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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (iii) math

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (iii) math

Answers (1)

Answer:

A=\left[\begin{array}{lll}-1 & 2 & 1\end{array}\right]

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

Given:

\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right] A=\left[\begin{array}{lll}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]

We know that two matrices B and C are eligible for the product BC only when number of columns of B is equal to number or rows of C.

So, from the given definition we can consider that the order of matrix A is 1 \times 3  i.e. we can assume \therefore\left[\begin{array}{l}4 \\ 1 \\ 3\end{array}\right]_{3 \times 1}\left[\begin{array}{lll}x_{1} & x_{2} & x_{3}\end{array}\right]_{1 \times 3}=\left[\begin{array}{ccc}-4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3\end{array}\right]_{3 \times 3}

\\\\ \Rightarrow\left[\begin{array}{lll} 4\left(x_{1}\right) & 4\left(x_{2}\right) & 4\left(x_{3}\right) \\ 1\left(x_{1}\right) & 1\left(x_{2}\right) & 1\left(x_{3}\right) \\ 3\left(x_{1}\right) & 3\left(x_{2}\right) & 3\left(x_{3}\right) \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} 4 x_{1} & 4 x_{2} & 4 x_{3} \\ x_{1} & x_{2} & x_{3} \\ 3 x_{1} & 3 x_{2} & 3 x_{3} \end{array}\right]=\left[\begin{array}{ccc} -4 & 8 & 4 \\ -1 & 2 & 1 \\ -3 & 6 & 3 \end{array}\right]

 

Equating the corresponding element of the two matrices, we have

x_{1}=-1, x_{2}=2, x_{3}=1

So, matrix  A=\left[\begin{array}{lll} -1 & 2 & 1\end{array}]\right.

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