Let A = [ \beginmatrix 0 & - 2 2 & 0 \endmatrix ] . If M and N are two matrices given by M = \sum\limits_k = 1^10 A^2k and N = \sum\limits_k = 1^10 A^2k - 1 then MN2 is :

Matrices and Determinants PYQ — JEE Main 2022 · 25 Jun · Shift 1 · Q31 with Solution

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Correct answer: A

$A=\begin{bmatrix}0&-2\\2&0\end{bmatrix}$

We need to compute

$M=\sum_{k=1}^{10} A^{2k}, \qquad N=\sum_{k=1}^{10} A^{2k-1}$

and determine the nature of $MN^2$ .

First,

=\begin{bmatrix}-4&0\\0&-4\end{bmatrix}=-4I$$ Hence, $$A^{2k}=(A^2)^k=(-4)^k I$$ and $$A^{2k-1}=A\cdot A^{2k-2}=A(A^2)^{k-1}=(-4)^{k-1}A$$ --- 3. **Compute $M$** $$M=\sum_{k=1}^{10} A^{2k}=\sum_{k=1}^{10} (-4)^k I =\left(\sum_{k=1}^{10}(-4)^k\right)I$$ So $M$ is a scalar multiple of the identity matrix. Let $$s=\sum_{k=1}^{10}(-4)^k$$ Then $$M=sI$$ --- 4. **Compute $N$** $$N=\sum_{k=1}^{10} A^{2k-1}=\sum_{k=1}^{10}(-4)^{k-1}A =\left(\sum_{k=1}^{10}(-4)^{k-1}\right)A$$ Let $$t=\sum_{k=1}^{10}(-4)^{k-1}$$ Then $$N=tA$$ --- 5. **Compute $N^2$** Since $N=tA$, $$N^2=t^2A^2=t^2(-4I)=-4t^2I$$ So $N^2$ is also a scalar multiple of the identity matrix. --- 6. **Compute $MN^2$** Since $M=sI$, $$MN^2=(sI)(-4t^2I)=-4st^2 I$$ Thus $MN^2$ is a scalar multiple of the identity matrix, hence it is **symmetric**. Now check whether it is the identity matrix. We have $$s=\sum_{k=1}^{10}(-4)^k=-4\sum_{j=0}^{9}(-4)^j=-4t$$ Therefore, $$MN^2=-4(-4t)t^2I=16t^3I$$ Clearly $t=\sum_{j=0}^9(-4)^j \neq 0$, so $MN^2$ is nonzero. Also there is no reason for $16t^3=1$; in fact it is a very large integer, not $1$. Hence $MN^2$ is **not** the identity matrix. So $MN^2$ is a **non-identity symmetric matrix**. --- 7. **Option check** - **A:** a non-identity symmetric matrix ✅ - **B:** a skew-symmetric matrix ❌ - **C:** neither symmetric nor skew-symmetric matrix ❌ - **D:** an identity matrix ❌ Therefore, the correct option is **A**.

View written solutionFree

Correct answer: A

$A=\begin{bmatrix}0&-2\\2&0\end{bmatrix}$

We need to compute

$M=\sum_{k=1}^{10} A^{2k}, \qquad N=\sum_{k=1}^{10} A^{2k-1}$

and determine the nature of $MN^2$ .

First,

=\begin{bmatrix}-4&0\\0&-4\end{bmatrix}=-4I$$ Hence, $$A^{2k}=(A^2)^k=(-4)^k I$$ and $$A^{2k-1}=A\cdot A^{2k-2}=A(A^2)^{k-1}=(-4)^{k-1}A$$ --- 3. **Compute $M$** $$M=\sum_{k=1}^{10} A^{2k}=\sum_{k=1}^{10} (-4)^k I =\left(\sum_{k=1}^{10}(-4)^k\right)I$$ So $M$ is a scalar multiple of the identity matrix. Let $$s=\sum_{k=1}^{10}(-4)^k$$ Then $$M=sI$$ --- 4. **Compute $N$** $$N=\sum_{k=1}^{10} A^{2k-1}=\sum_{k=1}^{10}(-4)^{k-1}A =\left(\sum_{k=1}^{10}(-4)^{k-1}\right)A$$ Let $$t=\sum_{k=1}^{10}(-4)^{k-1}$$ Then $$N=tA$$ --- 5. **Compute $N^2$** Since $N=tA$, $$N^2=t^2A^2=t^2(-4I)=-4t^2I$$ So $N^2$ is also a scalar multiple of the identity matrix. --- 6. **Compute $MN^2$** Since $M=sI$, $$MN^2=(sI)(-4t^2I)=-4st^2 I$$ Thus $MN^2$ is a scalar multiple of the identity matrix, hence it is **symmetric**. Now check whether it is the identity matrix. We have $$s=\sum_{k=1}^{10}(-4)^k=-4\sum_{j=0}^{9}(-4)^j=-4t$$ Therefore, $$MN^2=-4(-4t)t^2I=16t^3I$$ Clearly $t=\sum_{j=0}^9(-4)^j \neq 0$, so $MN^2$ is nonzero. Also there is no reason for $16t^3=1$; in fact it is a very large integer, not $1$. Hence $MN^2$ is **not** the identity matrix. So $MN^2$ is a **non-identity symmetric matrix**. --- 7. **Option check** - **A:** a non-identity symmetric matrix ✅ - **B:** a skew-symmetric matrix ❌ - **C:** neither symmetric nor skew-symmetric matrix ❌ - **D:** an identity matrix ❌ Therefore, the correct option is **A**.