Question

Let A = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right] . If ${\mu _1}$ and ${\mu _2}$ are column matrices such that A{\mu _1} = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right] and A{\mu _2} = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ 0 \end{array}} \right] then ${\mu _1} + {\mu _2}$ is equal to:
(A) \left[ {\begin{array}{*{20}{c}} { - 1} \\\ 1 \\\ 0 \end{array}} \right]
(B) \left[ {\begin{array}{*{20}{c}} { - 1} \\\ 1 \\\ { - 1} \end{array}} \right]
(C) \left[ {\begin{array}{*{20}{c}} { - 1} \\\ { - 1} \\\ 0 \end{array}} \right]
(D) \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 1} \\\ { - 1} \end{array}} \right]

Answer 1

Hint : In this we first calculate the order of ${\mu _1}$ and ${\mu _2}$ by using the concept of matrix multiplication. In this from order of matrix A and from order of matrix obtained as result after multiplication of A and ${\mu _1}$ we can find order of matrix ${\mu _1}$ and similar order of matrix ${\mu _2}$ . And then using it we can easily find the value of ${\mu _1}$ and ${\mu _2}$ so the value of ${\mu _1}$ + ${\mu _2}$ .

Complete step-by-step answer :
Given,
Matrix A = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right] , A{\mu _1} = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right] and {A_{{\mu _2}}} = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ 0 \end{array}} \right]
From the product of $A{\mu _1}$ we see that the order of the matrix obtained as a result is of $3 \times 1$ . But the order of matrix A is $3 \times 3$ .
Hence, from these two we can see that the order of the matrix ${\mu _1}$ will be of $3 \times 1$ , otherwise it is not possible to have a matrix of order $3 \times 1$ as a result.
Therefore, we can now let matrix ${\mu _1}$ = \left[ {\begin{array}{*{20}{c}} a \\\ b \\\ c \end{array}} \right] .
Now, substituting value of A and ${\mu _1}$ in given condition A{\mu _1} = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right] . We have,
\left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} a \\\ b \\\ c \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right]
Using laws of multiplication of matrix on left hand side we have
\left[ {\begin{array}{*{20}{c}} {1.a + 0 + 0.c} \\\ {2.a + 1.b + 0.c} \\\ {3.a + 2.b + 1.c} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right]
Now, on comparing above matrices.
$1.a + 0.b + 0.c = 1,\,\,2.a + 1.b + 0.c = 0\,\,and\,\,3.a + 2.b + 1.c = 0 \;$
On solving above formed equations we have
From first equation we have a = $1$
Using a = $1$ in $2a + b = 0$ to value of b
$\Rightarrow 2(1) + b = 0 \\\ \Rightarrow 2 + b = 0 \\\ \Rightarrow b = - 2 \;$
Now, using the value of both a and b in the third equation. We have,
$3a + 2b + c = 0 \\\ \Rightarrow 3(1) + 2( - 2) + c = 0 \\\ \Rightarrow 3 - 4 + c = 0 \\\ \Rightarrow - 1 + c = 0 \\\ \Rightarrow c = 1 \;$
Using value of a, b and c matrix ${\mu _1}$ becomes \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 2} \\\ 1 \end{array}} \right] .
Therefore, {\mu _1} = \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 2} \\\ 1 \end{array}} \right] …………………… (i)
Similarly we can find the value of ${\mu _2}$ .
From the product of $A{\mu _2}$ we see that the order of the matrix obtained as a result is of $3 \times 1$ . But the order of matrix A is $3 \times 3$ .
Hence, from these two we can see that the order of matrix ${\mu _2}$ will be of $3 \times 1$ , otherwise it is not possible to have a matrix of order $3 \times 1$ as a result.
Therefore, we can now let matrix ${\mu _1}$ = \left[ {\begin{array}{*{20}{c}} d \\\ e \\\ f \end{array}} \right] .
Now, substituting value of A and ${\mu _2}$ in given condition {A_{{\mu _2}}} = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ 0 \end{array}} \right] . We have,
\left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right]\left[ {\begin{array}{*{20}{c}} d \\\ e \\\ f \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ 0 \end{array}} \right]
Using laws of multiplication of matrix on left hand side we have
\left[ {\begin{array}{*{20}{c}} {1.d + 0.e + 0.f} \\\ {2.d + 1.e + 0.f} \\\ {3.d + 2.e + 1.f} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ 0 \end{array}} \right]

On comparing two matrices. We have
$1.d + 0.e + 0.f = 0 \\\ 2.d + 1.e + 0.f = 1 \\\ 3.d + 2.e + 1.f = 0 \\\$
Now, solving above formed equations for values of d, e and f.
From first equation we have
$1.d = 0 \\\ \Rightarrow d = 0 \\\$
Substituting value of d in second equation. We have
$2(0) + 1.e = 1 \\\ \Rightarrow 0 + e = 1 \\\ \Rightarrow e = 1 \\\$
Now, using the value of d and e in 3rd equation to find value of f.
$3(0) + 2(1) + f = 0 \\\ \Rightarrow 0 + 2 + f = 0 \\\ \Rightarrow f = - 2 \\\$
Hence, ${\mu _2}$ becomes \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ { - 2} \end{array}} \right]
Therefore, from above we have {\mu _2} = \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ { - 2} \end{array}} \right] ………………….(ii)
Adding (i) and (ii) we have
{\mu _1} + {\mu _2} = \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 2} \\\ 1 \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} 0 \\\ 1 \\\ { - 2} \end{array}} \right] \\\ \Rightarrow {\mu _1} + {\mu _2} = \left[ {\begin{array}{*{20}{c}} {1 + 0} \\\ { - 2 + 1} \\\ {1 - 2} \end{array}} \right] \\\ \Rightarrow {\mu _1} + {\mu _2} = \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 1} \\\ { - 1} \end{array}} \right] \\\
Therefore required value of ${\mu _1} + {\mu _2}$ is \left[ {\begin{array}{*{20}{c}} 1 \\\ { - 1} \\\ { - 1} \end{array}} \right] $$

Note : Solution of above problems can also be found by using the concept of inverse matrix multiplication. In this we first shift A to the left hand side and then calculate the inverse of the matrix and then find the product between the inverse matrix obtained and the result given to find the value of ${\mu _1}$ . In the same way we can calculate the value of ${\mu _2}$ and hence the required value of ${\mu _1} + {\mu _2}$ .