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  • Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question  48 sub question (vi)

Provide solution for rd sharma math class 12 chapter Algebra of matrices exercise 4.3 question  48 sub question (vi)

Answers (1)

Answer:

A=\left[\begin{array}{cc}1 & -2 \\ 2 & 0 \\ -5 & 4\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}{lll}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right]

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

Now, let

\\\\A=\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right] \\\\ \Rightarrow\left[\begin{array}{ll}a & b \\ c & d \\ e & f\end{array}\right]\left[\begin{array}{ccc}1 & 2 & 3 \\ 4 & 5 & 6\end{array}\right]=\left[\begin{array}{ccc}-7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9\end{array}\right] \\\\

\\\\\Rightarrow\left[\begin{array}{lll} a(1)+b(4) & a(2)+b(5) & a(3)+b(6) \\ c(1)+d(4) & c(2)+5(d) & c(3)+d(6) \\ e(1)+4(f) & e(2)+5(f) & 3(e)+6(f) \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\\\\\ \Rightarrow\left[\begin{array}{ccc} a+4 b & 2 a+5 b & 3 a+6 b \\ c+4 d & 2 c+5 d & 3 c+6 d \\ e+4 f & 2 e+5 f & 3 e+6 f \end{array}\right]=\left[\begin{array}{ccc} -7 & -8 & -9 \\ 2 & 4 & 6 \\ 11 & 10 & 9 \end{array}\right]\\

Equating the corresponding elements of the two matrices, we have

a + 4b = -7           ...(i)

2a +5b = -8       ...(ii)

Now multiply equation i by 2 and subtract equation ii from i

\begin{array}{l} 2 a+8 b=-14\\ 2 a+5 b=-8\\ 3 b=-6\\ \Rightarrow b=-2\\ \end{array} 

Put value of b=-2 in equation ii we get

2a + 5 (-2) = -8

\\\\ \Rightarrow 2 a-10=-8\\ \Rightarrow 2 a=-8+10\\ \Rightarrow 2 a=2\\ \Rightarrow a=1\\\

\\\\ c+4 d=2 ...(iii) \\\\ \Rightarrow 2 c+5 d=4 ... (iv)

Multiply equation iii by 2 and subtract equation iv from iii


\\\\ 2 c+8 d=4\\\\ \underline{2 c+5 d=4}\\\\

                      3 d=0 \Rightarrow d=0

Now, put d=0 in equation iv, we get

\\\\ \Rightarrow 2 c+5(0)=4\\\\ \Rightarrow 2 c=4\\\\ \Rightarrow c=2\\\\ \Rightarrow e+4 f=11 \cdots (v)\\\\ \Rightarrow 2 e+5 f=10 \cdots(vi) \\\\

Multiply equation v by 2 and subtract equation vi from v

\begin{array}{l} 2 e+8 f=22\\\\ \underline{2 e+5 f=10}\\\\ \Rightarrow 3 f=12\\ \\\Rightarrow f=4\\\\ \text { Put } f=4 \text { in equation }(v i)\\\\ \Rightarrow 2 e+5(4)=10\\\\ \Rightarrow 2 e+20=10\\\\ \Rightarrow 2 e=10-20\\\\ \Rightarrow 2 e=-10\\\\ \Rightarrow e=-5\\ \end{array}

 

Thus,

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

Hence,

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

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