How would I find the values to the matrix equation below?

Question

If there are any similar questions, please link them in the comments. I'm just a little confused as to how you get the answers (which are below).

Find the values of $a$, $b$, $c$, and $d$ in the matrix equation below. $$\begin{bmatrix}a&b\\c&d\end{bmatrix}+5\begin{bmatrix}-5&-6\\7&x\end{bmatrix}^T=\begin{bmatrix}-1&-3\\-3&6\end{bmatrix}$$ $$\begin{align}a&=24\\b&=-38\\c&=27\\d&=6-5x\end{align}$$

Answer 1

$$\begin{bmatrix}-5&-6\\7 & x\end{bmatrix}^T=\begin{bmatrix}-5 & 7 \\ -6 & x\end{bmatrix}$$ Therefore, $$\begin{bmatrix}a&b\\c&d\end{bmatrix}+5\begin{bmatrix}-5 & 7 \\ -6 & x\end{bmatrix}=\begin{bmatrix}a-25&b+35\\c-30 & d+5x\end{bmatrix}=\begin{bmatrix}-1 & -3 \\ -3 &6\end{bmatrix}.$$

Answer 2

The first step you'd want to take would be transposing the second matrix to get $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}+5 \begin{bmatrix} -5 & 7 \\ -6 & x \\ \end{bmatrix}=\begin{bmatrix} -1 & -3 \\ -3 & 6 \\ \end{bmatrix},$$ since transposing a matrix flips its entries around the main diagonal (upper-left to lower-right). Next, distribute the 5 inside that same matrix to get $$\begin{bmatrix} a & b \\ c & d \\ \end{bmatrix}+ \begin{bmatrix} -25 & 35 \\ -30 & 5x \\ \end{bmatrix}=\begin{bmatrix} -1 & -3 \\ -3 & 6 \\ \end{bmatrix}.$$ Now that you've got two matrices of the same size on the left side of the equation, you can add their corresponding entries: $$\begin{bmatrix} a-25 & b+35 \\ c-30 & d+5x \\ \end{bmatrix}=\begin{bmatrix} -1 & -3 \\ -3 & 6 \\ \end{bmatrix}$$ Since this statement says that the two resulting matrices are equal to each other, every one of their corresponding entries must be equal too, so we get the following four equations: \begin{array}{c} a - 25 = -1 \\ b + 35 = -3 \\ c - 30 = -3 \\ d + 5x = 6 \end{array} Those can then be solved for $a$, $b$, $c$, and $d$ to get the answers you included in the question.

Answer 3

The transpose operation in matrices is defined as switching the row and column indexes of matrix, that is, if you have $i^{th}$ row and $j^{th}$ column in A, then $i^{th}$ row of A is $i^{th}$ column of $A^T$ and $j^{th}$ column of A is $j^{th}$ row of $A^T$ for all i and j proper to the sizes of the matrix. Just to be systematic, the transpose of the matrix in your question is: $$ 5\begin{bmatrix} -5 & -6 \\ 7 & x \\ \end{bmatrix}^T = 5\begin{bmatrix} -5 & 7 \\ -6 & x \\ \end{bmatrix} $$ since first row is [-5 -6] and the second row is [7 x]. Notice that first column of $A^T$ is the same as first row of A, as it is for the second row.

So you have the equality $$ \begin{bmatrix} a-25 & b+35 \\ c-30 & d+5x \\ \end{bmatrix} = \begin{bmatrix} -1 & -3 \\ -3 & 6 \\ \end{bmatrix} $$ which is an equation system as follows:

$$ \begin{array}{c} a - 25 = -1 \\ b + 35 = -3 \\ c - 30 = -3 \\ d + 5x = 6 \end{array} $$ Then $$ a=24, b=-38, c=27, d=6-5x$$