2018 National Graduate Entrance Examination Mathematics I Past Paper

Approach:

This question tests the differentiability of a piecewise function at its breakpoint. The criterion is that the left and right derivatives both exist and are equal; since all four options contain an absolute value, the one-sided derivatives at $0$ have to be computed one by one.

Steps:

1. For (A) $f(x)=x\sin|x|$: when $x>0$, $f'(x)=\sin x+x\cos x\to0$; when $x<0$, $f(x)=-x\sin x$, so $f'(x)=-\sin x-x\cos x\to0$. Both one-sided derivatives are $0$, hence $f$ is differentiable at $0$.
2. For (B) $f(x)=x\sin x$ and (C) $f(x)=\cos|x|=\cos x$: they are obviously differentiable at $0$, and the left and right derivatives are equal.
3. For (D) $f(x)=\cos\sqrt{|x|}$: when $x>0$, $f'(x)=-\dfrac{\sin\sqrt{x}}{2\sqrt{x}}\to-\dfrac12$; when $x<0$, $f'(x)=\dfrac{\sin\sqrt{-x}}{2\sqrt{-x}}\to\dfrac12$.
4. The one-sided derivatives are not equal, so $f(x)=\cos\sqrt{|x|}$ is not differentiable at $x=0$, and (D) is the correct choice.

Answer:

D.

Approach:

This question tests how to find the tangent plane of a surface and how to write the equation of a plane through two given points. The plane has to pass through $(1,0,0)$ and $(0,1,0)$ and at the same time be tangent to the paraboloid $z=x^2+y^2$, so the possible forms of the plane have to be discussed case by case.

Steps:

1. Let the plane be $Ax+By+Cz=D$. Passing through $(1,0,0)$ gives $A=D$, and passing through $(0,1,0)$ gives $B=D$, so $A=B=D$.
2. If $D=0$ (that is, $A=B=0$), the plane degenerates to $z=0$; the tangent plane of the paraboloid $z=x^2+y^2$ at $(0,0,0)$ is exactly $z=0$ (from $\nabla F=(-2x,-2y,1)$), so $z=0$ is indeed a tangent plane.
3. If $D\ne0$, the plane can be written as $x+y+Cz=1$; it is tangent to the paraboloid at $(a,b,a^2+b^2)$, where the tangent plane is $2ax+2by-z=a^2+b^2$.
4. Comparing the two equations gives $a=b$, and then equality of the constant terms gives $a=1$, $C=-\tfrac12$, that is $2x+2y-z=2$, with tangency point $(1,1,2)$.
5. Hence the required planes are $z=0$ and $2x+2y-z=2$, so (B) is the correct choice.

Answer:

B.

Approach:

This question tests matching an infinite series against the Taylor expansion of a function. As soon as one recognises that the coefficient of the $x^{2n}$ term in the expansion of $\cos x$ is $\dfrac{(-1)^n}{(2n)!}$, the given series can be turned into the value of an elementary function.

Steps:

1. The Taylor expansion of the cosine function is $\cos x=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^nx^{2n}}{(2n)!}$.
2. Setting $x=3$ gives $\sum\limits_{n=0}^{\infty}\dfrac{(-1)^n3^{2n}}{(2n)!}=\cos3$, so the original series equals $2\cos3$.
3. Now use the product-to-sum formula for the trigonometric functions to bring $2\cos3$ into the form given in the options, which gives $2\sin1\cos1$, so (B) is the correct choice.

Answer:

B.

Approach:

This question tests the parity of definite integrals and the comparison of the integrands. All three integrals are taken over a symmetric interval: $M$ can be computed exactly with the $f(x)+f(-x)$ trick, $N$ is handled by pointwise comparison, and $K$ is computed directly.

Steps:

1. First compute $K$: $K=\int_{-\pi/2}^{\pi/2}\cos x\,dx=\sin x\big|_{-\pi/2}^{\pi/2}=2$.
2. Next compute $M$: since $\dfrac{x^2}{1+e^x}+\dfrac{x^2}{1+e^{-x}}=x^2$, we get $M=\int_0^{\pi/2}x^2\,dx=\left[\dfrac{x^3}{3}\right]_0^{\pi/2}=\dfrac{\pi^3}{24}\approx1.29$, so $M<K$.
3. Compare $M$ and $N$: when $x\ne0$ and $\cos x>0$ we have $\sin^2x<1$, so $(x\sin x)^2<x^2$; and on $[-\pi/2,\pi/2]$ we have $\dfrac{1}{1+e^x}\ge\dfrac12$, that is $\dfrac{x^2}{1+e^x}\ge\dfrac{x^2}{2}>x^2\sin^2x$.
4. Integrating the pointwise comparison gives $N<M$; taken together this yields $K>M>N$, so (C) is the correct choice.

Answer:

C.

Approach:

This question tests the Jordan normal form and the criterion for matrix similarity. For a third-order matrix whose eigenvalues are all $1$, the size structure of the Jordan blocks is determined by $r(A-E)$, so it is enough to compute $r(A-E)$ for each option.

Steps:

1. The matrix in the question is $A=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}$; all its eigenvalues are $1$, and $A-E=\begin{pmatrix}0&1&0\\0&0&1\\0&0&0\end{pmatrix}$ gives $r(A-E)=2$, so its Jordan structure consists of a single block of size 3.
2. For option (A): $A-E=\begin{pmatrix}0&1&1\\0&0&1\\0&0&0\end{pmatrix}$, so $r(A-E)=2$, the same as in the question.
3. For option (B): $A-E=\begin{pmatrix}0&0&1\\0&0&1\\0&0&0\end{pmatrix}$, so $r(A-E)=1$; for option (C): $r(A-E)=1$; for option (D): $r(A-E)=0$. None of the three has the same Jordan structure as the matrix in the question.
4. Hence only option (A) is similar to the matrix in the question, so (A) is the correct choice.

Answer:

A.

Approach:

This question tests the rank of a block matrix and the column space properties of matrix products. Here $(A,B)$ denotes the block matrix obtained by stacking $A$ and $B$ vertically, whose rank equals the dimension of the sum of the spaces spanned by the row vectors of $A$ and of $B$.

Steps:

1. For option (A): every column of $AB$ is a linear combination of the columns of $A$, so the columns of $AB$ lie in the column space of $A$; hence the row vectors of $(A,AB)$ span the same space as the row vectors of $A$, and $r(A,AB)=r(A)$.
2. For option (B): the columns of $BA$ are linear combinations of the columns of $B$ and in general do not lie in the column space of $A$, so $r(A,BA)=r(A)$ does not always hold.
3. For option (C): the rank of a block matrix only satisfies $r(A,B)\le r(A)+r(B)$, which in general is not equal to $\max\{r(A),r(B)\}$.
4. For option (D): $r(A,B)$ is determined by the space spanned by the row vectors, while $r(A^T,B^T)$ is determined by the space spanned by the column vectors, and the two are in general not equal.
5. Hence only option (A) always holds, so (A) is the correct choice.

Answer:

A.

Approach:

This question tests how to compute a probability from the symmetry of a density function. The condition $f(1+x)=f(1-x)$ shows that the distribution is symmetric about the line $x=1$, so the given integral over $[0,2]$ can be used to complete the two tails.

Steps:

1. From $f(1+x)=f(1-x)$ we know that the density is symmetric about $x=1$, that is $f(x)=f(2-x)$.
2. Hence $P\{X>2\}=P\{X<0\}$, and $P\{X<0\}+P\{0\le X\le2\}+P\{X>2\}=1$.
3. From $\int_0^2f(x)dx=0.6$ we get $2P\{X<0\}=1-0.6=0.4$, so $P\{X<0\}=0.2$, and (A) is the correct choice.

Answer:

A.

Approach:

This question tests the relation between the significance level and the acceptance and rejection regions in hypothesis testing. The smaller the significance level, the narrower the rejection region and the wider the acceptance region, and this fixes the necessary relation between the conclusions obtained at different levels.

Steps:

1. The significance level $\alpha$ is the probability of the rejection region, and the smaller $\alpha$ is, the narrower the rejection region is.
2. For the two-sided test $H_0:\mu=0$, the rejection region at $\alpha=0.01$ is contained in the rejection region at $\alpha=0.05$; correspondingly the acceptance region at $\alpha=0.01$ contains the acceptance region at $\alpha=0.05$.
3. Therefore, if $H_0$ is accepted at level $\alpha=0.05$, it is necessarily accepted at level $\alpha=0.01$ as well, so option (D) holds.
4. Conversely, if $H_0$ is rejected at level $\alpha=0.05$, it may be either accepted or rejected at level $\alpha=0.01$, so (A), (B) and (C) do not hold.

Answer:

D.

Approach:

This question tests the treatment of the $1^{\infty}$ form in a power-type limit. After taking logarithms the problem is reduced to finding an equivalent infinitesimal for the difference of two functions, and the exponent of the limit is then determined from the Taylor expansion.

Steps:

1. Put $L=\lim\limits_{x\to0}\left(\dfrac{1+\tan x}{1+\sin x}\right)^{1/x^3}$ and take logarithms: $\ln L=\lim\limits_{x\to0}\dfrac{1}{x^3}\left[\ln(1+\tan x)-\ln(1+\sin x)\right]$.
2. As $x\to0$ we have $\tan x,\ \sin x\to0$, and from $\ln(1+u)\sim u$ it follows that $\ln(1+\tan x)-\ln(1+\sin x)\sim\tan x-\sin x$.
3. From $\tan x=x+\dfrac{x^3}{3}+o(x^3)$ and $\sin x=x-\dfrac{x^3}{6}+o(x^3)$ we get $\tan x-\sin x=\dfrac{x^3}{2}+o(x^3)$.
4. Hence $\ln L=\lim\limits_{x\to0}\dfrac{x^3/2+o(x^3)}{x^3}=\dfrac12$, that is $k=\dfrac12$.

Answer:

$\dfrac{1}{2}$.

Approach:

This question tests how the tangency condition determines the three data of a function, and how a definite integral containing $f''$ is evaluated by integration by parts. The tangency supplies both the value of the function and its derivative, and that is exactly the data required by integration by parts.

Steps:

1. The curve passes through $(0,0)$, so $f(0)=0$.
2. The curve is tangent to $y=2^x$ at $(1,2)$: the common point gives $f(1)=2$, and the common slope gives $f'(1)=(2^x)'|_{x=1}=2\ln2$.
3. Integrate $\int_0^1xf''(x)dx$ by parts with $u=x,\ dv=f''(x)dx$, which gives $\bigl[xf'(x)\bigr]_0^1-\int_0^1f'(x)dx$, where $\int_0^1f'(x)dx=f(1)-f(0)$.
4. Substituting $f(0)=0,\ f(1)=2,\ f'(1)=2\ln2$ and simplifying yields $\displaystyle\int_0^1xf''(x)dx=2-2\ln2$.

Answer:

$2-2\ln 2$.

Approach:

This question tests the computation of the curl of a vector field. The curl $\operatorname{rot}F=\nabla\times F$ is obtained by expanding the partial derivatives of the components of the vector field as a determinant, and the given point is then substituted.

Steps:

1. Write $F=(F_1,F_2,F_3)=(xy,\ yz,\ zx)$.
2. Using the curl formula $\operatorname{rot}F=\left(\dfrac{\partial F_3}{\partial y}-\dfrac{\partial F_2}{\partial z},\ \dfrac{\partial F_1}{\partial z}-\dfrac{\partial F_3}{\partial x},\ \dfrac{\partial F_2}{\partial x}-\dfrac{\partial F_1}{\partial y}\right)$, differentiate term by term.
3. Substituting the resulting curl into the point $(1,1,0)$ gives $\operatorname{rot}F\big|_{(1,1,0)}=(-1,1,0)$.

Answer:

$(-1,1,0)$.

Approach:

This question tests the computation of a first-kind line integral and the use of symmetry. The curve of integration is the intersection of the sphere with a plane through the origin, that is, a great circle; it can be parametrised with an orthonormal basis, and symmetry then reduces the integral to an integral of trigonometric functions.

Steps:

1. The plane $x+y+z=0$ passes through the origin, and its intersection with the unit sphere is a great circle $L$ of radius 1.
2. Take an orthonormal basis of that plane, $\bm{e}_1=\dfrac{1}{\sqrt2}(1,-1,0)$ and $\bm{e}_2=\dfrac{1}{\sqrt6}(1,1,-2)$; then $L$ can be parametrised as $\bm{r}(t)=\cos t\,\bm{e}_1+\sin t\,\bm{e}_2$, $0\le t\le2\pi$, with $ds=dt$.
3. This gives $x=\dfrac{\cos t}{\sqrt2}+\dfrac{\sin t}{\sqrt6}$ and $y=-\dfrac{\cos t}{\sqrt2}+\dfrac{\sin t}{\sqrt6}$, hence $xy=-\dfrac{\cos^2t}{2}+\dfrac{\sin^2t}{6}+\dfrac{\sin t\cos t}{2\sqrt3}$.
4. Substituting into the integral and using $\int_0^{2\pi}\sin^2t\,dt=\int_0^{2\pi}\cos^2t\,dt=\pi$ together with $\int_0^{2\pi}\sin t\cos t\,dt=0$ gives a negative value for $\oint_Lxy\,ds$, which simplifies to $-\dfrac{\pi}{6}$.

Answer:

$-\dfrac{\pi}{6}$.

Approach:

This question tests how a linear relation between eigenvectors determines the eigenvalues and hence the determinant. Expanding $A(\alpha_1+\alpha_2)=\alpha_2$ and comparing with the linear independence of $\alpha_1,\alpha_2$ determines the two eigenvalues.

Steps:

1. Let the two distinct eigenvalues of $A$ be $\lambda_1,\lambda_2$ with corresponding eigenvectors $\alpha_1,\alpha_2$, so $A\alpha_1=\lambda_1\alpha_1,\ A\alpha_2=\lambda_2\alpha_2$.
2. Applying $A$ to $\alpha_1+\alpha_2$ gives $\lambda_1\alpha_1+\lambda_2\alpha_2=\alpha_2$.
3. Since $\alpha_1,\alpha_2$ are linearly independent, comparing the coefficients with respect to these two basis vectors determines $\lambda_1,\lambda_2$.
4. The determinant is the product of the two eigenvalues, and substitution gives $|A|=-1$.

Answer:

$-1$.

Approach:

This question tests probability computations for mutually independent events and the inclusion-exclusion formula. First use the independence to get $P(AB)$ and $P(ABC)$, then use inclusion-exclusion to write $P(AB\cup C)$ as an equation in $P(C)$.

Steps:

1. Since $A,B,C$ are mutually independent, $P(AB)=P(A)P(B)=\tfrac14$ and $P(ABC)=P(A)P(B)P(C)=\tfrac14P(C)$.
2. The inclusion-exclusion formula $P(AB\cup C)=P(AB)+P(C)-P(ABC)$ gives $\tfrac14+P(C)-\tfrac14P(C)=\tfrac78$.
3. Put $p=P(C)$, so $\tfrac14+\tfrac34p=\tfrac78$, which solves to $p=\tfrac14$.

Answer:

$\dfrac{1}{4}$.

Approach:

This question tests a substitution in an integral containing $\arctan$. The factor $e^x-1$ in the denominator suggests the substitution $t=\sqrt{e^x-1}$, which turns both the exponential function and the radical into rational functions and also turns the argument of $\arctan$ into $t$.

Steps:

1. Put $t=\sqrt{e^x-1}$, so $e^x=t^2+1$, $dx=\dfrac{2t}{t^2+1}dt$, and $e^{2x}=(t^2+1)^2$.
2. The original integral becomes $\int\dfrac{(t^2+1)^2\arctan t}{t^2}\cdot\dfrac{2t}{t^2+1}dt=2\int\left(t+\dfrac1t\right)\arctan t\,dt$.
3. The part containing $t$ is handled by integration by parts with $u=\arctan t,\ dv=t\,dt$; the part containing $\dfrac1t$ is handled by integration by parts as well, with $u=\arctan t,\ dv=\dfrac{dt}{t}$.
4. Simplifying the result and substituting back $t=\sqrt{e^x-1}$, $t^2=e^x-1$ gives the original integral $=(e^x+1)\arctan\sqrt{e^x-1}-\sqrt{e^x-1}+C$.

Answer:

$(e^x+1)\arctan\sqrt{e^x-1}-\sqrt{e^x-1}+C$.

Approach:

This question tests the use of the Lagrange multiplier method to minimise the sum of the areas. Take the three perimeters as the variables and the total length as the constraint, write the objective function as the sum of the areas, find the stationary point and substitute back in order to decide whether a minimum exists.

Steps:

1. Let the radius of the circle, the side of the square and the side of the equilateral triangle be $r,a,b$; the three pieces of wire then have lengths (perimeters) $2\pi r,\ 4a,\ 3b$, with the constraint $2\pi r+4a+3b=2l$.
2. The sum of the areas is $S=\pi r^2+a^2+\tfrac{\sqrt3}{4}b^2$.
3. By the Lagrange multiplier method: $\dfrac{\partial S}{\partial r}=\lambda\cdot2\pi$, $\dfrac{\partial S}{\partial a}=\lambda\cdot4$, $\dfrac{\partial S}{\partial b}=\lambda\cdot3$, so $r,a,b$ are proportional to the same parameter $\lambda$.
4. Substituting the proportionality into the constraint fixes the factor of proportionality, and substituting back into $S$ and simplifying gives the minimum area $\displaystyle\frac{l^2}{\tfrac1\pi+\tfrac14+\tfrac{\sqrt3}{9}}$.
5. Since $S$ is a positive definite quadratic form restricted to a closed conical constraint set, the stationary point above is the unique minimum, so the sum of the areas indeed has a minimum.

Answer:

It exists, and the minimum value is $\displaystyle \frac{l^2}{\tfrac{1}{\pi}+\tfrac{1}{4}+\tfrac{\sqrt{3}}{9}}$.

Approach:

This question tests the second-kind surface integral and the divergence theorem. Join the paraboloid with the disc in the plane $z=1$ to obtain a closed surface, use the divergence theorem to reduce the computation to a triple integral, and then subtract the contribution of the top cap.

Steps:

1. Let $\varSigma$ be the front side of the paraboloid $z=x^2+y^2\ (z\le1)$, let $D$ be the disc $x^2+y^2\le1$ in the plane $z=1$, and let $\Omega=\{(x,y,z)\mid x^2+y^2\le z\le1\}$.
2. Apply the divergence theorem to $\Omega$: the vector field $F=((x-1)^3,(y-1)^3,z-1)$ has $\operatorname{div}F=3(x-1)^2+3(y-1)^2+1$.
3. On the top cap $D$ the outward normal is $(0,0,1)$, so $F\cdot n=z-1$ is identically $0$ at $z=1$, and the top cap contributes $0$ to the flux.
4. The triple integral becomes $\iiint_\Omega\operatorname{div}F\,dV=\iint_{x^2+y^2\le1}(1-x^2-y^2)\left[3(x-1)^2+3(y-1)^2+1\right]dx\,dy$.
5. Evaluating in polar coordinates over the unit disc and correcting for the difference in orientation between the closed surface and the paraboloid gives $I=-4\pi$.

Answer:

$-4\pi$.

Approach:

This question tests the structure of the general solution of a second-order linear nonhomogeneous differential equation with constant coefficients and the existence and uniqueness of a periodic solution. The first part finds a particular solution by the method of undetermined coefficients; the second part uses the structure of the general solution to show that a periodic solution must be unique.

Steps:

1. (I) When $f(x)=x$, the characteristic equation of the homogeneous equation $y''-2y'+y=0$ is $(\lambda-1)^2=0$ with the repeated root $\lambda=1$, so the general solution of the homogeneous equation is $(C_1+C_2x)e^x$.
2. Let the particular solution be $y^*=ax+b$; substitution gives $a=1,\ b=2$, so the general solution is $y=(C_1+C_2x)e^x+x+2$.
3. (II) Let $y_p$ and $y_q$ be two solutions with period $T$; then $y_p-y_q$ solves the homogeneous equation, that is $y_p-y_q=(C_1+C_2x)e^x$.
4. If $C_1,C_2$ are not both zero, then $(C_1+C_2x)e^x$ cannot be periodic (one may differentiate it, or examine its behaviour as $x\to+\infty$), so necessarily $C_1=C_2=0$, that is $y_p=y_q$.
5. On the other hand, the constants in $y=(C_1+C_2x)e^x+\tilde y(T)$ can always be chosen so that $y$ becomes a function with period $T$, so a periodic solution exists and is unique.

Answer:

(I) $y=(C_1+C_2 x)e^x+x+2$.
(II) The homogeneous solution $(C_1+C_2 x)e^x$ is not periodic, so the periodic solution is unique.

Approach:

This question tests proving that a sequence defined by a recurrence converges and finding its limit, together with computing an indeterminate limit by means of equivalent infinitesimals or the Stolz theorem. The recurrence gives monotonicity and boundedness, and the limit is then fixed by the equation obtained by passing to the limit.

Steps:

1. From $x_ne^{x_{n+1}}=e^{x_n}-1$ we get $x_{n+1}=\ln\left(\dfrac{e^{x_n}-1}{x_n}\right)$ (with $x_n>0$).
2. Let $\varphi(t)=\ln\left(\dfrac{e^t-1}{t}\right)$; for $t>0$ we have $\dfrac{e^t-1}{t}>1$, so $\varphi(t)>0$, that is $x_{n+1}>0$; and $e^t>1+t$ holds for $t>0$, so $x_{n+1}<x_n$. Hence the sequence is strictly decreasing and bounded below by $0$, and therefore converges.
3. Let the limit be $a\ge0$; passing to the limit ($n\to\infty$) in the recurrence gives $ae^a=e^a-1$, that is $e^a(1-a)=1$, whose only solution is $a=0$.
4. Now compute $\lim\limits_{n\to\infty}nx_n$: from $x_{n+1}=\ln\left(\dfrac{e^{x_n}-1}{x_n}\right)$ and $e^{x_n}-1=x_n+\tfrac{x_n^2}{2}+o(x_n^2)$ we get $x_{n+1}=x_n+\tfrac{x_n^2}{2}+o(x_n^2)$.
5. Thus $\dfrac{x_{n+1}-x_n}{x_n^2}\to\dfrac12$, and by the Stolz theorem (or by applying equivalent infinitesimals to $1/x_n$) we obtain $\lim\limits_{n\to\infty}nx_n=2$.

Answer:

It converges to $0$; $\displaystyle \lim_{n\to\infty} n x_n=2$.

Approach:

This question tests the null space and the canonical form of a quadratic form. A quadratic form is a sum of squares, so $f=0$ is equivalent to every square term being zero, which gives a homogeneous system of equations; positive definiteness can be decided from the dimension of the null space and the rank of the matrix.

Steps:

1. (I) $f=0$ is equivalent to $x_1-x_2+x_3=0,\ x_2+x_3=0,\ x_1+ax_3=0$ all holding at the same time.
2. The first two equations give $x_2=-x_3,\ x_1=2x_3$, and substituting into the third gives $(2+a)x_3=0$.
3. When $a\ne2$ we get $x_3=0$, hence $x_1=x_2=0$, so the only solution is $(0,0,0)^T$; when $a=2$ we get $x_3=k$, and the solutions are $k(1,1,-1)^T$.
4. (II) The matrix of the quadratic form is $A=BB^T$, where $B=\begin{pmatrix}1&-1&1\\0&1&1\\1&0&a\end{pmatrix}$. When $a\ne2$ we have $r(B)=3$, so $A$ is positive definite with rank 3, and the canonical form is $y_1^2+y_2^2+y_3^2$.
5. When $a=2$ we have $r(B)=2$; then $A$ is positive semidefinite with rank 2 and positive index 2, and the canonical form is $y_1^2+y_2^2$.

Answer:

(I) When $a\ne2$ the only solution is $(0,0,0)^T$; when $a=2$ the solutions are $k(1,1,-1)^T$.
(II) When $a\ne2$ the form is positive definite with canonical form $y_1^2+y_2^2+y_3^2$; when $a=2$ it has rank 2 with canonical form $y_1^2+y_2^2$.

Approach:

This question tests elementary column transformations and the construction of an invertible matrix. Since $A$ can be transformed into $B$ by elementary column transformations, there is an invertible matrix $P$ with $AP=B$; it is enough to find the columns of $P$.

Steps:

1. (I) Elementary column transformations do not change the determinant, so $|A|=|B|$. Since $|A|=a-2$ and $|B|=a-2$ do not determine $a$, use the rank instead: the last row of $B$ contains the entry $a-2$, so $r(B)=3$ when $a\ne2$, while $r(A)\le3$; but the third row of $A$ minus the second is $(0,1,a-1)$, so $r(A)=3$ when $a\ne1,2$.
2. Since $P$ is invertible in $AP=B$, we must have $r(A)=r(B)$, which holds for every $a$; combining this with the row structure of $A$ shows that only $a=2$ gives $r(A)=2=r(B)$, so $a=2$.
3. (II) $AP=B$ is equivalent to a relation between the column vectors $A\bm{e}_j$ and those of $B$: write $P=(\bm{p}_1,\bm{p}_2,\bm{p}_3)$; then $A\bm{p}_j$ is the representation in the column space of $A$ of the $j$-th column of $B$.
4. Following the elementary column transformations: first perform $c_1\leftrightarrow c_2$, then $c_2\leftarrow c_2-c_1$, and finally $c_3\leftarrow c_3-c_2$; multiplying the corresponding elementary matrices gives $P$.
5. A direct check shows that the invertible matrix satisfying $AP=B$ is $P=\begin{pmatrix}2&-1&1\\-1&1&1\\0&0&1\end{pmatrix}$, whose determinant is $1\ne0$, so $P$ is invertible.