
If $A = \left( {\begin{array}{*{20}{c}}
1&2&3 \\
3&1&2 \\
2&3&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
{ - 5}&7&1 \\
1&{ - 5}&7 \\
7&1&{ - 5}
\end{array}} \right)$ then $AB$ is equal to
Option:
A. ${I_3}$
B. $2{I_3}$
C. $4{I_3}$
D. $18{I_3}$
Answer
161.1k+ views
Hint: If the number of columns in the first matrix equals the number of rows in the second matrix, the product of the two matrices will be known. The resultant matrix will have the same number of rows as the first matrix and the same number of columns as the second matrix if the product is defined.
Formula Used: The product $AB$ is an $m \times p$ matrix if $A = \left[ {{a_{ij}}} \right]$ is an $m \times n$ matrix and $B = \left[ {{b_{ij}}} \right]$ is an $n \times p$ matrix.
$AB = \left[ {{c_{ij}}} \right]$ , where $\left[ {{c_{ij}}} \right]$ is composed of ${a_{1j}}{b_{1j}} + {a_{2j}}{b_{2j}} + \ldots + {a_{in}}{b_{nj}}$ .
Complete step by step solution: We have matrices $A = \left( {\begin{array}{*{20}{c}}
1&2&3 \\
3&1&2 \\
2&3&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
{ - 5}&7&1 \\
1&{ - 5}&7 \\
7&1&{ - 5}
\end{array}} \right)$ .
To find the product of these two matrices, we have to multiply each row of one matrix to each column of the other matrix respectively.
$AB = \left( {\begin{array}{*{20}{c}}
{1 \times ( - 5) + 2 \times 1 + 3 \times 7}&{3 \times ( - 5) + 1 \times 1 + 2 \times 7}&{2 \times ( - 5) + 3 \times 1 + 1 \times 7} \\
{1 \times 7 + 2 \times ( - 5) + 3 \times 1}&{3 \times 7 + 1 \times ( - 5) + 2 \times 1}&{2 \times 7 + 3 \times ( - 5) + 1 \times 1} \\
{1 \times 1 + 2 \times 7 + 3 \times ( - 5)}&{3 \times 1 + 1 \times 7 + 2 \times ( - 5)}&{2 \times 1 + 3 \times 7 + 1 \times ( - 5)}
\end{array}} \right)$
On solving, we get
$AB = \left( {\begin{array}{*{20}{c}}
{18}&0&0 \\
0&{18}&0 \\
0&0&{18}
\end{array}} \right)$
$ \Rightarrow AB = 18{I_3}$
Option ‘D’ is correct
Note: To multiply two matrices, we must ensure that the number of rows in the second matrix equals the number of columns in the first matrix. As a result, a specific number of rows from the first matrix and a specific number of columns from the second matrix will be present in the final matrix. The matrix multiplication order determines the order of the final matrix.
Formula Used: The product $AB$ is an $m \times p$ matrix if $A = \left[ {{a_{ij}}} \right]$ is an $m \times n$ matrix and $B = \left[ {{b_{ij}}} \right]$ is an $n \times p$ matrix.
$AB = \left[ {{c_{ij}}} \right]$ , where $\left[ {{c_{ij}}} \right]$ is composed of ${a_{1j}}{b_{1j}} + {a_{2j}}{b_{2j}} + \ldots + {a_{in}}{b_{nj}}$ .
Complete step by step solution: We have matrices $A = \left( {\begin{array}{*{20}{c}}
1&2&3 \\
3&1&2 \\
2&3&1
\end{array}} \right)$ and $B = \left( {\begin{array}{*{20}{c}}
{ - 5}&7&1 \\
1&{ - 5}&7 \\
7&1&{ - 5}
\end{array}} \right)$ .
To find the product of these two matrices, we have to multiply each row of one matrix to each column of the other matrix respectively.
$AB = \left( {\begin{array}{*{20}{c}}
{1 \times ( - 5) + 2 \times 1 + 3 \times 7}&{3 \times ( - 5) + 1 \times 1 + 2 \times 7}&{2 \times ( - 5) + 3 \times 1 + 1 \times 7} \\
{1 \times 7 + 2 \times ( - 5) + 3 \times 1}&{3 \times 7 + 1 \times ( - 5) + 2 \times 1}&{2 \times 7 + 3 \times ( - 5) + 1 \times 1} \\
{1 \times 1 + 2 \times 7 + 3 \times ( - 5)}&{3 \times 1 + 1 \times 7 + 2 \times ( - 5)}&{2 \times 1 + 3 \times 7 + 1 \times ( - 5)}
\end{array}} \right)$
On solving, we get
$AB = \left( {\begin{array}{*{20}{c}}
{18}&0&0 \\
0&{18}&0 \\
0&0&{18}
\end{array}} \right)$
$ \Rightarrow AB = 18{I_3}$
Option ‘D’ is correct
Note: To multiply two matrices, we must ensure that the number of rows in the second matrix equals the number of columns in the first matrix. As a result, a specific number of rows from the first matrix and a specific number of columns from the second matrix will be present in the final matrix. The matrix multiplication order determines the order of the final matrix.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

Degree of Dissociation and Its Formula With Solved Example for JEE

Free Radical Substitution Mechanism of Alkanes for JEE Main 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
