7. Given ) x+y=0 x-3y+4z=0 ,which one of the ff alternatives is/are solution(s)of the given system of linear equation? A. (} -1 1 1 ) B. (} 1 -1 -1 ) C. (} 0.5 -0.5 -0.5 ) D. All E. None 8.Given matrix D = D=(} a_(11)&a_(12)&a_(13) a_(21)&a_(22)&a_(23) a_(31)&a_(32)&a_(33) ) ,which one is false about D? A. C_(11)=vert } a_(22)&a_(23) a_(32)&a_(33) vert B. dot (M)_(12)=-vert } a_(21)&a_(23) a_(31)&a_(33) vert C. M_(13)=vert } a_(21)&a_(22) a_(31)&a_(32) vert D. c_(32)=-vert } a_(11)&a_(13) a_(21)&a_(23) vert 9. Which one is true about an nxn invertible matrix A? det(A)=det(A^T) B. det(AA^T)=(det(A))^2 C. det(AA^-1)=1 D. det(A^-1)det(A^T)=1 . E 10. If M=(} -3&5 2&-3 is equal to; A. (} 3&5 2&3 ) B. (} -3&-5 -2&-3 ) C、 (} 3&-5 -2&3 ) (} -3&5 2&-3 ) What is the determinant of the matrix N N=(} -1&0&0 -1&-2&0 1&1&8 ) ? A. -32 B. 16 C、 -16 D. 5

### 7. Solution(s) of the given system of linear equations:<br />The system of equations is: <br />\[<br />\begin{aligned}<br />x + y &= 0 \\<br />x - 3y + 4z &= 0<br />\end{aligned}<br />\]<br /><br />Substitute each option into the equations to check if they satisfy both.<br /><br />- **Option A**: \((-1, 1, 1)\): <br /> Substituting into \(x + y = 0\): \(-1 + 1 = 0\) (True). <br /> Substituting into \(x - 3y + 4z = 0\): \(-1 - 3(1) + 4(1) = -1 - 3 + 4 = 0\) (True). <br /><br />- **Option B**: \((1, -1, -1)\): <br /> Substituting into \(x + y = 0\): \(1 - 1 = 0\) (True). <br /> Substituting into \(x - 3y + 4z = 0\): \(1 - 3(-1) + 4(-1) = 1 + 3 - 4 = 0\) (True). <br /><br />- **Option C**: \((0.5, -0.5, -0.5)\): <br /> Substituting into \(x + y = 0\): \(0.5 - 0.5 = 0\) (True). <br /> Substituting into \(x - 3y + 4z = 0\): \(0.5 - 3(-0.5) + 4(-0.5) = 0.5 + 1.5 - 2 = 0\) (True). <br /><br />Since all options satisfy the equations, the correct answer is: <br />**D. All**<br /><br />---<br /><br />### 8. False statement about matrix \(D\):<br />The cofactor and minor definitions are as follows: <br />- \(C_{ij} = (-1)^{i+j} M_{ij}\), where \(M_{ij}\) is the determinant of the submatrix obtained by removing the \(i\)-th row and \(j\)-th column. <br /><br />Check each option: <br />- **A**: \(C_{11} = \left| \begin{matrix} a_{22} & a_{23} \\ a_{32} & a_{33} \end{matrix} \right|\): True. <br />- **B**: \(\dot{M}_{12} = -\left| \begin{matrix} a_{21} & a_{23} \\ a_{31} & a_{33} \end{matrix} \right|\): True. <br />- **C**: \(M_{13} = \left| \begin{matrix} a_{21} & a_{22} \\ a_{31} & a_{32} \end{matrix} \right|\): True. <br />- **D**: \(c_{32} = -\left| \begin{matrix} a_{11} & a_{13} \\ a_{21} & a_{23} \end{matrix} \right|\): False. The cofactor \(C_{32}\) should be \((-1)^{3+2} M_{32} = (-1)^5 M_{32} = M_{32}\), not \(-M_{32}\). <br /><br />The correct answer is: <br />**D.**<br /><br />---<br /><br />### 9. True statement about an \(n \times n\) invertible matrix \(A\):<br />- **A**: \(\det(A) = \det(A^T)\): True. Determinants of a matrix and its transpose are equal. <br />- **B**: \(\det(AA^T) = (\det(A))^2\): True. For any square matrix \(A\), \(\det(AA^T) = \det(A) \cdot \det(A^T) = \det(A)^2\). <br />- **C**: \(\det(AA^{-1}) = 1\): True. Since \(AA^{-1} = I\), and \(\det(I) = 1\). <br />- **D**: \(\det(A^{-1}) \det(A^T) = 1\): False. \(\det(A^{-1}) = 1/\det(A)\), so \(\det(A^{-1}) \det(A^T) = \frac{\det(A^T)}{\det(A)} = 1\) only if \(\det(A) = 1\), which is not guaranteed for all invertible matrices. <br /><br />The correct answer is: <br />**E. All except D**<br /><br />---<br /><br />### 10. Inverse of matrix \(M\):<br />Matrix \(M = \begin{pmatrix} -3 & 5 \\ 2 & -3 \end{pmatrix}\). <br />The formula for the inverse of a \(2 \times 2\) matrix is: <br />\[<br />M^{-1} = \frac{1}{\det(M)} \begin{pmatrix} d & -b \\ -c & a \end{pmatrix},<br />\]<br />where \(\det(M) = ad - bc\). <br /><br />For \(M\): <br />\[<br />\det(M) = (-3)(-3) - (5)(2) = 9 - 10 = -1.<br />\]<br />Thus: <br />\[<br />M^{-1} = \frac{1}{-1} \begin{pmatrix} -3 & -5 \\ -2 & -3 \end{pmatrix} = \begin{pmatrix} 3 & 5 \\ 2 & 3 \end{pmatrix}.<br />\]<br /><br />The correct answer is: <br />**C. \(\begin{pmatrix} 3 & -5 \\ -2 & 3 \end{pmatrix}\)**<br /><br />---<br /><br />### Determinant of matrix \(N\):<br />Matrix \(N = \begin{pmatrix} -1 & 0 & 0 \\ -1 & -2 & 0 \\ 1 & 1 & 8 \end{pmatrix}\). <br />Using cofactor expansion along the first row: <br />\[<br />\det(N) = (-1) \cdot \det\begin{pmatrix} -2 & 0 \\ 1 & 8 \end{pmatrix} - 0 + 0.<br />\]<br />\[<br />\det\begin{pmatrix} -2 & 0 \\ 1 & 8 \end{pmatrix} = (-2)(8) - (0)(1) = -16.<br />\]<br />\[<br />\det(N) = (-1)(-16) = 16.<br />\]<br /><br />The correct answer is: <br />**B. 16**