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Q3.    If A = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix} and B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}, then verify

            (ii)    (A - B)' = A' - B'

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A' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}     B = \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix}

A=(A')' = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}

To prove: (A - B)' = A' - B'

L.H.S : (A - B)' = 

A-B = \begin{bmatrix} 3 & -1&0\\ 4 &2 & 1 \end{bmatrix}  - \begin{bmatrix} -1 & 2 & 1\\ 1 &2 &3 \end{bmatrix} 

A-B = \begin{bmatrix} 3-(-1) & -1-(2)&0-1\\ 4-1 &2-2 & 1-3 \end{bmatrix}

A-B = \begin{bmatrix} 4 & -3&-1\\ 3 &0 & -2 \end{bmatrix}

\therefore \, \, \, (A-B)' = \begin{bmatrix} 4 & 3\\ -3 &0\\-1 & -2 \end{bmatrix}

R.H.S:  A' - B'

A'-B' = \begin{bmatrix} 3 & 4\\ -1 &2 \\ 0 & 1 \end{bmatrix}  - \begin{bmatrix} -1 & 1\\ 2 &2 \\ 1 & 3 \end{bmatrix}

A'-B' = \begin{bmatrix} 4 & 3\\ -3 &0 \\ -1 & -2 \end{bmatrix}

Hence, L.H.S = R.H.S i.e. (A - B)' = A' - B'.

Posted by

seema garhwal

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