Browse papers
A

Section A: Long Answer Questions

Attempt any TWO questions.

3 questions·10 marks each
1long10 marks

Define a vector space. State and verify the axioms of a vector space. Show that the set of all 2x2 matrices forms a vector space over the field of real numbers.

Vector Space

A vector space VV over a field FF (here F=RF=\mathbb{R}) is a non-empty set equipped with two operations, vector addition (+:V×VV)(+:V\times V\to V) and scalar multiplication (:F×VV)(\cdot:F\times V\to V), satisfying the following axioms for all u,v,wVu,v,w\in V and α,βF\alpha,\beta\in F.

Axioms

Closure

  • u+vVu+v\in V
  • αuV\alpha u\in V

Additive axioms

  1. Commutativity: u+v=v+uu+v=v+u
  2. Associativity: (u+v)+w=u+(v+w)(u+v)+w=u+(v+w)
  3. Additive identity: there exists 0V0\in V with u+0=uu+0=u
  4. Additive inverse: for each uu there exists uV-u\in V with u+(u)=0u+(-u)=0

Scalar axioms 5. Distributivity over vector addition: α(u+v)=αu+αv\alpha(u+v)=\alpha u+\alpha v 6. Distributivity over scalar addition: (α+β)u=αu+βu(\alpha+\beta)u=\alpha u+\beta u 7. Associativity of scalars: α(βu)=(αβ)u\alpha(\beta u)=(\alpha\beta)u 8. Multiplicative identity: 1u=u1\cdot u=u

The set of all 2×22\times2 real matrices M2(R)M_{2}(\mathbb{R})

Let V=M2(R)={[abcd]:a,b,c,dR}V=M_{2}(\mathbb{R})=\left\{\begin{bmatrix}a&b\\c&d\end{bmatrix}: a,b,c,d\in\mathbb{R}\right\} with usual matrix addition and scalar multiplication.

Closure: Sum of two 2×22\times2 matrices is a 2×22\times2 matrix; a scalar multiple of a 2×22\times2 matrix is a 2×22\times2 matrix. \checkmark

Commutativity & Associativity: Entrywise addition of real numbers is commutative and associative, so matrix addition is too. \checkmark

Additive identity: The zero matrix 0=[0000]0=\begin{bmatrix}0&0\\0&0\end{bmatrix} satisfies A+0=AA+0=A. \checkmark

Additive inverse: For A=[abcd]A=\begin{bmatrix}a&b\\c&d\end{bmatrix}, A=[abcd]V-A=\begin{bmatrix}-a&-b\\-c&-d\end{bmatrix}\in V and A+(A)=0A+(-A)=0. \checkmark

Scalar axioms (5–8): Since each entry is a real number, the distributive, associative and identity laws of R\mathbb{R} give

α(A+B)=αA+αB,(α+β)A=αA+βA,\alpha(A+B)=\alpha A+\alpha B,\quad(\alpha+\beta)A=\alpha A+\beta A, α(βA)=(αβ)A,1A=A. \alpha(\beta A)=(\alpha\beta)A,\quad 1\cdot A=A.\ \checkmark

All eight axioms hold, so M2(R)M_{2}(\mathbb{R}) is a vector space over R\mathbb{R} (it has dimension 4).

vector-spaceaxioms
2long10 marks

Define linear dependence and independence of vectors. Determine whether a given set of vectors is linearly independent. Find a basis and the dimension of the subspace they span.

Linear Dependence and Independence

A set of vectors {v1,v2,,vn}\{v_1,v_2,\dots,v_n\} in a vector space VV is linearly independent if the only scalars c1,,cnc_1,\dots,c_n satisfying

c1v1+c2v2++cnvn=0c_1v_1+c_2v_2+\cdots+c_nv_n=0

are c1=c2==cn=0c_1=c_2=\cdots=c_n=0. If a non-trivial solution exists, the set is linearly dependent.

Worked Example

Test v1=(1,2,3), v2=(2,1,0), v3=(3,3,3)v_1=(1,2,3),\ v_2=(2,1,0),\ v_3=(3,3,3) in R3\mathbb{R}^3.

Form the matrix with these as rows and reduce:

[123210333]R22R1, R33R1[123036036]R3R2[123036000]\begin{bmatrix}1&2&3\\2&1&0\\3&3&3\end{bmatrix}\xrightarrow{R_2-2R_1,\ R_3-3R_1}\begin{bmatrix}1&2&3\\0&-3&-6\\0&-3&-6\end{bmatrix}\xrightarrow{R_3-R_2}\begin{bmatrix}1&2&3\\0&-3&-6\\0&0&0\end{bmatrix}

A zero row appears, so the vectors are linearly dependent. Indeed v3=v1+v2v_3=v_1+v_2.

Basis and Dimension of the spanned subspace

The non-zero rows of the echelon form are independent, so a basis of W=span{v1,v2,v3}W=\text{span}\{v_1,v_2,v_3\} is

{(1,2,3), (0,3,6)}equivalently{v1, v2}.\{(1,2,3),\ (0,-3,-6)\}\quad\text{equivalently}\quad\{v_1,\ v_2\}.

Hence dimW=2\dim W = 2 (the rank of the matrix). The subspace is a plane through the origin in R3\mathbb{R}^3.

linear-independencebasis
3long10 marks

Define a linear transformation. Show that a given mapping is a linear transformation and find its matrix representation with respect to standard bases.

Linear Transformation

Let VV and WW be vector spaces over a field FF. A map T:VWT:V\to W is a linear transformation if for all u,vVu,v\in V and scalar αF\alpha\in F:

T(u+v)=T(u)+T(v)(additivity)T(u+v)=T(u)+T(v)\quad(\text{additivity}) T(αu)=αT(u)(homogeneity)T(\alpha u)=\alpha\,T(u)\quad(\text{homogeneity})

Equivalently, T(αu+βv)=αT(u)+βT(v)T(\alpha u+\beta v)=\alpha T(u)+\beta T(v).

Example: verify linearity

Let T:R2R2T:\mathbb{R}^2\to\mathbb{R}^2 be T(x,y)=(x+2y, 3xy)T(x,y)=(x+2y,\ 3x-y).

Take u=(x1,y1), v=(x2,y2)u=(x_1,y_1),\ v=(x_2,y_2) and scalars α,β\alpha,\beta. Then αu+βv=(αx1+βx2, αy1+βy2)\alpha u+\beta v=(\alpha x_1+\beta x_2,\ \alpha y_1+\beta y_2) and

T(αu+βv)=((αx1+βx2)+2(αy1+βy2), 3(αx1+βx2)(αy1+βy2))T(\alpha u+\beta v)=\big((\alpha x_1+\beta x_2)+2(\alpha y_1+\beta y_2),\ 3(\alpha x_1+\beta x_2)-(\alpha y_1+\beta y_2)\big) =α(x1+2y1,3x1y1)+β(x2+2y2,3x2y2)=αT(u)+βT(v).=\alpha(x_1+2y_1,\,3x_1-y_1)+\beta(x_2+2y_2,\,3x_2-y_2)=\alpha T(u)+\beta T(v).

Hence TT is linear.

Matrix representation (standard bases)

Apply TT to the standard basis e1=(1,0), e2=(0,1)e_1=(1,0),\ e_2=(0,1):

T(e1)=(1,3),T(e2)=(2,1).T(e_1)=(1,3),\qquad T(e_2)=(2,-1).

These become the columns of the matrix:

[T]=[1231].[T]=\begin{bmatrix}1&2\\3&-1\end{bmatrix}.

Check: [T][xy]=[x+2y3xy][T]\begin{bmatrix}x\\y\end{bmatrix}=\begin{bmatrix}x+2y\\3x-y\end{bmatrix}, matching T(x,y)T(x,y). \checkmark

linear-transformation
B

Section B: Short Answer Questions

Attempt any EIGHT questions.

9 questions·5 marks each
4short5 marks

Define a vector space with two examples.

A vector space over a field FF is a set VV with vector addition and scalar multiplication satisfying the eight axioms (closure under both operations, commutativity and associativity of addition, additive identity 00 and inverses, the two distributive laws, associativity of scalars, and 1v=v1\cdot v=v).

Examples:

  1. Rn\mathbb{R}^n — the set of all real nn-tuples with componentwise addition and scalar multiplication.
  2. PnP_n — the set of all polynomials of degree n\le n with real coefficients, under ordinary polynomial addition and scalar multiplication. (Also valid: Mm×n(R)M_{m\times n}(\mathbb{R}), the space of m×nm\times n real matrices.)
vector-space
5short5 marks

Find the inverse of [[1,2],[3,4]].

For A=[1234]A=\begin{bmatrix}1&2\\3&4\end{bmatrix}, the determinant is

detA=(1)(4)(2)(3)=46=20,\det A = (1)(4)-(2)(3)=4-6=-2\neq 0,

so AA is invertible. Using A1=1detA[dbca]A^{-1}=\dfrac{1}{\det A}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}:

A1=12[4231]=[213212].A^{-1}=\frac{1}{-2}\begin{bmatrix}4&-2\\-3&1\end{bmatrix}=\begin{bmatrix}-2&1\\\tfrac{3}{2}&-\tfrac{1}{2}\end{bmatrix}.

Check: AA1=[1234][211.50.5]=[1001]. A A^{-1}=\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}-2&1\\1.5&-0.5\end{bmatrix}=\begin{bmatrix}1&0\\0&1\end{bmatrix}.\ \checkmark

matrix
6short5 marks

Define a linear transformation.

A linear transformation is a map T:VWT:V\to W between vector spaces over the same field FF that preserves the vector-space operations; that is, for all u,vVu,v\in V and every scalar αF\alpha\in F:

T(u+v)=T(u)+T(v)andT(αu)=αT(u).T(u+v)=T(u)+T(v)\qquad\text{and}\qquad T(\alpha u)=\alpha\,T(u).

Equivalently, T(αu+βv)=αT(u)+βT(v)T(\alpha u+\beta v)=\alpha T(u)+\beta T(v). A consequence is T(0V)=0WT(0_V)=0_W. Example: T:R2R2, T(x,y)=(x+y, xy)T:\mathbb{R}^2\to\mathbb{R}^2,\ T(x,y)=(x+y,\ x-y).

linear-transformation
7short5 marks

State the rank-nullity theorem.

Rank–Nullity Theorem

Let T:VWT:V\to W be a linear transformation where VV is a finite-dimensional vector space. Then

rank(T)+nullity(T)=dimV,\operatorname{rank}(T)+\operatorname{nullity}(T)=\dim V,

where rank(T)=dim(Im T)\operatorname{rank}(T)=\dim(\text{Im }T) is the dimension of the image (range) and nullity(T)=dim(kerT)\operatorname{nullity}(T)=\dim(\ker T) is the dimension of the kernel (null space).

For an m×nm\times n matrix AA representing TT, this reads rank(A)+nullity(A)=n\operatorname{rank}(A)+\operatorname{nullity}(A)=n (number of columns).

rank-nullity
8short5 marks

Define an orthonormal set of vectors.

An orthonormal set of vectors {v1,v2,,vn}\{v_1,v_2,\dots,v_n\} in an inner-product space is a set that is both orthogonal and normalized, i.e.

vi,vj=δij={1,i=j0,ij.\langle v_i,\,v_j\rangle=\delta_{ij}=\begin{cases}1,&i=j\\0,&i\neq j.\end{cases}

This means every pair of distinct vectors is orthogonal (vi,vj=0\langle v_i,v_j\rangle=0 for iji\neq j) and each vector is a unit vector (vi=1\|v_i\|=1).

Example: the standard basis {(1,0,0),(0,1,0),(0,0,1)}\{(1,0,0),(0,1,0),(0,0,1)\} of R3\mathbb{R}^3 is orthonormal.

orthogonal
9short5 marks

What is a characteristic equation of a matrix?

For a square matrix AA of order nn, the characteristic equation is

det(AλI)=0,\det(A-\lambda I)=0,

where λ\lambda is a scalar and II is the n×nn\times n identity matrix. The left side, p(λ)=det(AλI)p(\lambda)=\det(A-\lambda I), is the characteristic polynomial (a degree-nn polynomial in λ\lambda). Its roots are the eigenvalues of AA.

Example: for A=[2112]A=\begin{bmatrix}2&1\\1&2\end{bmatrix}, det(AλI)=(2λ)21=λ24λ+3=0\det(A-\lambda I)=(2-\lambda)^2-1=\lambda^2-4\lambda+3=0, giving eigenvalues λ=1,3\lambda=1,3.

eigenvalues
10short5 marks

Define a subspace with an example.

A subspace WW of a vector space VV over a field FF is a non-empty subset WVW\subseteq V that is itself a vector space under the operations of VV. Equivalently, WW is a subspace iff:

  1. 0W0\in W (it contains the zero vector),
  2. WW is closed under addition: u,vWu+vWu,v\in W\Rightarrow u+v\in W,
  3. WW is closed under scalar multiplication: αF, uWαuW\alpha\in F,\ u\in W\Rightarrow \alpha u\in W.

Example: W={(x,y,0):x,yR}W=\{(x,y,0):x,y\in\mathbb{R}\}, the xyxy-plane, is a subspace of R3\mathbb{R}^3 — it contains (0,0,0)(0,0,0) and is closed under addition and scalar multiplication.

subspace
11short5 marks

Find the trace of the matrix [[1,2,3],[4,5,6],[7,8,9]].

The trace of a square matrix is the sum of its main-diagonal (top-left to bottom-right) entries. For

A=[123456789],A=\begin{bmatrix}1&2&3\\4&5&6\\7&8&9\end{bmatrix},

the diagonal entries are 1,5,91,5,9, so

tr(A)=1+5+9=15.\operatorname{tr}(A)=1+5+9=15.
matrix
12short5 marks

State the Cayley-Hamilton theorem.

Cayley–Hamilton Theorem

Every square matrix satisfies its own characteristic equation. That is, if AA is an n×nn\times n matrix with characteristic polynomial

p(λ)=det(AλI)=λn+cn1λn1++c1λ+c0,p(\lambda)=\det(A-\lambda I)=\lambda^n+c_{n-1}\lambda^{n-1}+\cdots+c_1\lambda+c_0,

then substituting AA for λ\lambda gives the zero matrix:

p(A)=An+cn1An1++c1A+c0I=0.p(A)=A^n+c_{n-1}A^{n-1}+\cdots+c_1A+c_0 I = 0.

A useful consequence: if AA is invertible (c00c_0\neq0), then A1A^{-1} can be expressed as a polynomial in AA.

cayley-hamilton

Frequently asked questions

Where can I find the BSc CSIT (TU) Mathematics II (BSc CSIT, MTH163) question paper 2075?
The full BSc CSIT (TU) Mathematics II (BSc CSIT, MTH163) 2075 (regular) question paper is available free on Kekkei. You can read every question online and attempt the paper under timed exam conditions.
Does the Mathematics II (BSc CSIT, MTH163) 2075 paper come with solutions?
Yes. Every question on this Mathematics II (BSc CSIT, MTH163) past paper includes a step-by-step solution, plus instant AI feedback when you attempt it on Kekkei.
How many marks is the BSc CSIT (TU) Mathematics II (BSc CSIT, MTH163) 2075 paper?
The BSc CSIT (TU) Mathematics II (BSc CSIT, MTH163) 2075 paper carries 60 full marks and is meant to be completed in 180 minutes, across 12 questions.
Is practising this Mathematics II (BSc CSIT, MTH163) past paper free?
Yes — reading and attempting this Mathematics II (BSc CSIT, MTH163) past paper on Kekkei is completely free.