
What is the order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right]$ ?
A. $3 \times 1$
B. $1 \times 1$
C. \[1 \times 3\]
D. $3 \times 3$
Answer
163.8k+ views
Hint: Consider a matrix, $A$ of order $l \times m$ and another matrix, $B$ of order $m \times n$ . Let's say that matrix $C$ is formed as a result of the product of matrices $A$ and $B$, that is, $C = AB$. Then, the order will be $l \times n$.
Complete step by step Solution:
Given product of matrices:
$\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right]$
Let us first evaluate the order of each matrix in this product.
Order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right] = \left( {1 \times 3} \right)$
Order of $\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right] = \left( {3 \times 3} \right)$
Order of $\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right] = \left( {3 \times 1} \right)$
Now, we know that the order of a product of two matrices is the number of rows of the first matrix by the number of columns of the second matrix.
Therefore, order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right] = \left( {1 \times 3} \right)\left( {3 \times 3} \right) = \left( {1 \times 3} \right)$
And now,
Order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right] = \left( {1 \times 3} \right)\left( {3 \times 1} \right) = \left( {1 \times 1} \right)$
Hence, the order of final product is $\left( {1 \times 1} \right)$ .
Therefore, the correct option is (B).
Note: Matrix multiplication of two matrices is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix. This means that a matrix $A$ of order $\left( {j \times k} \right)$ and another matrix $B$ of order $\left( {l \times m} \right)$ can only be multiplied when $k = l$ .
Complete step by step Solution:
Given product of matrices:
$\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right]$
Let us first evaluate the order of each matrix in this product.
Order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right] = \left( {1 \times 3} \right)$
Order of $\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right] = \left( {3 \times 3} \right)$
Order of $\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right] = \left( {3 \times 1} \right)$
Now, we know that the order of a product of two matrices is the number of rows of the first matrix by the number of columns of the second matrix.
Therefore, order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right] = \left( {1 \times 3} \right)\left( {3 \times 3} \right) = \left( {1 \times 3} \right)$
And now,
Order of $\left[ {\begin{array}{*{20}{c}}
x&y&z
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&h&g \\
h&b&f \\
g&f&c
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
x \\
y \\
z
\end{array}} \right] = \left( {1 \times 3} \right)\left( {3 \times 1} \right) = \left( {1 \times 1} \right)$
Hence, the order of final product is $\left( {1 \times 1} \right)$ .
Therefore, the correct option is (B).
Note: Matrix multiplication of two matrices is only possible when the number of columns of the first matrix is equal to the number of rows of the second matrix. This means that a matrix $A$ of order $\left( {j \times k} \right)$ and another matrix $B$ of order $\left( {l \times m} \right)$ can only be multiplied when $k = l$ .
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