Daily Math Problems — June 27, 2026: Matrices, Systems of Linear Equations, Calculus

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Problems & worked solutions

RO M1

Problem 1 — Matrices

The trace of $A = (\begin{array}{ccc} 2 & 7 & 1 \\ 0 & 3 & 4 \\ 1 & 0 & 5 \end{array})$ is:

Show answer & worked solution

A. $5$
B. $9$
C. $10$ ✓ correct
D. $23$

$tr (A) = 2 + 3 + 5 = 10$ .

RO M1

Problem 2 — Systems of Linear Equations

By the Rouché–Capelli theorem, a system has solutions iff:

Show answer & worked solution

A. $\det (A) = 0$
B. $rank (A) = rank ([A ∣ b])$ ✓ correct
C. $rank (A) > rank ([A ∣ b])$
D. $The system is square$

A linear system is consistent (has at least one solution) iff $rank (A) = rank ([A ∣ b])$ — the rank of the coefficient matrix equals the rank of the augmented matrix.

RO M1

Problem 3 — Asymptotes

The function $f (x) = e^{- x}$ has at $x \to \infty$ :

Show answer & worked solution

A. $a horizontal asymptote at y = 0$ ✓ correct
B. $a horizontal asymptote at y = 1$
C. $a vertical asymptote at x = 0$
D. $no asymptote$

$\lim_{x \to \infty} e^{- x} = 0$ , so $y = 0$ is a horizontal asymptote at $+ \infty$ .

Problems & worked solutions

RO M1

Problem 1 — Matrices

The trace of $A = (\begin{array}{ccc} 2 & 7 & 1 \\ 0 & 3 & 4 \\ 1 & 0 & 5 \end{array})$ is:

Show answer & worked solution

A. $5$
B. $9$
C. $10$ ✓ correct
D. $23$

$tr (A) = 2 + 3 + 5 = 10$ .

RO M1

Problem 2 — Systems of Linear Equations

By the Rouché–Capelli theorem, a system has solutions iff:

Show answer & worked solution

A. $\det (A) = 0$
B. $rank (A) = rank ([A ∣ b])$ ✓ correct
C. $rank (A) > rank ([A ∣ b])$
D. $The system is square$

A linear system is consistent (has at least one solution) iff $rank (A) = rank ([A ∣ b])$ — the rank of the coefficient matrix equals the rank of the augmented matrix.

RO M1

Problem 3 — Asymptotes

The function $f (x) = e^{- x}$ has at $x \to \infty$ :

Show answer & worked solution

A. $a horizontal asymptote at y = 0$ ✓ correct
B. $a horizontal asymptote at y = 1$
C. $a vertical asymptote at x = 0$
D. $no asymptote$

$\lim_{x \to \infty} e^{- x} = 0$ , so $y = 0$ is a horizontal asymptote at $+ \infty$ .