Question

Let A=\left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right)and U1,U2,U3 be column matrices satisfying

1 \\\ 0 \\\ 0 \end{array}} \right)\,,\,A{U_2} = \left( {\begin{array}{*{20}{c}} 2 \\\ 3 \\\ 0 \end{array}} \right),\,A{U_3} = \left( {\begin{array}{*{20}{c}} 2 \\\ 3 \\\ 1 \end{array}} \right)$$. If U is 3*3 matrix whose column are U1,U2,U3 then |U|?

Answer 1

Hint : Here the given question is of matrix, in which the relation between matrices are given and in order to solve it we are going to use multiplication and addition property of matrices, and the data given in the question which describes the relation of all theses matrices.

Complete step-by-step answer :
The given question is of solving matrices, here first we have to find the matrix “U” in order to find the determinent of the matrix, on solving we get:
Here the relation is given for U1,U2,U3 in product with the matrix A, so we will here add all theses matrices and then take common the matrix A, in order to solve further, on solving we get:

\Rightarrow A{U_1} + A{U_2} + A{U_3} = \left( {\begin{array}{*{20}{c}} 1 \\\ 0 \\\ 0 \end{array}} \right)\, + \left( {\begin{array}{*{20}{c}} 2 \\\ 3 \\\ 0 \end{array}} \right) + \left( {\begin{array}{*{20}{c}} 2 \\\ 3 \\\ 1 \end{array}} \right) \\\ \Rightarrow A({U_1} + {U_2} + {U_3}) = \left( {\begin{array}{*{20}{c}} {1 + 2 + 2} \\\ {0 + 3 + 3} \\\ {0 + 0 + 1} \end{array}} \right) = \left( {\begin{array}{*{20}{c}} 5 \\\ 6 \\\ 1 \end{array}} \right) \\\

Now here we obtain a relation between matrices, and now to solve further for the matrix “U” which is summation of the three matrices U1,U2,U3, here to solve further we have to multiply the identity matrix on the right side of equation, we know the property of identity matrix that multiplying it by any matrix does not affect the values of matrix.
On solving we get:

1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right)(U) = \left( {\begin{array}{*{20}{c}} {5 \times 1 + 5 \times 0 + 5 \times 0}&{5 \times 0 + 5 \times 1 + 5 \times 0}&{5 \times 0 + 5 \times 0 + 5 \times 1} \\\ {6 \times 1 + 6 \times 0 + 6 \times 0}&{6 \times 0 + 6 \times 1 + 6 \times 0}&{6 \times 0 + 6 \times 0 + 6 \times 1} \\\ {1 \times 1 + 1 \times 0 + 1 \times 0}&{1 \times 0 + 1 \times 1 + 1 \times 0}&{1 \times 0 + 1 \times 0 + 1 \times 1} \end{array}} \right)$$ $$ \Rightarrow \left( {\begin{array}{*{20}{c}} 1&0&0 \\\ 2&1&0 \\\ 3&2&1 \end{array}} \right)(U) = \left( {\begin{array}{*{20}{c}} 5&5&5 \\\ 6&6&6 \\\ 1&1&1 \end{array}} \right)$$ Now using row and column arrangements we have to solve further:

\Rightarrow {C_1} \to {C_1} - 2{C_2} \\
\Rightarrow \left( {\begin{array}{{20}{c}}
{1 - 2(0)}&0&0 \\
{2 - 2(1)}&1&0 \\
{3 - 2(2)}&2&1
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{5 - 2(5)}&5&5 \\
{6 - 2(6)}&6&6 \\
{1 - 2(1)}&1&1
\end{array}} \right) \\
\Rightarrow \left( {\begin{array}{{20}{c}}
1&0&0 \\
0&1&0 \\
{ - 1}&2&1
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{ - 5}&5&5 \\
{ - 6}&6&6 \\
{ - 1}&1&1
\end{array}} \right) \\
\Rightarrow {C_2} \to {C_2} - 2{C_3} \\
\Rightarrow \left( {\begin{array}{{20}{c}}
1&{0 - 2(0)}&0 \\
0&{1 - 2(0)}&0 \\
{ - 1}&{2 - 2(1)}&1
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{ - 5}&{5 - 2(5)}&5 \\
{ - 6}&{6 - 2(6)}&6 \\
{ - 1}&{1 - 2(1)}&1
\end{array}} \right) \\
\Rightarrow \left( {\begin{array}{{20}{c}}
1&0&0 \\
0&1&0 \\
{ - 1}&0&1
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{ - 5}&{ - 5}&5 \\
{ - 6}&{ - 6}&6 \\
{ - 1}&{ - 1}&1
\end{array}} \right) \\
\Rightarrow {R_3} \to {R_3} + {R_1} \\
\Rightarrow \left( {\begin{array}{{20}{c}}
1&0&0 \\
0&1&0 \\
{ - 1 + 1}&{0 + 0}&{1 + 0}
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{ - 5}&{ - 5}&5 \\
{ - 6}&{ - 6}&6 \\
{ - 1 + ( - 5)}&{ - 1 + ( - 5)}&{1 + 5}
\end{array}} \right) \\
\Rightarrow \left( {\begin{array}{{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right)(U) = \left( {\begin{array}{{20}{c}}
{ - 5}&{ - 5}&5 \\
{ - 6}&{ - 6}&6 \\
{ - 6}&{ - 6}&6
\end{array}} \right) \\
\Rightarrow U = \left( {\begin{array}{{20}{c}}
{ - 5}&{ - 5}&5 \\
{ - 6}&{ - 6}&6 \\
{ - 6}&{ - 6}&6
\end{array}} \right)(\sin ce,\left( {\begin{array}{{20}{c}}
1&0&0 \\
0&1&0 \\
0&0&1
\end{array}} \right) = Identity,matrix) \\

Here we get the value of the matrix “U”, now determinant of matrix, $$ \Rightarrow |U| = \left| {\left( {\begin{array}{*{20}{c}} { - 5}&{ - 5}&5 \\\ { - 6}&{ - 6}&6 \\\ { - 6}&{ - 6}&6 \end{array}} \right)} \right| = 0$$ Here the value of determinant is zero because when two rows or columns of a determinant are the same then the value of determinant is zero. **Note** : Here the given question needs to find the value of the determinant given in the question, and to find that we have done a series of solutions needed to solve, as per instruction in the question. In solving a matrix you need to know the properties associated.