Semester 1 Mathematic 105/2017 Date: 06/05/2017 EtC. Name Grade &Section Roll No 1.Given M=[} x-4&6 4&x-6 ],ifvert Mvert =0 the value of x is? A. 0 B. 10 -10 D. A&B 2. If matrix B is a3times 3 non singular mat rix and vert Bvert =-2 ,then vert -3Bvert is equal to? A. 54 B. -6 C. -54 D. 6 E. -16 3. Given M = M=(} -1&-2&-3 0&-3&-4 1&1&0 ) ,which elementary row operation is most approprate to get (} -1&-2&-3 -1&1&1 1&1&0 ) from the given matrix M? C R_(2)arrow R_(1)-R_(2) D. R_(2)arrow R_(2)-R_(3) A. R_(2)arrow R_(2)+R_(1) B. R_(2)arrow R_(2)-R_(1) Which one of the ff matrix is in Reduced Row Echelon form? D (} 1&0&0 0&0&0 0&1&0 ) A. (} -1&0&1 0&0&0 0&1&0 ) B. (} 1&0&0 0&0&1 0&1&0 ) C. (} 1&0&12 0&1&0 0&0&0 ) A is an invertible 3times 3 matrix ,which one is false about matrix A? B. It is a singular 'matrix A. Its determinant is different from zero C. Its inverse is unique D vert A^2vert =vert Avert ast vert Avert E.All are true ch system of linear equation has No solution? C. ) x+2y+3z=5 3x+6y+9z=3 x+2y+8z=1 A. ) x+y+2z=1 y+2x+z=3 x+2y+2z=0 B. ) x-y+z=0 2y-z=3

1. To find the value of \( x \) for which the determinant of matrix \( M \) is zero, we calculate the determinant:<br /><br />\[<br />\text{det}(M) = (x-4)(x-6) - (6 \times 4) = 0<br />\]<br /><br />Simplifying,<br /><br />\[<br />(x-4)(x-6) - 24 = 0<br />\]<br /><br />Expanding,<br /><br />\[<br />x^2 - 10x + 24 - 24 = 0 \implies x^2 - 10x = 0<br />\]<br /><br />Factoring,<br /><br />\[<br />x(x - 10) = 0<br />\]<br /><br />Thus, \( x = 0 \) or \( x = 10 \). The correct answer is D. A&B.<br /><br />2. If matrix \( B \) is a \( 3 \times 3 \) non-singular matrix with \(\vert B\vert = -2\), then \(\vert -3B\vert = (-3)^3 \times \vert B\vert = -27 \times (-2) = 54\). The correct answer is A. 54.<br /><br />3. To transform the given matrix \( M \) to the desired form, we need to perform the operation \( R_{2} \rightarrow R_{2} + R_{1} \). The correct answer is A. \( R_{2} \rightarrow R_{2} + R_{1} \).<br /><br />4. A matrix is in Reduced Row Echelon Form if it satisfies certain conditions, including having leading 1s and zeros elsewhere in their columns. The matrix that meets these criteria is C. \(\begin{matrix} 1&0&12\\ 0&1&0\\ 0&0&0\end{matrix} \).<br /><br />5. For an invertible \( 3 \times 3 \) matrix \( A \), the statement that is false is B. It is a singular matrix. An invertible matrix cannot be singular by definition.<br /><br />6. The system of linear equations that has no solution is the one where the equations are inconsistent. This occurs when there is a contradiction, such as parallel planes that do not intersect. The correct answer is C. \(\{ \begin{matrix} x+2y+3z=5\\ 3x+6y+9z=3\\ x+2y+8z=1\end{matrix} \}\).