Q.Construct a $3 \times 4$ matrix, whose elements are given by:

Showing the 12 most recent of 16 on this concept.

1. TG EAPCET 2024Set eng-2024-05-10-AN1 markMCQ
   Q.$A = [a_{ij}]$ is a $3 \times 3$ matrix with positive integers as its elements. Elements of $A$ are such that the sum of all the elements of each row is equal to $6$ and $a_{22} = 2$. If $a_{ii} = \begin{cases} a_{ij} + a_{ji}, & j = i+1 \text{ when } i < 3 \\ a_{ij} + a_{ji}, & j = 4-i \text{ when } i = 3 \end{cases}$ for $i = 1, 2, 3$, then $|A| =$ (A) $6$ (B) $18$ (C) $3$ (D) $12$

   ›Reveal solutionSolution

   The problem gives a 3×3 matrix of positive integers with row sums 6, a fixed entry $a_{22}=2$, and a special relation linking symmetric pairs. Solving the constraints yields a unique matrix whose determinant is 12, so the correct option is (D).

   ---

   We start by understanding the structure. The matrix is $3\times3$ with positive integer entries. Each row sums to 6, and $a_{22}=2$. The condition on the diagonal entries is:

   $a_{ii} = \begin{cases} a_{ij} + a_{ji}, & j = i+1 \text{ when } i < 3 \\ a_{ij} + a_{ji}, & j = 4-i \text{ when } i = 3 \end{cases}$

   Let’s decode this:

   • For $i=1$: $j = i+1 = 2$, so $a_{11} = a_{12} + a_{21}$.
   • For $i=2$: $j = i+1 = 3$, so $a_{22} = a_{23} + a_{32}$.
   • For $i=3$: $j = 4-i = 1$, so $a_{33} = a_{31} + a_{13}$.

   Thus the diagonal entries are sums of symmetric off-diagonal pairs.

   ---

   Step 1: Use the given $a_{22}=2$.

   From $a_{22} = a_{23} + a_{32} = 2$, and since entries are positive integers, the only possibility is:

   $a_{23} = 1,\quad a_{32} = 1$

   (or swapped, but symmetry of the condition doesn’t matter — they are both 1).

   ---

   Step 2: Row sums give equations.

   Let the matrix be:

   $A = \begin{pmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & 2 & 1 \\ a_{31} & 1 & a_{33} \end{pmatrix}$

   Row sums = 6:

   • Row 1: $a_{11} + a_{12} + a_{13} = 6$
   • Row 2: $a_{21} + 2 + 1 = 6 \Rightarrow a_{21} = 3$
   • Row 3: $a_{31} + 1 + a_{33} = 6 \Rightarrow a_{31} + a_{33} = 5$

   ---

   Step 3: Diagonal conditions.

   We have:

   • $a_{11} = a_{12} + a_{21} = a_{12} + 3$
   • $a_{33} = a_{31} + a_{13}$

   From row 1: substitute $a_{11} = a_{12} + 3$ into $a_{11} + a_{12} + a_{13} = 6$:

   $(a_{12} + 3) + a_{12} + a_{13} = 6 \Rightarrow 2a_{12} + a_{13} = 3$

   Since entries are positive integers, $a_{12} \ge 1$, so:

   • If $a_{12}=1$, then $2(1) + a_{13}=3 \Rightarrow a_{13}=1$.
   • If $a_{12}=2$, then $4 + a_{13}=3$ impossible.
   • So only $a_{12}=1$, $a_{13}=1$.

   Then $a_{11} = a_{12}+3 = 4$.

   ---

   Step 4: Find remaining entries.

   From row 3: $a_{31} + a_{33} = 5$ and $a_{33} = a_{31} + a_{13} = a_{31} + 1$.

   Substitute: $a_{31} + (a_{31}+1) = 5 \Rightarrow 2a_{31}=4 \Rightarrow a_{31}=2$, then $a_{33}=3$.

   --- …

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2. TG EAPCET 2024Set eng-2024-05-09-FN1 markMCQ
   Q.A, B, C, D are square matrices such that A + B is symmetric, A - B is skew-symmetric and D is the transpose of C. If $A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix}$ and $C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix}$, then the matrix $B + D =$ (A) $\begin{bmatrix} -1 & 6 & 3 \\ 6 & 2 & -2 \\ 3 & -2 & 6 \end{bmatrix}$ (B) $\begin{bmatrix} -1 & 6 & 3 \\ 3 & 2 & -2 \\ 1 & -2 & 6 \end{bmatrix}$ (C) $\begin{bmatrix} 3 & 2 & -2 \\ 2 & 6 & 3 \\ -2 & 3 & 2 \end{bmatrix}$ (D) $\begin{bmatrix} 1 & -2 & 6 \\ -2 & 3 & 2 \\ 6 & 2 & 1 \end{bmatrix}$

   ›Reveal solutionSolution

   The key idea is to express $B$ in terms of $A$ using the symmetry conditions, and $D$ is simply $C^T$. Adding them gives the matrix in option (B).

   We are told that $A+B$ is symmetric and $A-B$ is skew-symmetric. This is a classic decomposition: any square matrix can be written uniquely as the sum of a symmetric and a skew-symmetric matrix. Here, $A$ is given, and the conditions let us solve for $B$ directly.

   Why this works:

   If $S = A+B$ is symmetric, then $S^T = S$. If $K = A-B$ is skew-symmetric, then $K^T = -K$. Adding these two equations gives a way to isolate $A$ and $B$. In fact, adding $S$ and $K$:

   $S + K = (A+B) + (A-B) = 2A$

   and subtracting:

   $S - K = (A+B) - (A-B) = 2B.$

   So we can find $B$ from $A$ alone, without ever needing $S$ or $K$ explicitly — just their symmetry properties.

   ---

   1. Find $B$ using the symmetry conditions. From $S = A+B$ symmetric, we have $(A+B)^T = A+B$. From $K = A-B$ skew-symmetric, we have $(A-B)^T = -(A-B)$. Add the two transposed equations:

   $(A+B)^T + (A-B)^T = (A+B) - (A-B)$

   The left side is $A^T + B^T + A^T - B^T = 2A^T$.

   The right side is $A+B - A + B = 2B$.

   Hence $2B = 2A^T$, so

   $B = A^T.$

   This is a neat result: when $A+B$ is symmetric and $A-B$ is skew-symmetric, $B$ must be the transpose of $A$.

   Tip

   A quick check: if $B = A^T$, then $A+B = A+A^T$ is symmetric, and $A-B = A-A^T$ is skew-symmetric. So the condition forces $B = A^T$ uniquely.

   Given

   $A = \begin{bmatrix} -1 & 2 & 3 \\ 4 & 3 & -2 \\ 3 & -4 & 5 \end{bmatrix},$

   we have

   $B = A^T = \begin{bmatrix} -1 & 4 & 3 \\ 2 & 3 & -4 \\ 3 & -2 & 5 \end{bmatrix}.$

   2. Find $D$ from $C$. We are told $D$ is the transpose of $C$. Given

   $C = \begin{bmatrix} 0 & 1 & -2 \\ 2 & -1 & 0 \\ 0 & 2 & 1 \end{bmatrix},$

   so

   $D = C^T = \begin{bmatrix} 0 & 2 & 0 \\ 1 & -1 & 2 \\ -2 & 0 & 1 \end{bmatrix}.$

   3. Add $B$ and $D$. $$$$

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3. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew symmetric matrix, then $AB =$ (A) $\begin{bmatrix} 48 & 27 & 48 \\ 52 & 19 & 22 \\ -59 & 43 & -67 \end{bmatrix}$ (B) $\begin{bmatrix} 48 & 26 & 36 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{bmatrix}$ (C) $\begin{bmatrix} 12 & 26 & 36 \\ 32 & 79 & 50 \\ -11 & 43 & -67 \end{bmatrix}$ (D) $\begin{bmatrix} 12 & 32 & 41 \\ 32 & 19 & 22 \\ -11 & 43 & -67 \end{bmatrix}$

   ›Reveal solutionSolution

   Using the definitions of symmetric and skew-symmetric matrices, we first determine the unknown entries $a,b,c,d$ by equating $A = A^T$ and $B = -B^T$. Then we compute the product $AB$ and match it to the given options. The result is option (B).

   We are given that $A$ is symmetric and $B$ is skew-symmetric. This gives us equations to solve for the unknowns $a, b, c, d$. Once we know the matrices, we multiply them and compare with the choices.

   Concept and intuition:

   A symmetric matrix equals its transpose; a skew-symmetric matrix equals the negative of its transpose. These conditions force certain entries to be equal or opposite, and also force diagonal entries of a skew-symmetric matrix to be zero. By applying these, we can fill in the missing numbers without any extra information.

   ---

   1. Use symmetry of $A$:

      For $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ to be symmetric, we need $A = A^T$.

      Compare entries:

      • $(1,2)$: $a = b$
      • $(1,3)$: $3 = 3$ (already satisfied)
      • $(2,3)$: $c = d$ So $b = a$ and $d = c$.

   2. Use skew-symmetry of $B$:

      For $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ to be skew-symmetric, we need $B = -B^T$.

      Compare entries:

      • $(1,3)$: $b = -6$ (since $B_{13} = b$ and $-B^T_{13} = -B_{31} = -6$)
      • $(2,3)$: $-7 = -c$ (since $B_{23} = -7$ and $-B^T_{23} = -B_{32} = -c$) → $c = 7$
      • $(3,1)$: $6 = -b$ (same as first condition, consistent)
      • $(3,2)$: $c = 7$ (already consistent)

      Thus $b = -6$, $c = 7$.

   3. Back-substitute into $A$:

      From step 1: $a = b = -6$, $d = c = 7$.

      So

   $A = \begin{bmatrix} 1 & -6 & 3 \\ -6 & 2 & 7 \\ 3 & 7 & 4 \end{bmatrix}.$

   4. Write $B$ with known values: $b = -6$, $c = 7$ gives

   $B = \begin{bmatrix} 0 & 5 & -6 \\ -5 & 0 & -7 \\ 6 & 7 & 0 \end{bmatrix}.$

   5. Compute $AB$:

      Multiply $A$ (3×3) by $B$ (3×3). Let $C = AB$.

      • $C_{11} = (1)(0) + (-6)(-5) + (3)(6) = 0 + 30 + 18 = 48$
      • $C_{12} = (1)(5) + (-6)(0) + (3)(7) = 5 + 0 + 21 = 26$ …

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4. TG EAPCET 2022Set eng-2022-07-18-AN1 markMCQ
   Q.Let $x = \alpha, y = \beta, z = \gamma$ be the unique solution of the system of simultaneous linear equations $2x + 3y - 2z + 4 = 0, 3x - 4y + 3z + 5 = 0, kx - 2y + z + 3 = 0$. If $\alpha = -2$ then $k =$ (A) $\begin{bmatrix} 1 & 2 \\ 3 & 5 \end{bmatrix}$ (B) $\begin{bmatrix} 5 & 3 \\ 1 & 2 \end{bmatrix}$ (C) $\begin{bmatrix} 3 & 5 \\ 1 & 2 \end{bmatrix}$ (D) $\begin{bmatrix} 3 & 5 \\ 2 & 1 \end{bmatrix}$

   ›Reveal solutionSolution

   We are given a system of three linear equations with an unknown coefficient $k$ and the value of one variable, $x = -2$, from its unique solution. By substituting $x = -2$ into the first two equations, we form a $2 \times 2$ system for $y$ and $z$. Solving this system gives $y=2$ and $z=3$. Substituting these values along with $x=-2$ into the third equation allows us to solve for $k$, which is $\boxed{1}$.

   Concept and Intuition

   A system of simultaneous linear equations has a unique solution if the determinant of its coefficient matrix is non-zero. In this problem, we are told that a unique solution $(x, y, z) = (\alpha, \beta, \gamma)$ exists, and we are given the value of $\alpha = -2$. This means we already know one component of the unique solution.

   The core idea is to use the given information to simplify the problem. Since $x = -2$ is part of the solution, it must satisfy all three equations. By substituting $x = -2$ into the first two equations, we reduce the problem from a $3 \times 3$ system to a $2 \times 2$ system involving only $y$ and $z$. Once we find $y$ and $z$, we will have the complete unique solution $(\alpha, \beta, \gamma)$. Finally, we can substitute all three values ($x, y, z$) into the third equation, which contains the unknown $k$, to determine its value.

   Step-by-step Solution

   1. Rewrite the system of equations in standard form: The given equations are:

   $2x + 3y - 2z + 4 = 0 \implies 2x + 3y - 2z = -4 \quad \text{(Eq. 1)}$

   $3x - 4y + 3z + 5 = 0 \implies 3x - 4y + 3z = -5 \quad \text{(Eq. 2)}$

   $kx - 2y + z + 3 = 0 \implies kx - 2y + z = -3 \quad \text{(Eq. 3)}$

   2. Substitute the given value of $x = \alpha = -2$ into Equations 1 and 2: Since $x = -2$ is part of the unique solution, it must satisfy these equations. Substitute $x = -2$ into (Eq. 1):

   $2(-2) + 3y - 2z = -4$

   $-4 + 3y - 2z = -4$

   $3y - 2z = 0 \quad \text{(Eq. A)}$

   Substitute $x = -2$ into (Eq. 2):
   

   $3(-2) - 4y + 3z = -5$

   $-6 - 4y + 3z = -5$

   $-4y + 3z = 1 \quad \text{(Eq. B)}$

   3. Solve the $2 \times 2$ system (Eq. A and Eq. B) for $y$ and $z$: We have the system:

   $3y - 2z = 0 \quad \text{(A)}$

   $-4y + 3z = 1 \quad \text{(B)}$

   From (A), we can express $y$ in terms of $z$: …
   

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5. TG EAPCET 2023Set eng-2023-05-13-AN1 markMCQ
   Q.If the inverse of the matrix $A = \begin{bmatrix} -1 & -3 & -2 \\ 0 & 1 & 2 \\ 3 & 4 & 5 \end{bmatrix}$ is $A^{-1} = \begin{bmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{bmatrix}$, then $a_1 + c_2 + b_3 =$ (A) $-6$ (B) $-\dfrac{2}{3}$ (C) $\dfrac{2}{3}$ (D) $6$

   ›Reveal solutionSolution

   Build $A^{-1}=\dfrac{1}{\det A}\,\operatorname{adj}A$. With $\det A=-9$, the required entries give $a_1+c_2+b_3=\dfrac{2}{3}$ — option (C).

   Setup. For $A=\begin{bmatrix}-1&-3&-2\\0&1&2\\3&4&5\end{bmatrix}$ we use $A^{-1}=\dfrac{1}{\det A}\operatorname{adj}A$, where $\operatorname{adj}A$ is the transpose of the cofactor matrix. Here $a_1$ is entry $(1,1)$, $b_3$ is entry $(2,3)$ and $c_2$ is entry $(3,2)$ of $A^{-1}$.

   Step 1 - Determinant. Expanding along column 1:

   $\det A=(-1)\begin{vmatrix}1&2\\4&5\end{vmatrix}+3\begin{vmatrix}-3&-2\\1&2\end{vmatrix}=(-1)(5-8)+3(-6+2)=3-12=-9.$

   Step 2 - Cofactors. The cofactor matrix $C_{ij}=(-1)^{i+j}M_{ij}$ is

   $C=\begin{bmatrix}-3&6&-3\\7&1&-5\\-4&2&-1\end{bmatrix},\qquad \operatorname{adj}A=C^{T}=\begin{bmatrix}-3&7&-4\\6&1&2\\-3&-5&-1\end{bmatrix}.$

   Step 3 - Inverse. …

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6. TG EAPCET 2024Set eng-2024-05-09-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i \leq 3, 1 \leq j \leq 3} a_{ij} =$ (A) $\dfrac{2}{3}$ (B) $\dfrac{1}{3}$ (C) $1$ (D) $17$

   ›Reveal solutionSolution

   The sum of all entries of $A^{-1}$ equals the sum of all entries of the adjugate matrix divided by $\det A$.

   For this $3\times3$ matrix, the sum is $\frac{1}{3}$, so the correct option is (B).

   ---

   Concept and intuition

   We want the sum of all nine entries of $A^{-1}$.

   A direct inversion is messy, but there’s a clever shortcut:

   For any invertible matrix $A$,

   $A^{-1} = \frac{1}{\det A} \cdot \operatorname{adj}(A)$

   where $\operatorname{adj}(A)$ is the adjugate (transpose of the cofactor matrix).

   So the sum of all entries of $A^{-1}$ is simply

   $\frac{1}{\det A} \times (\text{sum of all entries of } \operatorname{adj}(A)).$

   Now, each entry of $\operatorname{adj}(A)$ is a cofactor (a signed $2\times2$ determinant).

   If we can compute $\det A$ and the sum of all cofactors, we’re done — no need to invert the whole matrix.

   ---

   Step-by-step

   1. Compute $\det A$

   $A = \begin{bmatrix} 1 & 2 & 2 \\ 3 & 2 & 3 \\ 1 & 1 & 2 \end{bmatrix}$

   Expand along the first row:

   $\det A = 1\cdot\begin{vmatrix}2&3\\1&2\end{vmatrix} - 2\cdot\begin{vmatrix}3&3\\1&2\end{vmatrix} + 2\cdot\begin{vmatrix}3&2\\1&1\end{vmatrix}$

   $= 1\cdot(4-3) - 2\cdot(6-3) + 2\cdot(3-2) = 1\cdot1 - 2\cdot3 + 2\cdot1 = 1 - 6 + 2 = -3.$

   So $\det A = -3$.

   2. Sum of all cofactors (entries of the adjugate)

   The adjugate is the transpose of the cofactor matrix.

   Let $C_{ij}$ be the cofactor of entry $a_{ij}$ (i.e., $(-1)^{i+j}$ times the $2\times2$ determinant after removing row $i$, column $j$).

   Then $\operatorname{adj}(A)_{ji} = C_{ij}$.

   We need the sum of all entries of $\operatorname{adj}(A)$, which equals the sum of all cofactors $C_{ij}$ (since transposing doesn’t change the sum of all entries).

   Compute each cofactor:

   • $C_{11} = +\begin{vmatrix}2&3\\1&2\end{vmatrix} = 4-3 = 1$

   • $C_{12} = -\begin{vmatrix}3&3\\1&2\end{vmatrix} = -(6-3) = -3$

   • $C_{13} = +\begin{vmatrix}3&2\\1&1\end{vmatrix} = 3-2 = 1$

   • $C_{21} = -\begin{vmatrix}2&2\\1&2\end{vmatrix} = -(4-2) = -2$

   • $C_{22} = +\begin{vmatrix}1&2\\1&2\end{vmatrix} = 2-2 = 0$

   • $C_{23} = -\begin{vmatrix}1&2\\1&1\end{vmatrix} = -(1-2) = 1$

   • $C_{31} = +\begin{vmatrix}2&2\\2&3\end{vmatrix} = 6-4 = 2$

   • $C_{32} = -\begin{vmatrix}1&2\\3&3\end{vmatrix} = -(3-6) = 3$ …

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7. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
   Q.If $A + 2B = \begin{bmatrix} 1 & 2 & 0 \\ 6 & -3 & 3 \\ -5 & 3 & 1 \end{bmatrix}$ and $2A - B = \begin{bmatrix} 2 & -1 & 5 \\ 2 & -1 & 6 \\ 0 & 1 & 2 \end{bmatrix}$ then $\mathrm{tr}(A) - \mathrm{tr}(B) =$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Using traces: $\operatorname{tr}(A)=1$, $\operatorname{tr}(B)=-1$, so $\operatorname{tr}(A)-\operatorname{tr}(B)=2$ — option (B).

   Take the trace of each matrix equation. Let $a=\operatorname{tr}(A)$, $b=\operatorname{tr}(B)$.

   From $A+2B$: $\;a+2b = 1+(-3)+1 = -1.$

   From $2A-B$: $\;2a-b = 2+(-1)+2 = 3.$

   Solve the system:

   $\begin{cases} a+2b=-1 \ 2a-b=3 \end{cases}$ …

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8. TG EAPCET 2024Set eng-2024-05-10-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 9 & 3 & 0 \\ 1 & 5 & 8 \\ 7 & 6 & 2 \end{bmatrix}$ and $AA^T - A^2 = \begin{bmatrix} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{bmatrix}$, then $\sum_{1 \leq i \leq 3, 1 \leq j \leq 3} a_{ij} =$ (A) $35$ (B) $0$ (C) $33$ (D) $1$

   ›Reveal solutionSolution

   The sum of all entries in a matrix equals the sum of all entries in each row, which can be computed by multiplying the matrix by a column vector of ones. Using the property that $\mathbf{1}^T A^T = (A\mathbf{1})^T$, we find that $\sum a_{ij} = 0$.

   Concept: Sum of Matrix Entries

   When we need to find the sum of all entries in a matrix, there's an elegant trick: multiply the matrix by a column vector of all ones from the right (to sum rows), then sum those results. Equivalently, we can use the fact that:

   $\sum_{i,j} a_{ij} = \mathbf{1}^T M \mathbf{1}$

   where $\mathbf{1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$ and $M$ is our matrix.

   For our problem, we need to find the sum of all entries in $AA^T - A^2$.

   Solution

   1. Set up the sum formula

   We want to compute:

   $\sum_{i,j} a_{ij} = \mathbf{1}^T (AA^T - A^2) \mathbf{1} = \mathbf{1}^T AA^T \mathbf{1} - \mathbf{1}^T A^2 \mathbf{1}$

   2. Simplify the first term $\mathbf{1}^T AA^T \mathbf{1}$

   Using the property that $\mathbf{1}^T A = (A^T \mathbf{1})^T$, we can rewrite:

   $\mathbf{1}^T AA^T \mathbf{1} = (A^T \mathbf{1})^T (A^T \mathbf{1})$

   This is the dot product of the vector $A^T \mathbf{1}$ with itself. Let's compute $A^T \mathbf{1}$:

   $A^T \mathbf{1} = \begin{bmatrix} 9 & 1 & 7 \\ 3 & 5 & 6 \\ 0 & 8 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 17 \\ 14 \\ 10 \end{bmatrix}$

   Therefore:

   $\mathbf{1}^T AA^T \mathbf{1} = 17^2 + 14^2 + 10^2 = 289 + 196 + 100 = 585$

   3. Simplify the second term $\mathbf{1}^T A^2 \mathbf{1}$

   We can rewrite this as:

   $\mathbf{1}^T A^2 \mathbf{1} = \mathbf{1}^T A(A\mathbf{1}) = (A^T \mathbf{1})^T (A\mathbf{1})$

   First, compute $A\mathbf{1}$: …

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9. TG EAPCET 2022Set eng-2022-07-19-FN1 markMCQ
   Q.If the matrix $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$ satisfies the matrix equation $A^2 - 4A - 5I = 0$, then $A^{-1} =$ (A) $\dfrac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ -2 & 3 & -2 \\ 2 & 2 & -3 \end{bmatrix}$ (B) $\dfrac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$ (C) $\dfrac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ -2 & -2 & 3 \end{bmatrix}$ (D) $\dfrac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & 3 \end{bmatrix}$

   ›Reveal solutionSolution

   Rearranging $A^{2}-4A-5I=O$ gives $A(A-4I)=5I$, hence $A^{-1}=\frac{1}{5}(A-4I)$. Substituting $A$ produces the symmetric matrix of option (B).

   The concept: a matrix satisfying its own polynomial

   If a matrix satisfies a polynomial equation whose constant term is non-zero, you can read the inverse straight off it. That is the whole point of quoting the Cayley–Hamilton-type relation — it saves the entire adjoint/determinant computation.

   Step 1 — Rearrange

   $A^{2} - 4A - 5I = O \;\Longrightarrow\; A^{2} - 4A = 5I \;\Longrightarrow\; A\,(A - 4I) = 5I$

   Step 2 — Extract the inverse

   Multiply on the left by $A^{-1}$ (which exists since $\det A \neq 0$):

   $A - 4I = 5A^{-1} \qquad\Longrightarrow\qquad A^{-1} = \tfrac{1}{5}\,(A - 4I)$

   Step 3 — Substitute the given $A$

   $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}, \qquad A - 4I = \begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$

   $A^{-1} = \frac{1}{5}\begin{bmatrix} -3 & 2 & 2 \\ 2 & -3 & 2 \\ 2 & 2 & -3 \end{bmatrix}$

   Step 4 — Verify …

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10. TG EAPCET 2023Set eng-2023-05-12-AN1 markMCQ
    Q.If $\omega \neq 1$ is a cube root of unity, then

    [!FORMULA] $\left| \begin{array}{ccc} \omega + \omega^2 & \omega^2 + \omega^9 & \omega^9 + \omega \\ \omega^{27} + \omega^{31} & \omega^{31} + \omega^{17} & \omega^{17} + \omega^{27} \\ \omega^{30} + \omega^{41} & \omega^{41} + \omega^{19} & \omega^{19} + \omega^{30} \end{array} \right| =$

    (A) 3 (B) 2 (C) 1 (D) 0

    ›Reveal solutionSolution

    Using the property $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, every entry in the determinant simplifies to either $0$ or $-1$, making the entire determinant zero.

    The key to this problem is understanding the cube roots of unity. If $\omega \neq 1$ is a cube root of unity, then $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$. This means $\omega + \omega^2 = -1$. Also, any power of $\omega$ can be reduced modulo 3: $\omega^n = \omega^{n \bmod 3}$, because $\omega^3 = 1$.

    The determinant looks messy, but every entry is a sum of two powers of $\omega$. Our job is to simplify each sum using these two properties. Once we do, we will see a pattern that makes the determinant vanish.

    1. Simplify each exponent modulo 3.

       Since $\omega^3 = 1$, we have $\omega^9 = (\omega^3)^3 = 1$, $\omega^{27} = (\omega^3)^9 = 1$, $\omega^{31} = \omega^{30} \cdot \omega = (\omega^3)^{10} \cdot \omega = \omega$, $\omega^{17} = \omega^{15} \cdot \omega^2 = (\omega^3)^5 \cdot \omega^2 = \omega^2$, $\omega^{30} = (\omega^3)^{10} = 1$, $\omega^{41} = \omega^{39} \cdot \omega^2 = (\omega^3)^{13} \cdot \omega^2 = \omega^2$, and $\omega^{19} = \omega^{18} \cdot \omega = (\omega^3)^6 \cdot \omega = \omega$.

    2. Replace every entry with its simplified form.

       Let’s go row by row.

       First row:

       • Entry (1,1): $\omega + \omega^2 = -1$ (directly from $1+\omega+\omega^2=0$).
       • Entry (1,2): $\omega^2 + \omega^9 = \omega^2 + 1$. Since $1 + \omega + \omega^2 = 0$, we have $1 + \omega^2 = -\omega$. So this is $-\omega$.
       • Entry (1,3): $\omega^9 + \omega = 1 + \omega$. From $1+\omega+\omega^2=0$, we get $1+\omega = -\omega^2$. So this is $-\omega^2$.

       Second row:

       • Entry (2,1): $\omega^{27} + \omega^{31} = 1 + \omega$. That’s $-\omega^2$.
       • Entry (2,2): $\omega^{31} + \omega^{17} = \omega + \omega^2 = -1$.
       • Entry (2,3): $\omega^{17} + \omega^{27} = \omega^2 + 1 = -\omega$.

       Third row:

       • Entry (3,1): $\omega^{30} + \omega^{41} = 1 + \omega^2 = -\omega$.
       • Entry (3,2): $\omega^{41} + \omega^{19} = \omega^2 + \omega = -1$.
       • Entry (3,3): $\omega^{19} + \omega^{30} = \omega + 1 = -\omega^2$.

    3. Write the simplified determinant.

       The matrix becomes:

    $\begin{pmatrix} -1 & -\omega & -\omega^2 \\ -\omega^2 & -1 & -\omega \\ -\omega & -1 & -\omega^2 \end{pmatrix}$

    Factor $-1$ from each row (or from the whole matrix — careful: factoring $-1$ from each of the 3 rows gives $(-1)^3 = -1$ times the determinant of the matrix with entries $1, \omega, \omega^2$ etc.). But we can also check directly.

    4. Observe the pattern.

       Look at the first and third rows:

       Row 1: $(-1, -\omega, -\omega^2)$

       Row 3: $(-\omega, -1, -\omega^2)$

       These are not multiples, but notice that the second row is $(-\omega^2, -1, -\omega)$.

       A faster way: add all three rows. The sum of the three rows is:

    $(-1 - \omega^2 - \omega,\; -\omega -1 -1,\; -\omega^2 - \omega - \omega^2)$

    The first column sum: $-1 - \omega^2 - \omega = -(1+\omega+\omega^2) = 0$.

    The second column sum: $-\omega -1 -1 = -\omega -2$, which is not obviously zero. So row addition doesn’t give a zero row directly.

    Instead, note that each column is a cyclic shift of $(-1, -\omega^2, -\omega)$? Let’s check column 1: $(-1, -\omega^2, -\omega)$. Column 2: $(-\omega, -1, -1)$. That’s not a simple shift. …

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11. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.A and B are two non square matrices. If $P = A + B$, $Q = A^T B$, $R = AB^T$, then the matrices whose order is equal to the order of A are (A) PQ and QR (B) RQ and QP (C) PQ and RP (D) PQR and RPQ

    ›Reveal solutionSolution

    The key idea is that matrix multiplication is only defined when the inner dimensions match. By tracking the orders of $A$ and $B$ (which are non‑square), we can determine which products have the same order as $A$. The matrices that match $A$'s order are $PQ$ and $RP$, so the correct option is (C).

    The whole problem hinges on one simple rule: you can multiply two matrices only if the number of columns in the first equals the number of rows in the second. The order of the result is then (rows of first) × (columns of second). Since $A$ and $B$ are non‑square, their orders are different — let’s give them names so we can track everything cleanly.

    Let $A$ be of order $m \times n$ and $B$ be of order $p \times q$. Because they are non‑square, we know $m \neq n$ and $p \neq q$, but that’s not the main constraint here. The real constraint comes from the definitions of $P$, $Q$, and $R$.

    1. Order of $P = A + B$

       Addition is only defined if both matrices have the same order. So $A$ and $B$ must have the same order for $P$ to exist. That forces $m = p$ and $n = q$.

       Hence $P$ is also $m \times n$ — the same order as $A$.

    2. Order of $Q = A^T B$

       $A^T$ is $n \times m$. For $A^T B$ to be defined, the columns of $A^T$ (which is $m$) must equal the rows of $B$ (which is $p$). So $m = p$.

       The result $Q$ then has order $n \times q$.

       But from step 1 we already have $m = p$ and $n = q$, so $Q$ is $n \times n$ — a square matrix, not the same as $A$ ($m \times n$).

    3. Order of $R = A B^T$

       $B^T$ is $q \times p$. For $A B^T$ to be defined, columns of $A$ ($n$) must equal rows of $B^T$ ($q$). So $n = q$.

       The result $R$ has order $m \times p$. With $m = p$ from step 1, $R$ is $m \times m$ — again square, not the same as $A$.

    Now we check each product in the options, always asking: does the result have order $m \times n$ (the order of $A$)?

    • $PQ$: $P$ is $m \times n$, $Q$ is $n \times n$. Multiplication is defined (inner $n$ matches), and the result is $m \times n$ — same as $A$. ✓

    • $QR$: $Q$ is $n \times n$, $R$ is $m \times m$. For multiplication, we need $n = m$. Is that true? From step 1, $m = p$ and $n = q$; from step 3, $n = q$. So $n = q$ and $m = p$, but $m$ and $n$ are not necessarily equal (they are the dimensions of a non‑square matrix $A$, so $m \neq n$). Hence $QR$ is not defined. ✗ …

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12. TG EAPCET 2025Set eng-2025-05-02-AN1 markMCQ
    Q.A, C are $3 \times 3$ matrices. B, D are $3 \times 1$ matrices. If $AX = B$ has unique solution and $CX = D$ has infinite number of solutions, then (A) rank of $[A : D] =$ rank of $[C : B]$ (B) rank of $A =$ rank of $C$ (C) rank of $[A : B] <$ rank of $[B : D]$ (D) rank of $[A : D] \geq$ rank of $[C : B]$

    ›Reveal solutionSolution

    The unique solution of $AX = B$ forces $A$ to be invertible (rank 3), while infinite solutions of $CX = D$ mean $C$ is singular (rank $< 3$). Augmenting $A$ with any column preserves rank 3, while augmenting $C$ with any column gives rank at most 3, so rank of $[A : D] \geq$ rank of $[C : B]$.

    The heart of this problem lies in what the solution structure tells us about the coefficient matrices.

    When a linear system $AX = B$ has a unique solution, the coefficient matrix $A$ must be invertible. For a $3 \times 3$ matrix, this means $\text{rank}(A) = 3$. The system has exactly one solution because the columns of $A$ span all of $\mathbb{R}^3$ and are linearly independent.

    When a system $CX = D$ has infinitely many solutions, the coefficient matrix $C$ must be singular (non-invertible) and the system must be consistent. This means $\text{rank}(C) < 3$. The infinite solutions arise because the null space of $C$ is non-trivial—there are free variables.

    Now let's examine what happens when we form augmented matrices by appending columns.

    1. Analyze $[A : D]$ We know $\text{rank}(A) = 3$. When we append the column vector $D$ to form the $3 \times 4$ matrix $[A : D]$, the rank can be at most 3 (since we're in $\mathbb{R}^3$). Since $A$ already has rank 3, its columns span $\mathbb{R}^3$, so $D$ can be written as a linear combination of $A$'s columns. Therefore:

    $\text{rank}[A : D] = 3$

    2. Analyze $[C : B]$ We know $\text{rank}(C) < 3$ (either 0, 1, or 2). When we append $B$ to form $[C : B]$, the rank can increase by at most 1. The maximum possible rank is:

    $\text{rank}[C : B] \leq \min(3, \text{rank}(C) + 1) \leq 3$

    More precisely, since $\text{rank}(C) \leq 2$, we have $\text{rank}[C : B] \leq 3$.

    3. Compare the ranks

       From step 1: $\text{rank}[A : D] = 3$

       From step 2: $\text{rank}[C : B] \leq 3$

       Therefore:

    $\text{rank}[A : D] \geq \text{rank}[C : B]$

    4. Check the other options …

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