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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 65 sub question (iv) math

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 65 sub question (iv) math

Answers (1)

AnswerA=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right], such that A B=A C \ \ but \ \ \ B \neq C, A \neq 0

Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.

Solution: LetA=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] , \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]

Here,

A \neq 0, B \neq C

Consider LHS

\\A B=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 0 \\ -1 & 0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0+0 & 0+0 \end{array}\right]=\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]

Now consider RHS

\\A C=\left[\begin{array}{ll} 1 & 0 \\ 0 & 0 \end{array}\right] \ \left[\begin{array}{ll} 0 & 0 \\ 0 & 1 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0+0 & 0+0 \\ 0 +0 & 0+0 \end{array}\right] \\\\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right]

LHS=RHS

So, A B=A C \ \ for \ \ \ B \neq C, A \neq 0

We have A=\left[\begin{array}{ll}1 & 0 \\ 0 & 0\end{array}\right], B=\left[\begin{array}{cc}0 & 0 \\ -1 & 0\end{array}\right] and \ \ C=\left[\begin{array}{ll}0 & 0 \\ 0 & 1\end{array}\right]

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