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Question text 10Marks Consider the system of equations in the variables [math: x,y]x, y and [math: z]: [math: x+y+z=0x+(k+1)y+kz=−2−x+(k2−1)y+(2k−3)z=−4] \matrix{x+y+z=0 \cr x+(k+1)y+kz=-2 \cr -x+(k^2-1)y+(2k-3)z=-4} (a) When the parameter [math: k=−1]k=-1, the system has a solution [math: x=] Answer 1[input], [math: y=] Answer 2[input], [math: z=] Answer 3[input]. (b) When [math: k=0], the system has Answer 4[select: , a unique solution., no solution., infinitely many solutions.] (c) The system has infinitely many solutions when [math: k=] Answer 5[input] and [math: k=] Answer 6[input] (enter your answers in increasing order). (d) The system has no solution when [math: k=] Answer 7[input].Notes Report question issue Question 3 Notes

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We are given a system of three linear equations in x, y, z with a parameter k: x + y + z = 0 x + (k + 1) y + k z = -2 -x + (k^2 - 1) y + (2k - 3) z = -4 We will examine each part and evaluate the statements step by step, considering consistency and degrees of freedom as k varies. Part (a): When k = -1, the system has a solution with x = [Answer 1], y = [Answer 2], z = [Answer 3]. - Substitute k = -1 into the equations: Equation 1 remains x + y + z = 0. Equation 2 becomes x + (0) y + (-1) z = -2, i.e., x - z = -2. Equation 3 becomes -x + ((-1)^2 - 1) y + (2(-1) - 3) z = -4, which simplifies to -x + (0) y + (-5) z = -4, i.e., -x - 5z = -4, or x + 5z = 4. - From x - z = -2 we have x = z - 2. Plugging into x + 5z = 4 gives (z - 2) + 5z ......Login to view full explanation

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