(a) Find the rank of the matrix A = 1 & 2 & -1 & 0 -1 & 3 & 0 & -4 2 & 1 & 3 & -2 1 & 1 & 1 & -1 by reducing it to…

(a) Rank by row-reduced echelon form

Use Gaussian elimination to reduce A to row-reduced echelon form.

A = [1 2 -1 0; -1 3 0 -4; 2 1 3 -2; 1 1 1 -1].

Apply:

• R2 → R2 + R1 = (0,5,-1,-4)
• R3 → R3 − 2R1 = (0,-3,5,-2)
• R4 → R4 − R1 = (0,-1,2,-1)

Now swap R2 and R4: [1 2 -1 0; 0 -1 2 -1; 0 -3 5 -2; 0 5 -1 -4].

Then R2 → −R2: [1 2 -1 0; 0 1 -2 1; 0 -3 5 -2; 0 5 -1 -4].

Eliminate the second column:

• R1 → R1 − 2R2 = (1,0,3,-2)
• R3 → R3 + 3R2 = (0,0,-1,1)
• R4 → R4 − 5R2 = (0,0,9,-9)

So: [1 0 3 -2; 0 1 -2 1; 0 0 -1 1; 0 0 9 -9].

Now R3 → −R3 = (0,0,1,-1). Eliminate the third column:

• R1 → R1 − 3R3 = (1,0,0,1)
• R2 → R2 + 2R3 = (0,1,0,-1)
• R4 → R4 − 9R3 = (0,0,0,0)

Hence the row-reduced echelon form is [1 0 0 1; 0 1 0 -1; 0 0 1 -1; 0 0 0 0].

There are 3 non-zero rows. Therefore rank(A) = 3.

(b) Tracing the curve y²(x² − 1) = 2x − 1

The curve is y²(x² − 1) = 2x − 1.

Since only y² appears, the curve is symmetric about the x-axis.

Write y² = (2x − 1)/(x² − 1).

Intercepts:

• On the x-axis, y = 0 gives 2x − 1 = 0, so x = 1/2. Point: (1/2,0).
• On the y-axis, x = 0 gives −y² = −1, so y = ±1. Points: (0,±1).

Asymptotes:

• x = 1 and x = −1 make the denominator zero while the numerator is non-zero, so these are vertical asymptotes.
• As x → +∞,

y² = (2x − 1)/(x² − 1) → 0, so y = 0 is a horizontal asymptote in the positive x-direction.

Region of existence: Since y² ≥ 0, the quotient must be non-negative. It is so only for −1 < x ≤ 1/2 or x > 1. There is no real curve for x < −1 or 1/2 < x < 1.

Monotonicity: Let q(x) = (2x − 1)/(x² − 1). Then q′(x) = [2(x² − 1) − (2x − 1)2x]/(x² − 1)² = (−2x² + 2x − 2)/(x² − 1)² = −2(x² − x + 1)/(x² − 1)². But x² − x + 1 = (x − 1/2)² + 3/4 > 0, so q′(x) < 0 wherever defined. Thus |y| decreases on each interval.

Trace:

• For −1 < x ≤ 1/2, the curve comes down from x = −1 with y → ±∞, passes through (0,±1), and meets the x-axis at (1/2,0) with a vertical tangent, because x − 1/2 is proportional to y² near that point.
• For x > 1, the curve comes from x = 1 with y → ±∞ and tends to the x-axis y = 0 as x → +∞.
• The upper and lower halves are mirror images about the x-axis.
• The vertical lines x = −1 and x = 1 are asymptotes, and the x-axis is an asymptote for the right branch as x → +∞.

Thus the curve consists of two symmetric branches, one between x = −1 and x = 1/2, and the other for x > 1.

(c) Locus of the line meeting two lines and a circle

Let the variable line meet L1: y = mx, z = c at P = (p, mp, c) and L2: y = −mx, z = −c at Q = (q, −mq, −c).

Let R = (x,y,z) be any point on the line PQ. Then R = P + t(Q − P).

Since z = c(1 − 2t), we get t = (c − z)/(2c), 1 − t = (c + z)/(2c).

Put S = p + q and D = p − q. Then 2cx = cS + zD, 2cy/m = zS + cD.

Solving these two equations: S = 2c(cmx − zy)/(m(c² − z²)), D = 2c(cy − mzx)/(m(c² − z²)).

The line meets the circle x² + y² = a², z = 0 at the point where t = 1/2. At z = 0, the coordinates of intersection are x₀ = S/2, y₀ = mD/2. Since this point lies on the circle, (S/2)² + (mD/2)² = a², so S² + m²D² = 4a².

Substitute the values of S and D: 4c²(cmx − zy)²/[m²(c² − z²)²] + 4c²(cy − mzx)²/(c² − z²)² = 4a².

Multiplying by m²(c² − z²)² gives c²(cmx − zy)² + c²m²(cy − mzx)² = a²m²(c² − z²)².

Now (cmx − zy)² = (yz − cmx)² and (c² − z²)² = (z² − c²)², so this becomes c²m²(cy − mzx)² + c²(yz − cmx)² = a²m²(z² − c²)².

This is the required locus.