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  • If X =3 1 -1 and Y =2 1 -1 find (i) X + Y (ii) 2X - 3Y (iii) A matrix Z such that X + Y + Z is a zero matrix.

If

\begin{array}{l} X=\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right] \\ Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right] \end{array} find
(i) X + Y
(ii) 2X - 3Y
(iii) A matrix Z such that X + Y + Z is a zero matrix.

Answers (1)

If you want to add or subtract any two matrices, make sure these two matrices have the same order

That is,

If A and B are two matrices and to add them, if we have order of A as m × n, then order of B must be m × n.

We have matrices X and Y, where

\begin{array}{l} X=\left[\begin{array}{lll} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right] \\ Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right] \end{array}

 

According to convention,

If a matrix has M rows and N columns, the order of matrix is M $ \times $ N.

(i). We need to find the X + Y.

Let us first find the order of X and Y.

Order of X:

Number of rows = 2

$ \Rightarrow $  M = 2

Number of columns = 3

$ \Rightarrow $ N = 3

Then, order of matrix X = M $ \times $ N.

$ \Rightarrow $ Order of matrix X =2 $ \times $ 3

Order of Y:

Number of rows = 2

$ \Rightarrow $  M = 2

Number of columns = 3

$ \Rightarrow $  N = 3

Then, order of matrix Y = M $ \times $ N.

$ \Rightarrow $ Order of matrix Y = 2 $ \times $ 3

Since, order of matrix X = order of matrix Y

$ \Rightarrow $ Matrices X and Y can be added.

So,

\begin{aligned} &X+Y=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]+\left[\begin{array}{ccc} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\Rightarrow X+Y=\left[\begin{array}{lll} (3+2) & (1+1) & (-1-1) \\ (5+7) & (-2+2) & (-3+4) \end{array}\right]\\ &\Rightarrow \mathrm{X}+\mathrm{Y}=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\\ &\text { Thus, }\\ &X+Y=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right] \end{aligned}

(ii). We need to find 2X - 3Y.

Let us calculate 2X.

We have,

\begin{aligned} &X=\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]\\ &\text { Then, multiplying by } 2 \text { on both sides, we get }\\ &2 \mathrm{X}=2 \times\left[\begin{array}{ccc} 3 & 1 & -1 \\ 5 & -2 & -3 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}=\left[\begin{array}{lll} 2 \times 3 & 2 \times 1 & 2 \times-1 \\ 2 \times 5 & 2 \times-2 & 2 \times-3 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}=\left[\begin{array}{ccc} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right] \end{aligned}

Also,

\begin{aligned} &Y=\left[\begin{array}{lll} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\text { Multiplying by } 3 \text { on both sides, we get }\\ &3 Y=3 \times\left[\begin{array}{rrr} 2 & 1 & -1 \\ 7 & 2 & 4 \end{array}\right]\\ &\Rightarrow 3 \mathrm{Y}=\left[\begin{array}{lll} 3 \times 2 & 3 \times 1 & 3 \times-1 \\ 3 \times 7 & 3 \times 2 & 3 \times 4 \end{array}\right]\\ &\Rightarrow 3 Y=\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right] \end{aligned}

Now subtract 3Y from 2X.

\begin{aligned} &2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 6 & 2 & -2 \\ 10 & -4 & -6 \end{array}\right]-\left[\begin{array}{ccc} 6 & 3 & -3 \\ 21 & 6 & 12 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 6-6 & 2-3 & -2+3 \\ 10-21 & -4-6 & -6-12 \end{array}\right]\\ &\Rightarrow 2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right]\\ &\text { Thus, }\\ &2 \mathrm{X}-3 \mathrm{Y}=\left[\begin{array}{ccc} 0 & -1 & 1 \\ -11 & -10 & -18 \end{array}\right] \end{aligned}

We need to find matrix Z, such that X + Y + Z is a zero matrix.

That is,

X + Y + Z = 0

Or,

Z = -X - Y

Or,

Z = -(X + Y)

We have already found X + Y in part (i).

So, from part (i):

\begin{aligned} &X+Y=\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\\ &\text { Then, }\\ &Z=-\left(\left[\begin{array}{ccc} 5 & 2 & -2 \\ 12 & 0 & 1 \end{array}\right]\right)\\ &\Rightarrow \mathrm{Z}=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right]\\ &\mathrm{Thus},Z=\left[\begin{array}{ccc} -5 & -2 & 2 \\ -12 & 0 & -1 \end{array}\right] \end{aligned}

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