Question

2．已知矩陣A=$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$，B=$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$，求矩陣A^-1B．

Answer 1

分析 設(shè)矩陣A-1=$[\begin{array}{l}{a}&\\{c}&lckts2f\end{array}]$，通過AA^-1為單位矩陣可得A^-1，進(jìn)而可得結(jié)論．

解答 解：設(shè)矩陣A的逆矩陣為$[\begin{array}{l}{a}&\\{c}&hcm8lbx\end{array}]$，
則$[\begin{array}{l}{-1}&{0}\\{0}&{2}\end{array}]$$[\begin{array}{l}{a}&\\{c}&p81udek\end{array}]$=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$，即$[\begin{array}{l}{-a}&{-b}\\{2c}&{2d}\end{array}]$=$[\begin{array}{l}{1}&{0}\\{0}&{1}\end{array}]$，
故a=-1，b=0，c=0，d=$\frac{1}{2}$，
從而A^-1=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$，
∴A^-1B=$[\begin{array}{l}{-1}&{0}\\{0}&{\frac{1}{2}}\end{array}]$$[\begin{array}{l}{1}&{2}\\{0}&{6}\end{array}]$=$[\begin{array}{l}{-1}&{-2}\\{0}&{3}\end{array}]$．

點(diǎn)評(píng) 本題考查逆矩陣、矩陣的乘法，考查運(yùn)算求解能力，屬于基礎(chǔ)題．