- Recall the determinant rule for a \(2\times2\) matrix. - Multiply the entries on each diagonal. - Subtract the second diagonal product from the first.

For a \(2\times2\) matrix, \(\det(A)=3\cdot4-1\cdot2=10\).

\(10\)

- Think about what matrix must multiply \(D\) to produce the identity matrix. - The zero off-diagonal entries can stay zero. - Each nonzero diagonal entry needs a multiplicative reciprocal.

For a diagonal matrix with nonzero diagonal entries, invert each diagonal entry: \(D^{-1}=\begin{pmatrix}\frac{1}{2}&0\\0&\frac{1}{5}\end{pmatrix}\).

\(\begin{pmatrix}\frac{1}{2}&0\\0&\frac{1}{5}\end{pmatrix}\)

- For a \(2\times2\) matrix, use \(ad-bc\). - A square matrix is invertible exactly when its determinant is nonzero.

1. \(\det(A)=2(2)-1(4)=0\), so \(A\) is singular and not invertible. 2. \(\det(B)=2(1)-1(1)=1\ne0\), so \(B\) is invertible.

\(\det(A)=0\), \(\det(B)=1\), and only \(B\) is invertible.

- Compute the determinant first. - Swap the diagonal entries, negate the off-diagonal entries, and divide by the determinant.

1. \(\det(A)=1(1)-1(0)=1\). 2. Using the \(2\times2\) inverse formula, \(A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}\).

\(A^{-1}=\begin{pmatrix}1&-1\\0&1\end{pmatrix}\).

- Look at the entries below the main diagonal. - What simplified determinant rule applies to a triangular matrix?

1. The matrix is upper triangular, so its determinant is the product of its main diagonal entries. 2. \(\det(A)=2\cdot3\cdot(-1)=-6\).

\(-6\)

- Expanding along a row with zeros avoids unnecessary terms. - Delete the row and column containing the nonzero top entry to get a smaller determinant.

1. Only the first entry of the top row is nonzero, so the determinant equals its \(2\times2\) minor. 2. \(\det(B)=1\cdot\begin{vmatrix}2&1\\0&3\end{vmatrix}=2\cdot3-1\cdot0=6\).

\(6\)

- Think about what happens to a determinant when two rows repeat. - How does a zero determinant relate to invertibility?

1. Two identical rows are dependent, so the determinant is \(0\). 2. A square matrix with determinant \(0\) has no inverse.

Its determinant is \(0\), and it has no inverse.

- The determinant of a triangular or diagonal matrix uses its diagonal entries. - What happens when any compatible matrix is multiplied by the identity matrix?

1. The identity matrix is diagonal, so its determinant is \(1\cdot1\cdot1=1\). 2. Since \(II=I\), its inverse is itself.

\(\det(I)=1\) and \(I^{-1}=I\).

- Multiplication by this matrix undoes multiplication by \(A\). - What symbol denotes the inverse of a matrix?

1. The inverse of \(A\) is the matrix whose products with \(A\), in both orders, equal the identity. 2. Therefore \(B=A^{-1}\).

\(B\) is the inverse of \(A\): \(B=A^{-1}\).

- Arrange the three vectors as columns of a matrix. - What determinant value corresponds to linearly dependent columns? - Expand the determinant to obtain an expression in \(k\). - Solve the resulting equation.

1. The vectors are linearly dependent exactly when the determinant of the matrix with these vectors as columns is \(0\). 2. Compute \(\det\begin{pmatrix}2&1&1\\1&0&k\\3&2&1\end{pmatrix}=1-k\). 3. Set the determinant equal to \(0\): \(1-k=0\). Therefore, \(k=1\).

\(k=1\)

- What scalar triple product value indicates that three vectors are coplanar? - Write the vectors as columns of a \(3\times3\) determinant. - Expand and simplify the determinant as an expression in \(k\). - Set that expression equal to \(0\).

1. Three vectors in \(\mathbb{R}^3\) are linearly dependent exactly when their scalar triple product, or the corresponding determinant, is \(0\). 2. Compute \(\det\begin{pmatrix}2&4&2\\-1&1&5\\3&1&k\end{pmatrix}=2(k-5)-4(-k-15)+2\cdot(-4)=6k+42\). 3. Set the determinant equal to \(0\): \(6k+42=0\). Therefore, \(k=-7\).

\(k=-7\)

- How is the basis property related to linear independence? - Arrange the vectors as columns of a matrix. - What determinant value indicates linear dependence? - Solve the resulting equation in \(a\).

1. Three vectors form a basis of \(\mathbb{R}^3\) exactly when the determinant of the matrix with those vectors as columns is nonzero. 2. Compute \(\det\begin{pmatrix}1&2&8\\-2&1&a\\3&-1&3\end{pmatrix}=7a+7\). 3. The vectors fail to form a basis when \(7a+7=0\). Therefore, \(a=-1\).

\(\det[\mathbf x\ \mathbf y\ \mathbf z]=7(a+1)\). The vectors fail to form a basis when \(a=-1\).

- Arrange the three vectors as columns of a matrix. - What does a nonzero determinant say about the columns? - A vector can be written as a combination of the other two only when the three vectors are dependent. - Compute the determinant carefully.

1. Arrange the vectors as columns of a matrix and compute the determinant: \(\det\begin{pmatrix}1&0&2\\2&1&1\\1&-1&0\end{pmatrix}=-5\). 2. Because the determinant is nonzero, the vectors are linearly independent. Therefore, \(\mathbf{w}\) cannot be expressed as a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\).

\(\det[\mathbf u\ \mathbf v\ \mathbf w]=-5\ne0\). The vectors are linearly independent, so \(\mathbf w\) cannot be written as a linear combination of \(\mathbf u\) and \(\mathbf v\).

- A nonzero determinant certifies independence for three vectors in \(\mathbb R^3\). - Two proportional columns force determinant \(0\). - A zero column forces determinant \(0\).

1. For part a, \(\det[\mathbf a\ \mathbf b\ \mathbf c]=-2\ne0\), so the vectors are linearly independent. 2. In part b, \(\mathbf v=-2\mathbf u\), so the matrix with \(\mathbf u,\mathbf v,\mathbf w\) as columns has two proportional columns and determinant \(0\). The set is linearly dependent. 3. In part c, the matrix with \(\mathbf x,\mathbf y,\mathbf z\) as columns has a zero column, so its determinant is \(0\). The set is linearly dependent.

a) \(\det[\mathbf a\ \mathbf b\ \mathbf c]=-2\ne0\): linearly independent b) \(\det[\mathbf u\ \mathbf v\ \mathbf w]=0\) because \(\mathbf v=-2\mathbf u\): linearly dependent c) \(\det[\mathbf x\ \mathbf y\ \mathbf z]=0\) because one column is \(\mathbf0\): linearly dependent

- Arrange the vectors as columns of a matrix. - Set the determinant equal to \(0\). - After finding \(k\), solve for coefficients in a linear combination. - A nontrivial relation has at least one nonzero coefficient.

1. The vectors are linearly dependent when the determinant of the matrix with these vectors as columns is \(0\). 2. Compute \(\det\begin{pmatrix}2&1&5\\1&0&2\\0&3&k\end{pmatrix}=3-k\). Thus \(3-k=0\), so \(k=3\). 3. When \(k=3\), the vectors satisfy \(-2\mathbf{a}-\mathbf{b}+\mathbf{c}=\mathbf{0}\), which is nontrivial.

\(\det[\mathbf a\ \mathbf b\ \mathbf c]=3-k\). The determinant is zero when \(k=3\). For that value, \(-2\mathbf a-\mathbf b+\mathbf c=\mathbf0\) is a nontrivial relation.

- Express \(\mathbf{a}\), \(\mathbf{b}\), and \(\mathbf{c}\) by their coefficient columns relative to \(\mathbf{u},\mathbf{v},\mathbf{w}\). - What determinant value indicates that three coordinate columns are dependent? - After the determinant test, look for a simple nontrivial relation among \(\mathbf{a},\mathbf{b},\mathbf{c}\).

1. Relative to the independent vectors \(\mathbf{u},\mathbf{v},\mathbf{w}\), the coordinate columns of \(\mathbf{a},\mathbf{b},\mathbf{c}\) are \(\begin{pmatrix}1\\1\\0\end{pmatrix}\), \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\), and \(\begin{pmatrix}1\\0\\1\end{pmatrix}\). 2. Compute \(\det\begin{pmatrix}1&0&1\\1&1&0\\0&-1&1\end{pmatrix}=0\). Therefore, the three coordinate columns, and hence \(\mathbf{a},\mathbf{b},\mathbf{c}\), are linearly dependent. 3. Indeed, \(-\mathbf{a}+\mathbf{b}+\mathbf{c}=-(\mathbf{u}+\mathbf{v})+(\mathbf{v}-\mathbf{w})+(\mathbf{u}+\mathbf{w})=\mathbf{0}\), giving a nontrivial relation.

The coefficient determinant is \(\det\begin{pmatrix}1&0&1\\1&1&0\\0&-1&1\end{pmatrix}=0\). Thus the vectors are linearly dependent, with nontrivial relation \(-\mathbf a+\mathbf b+\mathbf c=\mathbf0\).

- Form three edge vectors from one common vertex. - Place those vectors as columns of a \(3\times3\) matrix. - The determinant gives the signed volume of the corresponding parallelepiped. - Divide its absolute value by \(6\) for a tetrahedron.

1. From vertex \(A\), the edge vectors are \(\overrightarrow{AB}=\langle 3,1,1\rangle\), \(\overrightarrow{AC}=\langle 1,4,-1\rangle\), and \(\overrightarrow{AD}=\langle 2,2,4\rangle\). 2. Place these vectors as the columns of a matrix: \(M=\begin{pmatrix}3&1&2\\1&4&2\\1&-1&4\end{pmatrix}\). 3. The determinant is \(\det(M)=42\). 4. A tetrahedron has one-sixth the volume of the corresponding parallelepiped, so \(V=\frac{1}{6}|\det(M)|=\frac{42}{6}=7\) cubic units.

\(7\) cubic units

- Form three edge vectors from one vertex. - A nonzero determinant shows that the vectors are linearly independent. - Divide the corresponding parallelepiped volume by \(6\).

1. From \(A\), \(\overrightarrow{AB}=\langle 4,0,0\rangle\), \(\overrightarrow{AC}=\langle 0,4,0\rangle\), and \(\overrightarrow{AD}=\langle 2,2,6\rangle\). 2. Their determinant is \(\det\begin{pmatrix}4&0&2\\0&4&2\\0&0&6\end{pmatrix}=96\ne0\). Therefore, the vectors are linearly independent, so the four points are not coplanar and form a tetrahedron. 3. Its volume is \(\frac{1}{6}|96|=16\) cubic units.

a) The edge-vector determinant is nonzero, so the points are not coplanar. b) \(16\) cubic units

- Use row-by-column multiplication for each product. - Track the negative signs carefully. - Recall the condition that identifies two matrices as inverses.

1. Compute \(MN=\begin{pmatrix}5&3\\3&2\end{pmatrix}\begin{pmatrix}2&-3\\-3&5\end{pmatrix}=\begin{pmatrix}10-9&-15+15\\6-6&-9+10\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). 2. In the reverse order, \(NM=\begin{pmatrix}2&-3\\-3&5\end{pmatrix}\begin{pmatrix}5&3\\3&2\end{pmatrix}=\begin{pmatrix}10-9&6-6\\-15+15&-9+10\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). 3. Since both products equal the identity matrix, \(M\) and \(N\) are inverses of each other.

a) \(MN=\begin{pmatrix}1&0\\0&1\end{pmatrix}\) and \(NM=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). b) \(M\) and \(N\) are inverse matrices.

- Compute the determinant before applying the inverse formula. - The two diagonal entries exchange positions in the numerator matrix. - The off-diagonal entries change signs.

1. For \(A=\begin{pmatrix}a&b\\c&d\end{pmatrix}\), \(A^{-1}=\frac{1}{ad-bc}\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\), provided \(ad-bc\ne0\). 2. Here, \(\det(A)=3\cdot1-1\cdot2=1\). 3. Therefore, \(A^{-1}=\begin{pmatrix}1&-1\\-2&3\end{pmatrix}\).

\(A^{-1}=\begin{pmatrix}1&-1\\-2&3\end{pmatrix}\)

- For \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\), the determinant is \(ad-bc\). - Invertibility is equivalent to a nonzero determinant.

1. \(\det(A_k)=6k-6=6(k-1)\). 2. A \(2\times2\) matrix is invertible exactly when its determinant is nonzero, so \(A_k\) is invertible for \(k\ne1\). 3. At \(k=1\), \(\det(A_1)=0\), so \(A_1\) is singular and has no inverse.

a) \(\det(A_k)=6(k-1)\) b) Invertible for \(k\ne1\) c) At \(k=1\), the matrix is singular (not invertible).

- Put the system coefficients into a square coefficient matrix. - Find the inverse of the coefficient matrix before multiplying the constant vector. - Multiplying by \(A^{-1}\) undoes multiplication by \(A\).

1. Write \(A=\begin{pmatrix}2&1\\5&3\end{pmatrix}\), \(\mathbf{x}=\begin{pmatrix}x\\y\end{pmatrix}\), and \(\mathbf{b}=\begin{pmatrix}7\\19\end{pmatrix}\). 2. The determinant is \(2\cdot3-1\cdot5=1\), so \(A^{-1}=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\). 3. Multiply by the inverse: \(\mathbf{x}=A^{-1}\mathbf{b}=\begin{pmatrix}3&-1\\-5&2\end{pmatrix}\begin{pmatrix}7\\19\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}\).

\((x,y)=(2,3)\)

- Compare the proposed numerator matrix with the standard \(2\times2\) inverse pattern. - Two changes happen to the original entries before dividing by the determinant. - A proposed inverse can be checked by multiplying it by the original matrix.

1. For a \(2\times2\) matrix \(\begin{pmatrix}a&b\\c&d\end{pmatrix}\), the inverse numerator is \(\begin{pmatrix}d&-b\\-c&a\end{pmatrix}\). 2. Noor changed the signs of the off-diagonal entries but did not interchange the two diagonal entries. 3. Therefore, \(A^{-1}=\frac12\begin{pmatrix}1&-1\\-2&4\end{pmatrix}\). 4. Multiplying \(A\) by this matrix gives the identity matrix, confirming the correction.

Noor failed to swap the diagonal entries. The correct inverse is \(A^{-1}=\frac12\begin{pmatrix}1&-1\\-2&4\end{pmatrix}\).

- Expansion along the first row gives alternating plus and minus signs. - The zero entry produces no term. - Evaluate each smaller determinant before combining the terms.

1. Expand using the first two entries of the first row: \(\det(A)=2\begin{vmatrix}3&1\\0&2\end{vmatrix}-1\begin{vmatrix}0&1\\1&2\end{vmatrix}\). 2. The minors equal \(6\) and \(-1\). 3. Thus \(\det(A)=2\cdot6-(-1)=13\).

\(13\)

- Attach an identity matrix to the right side of \(A\). - Which row scalings would make each diagonal entry on the left equal to \(1\)? - Apply each row operation to the whole augmented matrix.

1. Begin with \(\left[\begin{array}{ccc|ccc}2&0&0&1&0&0\\0&1&0&0&1&0\\0&0&3&0&0&1\end{array}\right]\). 2. Divide the first row by \(2\) and the third row by \(3\). 3. The left side becomes \(I\), while the right side becomes \(A^{-1}=\begin{pmatrix}\frac{1}{2}&0&0\\0&1&0\\0&0&\frac{1}{3}\end{pmatrix}\).

\(\left[\begin{array}{ccc|ccc}2&0&0&1&0&0\\0&1&0&0&1&0\\0&0&3&0&0&1\end{array}\right]\xrightarrow{R_1\leftarrow\frac{1}{2}R_1,\;R_3\leftarrow\frac{1}{3}R_3}\left[\begin{array}{ccc|ccc}1&0&0&\frac{1}{2}&0&0\\0&1&0&0&1&0\\0&0&1&0&0&\frac{1}{3}\end{array}\right]\). Thus \(A^{-1}=\begin{pmatrix}\frac{1}{2}&0&0\\0&1&0\\0&0&\frac{1}{3}\end{pmatrix}\).

- A zero determinant is the test for a square matrix that cannot be inverted. - The first row has two zeros, making expansion simple. - Set the resulting expression equal to zero.

1. Expanding along the first row gives \(\det(A)=\begin{vmatrix}t&1\\2&3\end{vmatrix}=3t-2\). 2. The matrix is not invertible when its determinant is zero. 3. Solve \(3t-2=0\) to obtain \(t=\frac{2}{3}\).

\(t=\frac{2}{3}\)

- Which row and column produce the upper-left entry of a matrix product? - Compute each first-row entry separately before checking the remaining rows. - What product identity characterizes an inverse?

1. For the first row of \(P^2\), the three entries are \(0\cdot0+1\cdot1+0\cdot0=1\), \(0\cdot1+1\cdot0+0\cdot0=0\), and \(0\cdot0+1\cdot0+0\cdot1=0\). 2. The other rows give \(P^2=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I\). 3. Since \(PP=I\), \(P^{-1}=P\).

First row of \(P^2\): \((0\cdot0+1\cdot1+0\cdot0,\;0\cdot1+1\cdot0+0\cdot0,\;0\cdot0+1\cdot0+0\cdot1)=(1,0,0)\). The complete product is \(P^2=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I\). Therefore \(P^{-1}=P\).

- Look for the entry above the main diagonal that prevents the left half from being the identity. - Apply every row operation to both halves of the augmented matrix. - Multiply the original matrix by your proposed inverse to check it.

1. Begin with \(\left[\begin{array}{ccc|ccc}1&2&0&1&0&0\\0&1&0&0&1&0\\0&0&1&0&0&1\end{array}\right]\). 2. Apply \(R_1\leftarrow R_1-2R_2\) to both halves. The result is \(\left[\begin{array}{ccc|ccc}1&0&0&1&-2&0\\0&1&0&0&1&0\\0&0&1&0&0&1\end{array}\right]\). 3. The left half is the identity, so \(A^{-1}=\begin{pmatrix}1&-2&0\\0&1&0\\0&0&1\end{pmatrix}\). 4. Direct multiplication gives \(AA^{-1}=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}=I\), confirming the inverse.

\(\left[\begin{array}{ccc|ccc}1&2&0&1&0&0\\0&1&0&0&1&0\\0&0&1&0&0&1\end{array}\right]\xrightarrow{R_1\leftarrow R_1-2R_2}\left[\begin{array}{ccc|ccc}1&0&0&1&-2&0\\0&1&0&0&1&0\\0&0&1&0&0&1\end{array}\right]\). Thus \(A^{-1}=\begin{pmatrix}1&-2&0\\0&1&0\\0&0&1\end{pmatrix}\) and \(AA^{-1}=I\).

- Arrange the vectors as columns of a square matrix. - What does a zero determinant indicate about the columns? - Compute each determinant carefully. - A nonzero determinant means the only linear combination producing the zero vector is the trivial one.

1. For part a, place the vectors in a matrix as columns. The determinant is \(\det\begin{pmatrix}4&-6\\-6&9\end{pmatrix}=4\cdot9-(-6)\cdot(-6)=36-36=0\). Therefore, the vectors are linearly dependent. 2. For part b, \(\det\begin{pmatrix}1&2&1\\2&0&1\\3&1&1\end{pmatrix}=3\ne0\). Therefore, the three vectors are linearly independent.

a) Linearly dependent b) Linearly independent

- What does it mean when one vector is a linear combination of the other two? - What geometric quantity does the scalar triple product represent? - Find \(\mathbf{w}\) before evaluating the determinant.

1. Because \(\mathbf{w}\) is a linear combination of \(\mathbf{u}\) and \(\mathbf{v}\), the three vectors are linearly dependent. Therefore, their scalar triple product must be \(0\). 2. First, \(\mathbf{w}=2\langle 1,0,2\rangle-\langle -1,3,1\rangle=\langle 3,-3,3\rangle\). 3. Then \(\det\begin{pmatrix}1&-1&3\\0&3&-3\\2&1&3\end{pmatrix}=12+6-18=0\), confirming the reasoning in part a.

a) The determinant must be \(0\) because the vectors are linearly dependent. b) \(\mathbf{w}=\langle 3,-3,3\rangle\) and \(\det(\mathbf{u},\mathbf{v},\mathbf{w})=0\).

- A nonzero determinant characterizes three independent vectors in \(\mathbb{R}^3\). - How does replacing a column by a linear combination affect the determinant? - How many linearly independent vectors can \(\mathbb{R}^3\) contain? - How does scaling determinant columns by nonzero constants affect the determinant?

1. Since \(\mathbf{u},\mathbf{v},\mathbf{w}\) are independent, \(\det(\mathbf{u},\mathbf{v},\mathbf{w})\ne0\). For part a, multilinearity gives \(\det(\mathbf{u},\mathbf{v},\mathbf{u}-\mathbf{v})=\det(\mathbf{u},\mathbf{v},\mathbf{u})-\det(\mathbf{u},\mathbf{v},\mathbf{v})=0\). Thus the set is linearly dependent. 2. Part b contains four vectors in \(\mathbb{R}^3\). At most three vectors can be linearly independent in a three-dimensional space, so the set is dependent. 3. If \(\mathbf{u}+\mathbf{v}\) were a scalar multiple of \(\mathbf{w}\), then \(\mathbf{u}\), \(\mathbf{v}\), and \(\mathbf{w}\) would satisfy a nontrivial linear relation. Therefore, \(\mathbf{u}+\mathbf{v}\) and \(\mathbf{w}\) are independent. 4. For part d, \(\det(2\mathbf{u},3\mathbf{v},4\mathbf{w})=2\cdot3\cdot4\det(\mathbf{u},\mathbf{v},\mathbf{w})=24\det(\mathbf{u},\mathbf{v},\mathbf{w})\ne0\). Therefore, \(2\mathbf{u},3\mathbf{v},4\mathbf{w}\) are independent.

a) Dependent: \(\mathbf{u}-\mathbf{v}\) is a combination of the first two vectors, so the determinant is \(0\). b) Dependent: four vectors in \(\mathbb{R}^3\) cannot all be independent. c) Independent: a relation between \(\mathbf{u}+\mathbf{v}\) and \(\mathbf w\) would contradict the independence of \(\mathbf u,\mathbf v,\mathbf w\). d) Independent: \(\det(2\mathbf u,3\mathbf v,4\mathbf w)=24\det(\mathbf u,\mathbf v,\mathbf w)\ne0\).

- For three vectors in \(\mathbb{R}^3\), connect linear independence with a nonzero determinant. - What happens to a determinant when one column is the zero vector? - Use linearity in one column to expand the determinant after replacing \(\mathbf{u}\) by \(\mathbf{u}+\mathbf{v}\). - Compare the number of vectors with the dimension of \(\mathbb{R}^3\).

1. Statement a is true. Three vectors in \(\mathbb{R}^3\) are linearly independent exactly when the determinant of the matrix having them as columns is nonzero. Therefore, they are dependent exactly when that determinant is \(0\). 2. Statement b is true. A matrix with the zero vector as one column has determinant \(0\), so its three column vectors are linearly dependent. 3. Statement c is true. Since \(\mathbf{u},\mathbf{v},\mathbf{w}\) are independent, \(\det(\mathbf{u},\mathbf{v},\mathbf{w})\ne0\). By linearity of the determinant in one column, \(\det(\mathbf{u}+\mathbf{v},\mathbf{v},\mathbf{w})=\det(\mathbf{u},\mathbf{v},\mathbf{w})+\det(\mathbf{v},\mathbf{v},\mathbf{w})=\det(\mathbf{u},\mathbf{v},\mathbf{w})\ne0\). Thus the new set is linearly independent. 4. Statement d is false. The dimension of \(\mathbb{R}^3\) is \(3\), so every set of four vectors in \(\mathbb{R}^3\) is linearly dependent.

a) True: three column vectors in \(\mathbb{R}^3\) are dependent exactly when their determinant is \(0\). b) True: the zero column makes the determinant \(0\). c) True: \(\det(\mathbf u+\mathbf v,\mathbf v,\mathbf w)=\det(\mathbf u,\mathbf v,\mathbf w)\ne0\). d) False: \(\mathbb{R}^3\) has dimension \(3\), so any four vectors are dependent.

- Form \(\overrightarrow{AB}\), \(\overrightarrow{AC}\), and \(\overrightarrow{AD}\) from the same starting point. - Put the three vectors into a \(3\times3\) determinant. Coplanarity requires that determinant to be \(0\). - After finding \(k\), use one plane point and two independent direction vectors for the parametric equation.

1. Direction vectors from \(A\) are \(\overrightarrow{AB}=\langle 2,-1,2\rangle\), \(\overrightarrow{AC}=\langle -1,2,-1\rangle\), and \(\overrightarrow{AD}=\langle k-2,1,3\rangle\). 2. The four points are coplanar exactly when the scalar triple product is \(0\). Using its determinant form, \(\det\begin{pmatrix}2&-1&k-2\\-1&2&1\\2&-1&3\end{pmatrix}=0\). 3. Expanding the determinant gives \(15-3k=0\), so \(k=5\). 4. A parametric equation of the plane through \(A\) with direction vectors \(\overrightarrow{AB}\) and \(\overrightarrow{AC}\) is \(\mathbf{r}=\langle 2,1,3\rangle+r\langle 2,-1,2\rangle+s\langle -1,2,-1\rangle\), where \(r,s\in\mathbb{R}\).

\(k=5\), and one plane equation is \(\mathbf{r}=\langle 2,1,3\rangle+r\langle 2,-1,2\rangle+s\langle -1,2,-1\rangle\), where \(r,s\in\mathbb{R}\).

- Form three displacement vectors from one common point and put them into a \(3\times3\) determinant. - A zero determinant means the three displacement vectors are linearly dependent and the four points are coplanar. - The absolute determinant is a parallelepiped volume; a tetrahedron has one-sixth of that volume.

1. Using \(P\) as the common starting point gives \(\overrightarrow{PQ}=\langle 2,1,3\rangle\), \(\overrightarrow{PR}=\langle -1,2,0\rangle\), and \(\overrightarrow{PS}=\langle 3,3,-1\rangle\). 2. The scalar triple product is the determinant \(\det\begin{pmatrix}2&-1&3\\1&2&3\\3&0&-1\end{pmatrix}=-32\). 3. Because this determinant is nonzero, the three displacement vectors are linearly independent, so the four points are not coplanar. 4. The absolute determinant, \(32\), is the volume of the corresponding parallelepiped. The tetrahedron has one-sixth that volume, so \(V=\frac{1}{6}\cdot32=\frac{16}{3}\) cubic units.

The points are not coplanar because the scalar triple product determinant is \(-32\ne0\). The tetrahedron's volume is \(\frac{16}{3}\) cubic units.

- What determinant value indicates dependent column vectors? - Arrange the vectors as columns of a \(3\times3\) matrix. - Simplify the determinant to a quadratic expression. - Factor or solve the quadratic equation.

1. The vectors are linearly dependent when the determinant of the matrix with these vectors as columns is \(0\). 2. Compute \(\det\begin{pmatrix}1&0&1\\2&k&1\\3&2&k\end{pmatrix}=k^2-3k+2\). 3. Factor: \(k^2-3k+2=(k-1)(k-2)\). 4. Therefore, the determinant is \(0\) when \(k=1\) or \(k=2\).

\(\det[\mathbf u\ \mathbf v\ \mathbf w]=k^2-3k+2=(k-1)(k-2)\). The vectors are dependent when \(k=1\) or \(k=2\).

- What determinant condition characterizes a basis of \(\mathbb{R}^3\)? - Compute the determinant as a polynomial in \(t\). - How could you factor the polynomial without guessing the final excluded parameter values? - Check your factorization by multiplying it back out.

1. The vectors form a basis exactly when the determinant of the matrix with these vectors as columns is nonzero. 2. Compute \(\det\begin{pmatrix}t&1&1\\1&t&1\\1&1&t\end{pmatrix}=t^3-3t+2\). 3. Factor the polynomial: \(t^3-3t+2=(t-1)^2(t+2)\). 4. The determinant is zero at \(t=1\) and \(t=-2\). Therefore, the vectors form a basis for every other real value of \(t\).

\(\det[\mathbf a\ \mathbf b\ \mathbf c]=t^3-3t+2=(t-1)^2(t+2)\). The vectors form a basis when \(t\in\mathbb{R}\setminus\{-2,1\}\).

- Place the three vectors as the columns of one matrix. - Expand the determinant using \(2\times2\) minors. - The absolute value of the determinant gives the volume scale factor in three dimensions.

1. The matrix is \(A=\begin{pmatrix}2&-1&1\\1&3&1\\2&0&5\end{pmatrix}\). 2. Expanding along the first row, \(\det(A)=2\begin{vmatrix}3&1\\0&5\end{vmatrix}-(-1)\begin{vmatrix}1&1\\2&5\end{vmatrix}+1\begin{vmatrix}1&3\\2&0\end{vmatrix}\). 3. Therefore, \(\det(A)=2\cdot15+(5-2)+(0-6)=27\). 4. The volume is the absolute value of the determinant: \(V=|\det(A)|=27\) cubic units.

a) \(\begin{pmatrix}2&-1&1\\1&3&1\\2&0&5\end{pmatrix}\) b) \(27\) c) \(27\) cubic units

- Use the dot-product area identity for the triangular base. - Put the three edge vectors into a \(3\times3\) matrix to find the volume. - A tetrahedron's volume is one-sixth the absolute value of that determinant. - Use \(V=\frac{1}{3}Gh\) to solve for the altitude.

1. For the base vectors, \(\|\mathbf{a}\|^2=9\), \(\|\mathbf{b}\|^2=9\), and \(\mathbf{a}\cdot\mathbf{b}=-4\). Thus, the base area is \(G=\frac{1}{2}\sqrt{9\cdot9-(-4)^2}=\frac{\sqrt{65}}{2}\). 2. Place the three edge vectors as columns: \(M=\begin{pmatrix}2&1&2\\-1&2&1\\2&-2&2\end{pmatrix}\). Since \(\det(M)=12\), \(V=\frac{1}{6}|\det(M)|=2\) cubic units. 3. Using \(V=\frac{1}{3}Gh\), \(h=\frac{3V}{G}=\frac{6}{\sqrt{65}/2}=\frac{12}{\sqrt{65}}\) units.

a) \(\frac{\sqrt{65}}{2}\) square units b) \(2\) cubic units c) \(\frac{12}{\sqrt{65}}\) units

- Write the three edge vectors from \(A\). - Check whether the two base vectors are perpendicular before finding the base area. - Use the three edge vectors as columns of a matrix for the volume. - Use the pyramid volume formula to solve for the altitude.

1. The edge vectors from \(A\) are \(\overrightarrow{AB}=\langle 3,0,4\rangle\), \(\overrightarrow{AC}=\langle 0,5,0\rangle\), and \(\overrightarrow{AD}=\langle 2,4,1\rangle\). 2. Because \(\overrightarrow{AB}\cdot\overrightarrow{AC}=0\), the base is a right triangle. Its area is \(G=\frac{1}{2}\cdot5\cdot5=\frac{25}{2}\) square units. 3. Form \(M=\begin{pmatrix}3&0&2\\0&5&4\\4&0&1\end{pmatrix}\). Since \(\det(M)=-25\), \(V=\frac{1}{6}|\det(M)|=\frac{25}{6}\) cubic units. 4. From \(V=\frac{1}{3}Gh_D\), \(h_D=\frac{3V}{G}=1\) unit.

a) \(\frac{25}{2}\) square units b) \(\frac{25}{6}\) cubic units c) \(1\) unit

- Put the three vectors into a \(3\times3\) matrix. - The volume is the absolute value of its determinant. - An absolute-value equation produces two cases. - Coplanar vectors form a parallelepiped with zero volume.

1. Place the vectors as columns: \(M_k=\begin{pmatrix}2&1&1\\1&3&k-1\\0&1&4\end{pmatrix}\). 2. The determinant is \(\det(M_k)=23-2k\). Therefore, \(V(k)=|23-2k|\). 3. For a volume of \(15\), \(|23-2k|=15\). Thus, \(23-2k=15\) gives \(k=4\), and \(23-2k=-15\) gives \(k=19\). 4. The points are coplanar when the volume is \(0\): \(23-2k=0\), so \(k=\frac{23}{2}\). At this value, the three vectors are linearly dependent.

a) \(V(k)=|23-2k|\) b) \(k=4\) or \(k=19\) c) \(k=\frac{23}{2}\); the volume is zero, so the vectors are coplanar.

- Use equal opposite-side vectors in the base parallelogram. - Every top vertex is obtained by the same translation vector \(\overrightarrow{AE}\). - Put three adjacent edge vectors into a matrix and take the absolute value of its determinant.

1. Since \(\overrightarrow{AD}=\overrightarrow{BC}\), \(\overrightarrow{BC}=\langle 1,3,0\rangle\), so \(D=A+\overrightarrow{BC}=(2, 4, 2)\). 2. The translation from the base to the top is \(\overrightarrow{AE}=\langle 1,1,4\rangle\). Therefore, \(F=B+\overrightarrow{AE}=(6, 2, 6)\), \(G=C+\overrightarrow{AE}=(7, 5, 6)\), and \(H=D+\overrightarrow{AE}=(3, 5, 6)\). 3. The three edge vectors from \(A\) are \(\overrightarrow{AB}=\langle 4,0,0\rangle\), \(\overrightarrow{AD}=\langle 1,3,0\rangle\), and \(\overrightarrow{AE}=\langle 1,1,4\rangle\). 4. Their matrix has determinant \(\det\begin{pmatrix}4&1&1\\0&3&1\\0&0&4\end{pmatrix}=48\). Thus, the volume is \(48\) cubic units.

a) \(D(2, 4, 2)\) b) \(F(6, 2, 6)\), \(G(7, 5, 6)\), \(H(3, 5, 6)\) c) \(48\) cubic units

- Use the base parallelogram's equal opposite-side vectors. - Put the three edge vectors from \(A\) into a \(3\times3\) matrix. - Find the base area from the two base vectors. - Use \(V=Gh\) to solve for the height.

1. In base parallelogram \(ABCD\), \(\overrightarrow{AD}=\overrightarrow{BC}=\langle -1,3,1\rangle\). Thus, \(D=A+\overrightarrow{BC}=(1, 3, 2)\). 2. The three edge vectors from \(A\) are \(\overrightarrow{AB}=\langle 3,1,0\rangle\), \(\overrightarrow{AD}=\langle -1,3,1\rangle\), and \(\overrightarrow{AE}=\langle 1,1,4\rangle\). 3. Their determinant is \(\det\begin{pmatrix}3&-1&1\\1&3&1\\0&1&4\end{pmatrix}=38\). Therefore, the volume is \(38\) cubic units. 4. For the base vectors, \(\|\overrightarrow{AB}\|^2=10\), \(\|\overrightarrow{AD}\|^2=11\), and \(\overrightarrow{AB}\cdot\overrightarrow{AD}=0\). Thus, the base area is \(G=\sqrt{10\cdot11}=\sqrt{110}\). 5. Since \(V=Gh\), \(h=\frac{38}{\sqrt{110}}\) units.

a) \(D(1, 3, 2)\) b) \(38\) cubic units c) \(\frac{38}{\sqrt{110}}\) units

- Use a dot product to test the angle at \(A\). - The two base sides are perpendicular. - Put the three edge vectors from \(A\) into a matrix. - Divide the determinant's absolute value by \(6\).

1. \(\overrightarrow{AB}=\langle 3,0,0\rangle\) and \(\overrightarrow{AC}=\langle 0,4,0\rangle\). Their dot product is \(0\), so \(\angle BAC=90^\circ\). 2. The base is a right triangle with leg lengths \(3\) and \(4\), so its area is \(\frac{1}{2}\cdot3\cdot4=6\) square units. 3. \(\overrightarrow{AD}=\langle 2,2,4\rangle\). The determinant of the edge-vector matrix is \(\det\begin{pmatrix}3&0&2\\0&4&2\\0&0&4\end{pmatrix}=48\). 4. Therefore, the tetrahedron's volume is \(\frac{1}{6}|48|=8\) cubic units.

a) \(90^\circ\) b) \(6\) square units c) \(8\) cubic units

- Notice that the three base points have the same \(z\)-coordinate. - Use a vector area identity for the base. - Use the three edge vectors from \(P\) as matrix columns. - Apply \(V=\frac{1}{3}Gh\).

1. \(\overrightarrow{PQ}=\langle 2,4,0\rangle\) and \(\overrightarrow{PR}=\langle -2,2,0\rangle\). Using the dot-product area identity gives \(G=\frac{1}{2}\sqrt{20\cdot8-4^2}=6\) square units. 2. \(\overrightarrow{PS}=\langle 1,1,6\rangle\), and \(\det\begin{pmatrix}2&-2&1\\4&2&1\\0&0&6\end{pmatrix}=72\). Thus, \(V=\frac{1}{6}|72|=12\) cubic units. 3. From \(V=\frac{1}{3}Gh\), \(h=\frac{3\cdot12}{6}=6\) units.

a) \(6\) square units b) \(12\) cubic units c) \(6\) units

- Put the edge vectors into a \(3\times3\) matrix. - Use dot products to check whether the edges are perpendicular. - Compare the three edge lengths.

1. Form the edge-vector matrix: \(M=\begin{pmatrix}4&-3&0\\0&0&5\\3&4&0\end{pmatrix}\). Its determinant is \(-125\), so the volume is \(125\) cubic units. 2. The pairwise dot products are \(\mathbf{a}\cdot\mathbf{b}=0\), \(\mathbf{a}\cdot\mathbf{c}=0\), and \(\mathbf{b}\cdot\mathbf{c}=0\). Thus, the edges are pairwise perpendicular. 3. Also, \(\|\mathbf{a}\|=\|\mathbf{b}\|=\|\mathbf{c}\|=5\). A rectangular prism with three equal edge lengths is a cube.

a) \(125\) cubic units b) A cube, because the three edge vectors are pairwise perpendicular and all have length \(5\).

- Write the parameterized coordinates of the fourth vertex. - Form three edge vectors from one base vertex. - Use a determinant and divide its absolute value by \(6\). - An absolute-value equation may have two solutions.

1. The fourth vertex is \(S_k=(2, 2, 1+k)\). 2. From \(A\), \(\overrightarrow{AB}=\langle 4,0,0\rangle\), \(\overrightarrow{AC}=\langle 0,3,0\rangle\), and \(\overrightarrow{AS_k}=\langle 1,1,1+k\rangle\). 3. The determinant is \(\det\begin{pmatrix}4&0&1\\0&3&1\\0&0&1+k\end{pmatrix}=12(1+k)\). 4. Therefore, \(V(k)=\frac{1}{6}|12(1+k)|=2|1+k|\). 5. Solve \(2|1+k|=10\), so \(|1+k|=5\). Thus, \(k=4\) or \(k=-6\).

The edge-matrix determinant is \(12(k+1)\). Since \(V(k)=2|k+1|\), the values are \(k=4\) or \(k=-6\).

- Use the three edge vectors from \(A\) in a determinant. - Use the dot-product area identity for the triangular base. - Solve \(V=\frac{1}{3}Gh\) for \(h\).

1. From \(A\), \(\overrightarrow{AB}=\langle 2,2,1\rangle\), \(\overrightarrow{AC}=\langle -1,4,2\rangle\), and \(\overrightarrow{AS}=\langle 4,5,10\rangle\). 2. The determinant is \(\det\begin{pmatrix}2&-1&4\\2&4&5\\1&2&10\end{pmatrix}=75\). Thus, \(V=\frac{1}{6}|75|=\frac{25}{2}\) cubic units. 3. For the base vectors, \(\|\overrightarrow{AB}\|^2=9\), \(\|\overrightarrow{AC}\|^2=21\), and \(\overrightarrow{AB}\cdot\overrightarrow{AC}=8\). Therefore, \(G=\frac{1}{2}\sqrt{9\cdot21-8^2}=\frac{5\sqrt{5}}{2}\) square units. 4. From \(V=\frac{1}{3}Gh\), \(h=\frac{3V}{G}=3\sqrt{5}\) units.

a) \(\frac{25}{2}\) cubic units b) Base area \(=\frac{5\sqrt{5}}{2}\) square units; \(h=3\sqrt{5}\) units

- Two square matrices are inverses when their products in both orders equal the identity matrix. - Write the relationship between the initial state, the next state, and the transition matrix. - Use the inverse matrix to undo multiplication by \(M\).

1. Verify both products: \(MA=\begin{pmatrix}0.6&0.2\\0.4&0.8\end{pmatrix}\begin{pmatrix}2&-0.5\\-1&1.5\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\), and \(AM=\begin{pmatrix}2&-0.5\\-1&1.5\end{pmatrix}\begin{pmatrix}0.6&0.2\\0.4&0.8\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). Therefore, \(A=M^{-1}\). 2. Since \(\mathbf{v}_1=M\mathbf{v}_0\), multiply by \(M^{-1}\): \(\mathbf{v}_0=M^{-1}\mathbf{v}_1=\begin{pmatrix}2&-0.5\\-1&1.5\end{pmatrix}\begin{pmatrix}40\\60\end{pmatrix}=\begin{pmatrix}80-30\\-40+90\end{pmatrix}=\begin{pmatrix}50\\50\end{pmatrix}\).

a) \(MA=AM=\begin{pmatrix}1&0\\0&1\end{pmatrix}\), so \(A=M^{-1}\). b) \(\mathbf{v}_0=\begin{pmatrix}50\\50\end{pmatrix}\).

- Multiply the two matrices using row-by-column multiplication. - Equal matrices have equal corresponding entries. - Separate the four equations into two systems with two unknowns each. - Solve each system by substitution or elimination.

1. Set up the equation \(\begin{pmatrix}4&7\\1&2\end{pmatrix}\begin{pmatrix}a&b\\c&d\end{pmatrix}=\begin{pmatrix}1&0\\0&1\end{pmatrix}\). 2. Equating corresponding entries gives two systems: \(4a+7c=1\) and \(a+2c=0\), \(4b+7d=0\) and \(b+2d=1\). 3. From \(a+2c=0\), \(a=-2c\). Substitution into \(4a+7c=1\) gives \(-8c+7c=1\), so \(c=-1\) and \(a=2\). 4. From \(b+2d=1\), \(b=1-2d\). Substitution into \(4b+7d=0\) gives \(4-8d+7d=0\), so \(d=4\) and \(b=-7\). Therefore, \(T^{-1}=\begin{pmatrix}2&-7\\-1&4\end{pmatrix}\).

The systems are \(\begin{cases}4a+7c=1\\a+2c=0\end{cases}\) and \(\begin{cases}4b+7d=0\\b+2d=1\end{cases}\). Their solutions are \(a=2\), \(c=-1\), \(b=-7\), and \(d=4\), so \(T^{-1}=\begin{pmatrix}2&-7\\-1&4\end{pmatrix}\).

- Invertibility of a square matrix can be decided from its determinant. - The determinant of a product is the product of the determinants. - Think about what multiplying a matrix by its inverse does to the determinant.

1. A square matrix is invertible when its determinant is nonzero. Since \(-2\ne0\) and \(5\ne0\), both \(A\) and \(B\) are invertible. 2. \(\det(AB)=\det(A)\det(B)=(-2)(5)=-10\), which is also nonzero, so \(AB\) is invertible. 3. For an invertible matrix \(M\), \(\det(M^{-1})=\frac{1}{\det(M)}\). 4. Therefore, \(\det\bigl((AB)^{-1}\bigr)=-\frac{1}{10}\).

a) All three are invertible because their determinants are nonzero. b) \(-10\) c) \(-\frac{1}{10}\)

- Work with the augmented matrix so every row operation affects both halves. - Look at the nonzero entries above the pivots. What must change for the left half to become the identity? - When the left half is the identity, check the right half by multiplication.

1. Begin with \(\left[\begin{array}{ccc|ccc}1&1&0&1&0&0\\0&1&1&0&1&0\\0&0&1&0&0&1\end{array}\right]\). 2. Apply \(R_1\leftarrow R_1-R_2\): \(\left[\begin{array}{ccc|ccc}1&0&-1&1&-1&0\\0&1&1&0&1&0\\0&0&1&0&0&1\end{array}\right]\). 3. Apply \(R_2\leftarrow R_2-R_3\) and \(R_1\leftarrow R_1+R_3\). The left side becomes \(I\), and the right side is \(A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}\). 4. Multiply to check \(AA^{-1}=I\).

\(R_1\leftarrow R_1-R_2\) gives \(\left[\begin{array}{ccc|ccc}1&0&-1&1&-1&0\\0&1&1&0&1&0\\0&0&1&0&0&1\end{array}\right]\). Then \(R_2\leftarrow R_2-R_3\) gives \(\left[\begin{array}{ccc|ccc}1&0&-1&1&-1&0\\0&1&0&0&1&-1\\0&0&1&0&0&1\end{array}\right]\). Finally \(R_1\leftarrow R_1+R_3\) gives \(\left[\begin{array}{ccc|ccc}1&0&0&1&-1&1\\0&1&0&0&1&-1\\0&0&1&0&0&1\end{array}\right]\). Thus \(A^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}\) and \(AA^{-1}=I\).

- Put the three vectors into a matrix and compute its determinant. - Consider which operations appear in a determinant of an integer matrix. - Use vector magnitudes and the dot product for the parallelogram area. - A zero determinant indicates linear dependence.

1. Place the vectors as columns: \(M_k=\begin{pmatrix}2&-1&1\\1&3&2\\0&1&k\end{pmatrix}\). 2. The determinant is \(\det(M_k)=7k-3\), so \(V(k)=|7k-3|\). 3. If \(k\) is an integer, then \(7k-3\) is an integer. Its absolute value is also an integer. 4. For \(\mathbf{a}\) and \(\mathbf{b}\), \(\|\mathbf{a}\|^2=5\), \(\|\mathbf{b}\|^2=11\), and \(\mathbf{a}\cdot\mathbf{b}=1\). Therefore, the parallelogram area is \(\sqrt{5\cdot11-1^2}=\sqrt{54}=3\sqrt{6}\). This is not an integer, so integer coordinates do not guarantee an integer area in three dimensions. 5. The volume is zero when \(7k-3=0\), so \(k=\frac{3}{7}\). At this value, the three vectors are linearly dependent and lie in one plane.

a) \(V(k)=|7k-3|\) b) For integer \(k\), \(7k-3\) and its absolute value are integers. c) \(3\sqrt{6}\) square units; no. d) \(k=\frac{3}{7}\); the vectors are coplanar and linearly dependent.

- Compare adjacent side lengths and their dot product. - Find a vector whose dot product with each base side vector is zero. - Use a determinant for the corresponding parallelepiped volume, then divide by \(3\). - Compare \(\overrightarrow{MS_k}\) with your perpendicular vector.

1. \(\overrightarrow{AB}=\langle 4,2,-4\rangle\) and \(\overrightarrow{AD}=\langle 4,-4,2\rangle\). Both have length \(6\), and their dot product is \(0\). Also, \(\overrightarrow{BC}=\overrightarrow{AD}\), so \(ABCD\) is a square. Its area is \(6^2=36\) square units. 2. The vector \(\mathbf{n}=\langle 1,2,2\rangle\) is perpendicular to both side vectors because \(\mathbf{n}\cdot\overrightarrow{AB}=0\) and \(\mathbf{n}\cdot\overrightarrow{AD}=0\). 3. Using the edge vectors as columns, \(\det\begin{pmatrix}4&4&4+k\\2&-4&-1+2k\\-4&2&-1+2k\end{pmatrix}=-108k\). A pyramid with a parallelogram base has one-third the corresponding parallelepiped's volume, so \(V(k)=\frac{1}{3}|-108k|=36|k|\). 4. \(\overrightarrow{MS_k}=\langle k,2k,2k\rangle=k\mathbf{n}\). For \(k\ne0\), this direction is normal to the base, so the line is perpendicular to the base. 5. When \(k=2\), \(h=\|\overrightarrow{MS_2}\|=\sqrt{2^2+4^2+4^2}=6\). Thus, \(V=\frac{1}{3}\cdot36\cdot6=72\), which agrees with \(V(2)=36|2|=72\).

a) \(ABCD\) is a square with area \(36\) square units. b) One perpendicular vector is \(\langle 1,2,2\rangle\). c) \(V(k)=36|k|\) d) \(\overrightarrow{MS_k}=k\langle 1,2,2\rangle\); for \(k=2\), \(V=72\) cubic units.

- Use the vectors as columns of a matrix. - A rectangular prism has pairwise perpendicular edge vectors. - Use dot products to impose the perpendicularity conditions. - Substitute the resulting parameter into the volume formula.

1. The determinant is \(\det\begin{pmatrix}2&2&t\\2&-1&6\\1&-2&-6\end{pmatrix}=72-3t\). Therefore, \(V(t)=|72-3t|\). 2. First, \(\mathbf{u}\cdot\mathbf{v}=0\). 3. For \(\mathbf{w}_t\) to be perpendicular to both, \(\mathbf{u}\cdot\mathbf{w}_t=2t+6=0\) and \(\mathbf{v}\cdot\mathbf{w}_t=2t+6=0\). Both equations give \(t=-3\). 4. Then \(V(-3)=|72-3(-3)|=81\) cubic units.

a) \(V(t)=|72-3t|\) b) \(t=-3\), and the volume is \(81\) cubic units.

- Form three vectors from \(A\) and use their determinant. - Use linearity of a determinant in one column. - Check the determinant formed with the movement direction. - A direction parallel to the base plane does not change perpendicular height.

1. From \(A\), \(\overrightarrow{AB}=\langle 3,-1,-2\rangle\), \(\overrightarrow{AC}=\langle 1,3,-1\rangle\), and \(\overrightarrow{AS}=\langle 5,7,7\rangle\). 2. The determinant is \(\det\begin{pmatrix}3&1&5\\-1&3&7\\-2&-1&7\end{pmatrix}=112\). Thus, \(V=\frac{1}{6}|112|=\frac{56}{3}\) cubic units. 3. For the moved vertex, \(\overrightarrow{AS_k}=\overrightarrow{AS}+k\mathbf{v}\). By linearity in the third column, \(\det(\overrightarrow{AB},\overrightarrow{AC},\overrightarrow{AS_k})=112+k\det(\overrightarrow{AB},\overrightarrow{AC},\mathbf{v})\). 4. The second determinant is \(\det\begin{pmatrix}3&1&1\\-1&3&-7\\-2&-1&0\end{pmatrix}=0\). Therefore, the determinant remains \(112\), so the volume remains \(\frac{56}{3}\). 5. A zero determinant with \(\overrightarrow{AB}\), \(\overrightarrow{AC}\), and \(\mathbf{v}\) means \(\mathbf{v}\) lies parallel to the base plane. Moving the vertex in that direction does not change its perpendicular distance from the base.

a) \(\frac{56}{3}\) cubic units b) The determinant remains \(112\), so the volume is constant. c) \(\mathbf{v}\) is parallel to the base plane, so the pyramid's height does not change.

- Write the general \(2\times2\) inverse before using the given inverse entry. - Identify which expression becomes the row-1, column-2 entry. - After solving for the parameter, check that the determinant is not zero.

1. The determinant is \(3a-2\), so for an invertible matrix \(A_a^{-1}=\frac{1}{3a-2}\begin{pmatrix}3&-2\\-1&a\end{pmatrix}\). 2. The row-1, column-2 entry is \(\frac{-2}{3a-2}\). Set \(\frac{-2}{3a-2}=-\frac12\). 3. Cross-multiplying gives \(-4=-(3a-2)\), so \(a=2\). 4. For \(a=2\), \(\det(A_a)=3(2)-2=4\ne0\), so the matrix is invertible.

\(a=2\); the determinant is \(4\), so the resulting matrix is invertible.