Let A be a 3 \times 3 real matrix such that A( \beginmatrix 1 1 0 \endmatrix ) = ( \beginmatrix 1 1 0 \endmatrix );A( \beginmatrix 1 0 1 \endmatrix ) = ( \beginmatrix - 1 0 1 \endmatrix ) and A( \beginmatrix 0 0 1 \endmatrix ) = ( \beginmatrix 1 1 2 \endmatrix ) . If X = (x_1,x_2,x_3)^T and I is an identity matrix of order 3, then the system (A - 2I)X = ( \beginmatrix 4 1 1 \endmatrix ) has :

Matrices and Determinants PYQ — JEE Main 2022 · 25 Jun · Shift 1 · Q26 with Solution

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Correct answer: B

Use the given action of $A$ to determine $A$ .

Let $v_1=\begin{pmatrix}1\\1\\0\end{pmatrix},\quad v_2=\begin{pmatrix}1\\0\\1\end{pmatrix},\quad v_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.$

We are given: $Av_1=v_1,\qquad Av_2=\begin{pmatrix}-1\\0\\1\end{pmatrix},\qquad Av_3=\begin{pmatrix}1\\1\\2\end{pmatrix}.$

First check whether $v_1,v_2,v_3$ are linearly independent.

Form the matrix with these as columns: $P=\begin{pmatrix}1&1&0\\1&0&0\\0&1&1\end{pmatrix}.$ Then $\det P=1\cdot\begin{vmatrix}0&0\\1&1\end{vmatrix}-1\cdot\begin{vmatrix}1&0\\0&1\end{vmatrix}+0=-1\neq 0.$ So $v_1,v_2,v_3$ form a basis of $\mathbb R^3$ .

Hence we can recover $A$ by finding its action on the standard basis vectors $e_1,e_2,e_3$ .

Express standard basis vectors in terms of $v_1,v_2,v_3$ .

We already have $e_3=v_3.$ Also, $v_2=e_1+e_3 \implies e_1=v_2-v_3.$ And $v_1=e_1+e_2 \implies e_2=v_1-e_1=v_1-v_2+v_3.$

Find $Ae_1,Ae_2,Ae_3$ .

Since $e_1=v_2-v_3$ , $Ae_1=Av_2-Av_3=\begin{pmatrix}-1\\0\\1\end{pmatrix}-\begin{pmatrix}1\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\-1\\-1\end{pmatrix}.$

Since $e_2=v_1-v_2+v_3$ ,

=\begin{pmatrix}1\\1\\0\end{pmatrix}-\begin{pmatrix}-1\\0\\1\end{pmatrix}+\begin{pmatrix}1\\1\\2\end{pmatrix} =\begin{pmatrix}3\\2\\1\end{pmatrix}.$$ And $$Ae_3=Av_3=\begin{pmatrix}1\\1\\2\end{pmatrix}.$$ Therefore, $$A=\begin{pmatrix}-2&3&1\\-1&2&1\\-1&1&2\end{pmatrix}.$$ 4. **Compute $A-2I$.** $$A-2I=\begin{pmatrix}-4&3&1\\-1&0&1\\-1&1&0\end{pmatrix}.$$ We need to solve $$(A-2I)X=\begin{pmatrix}4\\1\\1\end{pmatrix}.$$ So the augmented system is $$\left[\begin{array}{ccc|c} -4&3&1&4\\ -1&0&1&1\\ -1&1&0&1 \end{array}\right].$$ 5. **Solve the linear system.** The equations are: $$-4x_1+3x_2+x_3=4 \quad ...(1)$$ $$-x_1+x_3=1 \quad ...(2)$$ $$-x_1+x_2=1 \quad ...(3)$$ From (2), $$x_3=1+x_1.$$ From (3), $$x_2=1+x_1.$$ Substitute into (1): $$-4x_1+3(1+x_1)+(1+x_1)=4$$ $$-4x_1+3+3x_1+1+x_1=4$$ $$4=4.$$ So equation (1) is automatically satisfied. Thus there is one free variable, say $x_1=t$, and $$x_2=1+t,\qquad x_3=1+t.$$ Hence $$X=\begin{pmatrix}t\\1+t\\1+t\end{pmatrix},\qquad t\in\mathbb R.$$ Therefore, the system has **infinitely many solutions**. 6. **Check options.** - A: no solution $\to$ false - B: infinitely many solutions $\to$ true - C: unique solution $\to$ false - D: exactly two solutions $\to$ false So the correct option is **B**.

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View written solutionFree

Correct answer: B

Use the given action of $A$ to determine $A$ .

Let $v_1=\begin{pmatrix}1\\1\\0\end{pmatrix},\quad v_2=\begin{pmatrix}1\\0\\1\end{pmatrix},\quad v_3=\begin{pmatrix}0\\0\\1\end{pmatrix}.$

We are given: $Av_1=v_1,\qquad Av_2=\begin{pmatrix}-1\\0\\1\end{pmatrix},\qquad Av_3=\begin{pmatrix}1\\1\\2\end{pmatrix}.$

First check whether $v_1,v_2,v_3$ are linearly independent.

Hence we can recover $A$ by finding its action on the standard basis vectors $e_1,e_2,e_3$ .

Express standard basis vectors in terms of $v_1,v_2,v_3$ .

We already have $e_3=v_3.$ Also, $v_2=e_1+e_3 \implies e_1=v_2-v_3.$ And $v_1=e_1+e_2 \implies e_2=v_1-e_1=v_1-v_2+v_3.$

Find $Ae_1,Ae_2,Ae_3$ .

Since $e_1=v_2-v_3$ , $Ae_1=Av_2-Av_3=\begin{pmatrix}-1\\0\\1\end{pmatrix}-\begin{pmatrix}1\\1\\2\end{pmatrix}=\begin{pmatrix}-2\\-1\\-1\end{pmatrix}.$

Since $e_2=v_1-v_2+v_3$ ,

=\begin{pmatrix}1\\1\\0\end{pmatrix}-\begin{pmatrix}-1\\0\\1\end{pmatrix}+\begin{pmatrix}1\\1\\2\end{pmatrix} =\begin{pmatrix}3\\2\\1\end{pmatrix}.$$ And $$Ae_3=Av_3=\begin{pmatrix}1\\1\\2\end{pmatrix}.$$ Therefore, $$A=\begin{pmatrix}-2&3&1\\-1&2&1\\-1&1&2\end{pmatrix}.$$ 4. **Compute $A-2I$.** $$A-2I=\begin{pmatrix}-4&3&1\\-1&0&1\\-1&1&0\end{pmatrix}.$$ We need to solve $$(A-2I)X=\begin{pmatrix}4\\1\\1\end{pmatrix}.$$ So the augmented system is $$\left[\begin{array}{ccc|c} -4&3&1&4\\ -1&0&1&1\\ -1&1&0&1 \end{array}\right].$$ 5. **Solve the linear system.** The equations are: $$-4x_1+3x_2+x_3=4 \quad ...(1)$$ $$-x_1+x_3=1 \quad ...(2)$$ $$-x_1+x_2=1 \quad ...(3)$$ From (2), $$x_3=1+x_1.$$ From (3), $$x_2=1+x_1.$$ Substitute into (1): $$-4x_1+3(1+x_1)+(1+x_1)=4$$ $$-4x_1+3+3x_1+1+x_1=4$$ $$4=4.$$ So equation (1) is automatically satisfied. Thus there is one free variable, say $x_1=t$, and $$x_2=1+t,\qquad x_3=1+t.$$ Hence $$X=\begin{pmatrix}t\\1+t\\1+t\end{pmatrix},\qquad t\in\mathbb R.$$ Therefore, the system has **infinitely many solutions**. 6. **Check options.** - A: no solution $\to$ false - B: infinitely many solutions $\to$ true - C: unique solution $\to$ false - D: exactly two solutions $\to$ false So the correct option is **B**.

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