Determinants Matrices Question 158
Question:If $\begin{bmatrix} 3 & 0 & 5 \\0 & 2 & 0 \\1 & -3 & 2 \end{bmatrix} \begin{bmatrix} -y & 15y & 8y \\0 & 2 & 0 \\ y & -5y & -3y \end{bmatrix} $=$\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}$
then find the value of y
Options:
A) $\frac{1}{3}$
B) $\frac{1}{4}$
C) No unique value of ‘y’
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
We have $\begin{bmatrix}3 & 0 & 5 \\0 & 2 & 0 \\1 & -3 & 2 \end{bmatrix}\begin{bmatrix}-y & 15y & 8y \\0 & 2 & 0 \\y & -5y & -3y \end{bmatrix}$
=$\begin{bmatrix}6y & 0 & 0 \\0 & 2 & 0 \\0 & 15y-3 & 8y \end{bmatrix}$
=$\begin{bmatrix}1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & 1 \end{bmatrix}$
$\Rightarrow y=\frac{1}{3}$
BETA
