Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# If $A$=$\left[ \begin{matrix} 3 & x \\ 0 & 1 \\\end{matrix} \right]$ and $B$=$\left[ \begin{matrix} 9 & 16 \\ 0 & -y \\\end{matrix} \right]$. Find $x$ and $y$ when $A^2=B$

Last updated date: 23rd Jul 2024
Total views: 347.7k
Views today: 7.47k
Answer
Verified
347.7k+ views
Hint: We are given equations in the matrix form. So, we will first create a matrix on either side and then we will compare the elements. After doing that we will calculate the value of$x$and$y$. We need to find $A^2$ as well, which means we have to multiply the matrix $A$ by itself. The matrix multiplication is a bit of a complex process because it is not done like the real numbers. After making a matrix on both sides, we will compare the matrices element-wise and obtain the result.

Complete step by step answer:
To multiply the matrix $A$ by itself, we use the formula below for matrix multiplication:
If $A=[a_{ij}]$ is an $m\times n$ matrix and $B=[b_{ij}]$ is an $n\times p$ matrix,
The product AB is an $m\times p$ matrix.
$AB=[c_{ij}]$
Where$c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+...+a_{in}b_{nj}$
So, we have:
$A=\left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right]$
Using the formula we obtain:
\begin{align} & {{A}^{2}}=\left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3\times 3+\left( x\times 0 \right) & 3\times x+\left( x\times 1 \right) \\ 0\times 3+\left( 1\times 0 \right) & 0\times x+\left( 1\times 1 \right) \\ \end{matrix} \right] \\ \end{align}
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 9 & 4x \\ 0 & 1 \\ \end{matrix} \right]$
Hence, we have found $A^2$
Now, we plug these values in the equation given:
${{A}^{2}}=B$
$\Rightarrow \left[ \begin{matrix} 9 & 4x \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 16 \\ 0 & -y \\ \end{matrix} \right]$
Now, we compare the elements. After comparing the element at first row and second column we get:
$4x=16$
$\Rightarrow x=\dfrac{16}{4}$
$\Rightarrow x=4$
Now we compare the element at second row, second column:
$1=-y$
$\Rightarrow y=-1$
So, the values of $x$ and $y$ have been found.

Note: Make sure that you add the terms before giving the resultant value in each position of the resultant matrix. Look for any calculation mistake that might occur while doing multiplication. Always check the other elements to cross check if you have made any calculation mistakes.