Given X + Y = [tex]\left[\begin{array}{ccc}1&2\\1&1\\\end{array}\right][/tex] and X – Y = [tex]\left[\begin{array}{ccc}3&a

2 thoughts on “Given X + Y = [tex]\left[\begin{array}{ccc}1&2\\1&1\\\end{array}\right][/tex] and X – Y = [tex]\left[\begin{array}{ccc}3&a”

1. Answer:-

   Given:-

   \begin{gathered} \sf \: X + Y= \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \: \: \: – \: equation \: (1)\end{gathered}

   X+Y=[

   1

   1

   2

   1

   ]−equation(1)

   And,

   \begin{gathered} \sf \: X – Y = \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: – 1& \sf \: 1 \end{bmatrix} \: \: – \: \: equation \: (2)\end{gathered}

   X−Y=[

   3

   −1

   0

   1

   ]−equation(2)

   Add equations (1) & (2)

   \begin{gathered} \implies \sf \: X + Y + X – Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} + \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: – 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 1 + 3& \sf \: 2 + 0\\ \sf \: 1 – 1& \sf \: 1 + 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 4& \sf \: 2 \\ \sf \: 0& \sf \: 2\end{bmatrix} \\ \\ \\ \implies \sf \: X = \begin{bmatrix} \sf \: \frac{4}{2} & \sf \: \frac{2}{2} \\ \\ \sf \: \frac{0}{2} & \sf \: \frac{2}{2} \end{bmatrix} \\ \\ \\ \sf \implies \red{X = \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix}}\end{gathered}

   ⟹X+Y+X−Y=[

   1

   1

   2

   1

   ]+[

   3

   −1

   0

   1

   ]

   ⟹2X=[

   1+3

   1−1

   2+0

   1+1

   ]

   ⟹2X=[

   4

   0

   2

   2

   ]

   ⟹X=

   ⎣

   ⎢

   ⎡

   2

   4

   2

   0

   2

   2

   2

   2

   ⎦

   ⎥

   ⎤

   ⟹X=[

   2

   0

   1

   1

   ]

   Substitute the value of X in equation (1).

   \begin{gathered} \implies \sf \: \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} + Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} – \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} \\ \\ \\ \implies \sf \:Y =\begin{bmatrix} \sf \: 1 – 2& \sf \: 2 – 1 \\ \sf \: 1 – 0& \sf \: 1 – 1 \end{bmatrix} \\ \\ \\ \implies \sf \red{ \:Y =\begin{bmatrix} \sf \: – 1 & \sf \: 1 \\ \sf \: 1 & \sf \: 0 \end{bmatrix}}\end{gathered}

   ⟹[

   2

   0

   1

   1

   ]+Y=[

   1

   1

   2

   1

   ]

   ⟹Y=[

   1

   1

   2

   1

   ]−[

   2

   0

   1

   1

   ]

   ⟹Y=[

   1−2

   1−0

   2−1

   1−1

   ]

   ⟹Y=[

   −1

   1

   1

   0

   ]

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2. Answer:–

   Given:-

   [tex] \sf \: X + Y= \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \: \: \: – \: equation \: (1)[/tex]

   And,

   [tex] \sf \: X – Y = \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: – 1& \sf \: 1 \end{bmatrix} \: \: – \: \: equation \: (2)[/tex]

   Add equations (1) & (2)

   [tex] \implies \sf \: X + Y + X – Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} + \begin{bmatrix} \sf \: 3& \sf \: 0 \\ \sf \: – 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 1 + 3& \sf \: 2 + 0\\ \sf \: 1 – 1& \sf \: 1 + 1 \end{bmatrix} \\ \\ \\ \implies \sf \:2X = \begin{bmatrix} \sf \: 4& \sf \: 2 \\ \sf \: 0& \sf \: 2\end{bmatrix} \\ \\ \\ \implies \sf \: X = \begin{bmatrix} \sf \: \frac{4}{2} & \sf \: \frac{2}{2} \\ \\ \sf \: \frac{0}{2} & \sf \: \frac{2}{2} \end{bmatrix} \\ \\ \\ \sf \implies \red{X = \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix}}[/tex]

   Substitute the value of X in equation (1).

   [tex] \implies \sf \: \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} + Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} \\ \\ \\ \implies \sf \:Y = \begin{bmatrix} \sf \: 1& \sf \: 2 \\ \sf \: 1& \sf \: 1 \end{bmatrix} – \begin{bmatrix} \sf \: 2& \sf \: 1 \\ \sf \: 0& \sf \: 1\end{bmatrix} \\ \\ \\ \implies \sf \:Y =\begin{bmatrix} \sf \: 1 – 2& \sf \: 2 – 1 \\ \sf \: 1 – 0& \sf \: 1 – 1 \end{bmatrix} \\ \\ \\ \implies \sf \red{ \:Y =\begin{bmatrix} \sf \: – 1 & \sf \: 1 \\ \sf \: 1 & \sf \: 0 \end{bmatrix}}[/tex]

   Reply