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  • If find A–1. Using A–1, solve the system of linear equations x – 2y = 10, 2x – y – z = 8, –2y + z = 7.

If  A=\left[\begin{array}{ccc} 1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1 \end{array}\right] find A^{-1}. Using A^{-1}, solve the system of linear equations x-2y = 10, 2x -y -z = 8, -2y + z = 7

Answers (1)

\\A=\left|\begin{array}{ccc}1 & 2 & 0 \\ -2 & -1 & -2 \\ 0 & -1 & 1\end{array}\right|$ $\mathrm{A}^{-1}=\frac{\operatorname{adj} \mathrm{A}}{|\mathrm{A}|}$

Find IAI Expand IAl along Column 1

\\\begin{aligned} &\begin{array}{l} |\mathrm{A}|=\mathrm{a}_{11}(-1)^{1+1}\left|\begin{array}{cc} \mathrm{a}_{22} & \mathrm{a}_{23} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right|+\mathrm{a}_{21}(-1)^{2+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{32} & \mathrm{a}_{33} \end{array}\right| +\mathrm{a}_{31}(-1)^{3+1}\left|\begin{array}{cc} \mathrm{a}_{12} & \mathrm{a}_{13} \\ \mathrm{a}_{22} & \mathrm{a}_{23} \end{array}\right| \\ |\mathrm{A}|=(1)\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|-(-2)\left|\begin{array}{cc} 0 & 0 \\ -1 & 1 \end{array}\right|+0\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right| \\ =(-1+2)+2(0)+0 \\ =1 \end{array}\\ &\text { To find adj } A \end{aligned}

\\\begin{array}{l} a_{11}=\left|\begin{array}{cc} -1 & -2 \\ -1 & 1 \end{array}\right|=-1-2=-3 \\ a_{12}=-\left|\begin{array}{cc} -2 & -2 \\ 0 & 1 \end{array}\right|=-(-2+0)=2 \\ a_{13}=\left|\begin{array}{cc} -2 & -1 \\ 0 & -1 \end{array}\right|=2+0=2 \\ a_{21}=-\left|\begin{array}{cc} 2 & 0 \\ -1 & 1 \end{array}\right|=-(2+0)=-2 \\ a_{22}=\left|\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array}\right|=1 \\ a_{23}=-\left|\begin{array}{cc} 1 & 2 \\ 0 & -1 \end{array}\right|=-(-1)=1 \end{array}

\\\begin{array}{l} a_{31}=\left|\begin{array}{cc} 2 & 0 \\ -1 & -2 \end{array}\right|=-4 \\ a_{32}=-\left|\begin{array}{cc} 1 & 0 \\ -2 & -2 \end{array}\right|=-(-2)=2 \\ a_{33}=\left|\begin{array}{ccc} 1 & 2 \\ -2 & -1 \end{array}\right|=-1+4=3 \\ \therefore \text { adj } A=\left[\begin{array}{ccc} a_{11} & a_{12} & a_{13} \\ a_{21} & a_{22} & a_{23} \\ a_{31} & a_{32} & a_{33} \end{array}\right]^{T}=\left[\begin{array}{ccc} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]^{\mathrm{T}}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \\ \therefore A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 3 \end{array}\right]}{1}=\left[\begin{array}{ccc} -3 & -2 & -4 \\ 2 & 1 & 2 \\ 2 & 1 & 3 \end{array}\right] \end{array}

According to linear equation:

\\x -2y = 10\\ \\ \vspace{\baselineskip}2x -y -z = 8\\ \\ \vspace{\baselineskip}-2y + z = 7\\ \\

We know that, AX = B

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

Here,
So, transpose of \\ A^{-1}

\\ A^{-1}=\left[\begin{array}{lll}-3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3\end{array}\right]$

\\\begin{array}{l} \Rightarrow X=A^{-1} B \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{lll} -3 & 2 & 2 \\ -2 & 1 & 1 \\ -4 & 2 & 3 \end{array}\right]\left[\begin{array}{l} 10 \\ 8 \\ 7 \end{array}\right] \\ \Rightarrow\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} -30+16+14 \\ -20+8+7 \\ -40+16+21 \end{array}\right] \\ \therefore x=0, y=-5 \text { and } z=-3 \end{array}

 

 

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