If the matrix A = ( \beginmatrix 0 & 2 K & - 1 \endmatrix ) satisfies A(A^3 + 3I) = 2I , then the value of K is :

Matrices and Determinants PYQ — JEE Main 2021 · 27 Aug · Shift 1 · Q25 with Solution

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Correct answer: A

Given matrix

$A=\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$

and it satisfies

$A(A^3+3I)=2I.$

We need to find $K$ .

Rewrite the equation

Since matrix multiplication is associative,

$A(A^3+3I)=A^4+3A=2I.$

So we need to compute $A^2, A^3, A^4$ .

Compute $A^2$

$A^2=\begin{pmatrix}0&2\\ K&-1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$

Now multiply:

0\cdot 0+2K & 0\cdot 2+2(-1)\\ K\cdot 0+(-1)K & K\cdot 2+(-1)(-1) \end{pmatrix} =\begin{pmatrix} 2K & -2\\ -K & 2K+1 \end{pmatrix}.$$ --- 4. **Compute $A^3=A^2A$** $$A^3=\begin{pmatrix}2K&-2\\ -K&2K+1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$$ Multiplying, $$A^3=\begin{pmatrix} 2K\cdot 0+(-2)K & 2K\cdot 2+(-2)(-1)\\ (-K)\cdot 0+(2K+1)K & (-K)\cdot 2+(2K+1)(-1) \end{pmatrix}$$ $$A^3=\begin{pmatrix} -2K & 4K+2\\ 2K^2+K & -4K-1 \end{pmatrix}.$$ --- 5. **Compute $A^4=A^3A$** $$A^4=\begin{pmatrix}-2K&4K+2\\ 2K^2+K&-4K-1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$$ Multiplying, $$A^4=\begin{pmatrix} (-2K)\cdot 0+(4K+2)K & (-2K)\cdot 2+(4K+2)(-1)\\ (2K^2+K)\cdot 0+(-4K-1)K & (2K^2+K)\cdot 2+(-4K-1)(-1) \end{pmatrix}$$ $$A^4=\begin{pmatrix} 4K^2+2K & -8K-2\\ -4K^2-K & 4K^2+6K+1 \end{pmatrix}.$$ --- 6. **Use $A^4+3A=2I$** Now, $$3A=\begin{pmatrix}0&6\\ 3K&-3\end{pmatrix}.$$ Hence, $$A^4+3A=\begin{pmatrix} 4K^2+2K & -8K-2+6\\ -4K^2-K+3K & 4K^2+6K+1-3 \end{pmatrix}$$ $$A^4+3A=\begin{pmatrix} 4K^2+2K & 4-8K\\ -4K^2+2K & 4K^2+6K-2 \end{pmatrix}.$$ This must equal $$2I=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ So equating entries: ### From the $(1,2)$ entry: $$4-8K=0$$ $$8K=4$$ $$K=\frac12.$$ Check with another entry. ### From the $(1,1)$ entry: $$4K^2+2K=2.$$ Substitute $K=\frac12$: $$4\left(\frac14\right)+2\left(\frac12\right)=1+1=2,$$ which is correct. ### From the $(2,1)$ entry: $$-4K^2+2K=0.$$ Substitute $K=\frac12$: $$-4\left(\frac14\right)+1=-1+1=0,$$ correct. ### From the $(2,2)$ entry: $$4K^2+6K-2=2.$$ Substitute $K=\frac12$: $$4\left(\frac14\right)+6\left(\frac12\right)-2=1+3-2=2,$$ correct. Thus, $$\boxed{K=\frac12}.$$ --- 7. **Option matching** Option A is $$\boxed{\frac12}.$$ So the correct option is **A**. --- 8. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They match.

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View written solutionFree

Correct answer: A

Given matrix

$A=\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$

and it satisfies

$A(A^3+3I)=2I.$

We need to find $K$ .

Rewrite the equation

Since matrix multiplication is associative,

$A(A^3+3I)=A^4+3A=2I.$

So we need to compute $A^2, A^3, A^4$ .

Compute $A^2$

$A^2=\begin{pmatrix}0&2\\ K&-1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$

Now multiply:

0\cdot 0+2K & 0\cdot 2+2(-1)\\ K\cdot 0+(-1)K & K\cdot 2+(-1)(-1) \end{pmatrix} =\begin{pmatrix} 2K & -2\\ -K & 2K+1 \end{pmatrix}.$$ --- 4. **Compute $A^3=A^2A$** $$A^3=\begin{pmatrix}2K&-2\\ -K&2K+1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$$ Multiplying, $$A^3=\begin{pmatrix} 2K\cdot 0+(-2)K & 2K\cdot 2+(-2)(-1)\\ (-K)\cdot 0+(2K+1)K & (-K)\cdot 2+(2K+1)(-1) \end{pmatrix}$$ $$A^3=\begin{pmatrix} -2K & 4K+2\\ 2K^2+K & -4K-1 \end{pmatrix}.$$ --- 5. **Compute $A^4=A^3A$** $$A^4=\begin{pmatrix}-2K&4K+2\\ 2K^2+K&-4K-1\end{pmatrix}\begin{pmatrix}0&2\\ K&-1\end{pmatrix}$$ Multiplying, $$A^4=\begin{pmatrix} (-2K)\cdot 0+(4K+2)K & (-2K)\cdot 2+(4K+2)(-1)\\ (2K^2+K)\cdot 0+(-4K-1)K & (2K^2+K)\cdot 2+(-4K-1)(-1) \end{pmatrix}$$ $$A^4=\begin{pmatrix} 4K^2+2K & -8K-2\\ -4K^2-K & 4K^2+6K+1 \end{pmatrix}.$$ --- 6. **Use $A^4+3A=2I$** Now, $$3A=\begin{pmatrix}0&6\\ 3K&-3\end{pmatrix}.$$ Hence, $$A^4+3A=\begin{pmatrix} 4K^2+2K & -8K-2+6\\ -4K^2-K+3K & 4K^2+6K+1-3 \end{pmatrix}$$ $$A^4+3A=\begin{pmatrix} 4K^2+2K & 4-8K\\ -4K^2+2K & 4K^2+6K-2 \end{pmatrix}.$$ This must equal $$2I=\begin{pmatrix}2&0\\0&2\end{pmatrix}.$$ So equating entries: ### From the $(1,2)$ entry: $$4-8K=0$$ $$8K=4$$ $$K=\frac12.$$ Check with another entry. ### From the $(1,1)$ entry: $$4K^2+2K=2.$$ Substitute $K=\frac12$: $$4\left(\frac14\right)+2\left(\frac12\right)=1+1=2,$$ which is correct. ### From the $(2,1)$ entry: $$-4K^2+2K=0.$$ Substitute $K=\frac12$: $$-4\left(\frac14\right)+1=-1+1=0,$$ correct. ### From the $(2,2)$ entry: $$4K^2+6K-2=2.$$ Substitute $K=\frac12$: $$4\left(\frac14\right)+6\left(\frac12\right)-2=1+3-2=2,$$ correct. Thus, $$\boxed{K=\frac12}.$$ --- 7. **Option matching** Option A is $$\boxed{\frac12}.$$ So the correct option is **A**. --- 8. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** They match.

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