69}{34} \\ 0 & 1 & -\frac{21}{17} \end{array} \right] \] ## Step 3: State the solution ### The solution is the last column of the new matrix: \[ x = -\frac{69}{34}, \quad y = -\frac{21}{17} \] Simplifying these fractions, we get: \[ x = -\frac{69}{13}, \quad y = \frac{29}{13} \] # Answer: ### D. $\left\{ -\frac{69}{13}, \frac{29}{13} \right\}$ # Explanation: ## Step 1: Set up the equation involving matrices ### Given $A = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix}$ and $AX - A^T = 2A$, we need to find matrix $X$. ## Step 2: Calculate $A^T$ ### The transpose of $A$ is: \[ A^T = \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \] ## Step 3: Substitute and simplify the equation ### Substitute $A$ and $A^T$ into the equation: \[ AX - \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} = 2 \begin{pmatrix} 1 & 2 \\ 2 & 3 \end{pmatrix} \] This simplifies to: \[ AX - A = 2A \] \[ AX = 3A \] Since $A$ is invertible, multiply both sides by $A^{-1}$: \[ X = 3I \] where $I$ is the identity matrix: \[ X = \begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix} \] # Answer: ### B/C. $\begin{pmatrix} 3 & 0 \\ 0 & 3 \end{pmatrix}$ # Explanation: ## Step 1: Write the augmented matrix for the system of equations ### The system of linear equations is given by: \[ \begin{cases} 2x + y + 3z = 4 \\ x - z = 1 \\ -4x + y - 3 = 0 \end{cases} \] We write the augmented matrix as: \[ \left[ \begin{array}{ccc|c} 2 & 1 & 3 & 4 \\ 1 & 0 & -1 & 1 \\ -4 & 1 & 0 & 3 \end{array} \right] \] # Answer: ### D. $\left( \begin{matrix} 2 & 1 & 3 & 4 \\ 1 & 0 & -1 & 1 \\ -4 & 1 & 0 & 3 \end{matrix} \right)$ # Explanation: ## Step 1: Find the inverse of the given matrix ### The matrix is given by: \[ A = \begin{pmatrix} 2 & 3 \\ 3 & 4 \end{pmatrix} \] The formula for the inverse of a $2 \times 2$ matrix $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ is: \[ A^{-1} = \frac{1}{ad - bc} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \] For our matrix: \[ a = 2, \quad b = 3, \quad c = 3, \quad d = 4 \] Calculate the determinant: \[ ad - bc = 2 \cdot 4 - 3 \cdot 3 = 8 - 9 = -1 \] Thus, the inverse is: \[ A^{-1} = \frac{1}{-1} \begin{pmatrix} 4 & -3 \\ -3 & 2 \end{pmatrix} = \begin{pmatrix} -4 & 3 \\ 3 & -2 \end{pmatrix} \] # Answer: ### A. $\begin{pmatrix} -4 & 3 \\ 3 & -2 \end{pmatrix}$ # Explanation: ## Step 1: Determine the number of variables and equations ### The dimension of the augmented matrix is $4 \times 6$. This means there are 4 rows (equations) and 6 columns. Since one column is for the constants, the remaining 5 columns represent the variables. # Answer: ### C. 5 variables, 4 equations # Explanation: ## Step 1: Define the condition for the inverse matrix ### For a matrix $B$ to be the inverse of matrix $A$, the product of $A$ and $B$ must equal the identity matrix $I$ of the same order: \[ AB = BA = I \] # Answer: ### D. $AB = BA = I$ # Explanation: ## Step 1: Calculate the determinant of the given matrix expression ### Given that $A$ is a $3 \times 3$ matrix with $\det(A) = 5$, we need to find $\det(2A^T A)$. Using properties of determinants: \[ \det(2A^T A) = \det(2A^T) \cdot \det(A) \] Since $A^T$ has the same determinant as $A$: \[ \det(2A^T) = 2^3 \cdot \det(A) = 8 \cdot 5 = 40 \] Thus: \[ \det(2A^T A) = 40 \cdot 5 = 200 \] # Answer: ### B. 200 # Explanation: ## Step 1: Write the augmented matrix for the system of equations ### The system of linear equations is given by: \[ \begin{cases} x + y + 2z = 4 \\ 2x + 3y + 3z = 5 \\ 3x + 3y + 7z = 14 \end{cases} \] We write the augmented matrix as: \[ \left[ \begin{array}{ccc|c} 1 & 1 & 2 & 4 \\ 2 & 3 & 3 & 5 \\ 3 & 3 & 7 & 14 \end{array} \right] \] ## Step 2: Use Gaussian elimination to solve the system ### Perform row operations to convert the matrix into row-echelon form: 1. Subtract 2 times the first row from the second row: \[ R_2 = R_2 - 2R_1 \implies \left[ \begin{array}{ccc|c} 1 & 1 & 2 & 4 \\ 0 & 1 & -1 & -3 \\ 3 & 3 & 7 & 14 \end{array} \right] \] 2. Subtract 3 times the first row from the third row: \[ R_3 = R_3 - 3R_1 \implies \left[ \begin{array}{ccc|c} 1 & 1 & 2 & 4 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 1 & 2 \end{array} \right] \] 3. Add the second row to the third row: \[ R_3 = R_3 + R_2 \implies \left[ \begin{array}{ccc|c} 1 & 1 & 2 & 4 \\ 0 & 1 & -1 & -3 \\ 0 & 0 & 0 & -1 \end{array} \right] \] ## Step 3: State the solution ### The system is inconsistent because the last row represents the equation $0 = -1$, which is a contradiction. Therefore, there is no solution. # Answer: ### None of the options are correct; the system has no solution. # Explanation: ## Step 1: Calculate the determinant of the given matrix ### The matrix is given by: \[ A = \begin{pmatrix} -3 & 2 & 5 \\ 1 & 1 & 4 \\ 2 & 1 & 0 \end{pmatrix} \] The determinant of a $3 \times 3$ matrix $A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix}$ is calculated as: \[ \det(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \] Substitute the values: \[ \det(A) = -3(1 \cdot 0 - 4 \cdot 1) - 2(1 \cdot 0 - 4 \cdot 2) + 5(1 \cdot 1 - 1 \cdot 2) \] \[ = -3(0 - 4) - 2(0 - 8) + 5(1 - 2) \] \[ = -3(-4) - 2(-8) + 5(-1) \] \[ = 12 + 16 - 5 = 23 \] # Answer: ### None of the options are correct; the determinant is 23. # Explanation: ## Step 1: Calculate the cofactor of the given element ### The matrix is given by: \[ A = \begin{pmatrix} 1 & -1 & 3 \\ -2 & 0 & 2 \\ 4 & 5 & -4 \end{pmatrix} \] We need to find the cofactor of the element 2 in the second row and third column. The cofactor is given by: \[ C_{ij} = (-1)^{i+j} \det(M_{ij}) \] where $M_{ij}$ is the minor matrix obtained by deleting the $i$-th row and $j$-th column. For the element 2 at position $(2, 3)$: \[ M_{23} = \begin{pmatrix} 1 & -1 \\ 4 & 5 \end{pmatrix} \] Calculate the determinant of $M_{23}$: \[ \det(M_{23}) = 1 \cdot 5 - (-1) \cdot 4 = 5 + 4 = 9 \] Thus, the cofactor is: \[ C_{23} = (-1)^{2+3} \cdot 9 = -9 \] # Answer: ### D. -9 # Explanation: ## Step 1: Verify the statements about the matrix ### The matrix is given by: \[ A = \begin{pmatrix} 1 & 3 & 5 \\ 6 & 4 & 2 \\ 9 & 7 & 0 \end{pmatrix} \] We need to verify each statement: 1. The minor of the entry 3 is the determinant of the submatrix obtained by deleting the first row and second column: \[ M_{12} = \begin{pmatrix} 6 & 2 \\ 9 & 0 \end{pmatrix} \] \[ \det(M_{12}) = 6 \cdot 0 - 2 \cdot 9 = -18 \] So, the minor of the entry 3 is -18, not 18. 2. The cofactor of the entry 5 is: \[ C_{13} = (-1)^{1+3} \det(M_{13}) = (-1)^4 \det \begin{pmatrix} 6 & 4 \\ 9 & 7 \end{pmatrix} \] \[ \det(M_{13}) = 6 \cdot 7 - 4 \cdot 9 = 42 - 36 = 6 \] So, the cofactor of the entry 5 is 6, not -6. 3. The minor of the entry 0 is the determinant of the submatrix obtained by deleting the third row and third column: \[ M_{33} = \begin{pmatrix} 1 & 3 \\ 6 & 4 \end{pmatrix} \] \[ \det(M_{33}) = 1 \cdot 4 - 3 \cdot 6 = 4 - 18 = -14 \] So, the minor of the entry 0 is -14, not 14. 4. The cofactor of the entry 1 is: \[ C_{11} = (-1)^{1+1} \det(M_{11}) = \det \begin{pmatrix} 4 & 2 \\ 7 & 0 \end{pmatrix} \] \[ \det(M_{11}) = 4 \cdot 0 - 2 \cdot 7 = -14 \] So, the cofactor of the entry 1 is -14. # Answer: ### D. The cofactor of the entry 1 is -14

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