Q.Find the matrix $A$ satisfying the matrix equation: $\begin{bmatrix} 2 & 1 \\ 3 & 2 \end{bmatrix} A \begin{bmatrix} -3 & 2 \\ 5 & -3 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$.

Showing the 12 most recent of 32 on this concept.

1. AP EAPCET 2023Set eng-2023-05-19-FN1 markMCQ
   Q.If $A=\begin{bmatrix}2 & 1 & 3 & -1\\1 & -2 & 2 & -3\end{bmatrix}, B=\begin{bmatrix}2 & 1 & 0 & 3\\1 & -1 & 2 & 3\end{bmatrix}$ and $2A+3B-5C=O$, then $C=$ (A) $\begin{bmatrix}2 & 1 & 6/5 & 7/5\\1 & 7/5 & 2 & 3/5\end{bmatrix}$ (B) $\begin{bmatrix}-2 & 1 & 6/5 & 7/5\\1 & -7/5 & 2 & 3/5\end{bmatrix}$ (C) $\begin{bmatrix}-2 & 1 & 6/5 & 7/5\\1 & 7/5 & 2 & 3/5\end{bmatrix}$ (D) $\begin{bmatrix}2 & 1 & 6/5 & 7/5\\1 & -7/5 & 2 & 3/5\end{bmatrix}$

   ›Reveal solutionSolution

   Direct matrix arithmetic: $C=(2A+3B)/5$, computed entrywise, gives $\begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$.

   Concept and Intuition

   This is pure entrywise matrix algebra — scale each matrix, add, then divide by 5 (since $5C=2A+3B$).

   Step-by-Step Solution

   1. $2A = \begin{bmatrix}4&2&6&-2\\2&-4&4&-6\end{bmatrix}$.
   2. $3B = \begin{bmatrix}6&3&0&9\\3&-3&6&9\end{bmatrix}$.
   3. $2A+3B = \begin{bmatrix}10&5&6&7\\5&-7&10&3\end{bmatrix}$.
   4. $C = \dfrac{1}{5}(2A+3B) = \begin{bmatrix}2&1&6/5&7/5\\1&-7/5&2&3/5\end{bmatrix}$. …

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2. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 1 & 3 \\ 5 & 2 & 6 \\ -2 & -1 & -3 \end{bmatrix}$, then $A + A^3 + A^4 + A^5 + 3I =$ (A) $\begin{bmatrix} 4 & 2 & 1 \\ 2 & 5 & 6 \\ -3 & 2 & 3 \end{bmatrix}$ (B) $\begin{bmatrix} 4 & 1 & 3 \\ 5 & 5 & 6 \\ -2 & -1 & 0 \end{bmatrix}$ (C) $\begin{bmatrix} 3 & 1 & 4 \\ 3 & 1 & -2 \\ -1 & 2 & -1 \end{bmatrix}$ (D) $\begin{bmatrix} 4 & 1 & 3 \\ 2 & 3 & 5 \\ -3 & -2 & -3 \end{bmatrix}$

   ›Reveal solutionSolution

   A's characteristic polynomial is lambda^3=0, so by Cayley-Hamilton A^3=0 (hence A^4=A^5=0 too), collapsing the sum to A+3I, which matches option B.

   Concept and Intuition

   Rather than computing $A^3,A^4,A^5$ by repeated multiplication, use the Cayley-Hamilton theorem: every square matrix satisfies its own characteristic equation. If the characteristic polynomial turns out to be $\lambda^3=0$, we immediately get $A^3=0$, collapsing the higher powers.

   Step-by-Step Solution

   1. $\mathrm{tr}(A) = 1+2+(-3) = 0$.
   2. $\det(A) = 1[(2)(-3)-(6)(-1)] - 1[(5)(-3)-(6)(-2)] + 3[(5)(-1)-(2)(-2)] = 1(0) - 1(-3) + 3(-1) = 0+3-3=0$.
   3. Sum of principal 2x2 minors: $M_{11}=\begin{vmatrix}2&6\\-1&-3\end{vmatrix}=0$; $M_{22}=\begin{vmatrix}1&3\\-2&-3\end{vmatrix}=3$; $M_{33}=\begin{vmatrix}1&1\\5&2\end{vmatrix}=-3$. Sum $=0$. …

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3. AP EAPCET 2022Set eng-2022-07-06-FN1 markMCQ
   Q.If $A = \begin{pmatrix}3 & 4\\5 & 6\end{pmatrix}$ and $B = \begin{pmatrix}x & 0\\0 & y\end{pmatrix}$, $x,y\in N$, then (A) There is exactly one such matrix B such that AB = I (B) There is no matrix B such that AB = BA (C) There exist only a finite number of matrices B such that AB = BA (D) There exist infinite number of matrices B such that AB = BA

   ›Reveal solutionSolution

   Multiplying out $AB$ and $BA$ shows they're equal exactly when $x=y$, and since $x,y$ can be any of the infinitely many natural numbers with $x=y$, there are infinitely many commuting diagonal matrices B.

   Concept and Intuition

   For a diagonal matrix $B=\text{diag}(x,y)$ multiplying a general matrix $A$ on the left vs. right scales A's rows vs. columns differently — $AB$ scales A's columns by $x,y$ respectively, and $BA$ scales A's rows by $x,y$ respectively. So $AB=BA$ becomes a condition relating how each off-diagonal entry of $A$ gets scaled from each side, and typically forces the diagonal entries of B to be equal whenever the off-diagonal entries of A are both non-zero (as they are here).

   Step-by-Step Solution

   1. Compute $AB=\begin{pmatrix}3&4\\5&6\end{pmatrix}\begin{pmatrix}x&0\\0&y\end{pmatrix}=\begin{pmatrix}3x&4y\\5x&6y\end{pmatrix}$ (this scales each column of A by x then y).
   2. Compute $BA=\begin{pmatrix}x&0\\0&y\end{pmatrix}\begin{pmatrix}3&4\\5&6\end{pmatrix}=\begin{pmatrix}3x&4x\\5y&6y\end{pmatrix}$ (this scales each row of A by x then y).
   3. Set $AB=BA$ entrywise: $(1,1)$: $3x=3x$ (always true); $(1,2)$: $4y=4x\Rightarrow x=y$; $(2,1)$: $5x=5y\Rightarrow x=y$ (same condition); $(2,2)$: $6y=6y$ (always true).
   4. So the only requirement is $x=y$, with $x,y\in\mathbb{N}$. Since $x$ can be $1,2,3,\dots$ (infinitely many choices, each giving a valid B with $x=y$), there are infinitely many such matrices B. …

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4. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & 3 & 3 \\ 3 & 2 & 3 \\ 3 & 3 & 2 \end{bmatrix}$, then $A^2 - 8I =$ (A) $0$ (B) $8A$ (C) $7A$ (D) $5A$

   ›Reveal solutionSolution

   The key idea is to compute $A^2$ directly and then subtract $8I$; the result simplifies to $5A$, so the correct option is (D).

   We start with the matrix

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

   The problem asks for $A^2 - 8I$. Instead of guessing, we can compute $A^2$ by matrix multiplication and then subtract $8$ times the identity. This is a straightforward algebraic check — no need for eigenvalues or characteristic polynomials here, though those would also work.

   1. Compute $A^2$ Multiply $A$ by itself. The $(i,j)$ entry of $A^2$ is the dot product of row $i$ of $A$ with column $j$ of $A$.
      • For diagonal entries (e.g., $(1,1)$):

   $(2)(2) + (3)(3) + (3)(3) = 4 + 9 + 9 = 22.$

    By symmetry, all diagonal entries are $22$.
   

   • For off-diagonal entries (e.g., $(1,2)$):

   $(2)(3) + (3)(2) + (3)(3) = 6 + 6 + 9 = 21.$

    Again by symmetry, every off-diagonal entry is $21$.  
   

   So

   $A^2 = \begin{bmatrix} 22 & 21 & 21 \\ 21 & 22 & 21 \\ 21 & 21 & 22 \end{bmatrix}.$

   2. Subtract $8I$ The identity matrix $I$ has 1’s on the diagonal and 0’s elsewhere. Thus

   $A^2 - 8I = \begin{bmatrix} 22-8 & 21 & 21 \\ 21 & 22-8 & 21 \\ 21 & 21 & 22-8 \end{bmatrix} = \begin{bmatrix} 14 & 21 & 21 \\ 21 & 14 & 21 \\ 21 & 21 & 14 \end{bmatrix}.$

   3. Compare with multiples of $A$ Notice that $A$ is

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

   Multiply $A$ by 5:

   $$$$

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5. AP EAPCET 2024Set eng-2024-05-19-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$ and $\alpha A^2 + \beta A = 2I$ for some $\alpha, \beta \in \mathbb{R}$ then $\alpha + \beta =$ (A) 7 (B) 10 (C) 12 (D) 5

   ›Reveal solutionSolution

   Computing $A^2$ and matching the matrix equation $\alpha A^2+\beta A=2I$ entry-by-entry gives $\alpha=2,\ \beta=8$, so $\alpha+\beta=10$ — option (B).

   Concept and Intuition

   Given a $2\times2$ matrix satisfying a polynomial relation like $\alpha A^2+\beta A = 2I$, the cleanest approach is to compute $A^2$ directly, then equate corresponding entries of both sides of the matrix equation — this converts a single matrix equation into a small system of linear equations in the unknown scalars $\alpha,\beta$ (using just two independent entries is enough, and the remaining entries serve as a consistency check, since the relation must hold for the whole matrix, not just isolated numbers — this consistency is itself guaranteed for genuine matrix polynomial identities via Cayley–Hamilton-type reasoning, but verifying arithmetic keeps you safe under exam conditions).

   Step-by-Step Solution

   1. Compute $A^2$: $A^2=\begin{bmatrix}1&2\\-2&-5\end{bmatrix}\begin{bmatrix}1&2\\-2&-5\end{bmatrix} = \begin{bmatrix}1(1)+2(-2) & 1(2)+2(-5)\\-2(1)+(-5)(-2) & -2(2)+(-5)(-5)\end{bmatrix} = \begin{bmatrix}-3&-8\\8&21\end{bmatrix}$.
   2. Write $\alpha A^2+\beta A = 2I$ entry-wise:
      • $(1,1)$: $-3\alpha+\beta = 2$
      • $(1,2)$: $-8\alpha+2\beta = 0$ …

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6. AP EAPCET 2024Set eng-2024-05-22-AN1 markMCQ
   Q.$A=\begin{bmatrix}0 & 1 & 2\\ 2 & 3 & 0\\ 4 & 0 & 3\end{bmatrix}$ and B is a matrix such that $AB=BA$. If AB is not an identity matrix, then the matrix that can be taken as B is (A) $\begin{bmatrix}-9 & -3 & 6\\ -6 & 8 & -4\\ 12 & -4 & -2\end{bmatrix}$ (B) $\begin{bmatrix}9 & -3 & 6\\ -6 & 8 & -4\\ -12 & -4 & 2\end{bmatrix}$ (C) $\begin{bmatrix}9 & -3 & -6\\ -6 & 8 & -4\\ -12 & 4 & -2\end{bmatrix}$ (D) $\begin{bmatrix}9 & -3 & -6\\ -6 & -8 & 4\\ -12 & 4 & -2\end{bmatrix}$

   ›Reveal solutionSolution

   Direct computation is the only reliable way here — multiplying $A$ by each candidate $B$ shows option (D) is the unique one where $AB=BA$ (in fact $AB=BA=-30I$, a non-identity scalar matrix), so it satisfies every condition in the question.

   Concept and Intuition

   With no special structure making $B$ an obvious polynomial in $A$, the only certain way to check "does $B$ commute with $A$" among four numerically given candidates is to actually multiply both products, $AB$ and $BA$, and compare entry by entry. The question's extra clause — "if $AB$ is not an identity matrix" — is there to rule out a trivial commuting case ($B=A^{-1}$ scaled so $AB=I$), so once we find the candidate that truly commutes, we should double check its product isn't simply $I$.

   Step-by-Step Solution

   1. $A=\begin{bmatrix}0&1&2\\2&3&0\\4&0&3\end{bmatrix}$.
   2. Testing option (D), $B=\begin{bmatrix}9&-3&-6\\-6&-8&4\\-12&4&-2\end{bmatrix}$:
   3. Compute $AB$ row by row: row 1 of $A$ is $(0,1,2)$, so $(AB)_{1j}=0\cdot B_{1j}+1\cdot B_{2j}+2\cdot B_{3j}$, giving $(AB)_{1,\cdot}=(0-6-24,\,0-8+8,\,0+4-4)=(-30,0,0)$. Similarly rows 2 and 3 give $(0,-30,0)$ and $(0,0,-30)$.
   4. So $AB=\begin{bmatrix}-30&0&0\\0&-30&0\\0&0&-30\end{bmatrix}=-30I$.
   5. Compute $BA$ the same way (row of $B$ dotted into columns of $A$): it also comes out to $-30I$. …

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7. AP EAPCET 2023Set eng-2023-05-18-AN1 markMCQ
   Q.If $A+2B=\begin{pmatrix}1 & 2 & 0\\ 6 & -3 & 3\\ -5 & 3 & 1\end{pmatrix}$ and $2A-B=\begin{pmatrix}2 & -1 & 5\\ 2 & -1 & 6\\ 0 & 1 & 2\end{pmatrix}$, then $\mathrm{Tr}[A]-\mathrm{Tr}[B]=$ (A) $1$ (B) $2$ (C) $3$ (D) $4$

   ›Reveal solutionSolution

   Trace is linear, so instead of solving for $A$ and $B$ as full matrices, take the trace of each given equation and solve the resulting $2\times2$ linear system.

   Concept and Intuition

   Trace (sum of diagonal entries) is a linear functional: $\mathrm{Tr}[A+2B]=\mathrm{Tr}[A]+2\mathrm{Tr}[B]$. So applying trace to both matrix equations converts a matrix problem into a scalar linear system in $a=\mathrm{Tr}[A]$ and $b=\mathrm{Tr}[B]$.

   Step-by-Step Solution

   1. Trace of first matrix: $1+(-3)+1=-1$. So $a+2b=-1$.
   2. Trace of second matrix: $2+(-1)+2=3$. So $2a-b=3$.
   3. From equation 1: $a=-1-2b$.
   4. Substitute into equation 2: $2(-1-2b)-b=3 \Rightarrow -2-4b-b=3 \Rightarrow -5b=5 \Rightarrow b=-1$.
   5. Then $a=-1-2(-1)=1$. …

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8. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1 \end{bmatrix}$ is such that $A^2 = I$, then (A) $b = \dfrac{ac}{2}$ (B) $b = -\dfrac{ac}{2}$ (C) $b = \dfrac{a+c}{2}$ (D) $b = \sqrt{ac}$

   ›Reveal solutionSolution

   Multiplying the lower-triangular matrix $A$ by itself and forcing the result to equal $I$ gives a single condition on $a,b,c$: $b=-\dfrac{ac}{2}$.

   Concept and Intuition

   $A$ is lower triangular with $1$'s on the diagonal (except the $-1$ in the middle). Squaring a triangular matrix keeps it triangular, and the diagonal entries of $A^2$ are just the squares of the diagonal entries of $A$ ($1^2=1$, $(-1)^2=1$, $1^2=1$), which already match $I$'s diagonal automatically. The real content of the condition $A^2=I$ is in the off-diagonal (lower-triangular) entries.

   Step-by-Step Solution

   1. Write $A=\begin{bmatrix}1&0&0\\a&-1&0\\b&c&1\end{bmatrix}$.
   2. Compute $A^2 = A\cdot A$ row by row.
      • Row 1: $[1,0,0]\cdot A = [1,0,0]$ (matches $I$ automatically).
      • Row 2: $[a,-1,0]\cdot A$: first entry $= a(1)+(-1)(a)+0(b) = 0$; second entry $= a(0)+(-1)(-1)+0(c)=1$; third entry $=0$. So row 2 $=[0,1,0]$ (matches $I$ automatically, for any $a$).
      • Row 3: $[b,c,1]\cdot A$: first entry $= b(1)+c(a)+1(b) = 2b+ac$; second entry $= b(0)+c(-1)+1(c) = 0$; third entry $= 0+0+1=1$. So row 3 $=[2b+ac,\ 0,\ 1]$. …

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9. AP EAPCET 2021Set eng-2021-08-25-AN1 markMCQ
   Q.Let $A = \begin{bmatrix} 1 & 3 \\ 4 & -3 \end{bmatrix}$. Let $S = \left\{ \begin{bmatrix} x \\ y \end{bmatrix} \in R^2 \;/\; A\begin{bmatrix} x \\ y \end{bmatrix} = 3\begin{bmatrix} x \\ y \end{bmatrix} \right\}$ what is the cardinality of $S$? (A) 1 (B) Countably infinite (C) $|S| > 1$ but $S$ is finite (D) Uncountable

   ›Reveal solutionSolution

   $S$ is the eigenspace for eigenvalue $3$; since $\det(A-3I)=0$, it is a nontrivial line through the origin containing infinitely (uncountably) many vectors. Answer: (D).

   Concept and Intuition

   The condition $A\mathbf{v}=3\mathbf{v}$ means $\mathbf{v}$ is an eigenvector of $A$ for eigenvalue $3$ (or the zero vector). The solution set to $(A-3I)\mathbf{v}=0$ is either just $\{0\}$ (if $3$ is not an eigenvalue) or an entire subspace (a line, in 2D) if it is.

   Step-by-Step Solution

   1. $A - 3I = \begin{bmatrix}1-3 & 3\\ 4 & -3-3\end{bmatrix} = \begin{bmatrix}-2&3\\4&-6\end{bmatrix}$.
   2. $\det(A-3I) = (-2)(-6)-(3)(4) = 12-12=0$. …

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10. AP EAPCET 2022Set eng-2022-07-05-AN1 markMCQ
    Q.If $A = \begin{bmatrix} x & 0 \\ 1 & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 8 & 0 \\ 7 & 1 \end{bmatrix}$ and $A^3 = B$, then $x =$ (A) $-2$ or $3$ (B) $-2$ (C) $2$ or $-3$ (D) $2$

    ›Reveal solutionSolution

    Directly cubing the lower-triangular matrix $A$ and matching both entries to $B$ pins down $x=2$ uniquely. The answer is (D).

    Concept and Intuition

    $A$ is lower triangular, so its powers stay lower triangular, and the diagonal entries of $A^n$ are just the diagonal entries of $A$ raised to the $n$-th power — this makes computing $A^3$ by hand straightforward via repeated matrix multiplication.

    Step-by-Step Solution

    1. $A=\begin{bmatrix}x&0\\1&1\end{bmatrix}$.
    2. $A^2 = A\cdot A = \begin{bmatrix}x^2&0\\x+1&1\end{bmatrix}$ (bottom-left entry: $1\cdot x+1\cdot1=x+1$).
    3. $A^3 = A^2\cdot A = \begin{bmatrix}x^3&0\\(x+1)x+1&1\end{bmatrix} = \begin{bmatrix}x^3&0\\x^2+x+1&1\end{bmatrix}$.
    4. Set $A^3=B=\begin{bmatrix}8&0\\7&1\end{bmatrix}$: from the $(1,1)$ entry, $x^3=8\Rightarrow x=2$ (the unique real cube root). …

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11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.In solving a system of linear equations $AX=B$ by Cramer's rule, in the usual notation, if $\Delta_1=\begin{vmatrix}-11 & 1 & -7\\ -4 & 1 & -2\\ 5 & 1 & 1\end{vmatrix}$ and $\Delta_3=\begin{vmatrix}4 & 1 & -11\\ 1 & 1 & -4\\ 4 & 1 & 5\end{vmatrix}$, then $X=$ (A) $\begin{pmatrix}-1\\ 1\\ 2\end{pmatrix}$ (B) $\begin{pmatrix}2\\ 1\\ -1\end{pmatrix}$ (C) $\begin{pmatrix}1\\ -1\\ 2\end{pmatrix}$ (D) $\begin{pmatrix}1\\ 2\\ -1\end{pmatrix}$

    ›Reveal solutionSolution

    Reconstructing the coefficient matrix and RHS vector from the shared columns of $\Delta_1,\Delta_3$ and computing $\Delta,\Delta_1,\Delta_2,\Delta_3$ gives $X=(1,-1,2)$.

    Concept and Intuition

    In Cramer's rule for $AX=B$, $\Delta_i$ is formed by replacing the $i$-th column of $A$ with $B$, keeping the other columns unchanged. So $\Delta_1$'s columns 2,3 and $\Delta_3$'s columns 1,2 are literally the ORIGINAL matrix $A$'s corresponding columns — comparing the two given determinants lets us recover the full system.

    Step-by-Step Solution

    1. $\Delta_1=\begin{vmatrix}-11&1&-7\\-4&1&-2\\5&1&1\end{vmatrix}$: column 2 = original $A$'s column 2 $=(1,1,1)^T$; column 3 = original $A$'s column 3 $=(-7,-2,1)^T$; column 1 $=B=(-11,-4,5)^T$.
    2. $\Delta_3=\begin{vmatrix}4&1&-11\\1&1&-4\\4&1&5\end{vmatrix}$: column 1 = original $A$'s column 1 $=(4,1,4)^T$; column 2 = same $(1,1,1)^T$ (consistent); column 3 $=B=(-11,-4,5)^T$ (matches $\Delta_1$'s column 1, confirming $B$).
    3. Reconstructed system: $A=\begin{pmatrix}4&1&-7\\1&1&-2\\4&1&1\end{pmatrix}$, $B=\begin{pmatrix}-11\\-4\\5\end{pmatrix}$.
    4. $\Delta=\det(A)=4(1{\cdot}1-(-2){\cdot}1)-1(1{\cdot}1-(-2){\cdot}4)+(-7)(1{\cdot}1-1{\cdot}4)=4(3)-1(9)-7(-3)=12-9+21=24$. …

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12. AP EAPCET 2021Set eng-2021-08-19-AN1 markMCQ
    Q.$A = \begin{bmatrix} a^{2} & 15 & 31 \\ 12 & b^{2} & 41 \\ 35 & 61 & c^{2} \end{bmatrix}$ and $B = \begin{bmatrix} 2a & 3 & 5 \\ 2 & 2b & 8 \\ 1 & 4 & 2c-3 \end{bmatrix}$ are two matrices such that the sum of the principal diagonal elements of both $A$ and $B$ are equal, then the product of the principal diagonal elements of $B$ is ________ (A) 4 (B) 0 (C) $-4$ (D) $-12$

    ›Reveal solutionSolution

    Equating the traces of A and B forces $(a-1)^2+(b-1)^2+(c-1)^2=0$, so $a=b=c=1$; substituting into B's diagonal product gives $2\times2\times(-1)=-4$.

    Concept and Intuition

    When an equation reduces to a sum of squares equal to zero, each square must individually be zero (since squares of real numbers are never negative) — this is a very common and powerful trick to pin down exact values of multiple unknowns from a single scalar equation.

    Step-by-Step Solution

    1. Trace (sum of principal diagonal elements) of $A$: $a^2+b^2+c^2$.
    2. Trace of $B$: $2a+2b+(2c-3)$.
    3. Given these are equal: $a^2+b^2+c^2 = 2a+2b+2c-3$.
    4. Rearrange: $a^2-2a + b^2-2b + c^2-2c+3 = 0$.
    5. Complete the square for each variable: $(a^2-2a+1) + (b^2-2b+1) + (c^2-2c+1) = 0$, i.e., $(a-1)^2+(b-1)^2+(c-1)^2 = 0$.
    6. Since each square term is $\ge 0$ and they sum to zero, each must be exactly zero: $a=1,\ b=1,\ c=1$. …

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