Let A=aij be a 3×3 matrix such that A010=001,A413=010 and A212=100, then a23 equals :

Let A=a11 a12 a13a21 a22 a23a31 a32 a33A010=001⇒a12a22a32=010⇒a22=0;a12=0a32=1A413=010⇒4a11+a12+3a13=04a21+a22+3a23=1⇒4a21+3a23=14a31+a32+3a33=0A212=100⇒2a11+a12+2a13=12a21+a22+2a23=0⇒a21+a23=02a31+a32+2a33=0−4a23+3a23=1⇒a23=−1

Let A=aij be a 3×3 matrix such that A010=001,A413=010 and A212=100, then a23 equals :

Let A=a11 a12 a13a21 a22 a23a31 a32 a33A010=001⇒a12a22a32=010⇒a22=0;a12=0a32=1A413=010⇒4a11+a12+3a13=04a21+a22+3a23=1⇒4a21+3a23=14a31+a32+3a33=0A212=100⇒2a11+a12+2a13=12a21+a22+2a23=0⇒a21+a23=02a31+a32+2a33=0−4a23+3a23=1⇒a23=−1

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Let $A = [a_{ij}] be a 3 \times 3$ matrix such that $A [\begin{array}{l} 0 \\ 1 \\ 0 \end{array}] = [\begin{array}{l} 0 \\ 0 \\ 1 \end{array}], A [\begin{array}{l} 4 \\ 1 \\ 3 \end{array}] = [\begin{array}{l} 0 \\ 1 \\ 0 \end{array}] and A [\begin{array}{l} 2 \\ 1 \\ 2 \end{array}] = [\begin{array}{l} 1 \\ 0 \\ 0 \end{array}]$ , then $a_{23}$ equals :

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Detailed Solution

Let $A = [\begin{array}{lll} a_{11} a_{12} a_{13} \\ a_{21} a_{22} a_{23} \\ a_{31} a_{32} a_{33} \end{array}]$
$\begin{array}{l} A [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 1 \end{matrix}] \Rightarrow [\begin{matrix} a_{12} \\ a_{22} \\ a_{32} \end{matrix}] = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] \Rightarrow \begin{matrix} a_{22} = 0; a_{12} = 0 \\ a_{32} = 1 \end{matrix} \\ A [\begin{matrix} 4 \\ 1 \\ 3 \end{matrix}] = [\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}] \Rightarrow \begin{matrix} 4 a_{11} + a_{12} + 3 a_{13} = 0 \\ 4 a_{21} + a_{22} + 3 a_{23} = 1 \Rightarrow 4 a_{21} + 3 a_{23} = 1 \\ 4 a_{31} + a_{32} + 3 a_{33} = 0 \end{matrix} \\ A [\begin{matrix} 2 \\ 1 \\ 2 \end{matrix}] = [\begin{matrix} 1 \\ 0 \\ 0 \end{matrix}] \Rightarrow \begin{matrix} 2 a_{11} + a_{12} + 2 a_{13} = 1 \\ 2 a_{21} + a_{22} + 2 a_{23} = 0 \Rightarrow a_{21} + a_{23} = 0 \\ 2 a_{31} + a_{32} + 2 a_{33} = 0 \\ - 4 a_{23} + 3 a_{23} = 1 \Rightarrow a_{23} = - 1 \end{matrix} \end{array}$