Let S be the set of all column matrices such that b1, b2, b3 ∈ R and the system of equations (in real variables)– x + 2y + 5z = b12x – 4y + 3z = b2x – 2y + 2z = b3has at least one solution. Then which of the following system(s) (in real variables) has(have) at least one solution for each ∈ S?

Engineering

Mathematics

Inverse of a Matrix

Question

Let S be the set of all column matrices $[\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}]$ such that b₁, b₂, b₃ ∈ R and the system of equations (in real variables)

– x + 2y + 5z = b₁

2x – 4y + 3z = b₂

x – 2y + 2z = b₃

has at least one solution. Then which of the following system(s) (in real variables) has(have) at least one solution for each $[\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}]$ ∈ S?

x + y + 3z = b₁, 5x + 2y + 6z = b₂ and –2x – y – 3z = b₃

x + 2y + 5z = b₁, 2x + 3z = b₂ and x + 4y – 5z = b₃

–x + 2y – 5z = b₁, 2x – 4y + 10z = b₂ and x – 2y + 5z = b₃

x + 2y + 3z = b₁, 4y + 5z = b₂ and x + 2y + 6z = b₃

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$Let A = [\begin{matrix} - 1 & 2 & 5 \\ 2 & - 4 & 3 \\ 1 & - 2 & 2 \end{matrix}]$

$| A | = | \begin{matrix} - 1 & 2 & 5 \\ 2 & - 4 & 3 \\ 1 & - 2 & 2 \end{matrix} | = - 1 (- 8 + 6) - 2 (4 - 3) + 5 (0)$

= 2 – 2 = 0

$adj (A) \times B = [\begin{matrix} - 2 & - 14 & 26 \\ - 1 & - 7 & 13 \\ 0 & 0 & 0 \end{matrix}] [\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}] = 0$

$= [\begin{matrix} - 2 b_{1} - 14 b_{2} + 26 b_{3} \\ - b_{1} - 7 b_{2} + 13 b_{3} \\ 0 \end{matrix}] = 0$

⇒ b₁ + 7b₂ – 13 b₃ = 0

(A) |A| $\neq$ 0 ⇒ Unique solution

(B) |A| = 0 and adj (A) × $B = [\begin{matrix} 0 & 0 & 0 \\ 3 & 3 & 9 \\ - 1 & - 1 & - 3 \end{matrix}] [\begin{array}{l} b_{1} \\ b_{2} \\ b_{3} \end{array}] = [\begin{matrix} 0 \\ 3 b_{1} + 3 b_{2} + 9 b_{3} \\ - b_{1} - b_{2} - 3 b_{3} \end{matrix}]$

– x + 2y – 5z = $\frac{- b_{2}}{2}$

– x + 2y – 5z = – b₃

b₁ = $\frac{- b_{2}}{2}$ = – b₃ is not satisfy all the condition of b₁ + 7b₂ = 13b₃.

(D) |A| $\neq$ 0 ⇒ Unique solution.