Question

What is the value of the element in row 1, column 1 of matrix C?

Answer 1

Answer:

-3y

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 4

Step-by-step explanation:

I just took that and got it right

Answer 2

Answer:

Option A is correct.

Step-by-step explanation:

We given with $\begin{bmatrix}4&y\\0&1\end{bmatrix}X=\begin{bmatrix}y\\4\end{bmatrix}$

We have to find Element in row 1 and column 1 of Matrix C.

Matrix C is Matrix X.

So, we have

$X=\begin{bmatrix}4&y\\0&1\end{bmatrix}^{-1}\begin{bmatrix}y\\4\end{bmatrix}$

So first we find inverse of first matrix,

Inverse of matrix A is given by $\frac{1}{\left|A\right|}Adj\,A$

let, $A=\begin{bmatrix}4&y\\0&1\end{bmatrix}$

then,

$Adj\,A=\begin{bmatrix}1&-y\\0&4\end{bmatrix}$

|A| = 4 × 1 - 0 × y = 4

So, $X=\frac{1}{\left|A\right|}Adj\,A\times\begin{bmatrix}y\\4\end{bmatrix}$

$=\frac{1}{4}\begin{bmatrix}1&-y\\0&4\end{bmatrix}\begin{bmatrix}y\\4\end{bmatrix}$

$=\frac{1}{4}\begin{bmatrix}1\times y+(-y)\times4\\0\times y+4\times4\end{bmatrix}$

$=\frac{1}{4}\begin{bmatrix}y-4y\\16\end{bmatrix}$

$=\frac{1}{4}\begin{bmatrix}-3y\\16\end{bmatrix}$

$=\begin{bmatrix}\frac{-3y}{4}\\\frac{16}{4}\end{bmatrix}$

$=\begin{bmatrix}\frac{-3y}{4}\\4\end{bmatrix}$

Element at row 1 and column 1 of matrix is $\frac{-3y}{4}$

Therefore, Option A is correct.