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Given that A = $\begin{bmatrix} 2 & 1 \\ -5 & -4 \end{bmatrix} $ and B = $\begin{bmatrix} 3 & -1 \\ -1 & 0 \end{bmatrix} $

Find a 2 X 2 matrix C such that $CA= B$

I multiply both sides by $A^{-1}$

Since $A^{-1}A = I $

$ CI = BA^{-1}$

Since $CI = IC = C$

$ C = BA^{-1} $

However, when I carry on and find out the answer to matrix C, I can’t get the answer. Where have I gone wrong ?

user175089
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2 Answers2

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$C = BA^{-1}$ is o.k. What you have done wrong ? Show your computations !!

Fred
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The inverse matrix of $A$ is $$A^{-1}=\left[\begin{matrix}\frac 43&\frac 13\\ -\frac {5}3&-\frac {2}3\end{matrix}\right]$$ And then the matrix $C$ is $$C=BA^{-1}=\left[\begin{matrix}3&-1\\ -1&0\end{matrix}\right]\left[\begin{matrix}\frac 43&\frac 13\\ -\frac {5}3&-\frac {2}3\end{matrix}\right]=\left[\begin{matrix}\frac {17}3&\frac 32\\ -\frac {4}3&-\frac {1}3\end{matrix}\right]$$

Matrices aren’t a commutative group. Read this.

  • I did $A^{-1} \times B$ not $B \times A^{-1}$ why do they have different answers ? – user175089 Feb 16 '18 at 08:21
  • @user175089 $A^{-1}\times B$ is different from $B\times A^{-1}$ because matrices aren’t commutative group. –  Feb 16 '18 at 08:23