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Ncert Solutions Maths class 12th Maths Class 12th Matrices

Kindly Consider the following

60. $[\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}]$

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Vishal Baghel | Contributor-Level 10

4 months ago

Let A = $[\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}]$

We write A = IA

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}] = [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ A.

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ - 3 + 3 (1) & 0 + 3 (3) & - 5 + 3 (- 2) \\ 2 - 2 (1) & 5 - 2 (3) & 0 - 2 (- 2) \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 + 3 (1) & 1 + 0 & 0 + 0 \\ 0 - 2 (1) & 0 - 0 & 1 - 0 \end{matrix}]$ A. $(\begin{array}{l} R_{2} \to R_{2} + 3 R_{1} & R_{3} \to R_{3} - 2 R_{1} \end{array})$

$= [\begin{matrix} 1 & 3 & - 2 \\ 0 & 9 & - 11 \\ 0 & - 1 & . 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 0 \\ - 2 & 0 & 1 \end{matrix}]$ A.

$\to [\begin{matrix} 1 & 3 & - 2 \\ 0 & - 1 & 4 p \\ 0 & 9 & - 11 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ - 2 & 0 & 1 \\ 3 & 1 & 0 \end{matrix}]$ A. (R2 R3)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 9 & - 11 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ 3 & 1 & 0 \end{matrix}]$ A. (R2 (-1) R2)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 - 0 & 9 - 9 (1) & - 11 - 9 (- 4) \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ 3 - 9 (2) & 1 - 0 & 0 - 9 (- 1) \end{matrix}]$ A. (R3 R3--> 4R2)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 0 & 25 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ - 15 & 1 & 9 \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & - 4 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 2 & 0 & - 1 \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{5} \end{matrix}]$ A. $(R_{3} \to \frac{1}{25} R_{3})$

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 + 0 & 1 + 0 & - 4 + 4 (1) \\ 0 & 0 & 1 \end{matrix}] =$ $= [\begin{matrix} 1 & 0 \\ 2 + 4 (\frac{3}{5}) & 0 + 4 (\frac{1}{25}) & - 1 + 4 \times (\frac{9}{25}) \\ \frac{- 3}{25} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A. (R2 R2 + 4R3)

$\Rightarrow [\begin{matrix} 1 & 3 & - 2 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & 3 + 0 & - 2 + 2 (1) \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 + 2 (- \frac{3}{5}) & 0 + 2 (\frac{1}{25}) & 0 + 2 (\frac{9}{23}) \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A. (R1 R1 + 2R3)

$\Rightarrow [\begin{array}{l} 1 & 3 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{matrix} \frac{- 1}{5} & \frac{2}{25} & \frac{18}{25} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 + 0 & 3 - 3 (1) & 0 + 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] =$ $[\begin{matrix} \frac{- 1}{5} - 3 (\frac{- 2}{5}) & \frac{2}{25} - 3 (\frac{4}{25}) & \frac{18}{25} - 3 (\frac{11}{25}) \\ \frac{- 2}{5} & \frac{4}{5} & \frac{11}{25} \\ \frac{- 3}{5} & 25 & \frac{9}{25} \end{matrix}]$ A. (R1 R1 3R2)

$\Rightarrow [\begin{array}{l} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] = [\begin{matrix} 1 & \frac{- 10}{25} & \frac{- 15}{25} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{1}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

$\Rightarrow [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}] = [\begin{matrix} 1 & \frac{- 2}{5} & \frac{- 3}{5} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$ A.

∴A^-1 = $[\begin{matrix} 1 & \frac{- 2}{5} & \frac{- 3}{5} \\ \frac{- 2}{5} & \frac{4}{25} & \frac{11}{25} \\ \frac{- 3}{5} & \frac{1}{25} & \frac{9}{25} \end{matrix}]$

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If $A [\begin{matrix} 1 & 0 & 1 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 1 \end{matrix}]$ , $A [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = 4 [\begin{matrix} - 1 & 0 & 1 \end{matrix}]$ and $A [\begin{matrix} 0 & 1 & 0 \end{matrix}] = 2 [\begin{matrix} 1 & 0 & 0 \end{matrix}]$ where, A is a 3 × 3 matrix and (A –3I) $[\begin{matrix} x & y & z \end{matrix}] = [\begin{matrix} - 1 & 2 & 3 \end{matrix}]$ then the value of (x, y, z) is

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Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ ...(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

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Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

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Let $A = [\begin{matrix} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{matrix}]$

Given $A = [\begin{matrix} 1 & 0 & 1 \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$ ...(1)

$[\begin{matrix} x_{1} + z_{1} & x_{2} + z_{2} & x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 2 & 0 & 2 \end{matrix}]$

∴ x₁ + z₁ = 2 … (2)

x₂ + z₂ = 0 … (3)

x₃ + z₃ = 0 … (4)

Given $A = [\begin{matrix} - 1 & 0 & 1 \end{matrix}] = [\begin{matrix} - 4 & 0 & 4 \end{matrix}]$

$[\begin{matrix} - x_{1} + z_{1} & - x_{2} + z_{2} & - x_{3} + z_{3} \end{matrix}] = [\begin{matrix} 4 & 0 & 4 \end{matrix}]$

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Kindly Consider the following

60. $[\begin{matrix} 1 & 3 & - 2 \\ - 3 & 0 & - 5 \\ 2 & 5 & 0 \end{matrix}]$