Find the value of x, if $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)=\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$.
Answer
669.9k+ views
Hint: We solve this question starting with checking whether the matrices can be multiplied, by checking if the number of columns of the first matrix in the product is equal to number of rows in the second matrix. Then if the product is possible then we multiply the matrices given and we obtain another matrix. Then we equate the obtained matrix with the given result and by comparing all the elements in both matrices we can find the value of x.
Complete step by step answer:
We were given that the product of two matrices $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$ and $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$ is $\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$.
Let us go through the concept of multiplication of matrices, for solving the given question.
Consider two matrices ${{A}_{m\times n}}$ and ${{B}_{p\times q}}$, where m, n are number of rows and number of columns of the matrix A and p, q are number of rows and number of columns of the matrix B. Then product of matrices A and B is possible only when the number of columns of A is equal to number of columns of B, that is product of matrices A and B can be possible only if n=p.
Now let us check whether the given matrices are eligible for multiplication.
For the matrix $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$, number of columns are 2.
For the matrix $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$, number of rows are 2.
The number of columns of the first matrix in the product is equal to the number of rows in the second matrix. We can multiply the given matrices.
$\begin{align}
& \Rightarrow \left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
2\left( 1 \right)+3\left( -2 \right) & 2\left( -3 \right)+3\left( 4 \right) \\
5\left( 1 \right)+7\left( -2 \right) & 5\left( -3 \right)+7\left( 4 \right) \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
2+\left( -6 \right) & -6+12 \\
5+\left( -14 \right) & -15+28 \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
-4 & 6 \\
-9 & 13 \\
\end{matrix} \right) \\
\end{align}$
Given that the obtained product is equal to the matrix $\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$.
So, let us compare the obtained result with the above matrix.
$\Rightarrow \left( \begin{matrix}
-4 & 6 \\
-9 & 13 \\
\end{matrix} \right)=\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$
As we look at the two matrices above all the three known elements are equal and so let us equate the unknown element x. Then we get,
$\Rightarrow x=13$
So, the value of x is 13.
Hence the answer is 13.
Note:
While solving this question, checking whether the number of columns of the first matrix in the product is equal to the number of rows in the second matrix is necessary and one must not forget it because if the matrix does not satisfy that, the product of those is not possible. There is another possibility of mistake by multiplying the second matrix with the first matrix that is considering $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$ instead of $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$. But they do not give the same value when multiplied. $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)=\left( \begin{matrix}
1\left( 2 \right)+\left( -3 \right)\left( 5 \right) & 1\left( 3 \right)+\left( -3 \right)\left( 7 \right) \\
-2\left( 2 \right)+4\left( 5 \right) & -2\left( 3 \right)+4\left( 7 \right) \\
\end{matrix} \right)=\left( \begin{matrix}
-13 & -19 \\
16 & 22 \\
\end{matrix} \right)$, which is not equal to the value we got in the solution for the product of matrices. So, one should multiply the matrices in the given order and should not change them.
Complete step by step answer:
We were given that the product of two matrices $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$ and $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$ is $\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$.
Let us go through the concept of multiplication of matrices, for solving the given question.
Consider two matrices ${{A}_{m\times n}}$ and ${{B}_{p\times q}}$, where m, n are number of rows and number of columns of the matrix A and p, q are number of rows and number of columns of the matrix B. Then product of matrices A and B is possible only when the number of columns of A is equal to number of columns of B, that is product of matrices A and B can be possible only if n=p.
Now let us check whether the given matrices are eligible for multiplication.
For the matrix $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$, number of columns are 2.
For the matrix $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$, number of rows are 2.
The number of columns of the first matrix in the product is equal to the number of rows in the second matrix. We can multiply the given matrices.
$\begin{align}
& \Rightarrow \left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
2\left( 1 \right)+3\left( -2 \right) & 2\left( -3 \right)+3\left( 4 \right) \\
5\left( 1 \right)+7\left( -2 \right) & 5\left( -3 \right)+7\left( 4 \right) \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
2+\left( -6 \right) & -6+12 \\
5+\left( -14 \right) & -15+28 \\
\end{matrix} \right) \\
& \Rightarrow \left( \begin{matrix}
-4 & 6 \\
-9 & 13 \\
\end{matrix} \right) \\
\end{align}$
Given that the obtained product is equal to the matrix $\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$.
So, let us compare the obtained result with the above matrix.
$\Rightarrow \left( \begin{matrix}
-4 & 6 \\
-9 & 13 \\
\end{matrix} \right)=\left( \begin{matrix}
-4 & 6 \\
-9 & x \\
\end{matrix} \right)$
As we look at the two matrices above all the three known elements are equal and so let us equate the unknown element x. Then we get,
$\Rightarrow x=13$
So, the value of x is 13.
Hence the answer is 13.
Note:
While solving this question, checking whether the number of columns of the first matrix in the product is equal to the number of rows in the second matrix is necessary and one must not forget it because if the matrix does not satisfy that, the product of those is not possible. There is another possibility of mistake by multiplying the second matrix with the first matrix that is considering $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)$ instead of $\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)$. But they do not give the same value when multiplied. $\left( \begin{matrix}
1 & -3 \\
-2 & 4 \\
\end{matrix} \right)\left( \begin{matrix}
2 & 3 \\
5 & 7 \\
\end{matrix} \right)=\left( \begin{matrix}
1\left( 2 \right)+\left( -3 \right)\left( 5 \right) & 1\left( 3 \right)+\left( -3 \right)\left( 7 \right) \\
-2\left( 2 \right)+4\left( 5 \right) & -2\left( 3 \right)+4\left( 7 \right) \\
\end{matrix} \right)=\left( \begin{matrix}
-13 & -19 \\
16 & 22 \\
\end{matrix} \right)$, which is not equal to the value we got in the solution for the product of matrices. So, one should multiply the matrices in the given order and should not change them.
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