Mathematics - MCQ Practice Questions

Understanding:

We must find $|\vec{a}\times \vec{b}|$.

Formula:

Step 1: Evaluate the determinant

Step 2: Find the magnitude

Answer:

The magnitude of $\vec{a}\times \vec{b}$ is $\sqrt{3}$.

Quick Tip:

$|\vec{a}\times \vec{b}|$ gives the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$; here that area is $\sqrt{3}$.

Understanding:

We must find the angle $\theta$ between two vectors using the dot product formula.

Formula:

Step 1: Compute the dot product

Step 2: Compute the magnitudes

Step 3: Find $\cos \theta$

Step 4: Find $\theta$

Answer:

The angle between the vectors is ${\cos }^{-1}\left(\frac{1}{3}\right)$.

Understanding:

We must find the unit vector along $\vec{v}=3\hat{i}-4\hat{j}$.

Formula:

Step 1: Compute the magnitude

Step 2: Divide by the magnitude

Answer:

The unit vector in the direction of $\vec{v}$ is $\frac{3}{5}\hat{i}-\frac{4}{5}\hat{j}$.

Quick Tip:

A common trap is preserving the sign: the $j$-component is negative, matching the original vector's direction.

Understanding:

We must find the distance between points $A$ and $B$ given their position vectors.

Formula:

Step 1: Compute $\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}$

Step 2: Find the magnitude

Answer:

The length of $AB$ is $2\sqrt{6}$.

Understanding:

We must evaluate the scalar triple product (box product) $[\vec{a}\vec{b}\vec{c}]$.

Formula:

Step 1: Set up the determinant

Step 2: Evaluate

Answer:

The scalar triple product equals $1$.

Quick Tip:

For the standard basis vectors $\hat{i},\hat{j},\hat{k}$ in order, the scalar triple product is always $1$; reversing any two gives $-1$.

Understanding:

We must find the direction cosines of line $OP$.

Formula:

Step 1: Find $r=|OP|$

Step 2: Compute direction cosines

Verification:

Answer:

The direction cosines are $\frac{1}{3},\frac{2}{3},\frac{2}{3}$.

Quick Tip:

Always verify: the sum of squares of direction cosines must equal $1$.

Understanding:

We must find $\theta$ between $\vec{a}$ and $\vec{b}$.

Formula:

Step 1: Substitute values

Step 2: Find $\theta$

Answer:

The angle between the vectors is $\frac{\pi }{3}$.

Understanding:

We must find the area of triangle $ABC$ using the cross product formula.

Formula:

Step 1: Find $\overrightarrow{AB}$ and $\overrightarrow{AC}$

Step 2: Compute $\overrightarrow{AB}\times \overrightarrow{AC}$

Step 3: Find the magnitude

Step 4: Compute area

Answer:

The area of the triangle is $\frac{\sqrt{21}}{2}$.

Quick Tip:

Remember to halve $|\overrightarrow{AB}\times \overrightarrow{AC}|$; omitting the factor of $\frac{1}{2}$ is the most common error here.

Understanding:

We must find $\det (A^2-5A)$ where $A =

$.

Formula:

The Cayley–Hamilton theorem states that every square matrix satisfies its own characteristic equation. For a $2\times 2$ matrix:

so $A^2-(\text{tr}A)A+\det (A)I=0$.

Step 1: Compute the trace and determinant of $A$.

Step 2: Apply Cayley–Hamilton.

Step 3: Compute $A-5I$.

Step 4: Find the determinant.

Answer:

The determinant of $A^2-5A$ is $0$.

Quick Tip:

Whenever you see a polynomial in a matrix, apply Cayley–Hamilton first — it often collapses the expression to something simple.

Understanding:

We need $\det (B)$ where $B=A^{T}A$ and $A =

$.

Formula:

because $\det (A^T)=\det (A)$ for any square matrix.

Step 1: Use the multiplicative property.

Verification: Compute $A^{T}A$ directly.

Answer:

The determinant of $B=A^{T}A$ is $4$.

Quick Tip:

$\det (A^{T}A)=[\det (A){]}^2$ is always non-negative — a useful sanity check.

Understanding:

We must find $\det (\text{adj}A)$ given that $A$ is a $3\times 3$ matrix with $\det (A)=4$.

Formula:

For an $n\times n$ matrix $A$:

Step 1: Apply the formula with $n=3$ and $\det (A)=4$.

Answer:

The determinant of the adjugate of $A$ is $16$.

Quick Tip:

The companion formula $A\cdot \text{adj}(A)=\det (A)\cdot I$ gives $\det (A)\cdot \det (\text{adj}A)=[\det (A){]}^n$, which immediately yields the result.

Understanding:

A homogeneous system has a non-trivial solution if and only if the coefficient matrix is singular (determinant $=0$).

$

Formula:

Step 1: Expand the determinant along the first row.

Step 2: Evaluate each $2\times 2$ determinant.

Step 3: Substitute and simplify.

Step 4: Set equal to zero.

Answer:

The system has a non-trivial solution when $k=5$.

Quick Tip:

For a homogeneous system, always set $\det =0$; for a non-homogeneous system, use Cramer's rule or rank conditions instead.

Understanding:

We must find $A^{T}A$ for the rotation matrix $A =

$.

Formula:

A matrix is orthogonal when its rows (equivalently, columns) form an orthonormal set.

Step 1: Write $A^T$.

Step 2: Multiply $A^{T}A$.

using ${\cos }^2\theta +{\sin }^2\theta =1$.

Answer:

$A^{T}A=I$, the identity matrix.

Quick Tip:

Every rotation matrix is orthogonal, so $A^T=A^{-1}$ and $\det (A)=1$ — both facts are worth memorising for MCQs.

Understanding:

We must classify the matrix $A =

$.

Formula:

A matrix $A$ is skew-symmetric if and only if:

equivalently, $a_{ij}=-a_{ji}$ for all $i,j$, and all diagonal entries are $0$.

Step 1: Compute $A^T$.

Step 2: Check whether $A^T=-A$.

Since $A^T=-A$, the matrix is skew-symmetric.

Step 3: Verify it is not symmetric — clearly $A\ne A^T$.

Step 4: Check it is not orthogonal by noting $\det (A)=0$ (for any odd-order real skew-symmetric matrix, $\det =0$), so $A$ cannot be orthogonal.

Answer:

$A$ is a skew-symmetric matrix.

Quick Tip:

For any odd-order real skew-symmetric matrix, $\det (A)=0$ — this is a standard result worth remembering.

Understanding:

We identify the correct formula for the inverse of a product of two invertible matrices $A$ and $B$.

Formula:

For invertible matrices $A$ and $B$ of the same order:

This is the reversal (socks-and-shoes) rule.

Step 1: Verify by direct multiplication.

Step 2: Similarly check the other side.

Both products give $I$, confirming $B^{-1}{A}^{-1}$ is the inverse of $AB$.

Answer:

The correct identity is $(AB{)}^{-1}=B^{-1}{A}^{-1}$.

Quick Tip:

Matrix multiplication is not commutative, so the order reverses when taking inverses — just like putting on socks before shoes means removing shoes before socks.

Understanding:

We analyse the solution set of the system:

Formula:

For the system $AX=B$, the solution exists (and is unique or infinite) based on the rank condition:

Step 1: Form the coefficient matrix and augmented matrix.

Step 2: Apply $R_2\to R_2-2R_1$.

Step 3: Determine ranks.

Since both ranks are equal and less than $n=2$ (the number of unknowns), the system is consistent with infinitely many solutions.

Step 4: Note that equation (2) is simply $2\times$ equation (1), so the two equations represent the same line.

Answer:

The system has infinitely many solutions.

Quick Tip:

When one equation is a scalar multiple of the other (both the coefficients AND the RHS scale by the same factor), the two equations are dependent and the system has infinitely many solutions.

Understanding:

We must find the inverse of $A =

$.

Formula:

For a $2\times 2$ matrix $A =

$with$\det(A) \ne 0$:

Step 1: Compute the determinant.

Step 2: Apply the inverse formula.

Verification:

Answer:

The inverse of $A$ is $

$.

Quick Tip:

When $\det (A)=1$, the inverse is simply the adjugate — swap the diagonal entries and negate the off-diagonal entries.

Understanding:

We must identify which of the given logical statements is a tautology — a statement that is true for all possible truth values of its variables.

Formula:

A statement is a tautology if its truth table yields $\text{True}$ in every row:

Step 1: Evaluate each option

For $p\wedge (\sim p)$: when $p$ is True, $\sim p$ is False, so $p\wedge (\sim p)$ is False. This is a contradiction, not a tautology.

Step 2: Check $p\vee (\sim p)$

The statement is True in all cases — it is a tautology (Law of Excluded Middle).

Step 3: Eliminate remaining options

$p\wedge q$ is False when either $p$ or $q$ is False — not a tautology.

$p\Rightarrow q$ is False when $p$ is True and $q$ is False — not a tautology.

Answer:

The statement $p\vee (\sim p)$ is always true, making it a tautology.

Quick Tip:

The two fundamental results to memorise: $p\vee (\sim p)$ is always a tautology, and $p\wedge (\sim p)$ is always a contradiction.

Understanding:

We must find the negation of the universal statement: "All students are hardworking."

Step 1: Express the statement in logical form

Let $P(x)$ denote "$x$ is hardworking." The statement is:

Step 2: Apply the negation rule for universal quantifiers

The negation of "for all" is "there exists at least one for whom the predicate is false."

Step 3: Translate back to English

$\exists x,\sim P(x)$ means "There exists a student who is NOT hardworking," which is the same as "Some students are not hardworking."

Answer:

The negation of "All students are hardworking" is "Some students are not hardworking."

Quick Tip:

Negation of "All $A$ are $B$" is always "Some $A$ are not $B$" — never "No $A$ is $B$." A common exam trap is to select the stronger (but incorrect) negation.

Understanding:

We are given:

We must determine what can be concluded about $p$.

Formula:

The truth table of $p\Rightarrow q$:

Step 1: Use Modus Tollens

The rule of Modus Tollens states:

Since $p\Rightarrow q$ is True and $q$ is False, we must have $p$ is False.

Step 2: Verify by truth table

If $q$ is False and $p\Rightarrow q$ is True, only the row $p=F,q=F$ gives $p\Rightarrow q=T$. So $p$ must be False.

Step 3: Eliminate other options

$p\wedge q$ requires both to be True — False here since $q$ is False.

$\sim p\Rightarrow \sim q$: Since $p$ is False, $\sim p$ is True; $q$ is False, $\sim q$ is True; so $\sim p\Rightarrow \sim q=T$, meaning this option's claim of "false" is wrong.

Answer:

By Modus Tollens, $p$ must be false.

Quick Tip:

Modus Tollens: $(p\Rightarrow q)\wedge \sim q\Rightarrow \sim p$ is one of the most tested inference rules in Mathematical Reasoning.