The determinant of the unit lower triangular matrix $L$ is 1. There is one row swap, so $\det P=-1$, and the upper triangular matrix has determinant $1\cdot (-2)\cdot (3/2)\cdot 2=-6$. Therefore

First add row 1 to rows 2 and 3. This is the product of two elementary matrices, written compactly asSwap rows 2 and 4 with $P$, then subtract half of the new row 2 from row 3 with $E_2$:This gives the upper triangular matrixBecause $P^{-1}=P$, rearranging yields the convention $A=PLU$, whereThe conjugation by $P$ moves the earlier elimination multipliers to the appropriate rows. Ignoring this permutation when forming $L$ would give an incorrect PLU factorization. Here the same permutation is its own inverse, so $PA=LU$ also holds.

Undo each elementary row operation: divide row 3 by 3, add 3 times row 2 to row 4, and repeat the same swap. HenceMultiplying each inverse by its original matrix gives the identity matrix.

An elementary matrix acts on rows by left multiplication. In the stated order, the three matrices areThe first scales row 3, the second adds $-3$ times row 2 to row 4, and the third swaps rows 1 and 4. Right multiplication would instead perform column operations.

The first two columns are already orthogonal and have norms 15 and 5. Their normalized versions are $q_1=(3,4,0,0{)}^T/5$ and $q_2=(0,0,4,-3{)}^T/5$. Their inner products with the third column are 10 and 5. Subtracting these projections leaves $(20,-15,0,0{)}^T$, whose norm is 25. Thus an economy-size QR factorization isDirect multiplication gives $Q^{T}Q=I_3$ and $QR=A$. If a full square $Q$ is required, append $q_4=(0,0,3,4{)}^T/5$ and append a zero fourth row to $R$; the product is unchanged and the completed $Q$ is an orthogonal matrix.

In the column convention used for QR factorization, an $m\times k$ real matrix has orthonormal columns when $Q^{T}Q=I_k$. This says that each column has unit norm and distinct columns have zero inner product. If $Q$ is square, then $Q^{T}Q=I$ provides a left inverse; a square matrix with a left inverse is invertible, so $Q^{-1}=Q^T$, and also $Q{Q}^T=I$. Such a square matrix is an orthogonal matrix. A rectangular matrix with orthonormal columns generally has no two-sided inverse.

The Gram-Schmidt process constructs the vectors in order:The empty sum at $j=1$ is zero. Each subtraction removes the projection onto one previous unit vector, so $v_j$ is orthogonal to their span. Linear independence of the inputs ensures $v_j\ne 0$ in exact arithmetic, and normalization gives unit length.

In floating-point arithmetic, subtracting nearly equal projections can amplify roundoff and cause loss of orthogonality, especially for nearly dependent columns. Modified Gram-Schmidt, reorthogonalization, or Householder QR factorization can improve numerical behaviour; no implementation can remove the underlying sensitivity of nearly dependent input data.

As written, the assertion needs a nonzero-vector assumption: the singleton set containing the zero vector is mutually orthogonal but is linearly dependent. For the linear independence of nonzero orthogonal vectors, suppose ${\sum }_{i=1}^k{\alpha }_i{q}_i=0$. Taking the inner product with $q_j$ givesSince $q_j\ne 0$, positive definiteness gives $\Vert q_j{\Vert }^2>0$, hence ${\alpha }_j=0$. This holds for every $j$, proving linear independence under the necessary assumption.

Use binary units $1\mathrm{KB}=2^{10}$ bytes and $1\mathrm{MB}=2^{20}$ bytes, consistent with the given binary GB convention. Reading one average file takes seek plus rotation plus transfer; moving it requires the same costs again on the write:A half-full 16 GB disk contains $8\cdot 2^{30}$ used bytes, or $N=2^{20}$ average files. Under the stated per-file disk compaction model,A calculation with decimal KB and MB changes the unrounded figures, so the unit convention must be retained consistently. This model charges one seek and rotational delay on each read and each write; it does not assume a single streaming pass over the whole disk.

At 6000 revolutions per minute the disk makes 100 revolutions per second, so one turn takes 10 ms and the mean rotational latency is 5 ms. Adding seek time, data transfer time and controller overhead gives, using decimal MB,Using $1\mathrm{MB}=2^{20}$ bytes instead gives 19.09765625 ms, with the same rounded answer. The calculation assumes the stated rate applies to the sector transfer and that the seek and overhead are each paid once.

Conditioning on a cache hit or cache miss, the expected latency isThis is a straight line from $(0,45)$ to $(1,1)$; at hit rate $1/2$, the mean is 23 ms. The 45 ms is already the complete miss-service time in the question, so another cache-lookup millisecond is not added to it.

Device independence lets an application request input-output through a uniform interface without depending on a particular device's control registers, instruction protocol or geometry. A device driver translates the abstract requests into device-specific operations. Replacing one supported disk model with another can then preserve the application's file-reading interface.

Address a read-only memory with the two state bits and two input bits. Each word stores the two next-state bits and INC, so a direct replacement for all the combinational logic is $16\times 3$ bits, or 48 bits. Program its 16 words from the state table, duplicating a state's output over all four input addresses. The two D flip-flops and clock remain outside the ROM.

If INC retains its separate two-state-bit decoder, only the next-state logic needs a $16\times 2$-bit ROM, plus that output gate. Storing output in a separate state-only ROM would use $4\times 1$ further bits. These factored options have a different boundary from replacing all the combinational logic with one ROM.

Encode $(q_1{q}_0)$ by $A=00$, $B=01$, $C=11$, $I=10$, with reset to $A$. Two positive-edge-triggered D flip-flops store the state. Minimizing the Boolean functions from the complete table givesEquivalently, the minimal sum of products form of $D_1$ is ${\bar{x}}_0{q}_0{x}_1+{\bar{x}}_0{q}_0{q}_1$. The displayed factorization saves duplicated logic. The equations use only AND gates, OR gates and NOT gates; connect $D_j$ to the corresponding flip-flop data input and use one common clock.

For example, $C$ with input $00$ gives $(D_1{D}_0)=10$, entering the pulse state. In $I$, either $x_0=1$ gives $01$ or $x_0=0$ gives $00$, independently of $x_1$; the pulse cannot persist. The four states are distinguishable by their current output or future response to inputs $00$ and $10,00$, so two state bits are also necessary.

Use a four-state Moore machine. In state $A$ wait for an outer-sensor activation; $B$ combines the outer-only and both-sensors phases; $C$ records that the visitor has reached the inner-only phase; $I$ emits the one-cycle count pulse. Only $I$ has INC = 1.

The complete state transition table, with input order $x_1{x}_0$, isFor a normal entry the states are $A\to B\to B\to C\to I$; retreat after the first sensor instead gives $A\to B\to A$ with no pulse. Repeated samples at one sensor remain in the corresponding phase. State $I$ always exits on the next edge, so its output lasts exactly one clock period; an outer activation on that next edge goes directly to $B$, allowing the next visitor to start immediately.

This follows the question's single-visitor entry model and merges the two outer-active phases as its hint requests. Inputs are sampled synchronously; the table also specifies deterministic recovery for other sampled combinations. It is not a specification of a bidirectional occupancy counter for arbitrary overlapping entry and exit traffic.

Only now may the invalid codes 10–15 be marked as don't-care conditions. The $f$ Karnaugh map givesThe groups are $\{8,\dots ,15\}$, $\{4,5,12,13\}$, $\{4,6,12,14\}$ and $\{0,4,8,12\}$. They cover all required ones, avoid valid-digit zeros, and may assign either output to invalid codes. The valid inputs 0, 5, 6 and 9 cannot share a valid implicant with each other, so at least four product terms are necessary. The displayed four-term cover uses only seven literals.

Use $w,x,y$ as the three select bits of the 8-to-1 multiplexer, so address $i=4w+2x+y$ selects data input $D_i$. ConnectA single NOT gate produces $\bar{z}$, shared by $D_0$ and $D_3$; the other connections are wires and constant logic levels. For each fixed select address the data input reproduces the corresponding two rows of the truth table, including zeros for all invalid codes.