Q.(a) If $A = \begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}$, then show that $A^3 - 6A^2 + 7A + 2I = O$.

Showing the 12 most recent of 16 on this concept.

1. CBSE 2026Set 65/3/11 markMCQ
   Q.If $A^2 = 4A + 3I$ and $A^{-1} = xA + yI$, then the value of $(x+y)$ is: (A) $-1$ (B) $1$ (C) $\frac{5}{3}$ (D) $7$

   ›Reveal solutionSolution

   The key idea is to multiply the given matrix equation by $A^{-1}$ to express $A$ in terms of $I$, then compare coefficients with the given form of $A^{-1}$. The value of $(x+y)$ is $\boxed{\frac{5}{3}}$.

   Concept & Intuition

   When a matrix satisfies a polynomial equation like $A^2 = 4A + 3I$, it means $A$ behaves like a root of that polynomial. We can manipulate this equation algebraically just like we would with numbers — but with matrices, we must be careful about commutativity (here, $A$ commutes with itself and with $I$, so we're safe).

   The trick: if we multiply both sides by $A^{-1}$ (which exists, as we'll see), we get a linear expression for $A$ in terms of $I$. Then we can substitute that back into the given form $A^{-1} = xA + yI$ to find $x$ and $y$.

   Step-by-step solution

   1. Start with the given equation

      $A^2 = 4A + 3I$

      This is a matrix equation — every term is a $2\times2$ (or $n\times n$) matrix.

   2. Multiply both sides by $A^{-1}$ on the left

      Since $A^{-1}A = I$, we get:

      $A^{-1}A^2 = A^{-1}(4A + 3I)$

      $\Rightarrow (A^{-1}A)A = 4A^{-1}A + 3A^{-1}I$

      $\Rightarrow IA = 4I + 3A^{-1}$

      So:

      $A = 4I + 3A^{-1}$

   3. Rearrange to express $A^{-1}$ in terms of $A$ and $I$

      $3A^{-1} = A - 4I$

      $\Rightarrow A^{-1} = \frac{1}{3}A - \frac{4}{3}I$

   4. Compare with the given form

      We are told $A^{-1} = xA + yI$.

      Matching coefficients:

      $x = \frac{1}{3}$,

      $y = -\frac{4}{3}$ …

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2. CBSE 2026Set A1 markMCQ
   Q.If $A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$, then $A^3 =$

   (a) $3A$

   (b) $4A$

   (c) $2A$

   (d) None of these

   ›Reveal solutionSolution

   $A^2 = 2A$, so $A^3 = 2A^2 = 4A$.

   With $A = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$, first compute $A^2$: …

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3. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A$ is an invertible matrix of order 2, then $\det(A^{-1})$ is equal to:

   (a) $\det(A)$

   (b) $\frac{1}{\det(A)}$

   (c) $1$

   (d) $0$

   ›Reveal solutionSolution

   $\det(A^{-1})=\dfrac{1}{\det A}$.

   From $AA^{-1}=I$, $\det(A)\det(A^{-1})=\det(I)=1$, so $\det(A^{-1})=\dfrac{1}{\det(A)}$. (This holds for a …

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4. CBSE 2026Set ANNUAL1 mark
   Q.Fill in the blank: Unique solution of equation $AX = B$ is given by $X =$ ______, where $|A| \neq 0$.

   ›Reveal solutionSolution

   The unique solution of $AX=B$ (when $|A| e0$) is $X=A^{-1}B$.

   …

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5. CBSE 2026Set ANNUAL1 markMCQ
   Q.If $A=\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$, then $A^{2026}$ is equal to

   (a) $\begin{bmatrix} 0 & 1 \\ 0 & 0 \end{bmatrix}$

   (b) $\begin{bmatrix} 0 & 2026 \\ 0 & 0 \end{bmatrix}$

   (c) $\begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix}$

   (d) $\begin{bmatrix} 2026 & 0 \\ 0 & 2026 \end{bmatrix}$

   ›Reveal solutionSolution

   $A^2=O$, hence $A^{2026}=O$ (zero matrix).

   Compute $A^{2}$:

   $A^{2}=\begin{bmatrix}0&1\\0&0\end{bmatrix}\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0\cdot0+1\cdot0 & 0\cdot1+1\cdot0\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}=O.$

   …

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6. CBSE 2025Set 65/2/11 markMCQ
   Q.If $A$ and $B$ are square matrices of order $m$ such that $A^2 - B^2 = (A - B)(A + B)$, then which of the following is always correct? (A) $A = B$ (B) $AB = BA$ (C) $A = 0$ or $B = 0$ (D) $A = I$ or $B = I$

   ›Reveal solutionSolution

   The given matrix identity $A^2 - B^2 = (A - B)(A + B)$ holds true if and only if the matrices $A$ and $B$ commute, meaning $AB = BA$. The correct option is (B).

   Concept and Intuition

   In scalar algebra, we are accustomed to the identity $a^2 - b^2 = (a - b)(a + b)$. This identity relies on the commutative property of multiplication, where $ab = ba$. For example, when we expand $(a - b)(a + b)$, we get $a^2 + ab - ba - b^2$, and since $ab = ba$, the middle terms cancel out, leaving $a^2 - b^2$.

   However, matrix multiplication is generally not commutative. That is, for two matrices $A$ and $B$, it is usually the case that $AB \neq BA$. This non-commutativity is the crucial difference that this problem tests. If we blindly apply scalar algebra rules to matrices without considering the order of multiplication, we might make an error. The problem statement essentially gives us a condition under which the scalar identity does hold for matrices, and we need to find out what that condition implies about $A$ and $B$.

   Step-by-step Derivation

   1. Start with the given equation: We are given that $A$ and $B$ are square matrices of order $m$ such that:

   $A^2 - B^2 = (A - B)(A + B)$

   2. Expand the right-hand side carefully: When multiplying matrices, we must maintain the order of multiplication. We distribute $(A - B)$ over $(A + B)$:

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

   Now, distribute $A$ and $B$ into their respective parentheses:
   

   $A(A + B) - B(A + B) = A \cdot A + A \cdot B - B \cdot A - B \cdot B$

   This simplifies to:
   

   $A^2 + AB - BA - B^2$

   > [!WARNING]
   > A common mistake is to assume $AB = BA$ from the start, which would incorrectly simplify $AB - BA$ to $0$. Remember that matrix multiplication is not generally commutative.
   

   3. Substitute the expanded form back into the original equation:

   Now we equate the left-hand side of the given equation with our expanded right-hand side:

   $A^2 - B^2 = A^2 + AB - BA - B^2$

   4. Simplify the equation: We can subtract $A^2$ from both sides of the equation:

   $-B^2 = AB - BA - B^2$

   Next, add $B^2$ to both sides:
   

   $0 = AB - BA$

   Rearranging this equation, we get:
   

   $AB = BA$

   5. Interpret the result:

      The derivation shows that for the identity $A^2 - B^2 = (A - B)(A + B)$ to hold true for matrices $A$ and $B$, it is necessary that $AB = BA$. This means that matrices $A$ and $B$ must commute.

   6. Evaluate the given options: …

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7. CBSE 2025Set 65/4/11 markMCQ
   Q.If A and B are square matrices of same order such that $AB = A$ and $BA = B$, then $A^2 + B^2$ is equal to : (A) $A + B$ (B) $BA$ (C) $2(A + B)$ (D) $2BA$

   ›Reveal solutionSolution

   When $AB = A$ and $BA = B$, the matrices satisfy idempotent-like relations that allow us to express higher powers in terms of the originals; computing $A^2 + B^2$ using these relations yields $A + B$.

   The key insight here is to use the given relations to simplify powers of $A$ and $B$. We're told that $AB = A$ and $BA = B$, which means each matrix "absorbs" the other in a specific order. These relations are our tools to reduce any product back to simpler forms.

   Let's compute $A^2$ and $B^2$ separately, then add them.

   Finding $A^2$:

   1. Write $A^2 = A \cdot A$.

   2. We need to express this using our given relations. Notice that from $AB = A$, we can write $A = AB$.

   3. Substitute this into $A^2$:

   $A^2 = A \cdot A = (AB) \cdot A = A(BA)$

   4. But we know $BA = B$, so:

   $A^2 = A(BA) = AB = A$

   Finding $B^2$:

   1. Write $B^2 = B \cdot B$.

   2. From $BA = B$, we have $B = BA$.

   3. Substitute:

   $B^2 = B \cdot B = (BA) \cdot B = B(AB)$

   4. Using $AB = A$: $B^2 = B(AB) = BA = B$ …

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8. CBSE 2025Set ANNUAL1 markMCQ
   Q.If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ be such that $A^{-1} = kA$, then $k =$

   (a) $19$

   (b) $\dfrac{1}{19}$

   (c) $-19$

   (d) $-\dfrac{1}{19}$

   ›Reveal solutionSolution

   Compute |A| and adj(A), form A⁻¹, and compare it entry-by-entry with kA.

   $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$, $|A| = 2(-2) - 3(5) = -4-15 = -19$

   $\text{adj}(A) = \begin{bmatrix} -2 & -3 \\ -5 & 2 \end{bmatrix}$

   …

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9. CBSE 2024Set 65/1/11 markMCQ
   Q.If $A$ and $B$ are two non-zero square matrices of the same order such that $(A + B)^2 = A^2 + B^2$, then: (A) $AB = O$ (B) $AB = -BA$ (C) $BA = O$ (D) $AB = BA$

   ›Reveal solutionSolution

   The key idea is to expand $(A+B)^2$ and compare it with $A^2+B^2$ — the cross terms must cancel, which forces $AB = -BA$. The correct option is (B).

   Concept and Intuition

   When you square a sum of matrices, you get the same expansion as with numbers: $(A+B)^2 = A^2 + AB + BA + B^2$. The only difference is that matrix multiplication is not commutative — $AB$ and $BA$ are generally different. The given condition says this sum equals $A^2 + B^2$, so the two middle terms $AB$ and $BA$ must add up to the zero matrix. That means $AB + BA = O$, which rearranges to $AB = -BA$. This is the definition of anti-commuting matrices.

   Watch out

   A common mistake is to assume $AB = O$ or $BA = O$ individually. The condition only forces their sum to be zero, not each term separately. For example, if $A = \begin{pmatrix}0 & 1 \\ 0 & 0\end{pmatrix}$ and $B = \begin{pmatrix}0 & 0 \\ 1 & 0\end{pmatrix}$, then $AB \neq O$ and $BA \neq O$, but $AB = -BA$ holds.

   Step-by-Step Solution

   1. Expand the square Since matrix multiplication is distributive, we have:

   $(A+B)^2 = (A+B)(A+B) = A^2 + AB + BA + B^2.$

   This is exactly like the binomial expansion for numbers, but we must keep the order of multiplication.

   2. Apply the given condition The problem states:

   $(A+B)^2 = A^2 + B^2.$

   Substituting the expansion:

   $A^2 + AB + BA + B^2 = A^2 + B^2.$

   3. Cancel the common terms Subtract $A^2 + B^2$ from both sides:

   $AB + BA = O.$

   4. Rearrange to the required form The equation $AB + BA = O$ is equivalent to:

   $AB = -BA.$

   This is the defining relation for matrices that anti-commute. …

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10. CBSE 2024Set D1 markMCQ
    Q.If $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$ then $A^{-1} =$

    (a) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ \frac{1}{6} & \frac{1}{12} \end{bmatrix}$

    (b) $\begin{bmatrix} \frac{1}{4} & \frac{1}{8} \\ -\frac{1}{6} & \frac{1}{12} \end{bmatrix}$

    (c) $\begin{bmatrix} 4 & 8 \\ 6 & 12 \end{bmatrix}$

    (d) $\begin{bmatrix} 4 & 8 \\ -6 & 12 \end{bmatrix}$

    ›Reveal solutionSolution

    For a $2\times2$ matrix $A^{-1} = \frac{1}{\det A}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$.

    Here $A = \begin{bmatrix} 2 & -3 \\ 4 & 6 \end{bmatrix}$, so $\det A = (2)(6) - (-3)(4) = 12 + 12 = 24.$

    The adjoint (swap diagonal, negate off-diagonal) is $\begin{bmatrix} 6 & 3 \\ -4 & 2 \end{bmatrix}$. Hence …

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11. CBSE 2024Set D1 markMCQ
    Q.$A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} \Rightarrow A^5 =$

    (a) $\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}$

    (b) $\begin{bmatrix} 0 & 1 \\ 0 & 1 \end{bmatrix}$

    (c) $\begin{bmatrix} 0 & 5 \\ 5 & 0 \end{bmatrix}$

    (d) $\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

    ›Reveal solutionSolution

    $A^2 = I \Rightarrow A^5 = A$.

    Compute $A^2$:

    $A^2 = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix} = I.$

    Therefore …

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12. CBSE 2023Set 65/1/11 markMCQ
    Q.If $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$ and $(3I + 4A)(3I - 4A) = x^2I$, then the value(s) x is/are : (A) $\pm \sqrt{7}$ (B) 0 (C) $\pm 5$ (D) 25

    ›Reveal solutionSolution

    The key idea is to treat the matrix expression $(3I+4A)(3I-4A)$ as a polynomial in $A$, then use the fact that $A^2 = -I$ to simplify it to a scalar multiple of $I$. The result is $25I$, so $x^2 = 25$ and $x = \pm 5$.

    We start with the matrix $A = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix}$. Notice that $A$ is a special matrix — it behaves like the imaginary unit $i$ in complex numbers because $A^2 = -I$. Let’s verify:

    $A^2 = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} = -I$.

    This property is the heart of the problem. When we multiply two linear combinations of $I$ and $A$, the result will be a combination of $I$ and $A$ again, but because $A^2 = -I$, any $A^2$ term collapses back to a multiple of $I$. So the product $(3I+4A)(3I-4A)$ should simplify to something like $(\text{number})I + (\text{number})A$. Let’s find out exactly.

    1. Expand the product carefully — but treat $I$ and $A$ as commuting matrices (they do, since $I$ commutes with everything).

       $(3I+4A)(3I-4A) = 3I \cdot 3I + 3I \cdot (-4A) + 4A \cdot 3I + 4A \cdot (-4A)$

       $= 9I^2 -12IA + 12AI -16A^2$.

       Since $I^2 = I$, $IA = A$, and $AI = A$, the middle terms $-12A + 12A$ cancel exactly. So we get:

       $= 9I - 16A^2$.

    2. Now use $A^2 = -I$ to replace $A^2$:

       $9I - 16(-I) = 9I + 16I = 25I$.

       So the product simplifies to $25I$, a pure scalar multiple of the identity matrix.

    3. The problem states that this product equals $x^2 I$. Therefore: …

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