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can anyone help me with this :) ?

Solve matrix equation: $(A-I)(X+I)B^T = 6B$ where:

$$ A= \begin{bmatrix} 3 & 2 \\ 5 & 3 \\ \end{bmatrix} $$ $$ B= \begin{bmatrix} 3 & 2 \\ 7 & 5 \\ \end{bmatrix} $$ where $B^T$ is transposition of $B$

3 Answers 3

5

$X=(A-I)^{-1} \times 6B \times (B^T)^{-1}-I$

2

$$(A-I)(X+I)B^T = 6B$$ Note that $B$ and $B^T$ are square matrices with determinant non-zero ($1$) and hence are invertible.

So, $$(A-I)(X+I) = 6B{\left(B^T\right)}^{-1}$$ $$(A-I)X + (A-I) = 6B{\left(B^T\right)}^{-1}$$ $$(A-I)X=I-A+6B{\left(B^T\right)}^{-1}$$

Since $A-I$ is also invertible, $$X = (A-I)^{-1}\left(I-A+6B{\left(B^T\right)}^{-1}\right)$$ $$= -I+6(A-I)^{-1}B{\left(B^T\right)}^{-1}$$

0

This is a system of $4$ linear equations in the variables $x_1,\ldots ,x_4$ of $X$, which has a unique solution, namely $$ X=\begin{pmatrix} 27 & -38 \cr 5 & -8\end{pmatrix}. $$ So it is only solving $4$ linear equations in $4$ unknowns.

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    Could you expand your thought? – 2017-02-17
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    There is nothing to expand. Write $X$ with $4$ unknowns $x_1,x_2,x_3,x_4$ and write down the matrix equation. This gives four linear equations, one for each coefficient. Then solve it, say, with Gauss elimination. – 2017-02-18