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Ncert Solutions Maths class 12th Maths Class 12th Matrices

17.Find X and Y, if

(i) $X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ and $X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

(ii) $2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ and $3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$

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Answered by

Vishal Baghel | Contributor-Level 10

4 months ago

(i) x + y = $[\begin{array}{l} 7 & 0 \\ 2 & 5 \end{array}]$

x - y = $[\begin{array}{l} 3 0 \\ 0 3 \end{array}]$

$\Rightarrow Y = \frac{1}{2} [\begin{array}{l} 4 & 0 \\ 2 & 2 \end{array}] = [\begin{array}{l} \frac{1}{2} \times 4 & \frac{1}{2} \times 0 \\ \frac{1}{2} \times 2 & \frac{1}{2} \times 2 \end{array}]$ $= [\begin{array}{l} 2 0 \\ 1 1 \end{array}]$

(ii) 2x + 3y = $[\begin{array}{l} 2 & 3 \\ 4 & 0 \end{array}]$ ___ (1)

2x + 2y = $[\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$ _____ (2)

Multiplying eqⁿ by 2 and $q^{n} (u)$

2 × (2x+ 3y) = 2 × $[\begin{array}{l} 2 3 \\ 4 0 \end{array}]$

$\Rightarrow$ 4x + 6y = $[\begin{array}{l} 2 \times 2 & 2 \times 3 \\ 2 \times 4 & 2 \times 0 \end{array}]$

$\Rightarrow$ 4x + 6y = $[\begin{array}{l} 4 & 6 \\ 8 & 0 \end{array}]$ $\to (i i i) .$

3 × (3x+ 2y) = 3 × $[\begin{array}{l} 2 - 2 \\ - 1 5 \end{array}]$

$\Rightarrow$ 9x + 6y = $[\begin{matrix} 3 \times (2) & 3 \times (- 2) \\ 3 \times (- 1) & 3 \times 5 \end{matrix}] = [\begin{matrix} 6 & - 6 \\ - 3 & 15 \end{matrix}]$ $\to (i v)$

Subtracting eqⁿ (iii) from (iv) we get,

4x+ 6y - (4x+ 6y) = $[\begin{array}{l} 6 - 6 \\ - 3 15 \end{array}] -$ $[\begin{array}{l} 4 6 \\ 8 0 \end{array}]$

$\Rightarrow$ 5x = $[\begin{array}{l} 6 - 4 - 6 - 6 \\ - 3 - 8 15 - 0 \end{array}] =$ $[\begin{array}{l} 2 - 12 \\ - 11 15 \end{array}]$

$\Rightarrow x = \frac{1}{5} [\begin{array}{r} 2 & - 12 \\ - 11 & 15 \end{array}] = [\begin{array}{r} 2 \times \frac{1}{5} & - 12 \times \frac{1}{5} \\ - 11 \times \frac{1}{5} & 15 \times \frac{1}{5} \end{array}] =$ $[\begin{array}{l} \frac{2}{5} \frac{- 12}{5} \\ \frac{- 11}{5 3} \end{array}]$

From eqⁿ (u);

$3 y = [\begin{array}{l} 2 & 3 \\ 4 & 0 \end{array}] - 2 x = [\begin{array}{l} 2 & 3 \\ 4 & 0 \end{array}] - [\begin{matrix} 2 \times \frac{2}{5} & 2 \times (- \frac{12}{5}) \\ 2 \times (- 11 / 5) & 2 \times 3 \end{matrix}]$

$= [\begin{array}{l} 2 3 \\ 4 0 \end{array}] -$ $[\begin{array}{l} \frac{4}{5} \frac{- 24}{5} \\ \frac{- 22}{5 6} \end{array}]$

$= [\begin{array}{l} 2 \frac{- 4}{5} 3 - (\frac{- 24}{5}) \\ 4 - (\frac{- 22}{5}) 0 - 6 \end{array}]$

$= [\begin{matrix} \frac{5 \times 2 - 4}{5} & \frac{3 \times 5 + 24}{5} \\ \frac{4 \times 5 + 22}{5} & - 6 \end{matrix}]$

$= [\begin{matrix} \frac{10 - 4}{5} & \frac{15 + 24}{5} \\ \frac{20 + 22}{5} & - 6 \end{matrix}]$

y $= \frac{1}{3} [\begin{matrix} \frac{6}{5} & \frac{3 q}{5} \\ \frac{42}{5} & - 6 \end{matrix}] =$ $[\begin{array}{l} \frac{1}{3} \times \frac{6}{5} \frac{1}{3} \times \frac{39}{5} \\ \frac{1}{3} \times \frac{42}{5} \frac{1}{3} \times (- 6) \end{array}]$

$= [\begin{array}{l} \frac{2}{5} \frac{13}{5} \\ \frac{14}{5 - 2} \end{array}]$

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17.Find X and Y, if

(i) $X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ and $X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

(ii) $2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ and $3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$

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17.Find X and Y, if (i) X+Y=[7025] and X−Y=[3003] (ii) 2X+3Y=[2340] and 3X+2Y=[2−2−15]

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17.Find X and Y, if

(i) $X + Y = [\begin{matrix} 7 & 0 \\ 2 & 5 \end{matrix}]$ and $X - Y = [\begin{matrix} 3 & 0 \\ 0 & 3 \end{matrix}]$

(ii) $2 X + 3 Y = [\begin{matrix} 2 & 3 \\ 4 & 0 \end{matrix}]$ and $3 X + 2 Y = [\begin{matrix} 2 & - 2 \\ - 1 & 5 \end{matrix}]$