Let A=[\beginarraylll0 & 1 & 2 a & 0 & 3 1 & c & 0\endarray] , where a, c \in \mathbbR . If A^3=A and the positive value of a belongs to the interval (n-1, n] , where n \in \mathbbN , then n is equal to \underline\hspace2cm .

Matrices and Determinants PYQ — JEE Main 2023 · 11 Apr · Shift 1 · Q41 with Solution

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Correct answer: 2

We are given $A=\begin{bmatrix}0&1&2\\ a&0&3\\ 1&c&0\end{bmatrix}$ and the condition $A^3=A.$

This implies $A(A^2-I)=0,$ but the easiest way is to compute $A^2$ and then $A^3$ .

Compute $A^2$ .

Using matrix multiplication, $A^2=A\cdot A.$

Now compute each entry:

First row:

\begin{aligned} (1,1)&=0\cdot 0+1\cdot a+2\cdot 1=a+2,\\ (1,2)&=0\cdot 1+1\cdot 0+2\cdot c=2c,\\ (1,3)&=0\cdot 2+1\cdot 3+2\cdot 0=3. \end{aligned}

Second row:

\begin{aligned} (2,1)&=a\cdot 0+0\cdot a+3\cdot 1=3,\\ (2,2)&=a\cdot 1+0\cdot 0+3\cdot c=a+3c,\\ (2,3)&=a\cdot 2+0\cdot 3+3\cdot 0=2a. \end{aligned}

Third row:

\begin{aligned} (3,1)&=1\cdot 0+c\cdot a+0\cdot 1=ac,\\ (3,2)&=1\cdot 1+c\cdot 0+0\cdot c=1,\\ (3,3)&=1\cdot 2+c\cdot 3+0\cdot 0=2+3c. \end{aligned}

So, $A^2=\begin{bmatrix}a+2&2c&3\\ 3&a+3c&2a\\ ac&1&2+3c\end{bmatrix}.$

Compute $A^3=A^2A$ .

We only need enough entries to compare with $A$ , but let us compute systematically.

Using $A^3=A^2A,$ with

\quad A=\begin{bmatrix}0&1&2\\ a&0&3\\ 1&c&0\end{bmatrix},$$ we get: - First row of $A^3$:

\begin{aligned} (1,1)&=(a+2)\cdot 0+(2c)\cdot a+3\cdot 1=2ac+3,\ (1,2)&=(a+2)\cdot 1+(2c)\cdot 0+3\cdot c=a+2+3c,\ (1,3)&=(a+2)\cdot 2+(2c)\cdot 3+3\cdot 0=2a+4+6c. \end{aligned}

- Second row of $A^3$:

\begin{aligned} (2,1)&=3\cdot 0+(a+3c)\cdot a+(2a)\cdot 1=a^2+3ac+2a,\ (2,2)&=3\cdot 1+(a+3c)\cdot 0+(2a)\cdot c=3+2ac,\ (2,3)&=3\cdot 2+(a+3c)\cdot 3+(2a)\cdot 0=6+3a+9c. \end{aligned}

- Third row of $A^3$:

\begin{aligned} (3,1)&=(ac)\cdot 0+1\cdot a+(2+3c)\cdot 1=a+2+3c,\ (3,2)&=(ac)\cdot 1+1\cdot 0+(2+3c)\cdot c=ac+2c+3c^2,\ (3,3)&=(ac)\cdot 2+1\cdot 3+(2+3c)\cdot 0=2ac+3. \end{aligned}

Thus, $$A^3=\begin{bmatrix} 2ac+3 & a+2+3c & 2a+4+6c\\ a^2+3ac+2a & 3+2ac & 6+3a+9c\\ a+2+3c & ac+2c+3c^2 & 2ac+3 \end{bmatrix}.$$ --- 4. Since $A^3=A$, compare corresponding entries. From the $(1,1)$ entry: $$2ac+3=0$$ so $$ac=-\frac{3}{2}. \qquad (1)$$ From the $(1,2)$ entry: $$a+2+3c=1$$ so $$a+3c=-1. \qquad (2)$$ From the $(1,3)$ entry: $$2a+4+6c=2$$ so $$a+3c=-1,$$ which is the same as (2). From the $(2,2)$ entry: $$3+2ac=0$$ which again gives $$ac=-\frac{3}{2},$$ consistent with (1). From the $(3,1)$ entry: $$a+2+3c=1,$$ again same as (2). So we solve the system $$ac=-\frac{3}{2}, \qquad a+3c=-1.$$ --- 5. Solve for $a$. From $$a+3c=-1,$$ we get $$c=\frac{-1-a}{3}.$$ Substitute into $ac=-\frac{3}{2}$: $$a\left(\frac{-1-a}{3}\right)=-\frac{3}{2}.$$ Multiply by $3$: $$-a-a^2=-\frac{9}{2}.$$ So, $$a^2+a-\frac{9}{2}=0.$$ Multiply by $2$: $$2a^2+2a-9=0.$$ Now solve: $$a=\frac{-2\pm\sqrt{4+72}}{4}=\frac{-2\pm\sqrt{76}}{4}=\frac{-1\pm\sqrt{19}}{2}.$$ The positive value is $$a=\frac{-1+\sqrt{19}}{2}.$$ Numerically, $$\sqrt{19}\approx 4.359,$$ so $$a\approx \frac{-1+4.359}{2}=\frac{3.359}{2}\approx 1.6795.$$ Thus, $$a\in(1,2].$$ Hence, comparing with $(n-1,n]$, we get $$n=2.$$ --- 6. Verification with stored answer. Our derived answer is $2$, which matches the stored correct answer.

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View written solutionFree

Correct answer: 2

We are given $A=\begin{bmatrix}0&1&2\\ a&0&3\\ 1&c&0\end{bmatrix}$ and the condition $A^3=A.$

This implies $A(A^2-I)=0,$ but the easiest way is to compute $A^2$ and then $A^3$ .

Compute $A^2$ .

Using matrix multiplication, $A^2=A\cdot A.$

Now compute each entry:

First row:

\begin{aligned} (1,1)&=0\cdot 0+1\cdot a+2\cdot 1=a+2,\\ (1,2)&=0\cdot 1+1\cdot 0+2\cdot c=2c,\\ (1,3)&=0\cdot 2+1\cdot 3+2\cdot 0=3. \end{aligned}

Second row:

\begin{aligned} (2,1)&=a\cdot 0+0\cdot a+3\cdot 1=3,\\ (2,2)&=a\cdot 1+0\cdot 0+3\cdot c=a+3c,\\ (2,3)&=a\cdot 2+0\cdot 3+3\cdot 0=2a. \end{aligned}

Third row:

\begin{aligned} (3,1)&=1\cdot 0+c\cdot a+0\cdot 1=ac,\\ (3,2)&=1\cdot 1+c\cdot 0+0\cdot c=1,\\ (3,3)&=1\cdot 2+c\cdot 3+0\cdot 0=2+3c. \end{aligned}

So, $A^2=\begin{bmatrix}a+2&2c&3\\ 3&a+3c&2a\\ ac&1&2+3c\end{bmatrix}.$

Compute $A^3=A^2A$ .

We only need enough entries to compare with $A$ , but let us compute systematically.

Using $A^3=A^2A,$ with

\quad A=\begin{bmatrix}0&1&2\\ a&0&3\\ 1&c&0\end{bmatrix},$$ we get: - First row of $A^3$:

\begin{aligned} (1,1)&=(a+2)\cdot 0+(2c)\cdot a+3\cdot 1=2ac+3,\ (1,2)&=(a+2)\cdot 1+(2c)\cdot 0+3\cdot c=a+2+3c,\ (1,3)&=(a+2)\cdot 2+(2c)\cdot 3+3\cdot 0=2a+4+6c. \end{aligned}

- Second row of $A^3$:

\begin{aligned} (2,1)&=3\cdot 0+(a+3c)\cdot a+(2a)\cdot 1=a^2+3ac+2a,\ (2,2)&=3\cdot 1+(a+3c)\cdot 0+(2a)\cdot c=3+2ac,\ (2,3)&=3\cdot 2+(a+3c)\cdot 3+(2a)\cdot 0=6+3a+9c. \end{aligned}

- Third row of $A^3$:

\begin{aligned} (3,1)&=(ac)\cdot 0+1\cdot a+(2+3c)\cdot 1=a+2+3c,\ (3,2)&=(ac)\cdot 1+1\cdot 0+(2+3c)\cdot c=ac+2c+3c^2,\ (3,3)&=(ac)\cdot 2+1\cdot 3+(2+3c)\cdot 0=2ac+3. \end{aligned}

Thus, $$A^3=\begin{bmatrix} 2ac+3 & a+2+3c & 2a+4+6c\\ a^2+3ac+2a & 3+2ac & 6+3a+9c\\ a+2+3c & ac+2c+3c^2 & 2ac+3 \end{bmatrix}.$$ --- 4. Since $A^3=A$, compare corresponding entries. From the $(1,1)$ entry: $$2ac+3=0$$ so $$ac=-\frac{3}{2}. \qquad (1)$$ From the $(1,2)$ entry: $$a+2+3c=1$$ so $$a+3c=-1. \qquad (2)$$ From the $(1,3)$ entry: $$2a+4+6c=2$$ so $$a+3c=-1,$$ which is the same as (2). From the $(2,2)$ entry: $$3+2ac=0$$ which again gives $$ac=-\frac{3}{2},$$ consistent with (1). From the $(3,1)$ entry: $$a+2+3c=1,$$ again same as (2). So we solve the system $$ac=-\frac{3}{2}, \qquad a+3c=-1.$$ --- 5. Solve for $a$. From $$a+3c=-1,$$ we get $$c=\frac{-1-a}{3}.$$ Substitute into $ac=-\frac{3}{2}$: $$a\left(\frac{-1-a}{3}\right)=-\frac{3}{2}.$$ Multiply by $3$: $$-a-a^2=-\frac{9}{2}.$$ So, $$a^2+a-\frac{9}{2}=0.$$ Multiply by $2$: $$2a^2+2a-9=0.$$ Now solve: $$a=\frac{-2\pm\sqrt{4+72}}{4}=\frac{-2\pm\sqrt{76}}{4}=\frac{-1\pm\sqrt{19}}{2}.$$ The positive value is $$a=\frac{-1+\sqrt{19}}{2}.$$ Numerically, $$\sqrt{19}\approx 4.359,$$ so $$a\approx \frac{-1+4.359}{2}=\frac{3.359}{2}\approx 1.6795.$$ Thus, $$a\in(1,2].$$ Hence, comparing with $(n-1,n]$, we get $$n=2.$$ --- 6. Verification with stored answer. Our derived answer is $2$, which matches the stored correct answer.

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