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  • Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (v) maths textbook solution

Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 48 sub question (v) maths textbook solution

Answers (1)

Answer:

A=\left[\begin{array}{ccc}1 & -2 & -5 \\ 3 & 4 & 0\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}{cc}2 & -1 \\ 1 & 0 \\ -3 & 4\end{array}\right] A=\left[\begin{array}{ccc}-1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15\end{array}\right]

The matrix given on the LHS of the equation is a 3 \times 2 matrix and the one given on the RHS of the equation is 3 \times 3 matrix. So,  A has to be 2 \times 3 matrixNow, let

\begin{array}{l} \text { Now, let } A=\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{cc} 2 & -1 \\ 1 & 0 \\ -3 & 4 \end{array}\right]\left[\begin{array}{lll} a & b & c \\ d & e & f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\ \end{array}

\begin{array}{l} \Rightarrow\left[\begin{array}{ccc} 2(a)+(-1) d & 2(b)+(-1)(e) & 2(c)+(-1) f \\ 1(a)+0(d) & 1(b)+0(e) & 1(c)+0(f) \\ -3(a)+4(d) & -3(b)+4(e) & -3(c)+4(f) \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a+0 & b+0 & c+0 \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ccc} 2 a-d & 2 b-e & 2 c-f \\ a & b & c \\ -3 a+4 d & -3 b+4 e & -3 c+4 f \end{array}\right]=\left[\begin{array}{ccc} -1 & -8 & -10 \\ 1 & -2 & -5 \\ 9 & 22 & 15 \end{array}\right] \end{array}

Equating the corresponding elements of the two matrices, we have

\\\\ a = 1, b = -2 , c = -5 \\\\ then \ \ \, 2a - d = -1

\begin{array}{l} \Rightarrow 2(1)-d=-1 \\\\ \Rightarrow 2-d=-1 \\\\ \Rightarrow d=2+1=3 \\\\ \Rightarrow 2 b-e=-8 \\\\ \Rightarrow 2(-2)-e=-8 \\\\ \Rightarrow-4-e=-8 \\\\ \Rightarrow e=-4+8=4 \\\\ \Rightarrow e=4 \end{array}

\begin{array}{l} \Rightarrow 2 c-f=-10 \\\\ \Rightarrow 2(-5)-f=-10 \\\\ \Rightarrow-10-f=-10 \\\\ \Rightarrow f=-10+10 \\\\ \Rightarrow f=0 \\ \end{array}

Hence, a=1, b=-2, c=-5, d=3, e=4 \text { and } f=0 \\

Therefore, matrix

A=\left[\begin{array}{ccc} 1 & -2 & -5 \\ 3 & 4 & 0 \end{array}\right]

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