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Paper 1 Answers

Time2 hours 30 minutes

Marks90

SetSet 1

PaperLevel 1 - Linear Algebra Paper 1

Information

Section A: Vectors In Rn
Section B: Matrix Algebra
Section C: Linear Systems
Section D: Determinants
Section E: Subspaces and Dimension
Section F: Linear Maps and Eigenvalues

Marking Guidance

These worked answers show the expected method and final result. Equivalent correct reasoning should receive credit.

Section A: Vectors In Rn

Answer 1. Vector geometry and planes

[15 marks]

This question concerns vector operations, scalar products, vector products, and planes in \(\mathbb R^3\).

a) Let \(\mathbf u=(2,-1,3)\) and \(\mathbf v=(-1,4,2)\). Compute \(2\mathbf u-\mathbf v\).

[3 marks]

Compute the scalar multiple first: \(2\mathbf u=(4,-2,6)\). Then subtract component by component: \(2\mathbf u-\mathbf v=(4,-2,6)-(-1,4,2)=(5,-6,4)\). Therefore \(2\mathbf u-\mathbf v=(5,-6,4)\).

Answer 1. Vector geometry and planes

[15 marks]

b) Use a scalar product to decide whether \(\mathbf u\) and \(\mathbf v\) from part (a) are orthogonal.

[3 marks]

Use the scalar product: \(\mathbf u\cdot\mathbf v=2(-1)+(-1)4+3(2)=-2-4+6=0\). Since both vectors are non-zero and their scalar product is zero, \(\mathbf u\) and \(\mathbf v\) are orthogonal. The angle between them is \(\pi/2\).

Answer 1. Vector geometry and planes

[15 marks]

c) Compute \(\mathbf u\times\mathbf v\) for the vectors from part (a), and verify that the result is perpendicular to both \(\mathbf u\) and \(\mathbf v\).

[5 marks]

Use the component formula for the vector product. For \(\mathbf u=(2,-1,3)\) and \(\mathbf v=(-1,4,2)\), \(\mathbf u\times\mathbf v=(( -1)2-3(4),\,3(-1)-2(2),\,2(4)-(-1)(-1))=(-14,-7,7)\). Check with \(\mathbf u\): \((-14,-7,7)\cdot(2,-1,3)=-28+7+21=0\). Check with \(\mathbf v\): \((-14,-7,7)\cdot(-1,4,2)=14-28+14=0\). Therefore \(\mathbf u\times\mathbf v=(-14,-7,7)\), perpendicular to both input vectors.

Answer 1. Vector geometry and planes

[15 marks]

d) Use your vector product from part (c) as a normal vector to find an equation of the plane through \(P=(1,0,-2)\).

[4 marks]

A plane with normal vector \(\mathbf n=(-14,-7,7)\) through \(P=(1,0,-2)\) satisfies \(\mathbf n\cdot((x,y,z)-P)=0\). Thus \(-14(x-1)-7(y-0)+7(z+2)=0\). Expanding gives \(-14x+14-7y+7z+14=0\), so \(-14x-7y+7z+28=0\). Dividing by \(7\) gives \(-2x-y+z+4=0\), equivalently \(2x+y-z=4\).

Section B: Matrix Algebra

Answer 2. Matrix products and inverses

[15 marks]

This question concerns matrix multiplication, transpose rules, inverse matrices, and non-commutativity.

a) Let \(A=\begin{pmatrix}1&0&-1\\2&1&3\end{pmatrix}\) and \(B=\begin{pmatrix}2&3\\1&-2\\0&4\end{pmatrix}\). Compute \(AB\).

[4 marks]

The product is defined because \(A\) is \(2\) by \(3\) and \(B\) is \(3\) by \(2\). The result is \(2\) by \(2\). Compute row-column products: entry \((1,1)\) is \(1(2)+0(1)+(-1)(0)=2\); entry \((1,2)\) is \(1(3)+0(-2)+(-1)(4)=-1\); entry \((2,1)\) is \(2(2)+1(1)+3(0)=5\); entry \((2,2)\) is \(2(3)+1(-2)+3(4)=16\). Therefore \(AB=\begin{pmatrix}2&-1\\5&16\end{pmatrix}\).

Answer 2. Matrix products and inverses

[15 marks]

b) State the transpose rule for a product and use part (a) to find \((AB)^T\).

[3 marks]

First compute \(B^T\), which is \(\begin{pmatrix}2&1&0\\3&-2&4\end{pmatrix}\). Also \(A^T=\begin{pmatrix}1&2\\0&1\\-1&3\end{pmatrix}\). The transpose rule gives \((AB)^T=B^TA^T\). Since \(AB=\begin{pmatrix}2&-1\\5&16\end{pmatrix}\), \((AB)^T=\begin{pmatrix}2&5\\-1&16\end{pmatrix}\). This matches \(B^TA^T\), confirming the reversed order in the transpose rule.

Answer 2. Matrix products and inverses

[15 marks]

c) Use an inverse matrix to solve \(\begin{pmatrix}3&1\\2&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}4\\3\end{pmatrix}\).

[5 marks]

Let \(C=\begin{pmatrix}3&1\\2&1\end{pmatrix}\). Its determinant is \(3(1)-1(2)=1\), so \(C\) is invertible. The inverse is \(C^{-1}=\begin{pmatrix}1&-1\\-2&3\end{pmatrix}\). For \(C\mathbf x=\begin{pmatrix}4\\3\end{pmatrix}\), multiply by \(C^{-1}\): \(\mathbf x=C^{-1}\begin{pmatrix}4\\3\end{pmatrix}=\begin{pmatrix}1(4)-1(3)\\-2(4)+3(3)\end{pmatrix}=\begin{pmatrix}1\\1\end{pmatrix}\). Therefore \(x=1\) and \(y=1\).

Answer 2. Matrix products and inverses

[15 marks]

d) Give a concrete example showing that matrix multiplication is not commutative.

[3 marks]

Matrix multiplication is not generally commutative. For example, with \(D=\begin{pmatrix}1&1\\0&1\end{pmatrix}\) and \(E=\begin{pmatrix}1&0\\2&1\end{pmatrix}\), \(DE=\begin{pmatrix}3&1\\2&1\end{pmatrix}\), while \(ED=\begin{pmatrix}1&1\\2&3\end{pmatrix}\). Since these products are different, \(DE\ne ED\).

Section C: Linear Systems

Answer 3. Solving and classifying systems

[15 marks]

This question uses augmented matrices, row reduction, and solution-set interpretation.

a) Write the augmented matrix for the system \(x+y+z=3\), \(2x+3y+5z=8\), \(x+2y+4z=6\).

[3 marks]

The coefficient matrix is \(A=\begin{pmatrix}1&1&1\\2&3&5\\1&2&4\end{pmatrix}\), the unknown vector is \(\mathbf x=\begin{pmatrix}x\\y\\z\end{pmatrix}\), and the right-hand side is \(\mathbf b=\begin{pmatrix}3\\8\\6\end{pmatrix}\). The augmented matrix is \(\left(\begin{array}{ccc|c}1&1&1&3\\2&3&5&8\\1&2&4&6\end{array}\right)\).

Answer 3. Solving and classifying systems

[15 marks]

b) Row-reduce the augmented matrix from part (a) enough to classify the system.

[5 marks]

Start from \(\left(\begin{array}{ccc|c}1&1&1&3\\2&3&5&8\\1&2&4&6\end{array}\right)\). Use \(R_2\leftarrow R_2-2R_1\) and \(R_3\leftarrow R_3-R_1\) to get \(\left(\begin{array}{ccc|c}1&1&1&3\\0&1&3&2\\0&1&3&3\end{array}\right)\). Then \(R_3\leftarrow R_3-R_2\) gives \(\left(\begin{array}{ccc|c}1&1&1&3\\0&1&3&2\\0&0&0&1\end{array}\right)\). The last row means \(0=1\), a contradiction. Therefore the system is inconsistent and has no solution.

Answer 3. Solving and classifying systems

[15 marks]

c) Replace the third equation by \(x+2y+4z=5\). Find the solution set of the new system.

[5 marks]

For the new system, row reduction gives \(\left(\begin{array}{ccc|c}1&1&1&3\\0&1&3&2\\0&0&0&0\end{array}\right)\). The variables \(x\) and \(y\) are pivot variables, and \(z\) is free. Let \(z=t\). The second row gives \(y+3t=2\), so \(y=2-3t\). The first row gives \(x+y+t=3\), so \(x=3-(2-3t)-t=1+2t\). Therefore the solution set is \((x,y,z)=(1,2,0)+t(2,-3,1)\), where \(t\in\mathbb R\).

Answer 3. Solving and classifying systems

[15 marks]

d) Explain the difference between a pivot variable and a free variable.

[2 marks]

A pivot column is a column containing a leading entry in echelon or reduced echelon form. A variable whose column is a pivot column is a pivot variable. Any non-pivot variable is free, meaning it can be chosen as a parameter. In part (c), the column for \(z\) has no pivot, so \(z\) is free.

Section D: Determinants

Answer 4. Determinants and their consequences

[15 marks]

This question concerns determinant calculation, determinant properties, geometry, and invertibility.

a) Compute \(\det A\) for \(A=\begin{pmatrix}2&1&0\\0&3&1\\1&2&4\end{pmatrix}\).

[4 marks]

Expand along the first row: \(\det A=2\det\begin{pmatrix}3&1\\2&4\end{pmatrix}-1\det\begin{pmatrix}0&1\\1&4\end{pmatrix}+0\). The first minor is \(3(4)-1(2)=10\). The second minor is \(0(4)-1(1)=-1\). Therefore \(\det A=2(10)-1(-1)=21\).

Answer 4. Determinants and their consequences

[15 marks]

b) Use your determinant from part (a) to decide whether \(A\) is invertible and whether its columns are linearly independent.

[3 marks]

A square matrix is invertible exactly when its determinant is non-zero. Since \(\det A=21\ne0\), the matrix \(A\) is invertible. The columns of \(A\) are also linearly independent because a non-zero determinant means the column vectors span non-zero three-dimensional volume.

Answer 4. Determinants and their consequences

[15 marks]

c) Let \(B\) be obtained from \(A\) by swapping two rows, and let \(C\) be obtained from \(A\) by multiplying one row by \(5\). Find \(\det B\) and \(\det C\).

[4 marks]

If \(B\) is obtained by swapping two rows of \(A\), then \(\det B=-\det A\). Since \(\det A=21\), \(\det B=-21\). If \(C\) is obtained by multiplying one row of \(A\) by \(5\), then \(\det C=5\det A=105\).

Answer 4. Determinants and their consequences

[15 marks]

d) Find the area of the parallelogram spanned by \(\mathbf p=(3,-1)\) and \(\mathbf q=(1,4)\).

[4 marks]

The parallelogram area is the absolute value of the determinant with the vectors as columns: \(\left|\det\begin{pmatrix}3&1\\-1&4\end{pmatrix}\right|\). The determinant is \(3(4)-1(-1)=12+1=13\). Therefore the area is \(13\) square units.

Section E: Subspaces and Dimension

Answer 5. Subspaces, bases, and coordinates

[15 marks]

This question concerns subspaces, spans, independence, bases, and coordinates.

a) Show that \(W=\{(x,y,z)\in\mathbb R^3:x-2y+z=0\}\) is a subspace of \(\mathbb R^3\).

[5 marks]

The zero vector satisfies \(0-2(0)+0=0\), so \(\mathbf0\in W\). If \(\mathbf u=(x_1,y_1,z_1)\in W\) and \(\mathbf v=(x_2,y_2,z_2)\in W\), then \(x_1-2y_1+z_1=0\) and \(x_2-2y_2+z_2=0\). For scalars \(\lambda,\mu\), the defining expression for \(\lambda\mathbf u+\mu\mathbf v\) is \(\lambda(x_1-2y_1+z_1)+\mu(x_2-2y_2+z_2)=0\). Therefore every linear combination stays in \(W\), so \(W\) is a subspace of \(\mathbb R^3\).

Answer 5. Subspaces, bases, and coordinates

[15 marks]

b) Find a basis for \(W\) and state \(\dim W\).

[5 marks]

Solve the defining equation for \(x\): \(x=2y-z\). Let \(y=s\) and \(z=t\). Then \((x,y,z)=(2s-t,s,t)=s(2,1,0)+t(-1,0,1)\). Thus \(W=\operatorname{span}((2,1,0),(-1,0,1))\). These two vectors are not scalar multiples, so they are linearly independent. Therefore they form a basis for \(W\), and \(\dim W=2\).

Answer 5. Subspaces, bases, and coordinates

[15 marks]

c) Decide whether \((3,2,-1)\) lies in the span of your basis from part (b).

[3 marks]

A vector lies in the span exactly when it can be written as \(a(2,1,0)+b(-1,0,1)\). This gives \((2a-b,a,b)=(3,2,-1)\). From the second component, \(a=2\). From the third component, \(b=-1\). The first component then gives \(2(2)-(-1)=5\), not \(3\). The equations are inconsistent, so \((3,2,-1) otin W\). Equivalently, \(3-2(2)+(-1)=-2\ne0\).

Answer 5. Subspaces, bases, and coordinates

[15 marks]

d) Explain why \(S=\{(x,y,z):x-2y+z=1\}\) is not a subspace of \(\mathbb R^3\).

[2 marks]

The set \(S\) is not a subspace because a subspace must contain the zero vector. For \(\mathbf0=(0,0,0)\), the defining expression gives \(0-2(0)+0=0\), not \(1\). Therefore \(\mathbf0 otin S\), so \(S\) is not a subspace.

Section F: Linear Maps and Eigenvalues

Answer 6. Linear maps and eigenvalue structure

[15 marks]

This question concerns matrix representations, kernels, images, rank-nullity, and diagonalisation.

a) Let \(T:\mathbb R^2\to\mathbb R^2\) be defined by \(T(x,y)=(2x-y,x+3y)\). Find the standard matrix of \(T\).

[3 marks]

The standard matrix has columns \(T(1,0)\) and \(T(0,1)\). We have \(T(1,0)=(2,1)\) and \(T(0,1)=(-1,3)\). Therefore the matrix representation is \(A=\begin{pmatrix}2&-1\\1&3\end{pmatrix}\).

Answer 6. Linear maps and eigenvalue structure

[15 marks]

b) Find \(\ker T\), \(\operatorname{im}T\), the rank, and the nullity.

[5 marks]

The kernel solves \(A\begin{pmatrix}x\\y\end{pmatrix}=\mathbf0\), so \(2x-y=0\) and \(x+3y=0\). From the first equation, \(y=2x\). Substitute into the second: \(x+6x=7x=0\), so \(x=0\) and \(y=0\). Thus \(\ker T=\{\mathbf0\}\). The determinant of \(A\) is \(2(3)-(-1)(1)=7\ne0\), so \(A\) is invertible and \(\operatorname{im}T=\mathbb R^2\). Therefore rank is \(2\) and nullity is \(0\).

Answer 6. Linear maps and eigenvalue structure

[15 marks]

c) Find the eigenvalues of \(M=\begin{pmatrix}4&1\\0&2\end{pmatrix}\) and one eigenvector for each eigenvalue.

[5 marks]

For \(M=\begin{pmatrix}4&1\\0&2\end{pmatrix}\), the characteristic polynomial is \(\det(M-\lambda I)=(4-\lambda)(2-\lambda)\). Therefore the eigenvalues are \(\lambda=4\) and \(\lambda=2\). For \(\lambda=4\), \(M-4I=\begin{pmatrix}0&1\\0&-2\end{pmatrix}\), so \(y=0\); one eigenvector is \((1,0)\). For \(\lambda=2\), \(M-2I=\begin{pmatrix}2&1\\0&0\end{pmatrix}\), so \(2x+y=0\); one eigenvector is \((1,-2)\).

Answer 6. Linear maps and eigenvalue structure

[15 marks]

d) Decide whether \(M\) from part (c) is diagonalizable, and give suitable matrices \(P\) and \(D\) if it is.

[2 marks]

The two eigenvalues are distinct, so their eigenvectors are linearly independent. A \(2\) by \(2\) matrix with two linearly independent eigenvectors is diagonalizable. With \(P=\begin{pmatrix}1&1\\0&-2\end{pmatrix}\) and \(D=\begin{pmatrix}4&0\\0&2\end{pmatrix}\), using the eigenvectors in matching order, \(M=PDP^{-1}\).