Let A, B and C be three 2 × 2 matrices with real entries such that B = ( I + A ) - 1 a n d A + C = I . If B C = [ 1 - 5 - 1 2 ] and C B [ x 1 x 2 ] = [ 12 - 6 ] , then x 1 + x 2 is. [2026]

Question

Let A, B and C be three $2 \times 2$ matrices with real entries such that $B = {(I + A)}^{- 1} and A + C = I .$ If $B C = [[\begin{matrix} 1 & - 5 \\ - 1 & 2 \end{matrix}]] and C B [[\begin{matrix} x_{1} \\ x_{2} \end{matrix}]] = [[\begin{matrix} 12 \\ - 6 \end{matrix}]],$ then $x_{1} + x_{2}$ is. [2026]

Smart Achievers · Accepted Answer

(4)

$B = {(I + A)}^{- 1}, A + C = I$

$\Rightarrow B (I + A) = (I + A) B = I$

$\Rightarrow B + B A = B + A B$

$\Rightarrow B + B (I - C) = B + (I - C) B$

$\Rightarrow 2 B - B C = 2 B - C B$

$\Rightarrow B C = C B$

$∴$ $C B [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 1 & - 5 \\ - 1 & 2 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 12 \\ - 6 \end{matrix}]$

$\Rightarrow [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = {[\begin{matrix} 1 & - 5 \\ - 1 & 2 \end{matrix}]}^{- 1} [\begin{matrix} 12 \\ - 6 \end{matrix}] = \frac{- 1}{3} [\begin{matrix} 2 & 5 \\ 1 & 1 \end{matrix}] [\begin{matrix} 32 \\ - 6 \end{matrix}]$

$\Rightarrow [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = [\begin{matrix} 2 \\ - 2 \end{matrix}]$

$∴$ $x_{1} + x_{2} = 0$

Solution

1 -2

2 4

3 2

4 0

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Solution

Q. Let A, B and C be three 2×2 matrices with real entries such that B=(I+A)-1 and A+C=I. If BC=[1-5-12] and CB[x1x2]=[12-6], then x1+x2 is. [2026]

1 -2

2 4

3 2

4 0

Ans. (4) B=(I+A)-1, A+C=I ⇒B(I+A)=(I+A)B=I ⇒B+BA=B+AB ⇒B+B(I-C)=B+(I-C)B ⇒2B-BC=2B-CB ⇒BC=CB ∴ CB[x1x2]=[1-5-12][x1x2]=[12-6] ⇒ [x1x2]=[1-5-12]-1[12-6]=-13[2511][32-6] ⇒[x1x2]=[2-2] ∴ x1+x2=0

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