Thank you @LeenSharma Veteran.

You made it look so simple.

0 votes

Consider a system of equations (λ – a)x + 2y +3z = 0, x +2(λ – b) y + 3z = 0, x + 2y + 3(λ – c) z = 0, which has a non-trivial solution.

Product of all values of λ for above system is

(1) | abc + a + b + c + 2 |

(2) | abc + a + b + c – 2 |

(3) | abc – a – b – c – 2 |

(4) | abc – a – b – c + 2 |

Ans given is Option C.

Can anyone explain the complete solution?

2 votes

$A=\begin{bmatrix} \lambda- a &2 &3 \\ 1& 2\lambda-2b &3 \\ 1&2 & 3\lambda-3c \end{bmatrix}$

$=\begin{bmatrix} -a+\lambda &2 &3 \\ 1&-2b+2\lambda &3 \\ 1&2 & -3c+3\lambda \end{bmatrix}$

We find the values of $\lambda$ which satisfy the characteristic equation of the matrix $A$, namely those values of $\lambda$ for which

$|A-\lambda I|=0$

$\begin{vmatrix} -a+\lambda &2 &3 \\ 1&-2b+2\lambda &3 \\ 1&2 & -3c+3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a+\lambda &2+0 &3+0 \\ 1+0&-2b+2\lambda &3+0 \\ 1+0&2+0 & -3c+3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{vmatrix}+\begin{vmatrix} \lambda &0 &0 \\ 0& 2\lambda &0 \\ 0&0 & 3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{vmatrix}+6\lambda ^{3}\begin{vmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0&0 & 1 \end{vmatrix}=0$

$-6abc+6a+6b+6c+12+6\lambda ^{3}=0$

$6\lambda ^{3}=6abc-6a-6b-6c-12$

$\lambda ^{3}=abc-a-b-c-2$

Hence,Option$(3)abc-a-b-c-2$.

$=\begin{bmatrix} -a+\lambda &2 &3 \\ 1&-2b+2\lambda &3 \\ 1&2 & -3c+3\lambda \end{bmatrix}$

We find the values of $\lambda$ which satisfy the characteristic equation of the matrix $A$, namely those values of $\lambda$ for which

$|A-\lambda I|=0$

$\begin{vmatrix} -a+\lambda &2 &3 \\ 1&-2b+2\lambda &3 \\ 1&2 & -3c+3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a+\lambda &2+0 &3+0 \\ 1+0&-2b+2\lambda &3+0 \\ 1+0&2+0 & -3c+3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{vmatrix}+\begin{vmatrix} \lambda &0 &0 \\ 0& 2\lambda &0 \\ 0&0 & 3\lambda \end{vmatrix}=0$

$\begin{vmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{vmatrix}+6\lambda ^{3}\begin{vmatrix} 1 &0 &0 \\ 0& 1 &0 \\ 0&0 & 1 \end{vmatrix}=0$

$-6abc+6a+6b+6c+12+6\lambda ^{3}=0$

$6\lambda ^{3}=6abc-6a-6b-6c-12$

$\lambda ^{3}=abc-a-b-c-2$

Hence,Option$(3)abc-a-b-c-2$.

0 votes

**M**=$\begin{pmatrix} -a+\lambda &2 &3 \\ 1&-2b+2\lambda &3 \\ 1&2 & -3c+3\lambda \end{pmatrix}$=0

can be written as:-

**M=**$\begin{pmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{pmatrix}$ **/ -6** - $\lambda$/-6 $\begin{pmatrix} -1 &0 &0 \\ 0&-2 &0 \\ 0&0 & -3 \end{pmatrix}$=0

**|M'-$\lambda$I|=0**

**:represent** $\lambda$ is eigen values of given matrix.....and product of all the values of is** equal to the determinent of matrix M' /-6**

M'=$\begin{pmatrix} -a &2 &3 \\ 1&-2b &3 \\ 1&2 & -3c \end{pmatrix}$ /-6

**Det(M')**devided by -6**=option C**