Mathematics - MCQ Practice Questions

Understanding:

We need to find the inverse of the $2\times 2$ matrix $A =

$.

Formula:

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

$,

Step 1: Compute the determinant.

Step 2: Apply the inverse formula.

Verification:

Answer:

The inverse of $A$ is $

$.

Quick Tip:

When $\det (A)=1$, simply swap the main diagonal entries and negate the off-diagonal entries — no division needed.

Understanding:

We must evaluate the determinant of the given $3\times 3$ matrix involving cube roots of unity $\omega$, where $1+\omega +{\omega }^2=0$ and ${\omega }^3=1$.

Formula:

For a determinant $D$, if we add all rows (or columns) and the resulting row (or column) is a zero vector, then $D=0$.

Step 1: Add all three columns together.

Adding $C_1+C_2+C_3$ gives a new first column with entries:

Step 2: Conclude from the zero column.

Since the sum of all three columns yields a column of zeros, and this operation does not change the value of the determinant, the resulting determinant has an entire column of zeros.

A determinant with a column (or row) of all zeros equals $0$.

Answer:

The value of the determinant is $0$.

Quick Tip:

Whenever you see a matrix built from $1,\omega ,{\omega }^2$ in a cyclic pattern, immediately check $C_1+C_2+C_3$; it almost always collapses to zero using $1+\omega +{\omega }^2=0$.

Understanding:

We need to find $\det (3A)$ given that $A$ is a $3\times 3$ matrix and $\det (A)=5$.

Formula:

For an $n\times n$ matrix $A$ and a scalar $k$,

Step 1: Identify $n$ and $k$.

Step 2: Apply the scalar multiplication property.

Answer:

The value of $\det (3A)$ is $135$.

Quick Tip:

A very common mistake is writing $\det (kA)=k\cdot \det (A)$. Remember: the scalar $k$ is pulled out once per row, so for an $n\times n$ matrix the factor is $k^n$, not $k$.

Understanding:

We need to find $x$ by equating the two $2\times 2$ determinants.

Formula:

Step 1: Evaluate the left-hand side determinant.

Step 2: Evaluate the right-hand side determinant.

Step 3: Set the two expressions equal and solve.

Answer:

The values of $x$ satisfying the equation are $x=\pm 4$.

Quick Tip:

After computing both determinants as polynomials in $x$, the equation is a simple quadratic — remember to include both the positive and negative roots.

Understanding:

We need to compute $\det (AB)$ where $A =

$and$B =

$.

Formula:

For square matrices of the same order,

Step 1: Compute $\det (A)$.

Step 2: Compute $\det (B)$.

Step 3: Apply the product rule.

Verification:

Answer:

The value of $\det (AB)$ is $-12$.

Quick Tip:

Using $\det (AB)=\det (A)\cdot \det (B)$ is far faster than computing the full matrix product when only the determinant is required.

Understanding:

We need to determine the nature of the solution set of the system:

Formula:

For a system $AX=B$, the solution type depends on the rank of the coefficient matrix $A$ and the augmented matrix $[A|B]$:

Step 1: Write the coefficient matrix and compute its determinant.

Step 2: Observe the relationship between the equations.

Multiplying Equation 1 by $2$: $2x+2y=6$, which is exactly Equation 2.

So both equations represent the same line, meaning every point on $x+y=3$ is a solution.

Step 3: Confirm using rank.

This confirms infinitely many solutions.

Answer:

The system has infinitely many solutions (the two equations are identical lines).

Quick Tip:

Whenever $\det (A)=0$, the system is either inconsistent (no solution) or dependent (infinitely many solutions). Check the augmented matrix to distinguish between the two cases.

Understanding:

We need to find $\det (\text{adj}A)$ for a $3\times 3$ matrix $A$ with $\det (A)=-4$.

Formula:

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

Step 1: Identify the values.

Step 2: Apply the formula.

Answer:

The value of $\det (\text{adj}A)$ is $16$.

Quick Tip:

Two results worth memorising: $\det (\text{adj}A)=(\det A{)}^{n-1}$ and $\text{adj}(\text{adj}A)=(\det A{)}^{n-2}A$. These appear repeatedly in competitive exams.

Understanding:

We need to find the value of $k$ for which the linear system has no solution.

Formula:

A system $AX=B$ has no solution when $\det (A)=0$ but the system is inconsistent (i.e., $\rho ([A|B])>\rho (A)$).

Step 1: Write the coefficient matrix and set its determinant to zero.

Step 2: Check $k=1$.

Equations become $x+y=4$ and $x+y=4$ — identical lines, so infinitely many solutions. Rejected.

Step 3: Check $k=-1$.

Equations become $x-y=4$ and $-x+y=4$, i.e., $x-y=-4$.

These are parallel lines ($x-y=4$ and $x-y=-4$) — no common solution.

Answer:

The system has no solution when $k=-1$.

Quick Tip:

For a $2\times 2$ system, $\det (A)=0$ gives candidate values of $k$. Always substitute each back to distinguish between the "no solution" case (parallel, inconsistent) and the "infinitely many" case (coincident lines).

Understanding:

We need to find $\det \left(\frac{1}{2}A\right)$ where $A$ is $3\times 3$ and $\det (A)=6$.

Formula:

For an $n\times n$ matrix and scalar $k$:

Step 1: Identify values.

Step 2: Apply the formula.

Answer:

The value of $\det \left(\frac{1}{2}A\right)$ is $\frac{3}{4}$.

Quick Tip:

Note that option $\frac{6}{8}$ is the same as $\frac{3}{4}$ but left unsimplified — always simplify fractions in your final answer to match the standard form given in exam options.

Understanding:

We must find the cofactor $C_{23}$ of the matrix $A$, i.e., the cofactor of the element in row 2, column 3.

Formula:

The cofactor $C_{ij}$ is defined as:

where $M_{ij}$ is the minor obtained by deleting row $i$ and column $j$.

Step 1: Find the minor $M_{23}$ by deleting row 2 and column 3.

Step 2: Apply the sign factor.

Answer:

The cofactor $C_{23}$ is $6$.

Quick Tip:

The sign pattern for cofactors forms a checkerboard: $

$.Position$(2,3)$carriesa$-$sign,so$C_{23} = -M_{23}$.

Understanding:

We must find the perimeter of a rhombus whose diagonals are given.

Formula:

The diagonals of a rhombus bisect each other at right angles. The side $a$ of the rhombus is:

and the perimeter is $P=4a$.

Step 1: Find the half-diagonals.

Step 2: Find the side using Pythagoras.

Step 3: Find the perimeter.

Answer:

The perimeter of the rhombus is 40 cm.

Quick Tip:

The half-diagonals $6$ and $8$ form a classic $3$-$4$-$5$ right triangle (scaled by $2$), so the hypotenuse $=10$ can be spotted instantly without full calculation.

Understanding:

We must find the length of a tangent from an external point to a circle.

Formula:

The tangent from an external point is perpendicular to the radius at the point of contact, so by Pythagoras:

Step 1: Substitute the given values.

Answer:

The length of the tangent is 12 cm.

Quick Tip:

The numbers $5$, $12$, $13$ form a Pythagorean triple — recognising it saves computation time in exams.

Understanding:

We must find the height of a trapezium given its area and parallel sides.

Formula:

Step 1: Substitute the known values and solve for $h$.

Answer:

The height of the trapezium is 10 cm.

Understanding:

Two chords intersect inside a circle. We must find the length of one segment using the intersecting chords theorem.

Formula:

When two chords intersect inside a circle at point $P$:

Step 1: Substitute and solve for $PD$.

Answer:

The length of $PD$ is 12 cm.

Quick Tip:

The intersecting chords theorem states that the products of the two segments of each chord are equal. A common error is to add instead of multiply the segments.

Understanding:

We must find the area of a triangle with given side lengths.

Formula:

First check whether the triangle is right-angled:

If yes, use:

Step 1: Check for right angle.

So the triangle is right-angled with legs $7$ cm and $24$ cm.

Step 2: Compute the area.

Answer:

The area of the triangle is 84 square centimetres.

Quick Tip:

The triple $7$-$24$-$25$ is a standard Pythagorean triple worth memorising alongside $3$-$4$-$5$ and $5$-$12$-$13$.

Understanding:

We must find the length of a chord given the radius and its distance from the centre.

Formula:

The perpendicular from the centre bisects the chord. Using Pythagoras on the right triangle formed:

Step 1: Substitute the values.

Answer:

The length of the chord is 16 cm.

Quick Tip:

The numbers $6$, $8$, $10$ form a scaled $3$-$4$-$5$ triple, making the half-chord $=8$ easy to spot.

Understanding:

We must find the area of a right triangle given the hypotenuse and one leg.

Formula:

First find the missing leg, then use:

where $a$ and $b$ are the two legs.

Step 1: Find the other leg $b$.

Step 2: Compute the area.

Answer:

The area of the right triangle is 120 square centimetres.

Quick Tip:

The triple $10$-$24$-$26$ is simply $2\times (5\text{-}12\text{-}13)$. Recognising such scaled triples avoids lengthy square-root computation.

Understanding:

We must identify the measure of an angle inscribed in a semicircle.

Formula:

The inscribed angle theorem states:

Step 1: Identify the central angle for a semicircle.

A semicircle subtends an arc of $180^{\circ }$ at the centre.

Step 2: Apply the inscribed angle theorem.

Answer:

The angle inscribed in a semicircle is always $90^{\circ }$.

Quick Tip:

This is Thales' theorem — any angle inscribed in a semicircle (i.e., the angle at the circumference subtended by the diameter) is a right angle. It is one of the most frequently tested circle theorems.

Understanding:

We must find the area of a rectangle given its perimeter and diagonal.

Formula:

Let the length and width be $l$ and $w$. Then:

Step 1: Find $l+w$.

Step 2: Find $l^2+w^2$.

Step 3: Compute the area.

Verification:

$l$ and $w$ satisfy $t^2-35t+300=0$.

So $l=20\text{ cm}$ and $w=15\text{ cm}$. Check: $20+15=35$ ✓, $\sqrt{20^2+15^2}=\sqrt{625}=25$ ✓, Area $=20\times 15=300$ ✓.

Answer:

The area of the rectangle is 300 square centimetres.

Quick Tip:

The identity $\text{Area}=\frac{(l+w{)}^2-(l^2+w^2)}{2}$ converts the perimeter and diagonal directly into the area without solving for $l$ and $w$ individually.

Understanding:

We must compute the dot product of two vectors.

Formula:

Step 1: Multiply corresponding components

Answer:

The dot product $\vec{a}\cdot \vec{b}$ equals $-8$.

Quick Tip:

A common error is to forget the sign of the third component; always carry the minus sign through each multiplication carefully.