Use multiplicativity of the determinant and the determinant criterion for an invertible matrix:Neither factor can therefore be zero, so both matrices are nonsingular. The determinant of a triangular matrix is the product of its diagonal entries:Each product is nonzero, forcing every diagonal factor to be nonzero. These explicitly stated determinant properties justify both conclusions.

Substitute the QR decomposition into the normal equations:The diagonal of $R$ is nonzero, so $R^T$ is invertible. Cancel it algebraically to obtainCompute the three inner products in $Q^{T}b$, then use back substitution in this upper triangular matrix. There is no need to form an inverse or explicitly construct $A^{T}A$. The orthogonal projection of $b$ onto the column space is $Q{Q}^{T}b$; its orthogonal residual proves that this solution minimizes $\Vert Ay-b{\Vert }_2$.

Collect the orthonormal vectors as columns and compute $R=Q^{T}A$. This gives the thin QR decompositionThe inner products in part (e) prove $Q^{T}Q=I_3$; the column identities $a_1=2q_1$, $a_2=4q_1+2q_2$, $a_3=2q_1+2q_2+2q_3$ prove $A=QR$.

The Gram-Schmidt process first gives $u_1=a_1=(1,1,1,1{)}^T$, of norm two. Next $a_1^T{a}_2/(a_1^T{a}_1)=2$, soFor the third column, both coefficients onto $u_1,u_2$ equal one, givingThese three nonzero vectors are mutually orthogonal and each has norm two. The resulting orthonormal basis is

If $Ax=b$ has a solution, the multiplication identity from part (c) expresses $b$ as a linear combination of the columns, so $b\in \mathrm{Col}(A)$.

Conversely, if $b$ lies in that column space, its definition supplies scalars $x_1,x_2,x_3$ with $b=x_1{a}_1+x_2{a}_2+x_3{a}_3$. Put those scalars into the column vector $x$. Then $Ax=b$. This proves both directions of the claimed equivalence.

Writing the columns as $a_1,a_2,a_3$, matrix multiplication givesThe column space is the linear span of these columns, so this particular linear combination belongs to it. The assertion holds for every vector of coefficients $y$.

The columns are linearly independent. A linear combination of them with coefficients $x_1,x_2,x_3$ is exactly $Ax$. Part (a) shows that this vanishes only when all three coefficients vanish, which is the definition of column independence. Hence $A$ has column rank three.

No: the only solution is the zero vector. Subtract the fourth row equation from the first to get $2x_3=0$. Subtract the third from the fourth to get $2x_2=0$. With $x_2=x_3=0$, any row then gives $x_1=0$. Thus this homogeneous linear system has trivial null space.

Use $k$ clocked processor registers, each storing an $n$-bit word, two independent combinational read ports and one synchronous write port. Each address requires $a=\lceil {\log }_{2}k\rceil$ bits. The read addresses select two $k$-to-one multiplexers, each $n$ bits wide. A write-address binary decoder combines its selected line with the global write enable to form the individual enables:At the active clock edge, register $i$ captures the common write data if $w{e}_i=1$, otherwise retaining its word. Reads return the currently stored words selected by their respective addresses. For non-power-of-two $k$, unused address codes need a specified harmless implementation. A simultaneous read and write of one address follows the chosen edge/read timing; an additional forwarding circuit would be needed if a different bypass convention were required.

The preemptive policy improves B's scheduling turnaround time from 17 to 10 ms and the mean from 16.5 to 14 ms. A finishes later, at 18 rather than 16 ms, and the total completion time rises from 17 to 18 ms. Thus preemption improves mean turnaround here; nonpreemption gives the shorter makespan here.

The result illustrates different scheduling objectives, not a universal winner. These calculations assume negligible context switch overhead; charging for the additional switches could change the comparison. Both policies perform the same 15 ms of useful CPU work.

Use round-robin scheduling with scheduling quantum 2 ms. A's first burst is split by B, and A's second burst is also interrupted once. The resulting CPU intervals areA blocks for input-output during $[5,7)$ and $[12,14)$; B blocks during $[3,5)$ and $[6,8)$. At time 8, B joins the ready queue while A uses the remainder of its current quantum; B runs at time 9. No other process is runnable during A's final four milliseconds, so A can continue across the time-16 quantum boundary. Consequently

With the given ready-order rule and A first, the CPU timeline in milliseconds isA performs input-output during $[4,6)$ and $[10,12)$; B does so during $[5,7)$ and $[11,13)$. In particular, B becoming ready at time 7 does not interrupt A's ongoing nonpreemptive burst. The completion times are $C_A=16$ and $C_B=17$, givingThere are 15 ms of useful CPU execution and 2 ms of idle time on this particular timeline.