studypandit.in

GATE [EC, 2017, 1 mark] Question : Consider the $5\times 5$ matrix $$A=\begin{bmatrix} 1 & 2 & 3 & 4 & 5\\ 5 & 1 & 2 & 3 & 4\\ 4 & 5 & 1 & 2 & 3\\ 3 & 4 & 5 & 1 & 2\\ 2 & 3 & 4 & 5 & 1 \end{bmatrix}$$ It is given that $A$ has only one real eigen value. Then the real eigen value of $A$ is (a) $-2.5$ (b) $0$ (c) $15$ (d) $25$ Solution : The characteristic equation is $|A-\lambda I|=0$ $\implies\begin{vmatrix} 1-\lambda & 2 & 3 & 4 & 5\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda & 2\\ 2 & 3 & 4 & 5 & 1-\lambda \end{vmatrix}=0$ $R_1\to R_1+R_2+R_3+R_4+R_5$ $\implies\begin{vmatrix} 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda & 15-\lambda\\ 5 & 1-\lambda & 2 & 3 & 4\\ 4 & 5 & 1-\lambda & 2 & 3\\ 3 & 4 & 5 & 1-\lambda ...

GATE [EE, 2007, 2 marks] Question : The linear operation $L(x)$ is defined by the cross product $L(x)=b\times X$, where $b=\begin{bmatrix} 0 & 1 & 0 \end{bmatrix}^T$ and $X=\begin{bmatrix} x_1 & x_2 & x_3 \end{bmatrix}^T$ are three dimensional vectors. The $3\times 3$ matrix $M$ of this operation sstisfies $$L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$$ Then the eigen values of $M$ are (a) $0,+1,-1$ (b) $1,-1,1$ (c) $i,-i,1$ (d) $i,-i,0$ Solution : As $L(x)=M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}$ and $L(x)=b\times X$ $\therefore M\begin{bmatrix} x_1\\ x_2\\ x_3 \end{bmatrix}=b\times X \quad\quad\ldots(1)$ Since $M$ is $3\times 3$ matrix, let $M=\begin{bmatrix} a_1 & a_2 & a_3\\ a_4 & a_5 & a_6\\ a_7 & a_8 & a_9 \end{bmatrix}$ and $b\times X=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 0 & 1 & 0\\ x_1 & x_2 & x_3 \end{vmatrix}$ $=x_3\hat{i}+0\hat{j}-x_1\hat{k}$ $\begin{bmatrix} x_3 & 0 ...

UPSC [CSE 2019 P1] Question 1(d) : If $$A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$$ then show that $AB=6I_3$. Use this result to solve the following system of equations : $$\begin{array}{r} 2x+y+z=5\\ x-y=0\\ 2x+y-z=1\\ \end{array}$$ Solution : Given $A=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix} \text{and } B=\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 1 & 2 & 1\\ 1 & -4 & 1\\ 3 & 0 & -3 \end{bmatrix}\begin{bmatrix} 2 & 1 & 1\\ 1 & -1 & 0\\ 2 & 1 & -1 \end{bmatrix}$ $\implies AB=\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 & 0\\ 0 & 0 & 6 \end{bmatrix}=6\begin{bmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1 \end{bmatrix}=6I...