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Mathematics

If 2[3x01]+3[13y2]=[z−7158]2\begin{bmatrix}[r] 3 & x \ 0 & 1 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 3 \ y & 2 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix}, the values of x, y and z are :

  1. x = 8, y = -5 and z = 9

  2. x = -8, y = 5 and z = 9

  3. x = -8, y = -5 and z = -9

  4. x = -8, y = 5 and z = -9

Matrices

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Answer

Given,

⇒2[3x01]+3[13y2]=[z−7158]⇒[62x02]+[393y6]=[z−7158]⇒[6+32x+90+3y2+6]=[z−7158]⇒[92x+93y8]=[z−7158]\Rightarrow 2\begin{bmatrix}[r] 3 & x \ 0 & 1 \end{bmatrix} + 3\begin{bmatrix}[r] 1 & 3 \ y & 2 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6 & 2x \ 0 & 2 \end{bmatrix} + \begin{bmatrix}[r] 3 & 9 \ 3y & 6 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 6 + 3 & 2x + 9 \ 0 + 3y & 2 + 6 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix} \\[1em] \Rightarrow \begin{bmatrix}[r] 9 & 2x + 9 \ 3y & 8 \end{bmatrix} = \begin{bmatrix}[r] z & -7 \ 15 & 8 \end{bmatrix} \\[1em]

From above equation we get :

⇒ z = 9, 3y = 15 and 2x + 9 = -7

⇒ z = 9, y = 153\dfrac{15}{3} and 2x = -7 - 9

⇒ z = 9, y = 5 and 2x = -16

⇒ z = 9, y = 5 and x = −162-\dfrac{16}{2}

⇒ z = 9, y = 5 and x = -8.

Hence, Option 2 is the correct option.

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