Leave a Comment / Practice Questions, Linear Algebra / By Rsquared Mathematics Linear Algebra MCQs and MSQs with Solutions for CSIR NET, IIT JAMSystem of Equations – IV Practice Questions for NET JRF Linear Algebra <br> Assignment: System of Equations 31. The solutions to the system of equations : \((1-i) x_{1}-i x_{2}=0\) \(2 x_{1}+(1-i) x_{2}=0\) is given by: (a.) \(\left(x_{1}, x_{2}\right)=(0,0)\) (b.) \(\left(x_{1}, x_{2}\right)=(1,1)\) (c.) \(\left(x_{1}, x_{2}\right)=c\left(1, \cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)\) where \(c\) is any complex number. (d.) \(\left(x_{1}, x_{2}\right)=c\left(\cos \frac{3 \pi}{4}, i \sin \frac{3 \pi}{4}\right)\) where \(c\) is any complex number. Explanation for Question 31: The matrix corresponding to the system is non-singular; thus, \(0\) is the only solution for a homogeneous system. Submit 32. Let \(A\) be an \(m \times n\) matrix where \(m (a.) The system of equations has no solution. (b.) The system of equations has a solution if and only if it has infinitely many solutions. (c.) The system of equations has a unique solution. (d.) The system of equations has at least one solution. Explanation for Question 32: Since \(A\) cannot have full rank, thus if solution exists, it is not unique. Submit 33. The system \(A x=0\), where \(A\) is an \(n \times n\) matrix, (a.) May not have a nonzero solution (b.) Always has a nonzero solution (c.) Always has at least 2 linearly independent solutions (d.) Always has at least \(n\) linearly independent solutions Explanation for Question 33: NA Submit 34. Let \(\mathrm{P}\) be a matrix of order \(m \times n\) and \(\mathrm{Q}\) be a matrix of order \(n \times p\), \(n \neq p\). If \(\operatorname{rank}(\mathrm{P})=n\) and \(\operatorname{rank}(\mathrm{Q})=p\), then \(\operatorname{rank}(\mathrm{P Q})\) is (a.) \(n\) (b.) \(p\) (c.) \(np\) (d.) \(n+p\) Explanation for Question 34: Note that the composition of a 1-1 map is again one-one. Since LT corresponding to \(P\) and \(Q\) are one-one maps (full column rank), matrix multiplication is nothing but the composition of their corresponding LT. Therefore, it will also have a full column rank.(one-one) Submit 35. Let \(\mathrm{A}\) be a \(m \times n\) matrix with row rank \(r=\) column rank. The dimension of the space of solutions of the system of linear equations \(\mathrm{AX}=0\) is (a.) \(r\) (b.) \(n-r\) (c.) \(m-r\) (d.) \(\min .(m, n)-r\) Explanation for Question 35: If the column rank is \(r\) and there are \(n\) columns, then the null space has a dimension \(n-r\) (Rank Nullity Theorem). Submit 36. Let \(\mathrm{M}\) be a \(m \times n(m < n)\) matrix with rank \(m\) Then (a.) For every \(\mathrm{b}\) in \(\mathbb{R}^{m} \quad \mathrm{M} x=\mathrm{b}\) has a unique solution (b.) For every \(\mathrm{b}\) in \(\mathbb{R}^{m} \quad \mathrm{M} x=\mathrm{b}\) has \(\mathrm{a}\) solution but it is not unique (c.) There exists \(\mathrm{b} \in \mathbb{R}^{m}\) for which \(\mathrm{M} x=\mathrm{b}\) has not solution (d.) None of the above Explanation for Question 36: Since \(M\) is full row rank, it corresponds to an onto LT. Hence, for every \(b\), it will be consistent. Also, \(m < n\), thus \(M\) does not have full column rank; hence, there will be more than one solution for any \(b\). Submit 37. Consider the system of linear equations \(a_{1} x+b_{1} y+c_{1} z=d_{1}\), \(a_{2} x+b_{2} y+c_{2} z=d_{2}\), \(a_{3} x+b_{3} y+c_{3} z=d_{3}\), Where \(a_{i}, b_{i}, c_{i}, d_{i}\) are real numbers for \(1 \leq i \leq 3\). If \(\left|\begin{array}{lll}b_{1} & c_{1} & d_{1} \\ b_{2} & c_{2} & d_{2} \\ b_{3} & c_{3} & d_{3}\end{array}\right| \neq 0\) then the above system has (a.) At most one solution (b.) Always exactly one solution (c.) More than one but finitely many solutions (d.) Infinitely many solutions. Explanation for Question 37: NA Submit 38. Let \(A=\left(\begin{array}{ccc}12 & 24 & 5 \\ x & 6 & 2 \\ -1 & -2 & 3\end{array}\right)\). The value of \(x\) for which the matrix \(A\) is not invertible is (a.) 6 (b.) 12 (c.) 3 (d.) 2 Explanation for Question 38: NA Submit 39. Let \(A=\left(\begin{array}{ll}\pi & p \\ q & r\end{array}\right)\) where \(p, q, r\) are rational numbers. If \(\operatorname{det} A=0\) and \(p \neq 0\) then the value of \(q^{2}+r^{2}\) (a.) is 2 (b.) is 1 (c.) is 0 (d.) cannot be determined using the given information Explanation for Question 39: \(|A|=0 \implies \pi r = pq\). As \(p, q, r\) are rational, it is not possible unless \(r\) and \(q\) are zero. \(\implies r^2+q^2= 0\). Submit 40. Let \(A=\left(\begin{array}{ll}a & \pi \\ \pi & \frac{1}{49} \end{array}\right)\), where \(a\) is a real number. Then, \(\mathrm{A}\) is invertible (a.) for all \(a \neq 22^{2}\) (b.) for all \(a \neq 180^{2} \times 49\) (c.) for all \(a \neq 22^{2}\) or \(a \neq 180^{2} \times 49\) (d.) for all rational \(a\) Explanation for Question 40: Note that \(\pi^2\) is never equal to \(\frac{a}{49}\) for any rational \(a\). Submit 41. Let \(\mathrm{A}\) be an \(n \times n\) invertible matrix with integer entries and assume that \(A^{-1}\) also only integer entries. Then, (a.) \(\operatorname{det} A=n\) (b.) \(\operatorname{det} A= \pm 1\) (c.) \(\operatorname{det} A=n^{2}\) (d.) det \(A\) will depend on the entries \(A\) and \(A^{-1}\) Explanation for Question 41: We know that \(|A^{-1}A|=|I|=1\). We also have that \(|A^{-1}A|=|A|^{-1}|A|=1\). Note that if \(A\) has integer entries, \(|A|\) is also an integer. Now, what can we multiply to an integer to get \(1\) or \(-1\). Submit 42. The determinant \(\left|\begin{array}{cccccc}x_{0} & x & x_{2} & x_{3} & x_{4} \\ x_{0} & x_{1} & x & x_{3} & x_{4} \\ x_{0} & x_{1} & x_{2} & x & x_{4} \\ x_{0} & x_{1} & x_{2} & x_{3} & x\end{array}\right|\) equal to (a.) \(\left[x_{0}\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\right]^{4}\) (b.) \(x_{0}\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\) (c.) \(x_{0}\left[\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)\right]^{4}\) (d.) \(x_{0} x_{1} x_{2} x_{3} x_{4}\) Explanation for Question 42: NA Submit 43. For real numbers \(a, b, c\), the following linear system of equations \(x+y+z=1\) \(a x+b y+c z=1\) \(a^{2} x+b^{2} y+c^{2} z=1\) has a unique solution if and only if (a.) \(b=c\) and \(b \neq a\). (b.) \(a=b\) and \(a \neq c\). (c.) \(a=c\) and \(a \neq b\). (d.) \(a \neq b, b \neq c\) and \(a \neq c\). Explanation for Question 43: Hint – Solve the determinant to be non-zero. Submit 44. Let \(M=\left[\begin{array}{ccc}1 & 1 & 0 \\ -1 & 1 & 2 \\ 2 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]\). Then the rank of \(M\) is (a.) 3 (b.) 4 (c.) 2 (d.) 1 Explanation for Question 44: Use row echelon form. Submit 45. Let \(D_{1}=\operatorname{det}\left(\begin{array}{lll}a & b & c \\ x & y & z \\ p & q & r\end{array}\right)\) and \(D_{2}=\operatorname{det}\left(\begin{array}{ccc}-x & a & -p \\ y & -b & q \\ z & -c & r\end{array}\right)\). Then (a.) \(D_{1}=D_{2}\) (b.) \(D_{1}=2 D_{2}\) (c.) \(D_{1}=-D_{2}\) (d.) \(2 D_{1}=D_{2}\) Explanation for Question 45: Majduri ka kaam. Submit Leave a Comment Cancel ReplyYour email address will not be published. Required fields are marked *Type here..Name* Email* Website Save my name, email, and website in this browser for the next time I comment.
Practice Questions for NET JRF Linear Algebra : Vector Space and Subspaces – I Practice Questions | By Rsquared Mathematics | Linear Algebra, Vector Space and Subspaces
Practice Questions for NET JRF Linear Algebra : Vector Space and Subspaces – II Practice Questions, Linear Algebra | By Rsquared Mathematics | Linear Algebra, Vector Space and Subspaces
