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  • Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 49

Need solution for RD Sharma maths class 12 chapter Algebra of matrices exercise 4.3 question 49

Answers (1)

Answer:

A=\left[\begin{array}{cc}4 & 2 \\ -1 & 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:

A\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6 I_{2}

 is identity matrix of order 2

I_{2}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]

Now, let

A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]

\\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]\left[\begin{array}{cc}1 & -2 \\ 1 & 4\end{array}\right]=6\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ll}a+b & -2 a+4 b \\ c+d & -2 c+4 d\end{array}\right]=\left[\begin{array}{ll}6 & 0 \\ 0 & 6\end{array}\right]

Since, corresponding entries of equal matrices are equal, so

\\\\ \Rightarrow a+b=6 \quad \ldots(i) \\\\ \Rightarrow-2 a+4 b=0 \quad \ldots(i i)\\\\ \Rightarrow c+d=0 \quad \ldots(iii)\\\\ \Rightarrow-2 c+4 d=6 \quad \ldots(iv) \\

Multiply equation i by 4 and subtract equation ii from i

\\\\ 4 a+4 b=24\\\\ -2 a+4 b=0\\\\ 6 a=24\\\\ a=\frac{24}{6}\\\\ \Rightarrow a=4

 

Put a=4 in equation (i)

\\\\ \Rightarrow a+b=6\\\\ \Rightarrow 4+b=6\\\\ \Rightarrow b=6-4=2\\\\ \Rightarrow b=2

Multiply equation iii by 2 and add equation iii and iv

\\\\ 2 c+2 d=0\\\\ -2 c+4 d=6\\\\ 6 d=6\\\\ \Rightarrow d=1

Put d=1 in equation iii

\\\\ \Rightarrow c+d=0\\ \Rightarrow c=-1

Hence 

A=\left[\begin{array}{cc} 4 & 2 \\ -1 & 1 \end{array}\right]

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