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 $\det (A^2-5A)$ 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.