Question

Find the value of x, y and z if A = \left[ {\begin{array}{*{20}{c}} 0&{2y}&z; \\\ x&y;&{ - z} \\\ x&{ - y}&z; \end{array}} \right] satisfies $A' = {A^{ - 1}}$

Answer 1

We can find the transpose of the matrix by interchanging the rows and columns. Then we can equate the inverse to the transpose. We can take multiply the matrix with its inverse and equate it to the identity matrix. After multiplication, we can equate the elements of the matrix and solve for x, y, and z.

Complete step by step answer:

We have a square matrix A with order 3.
A = \left[ {\begin{array}{*{20}{c}} 0&{2y}&z; \\\ x&y;&{ - z} \\\ x&{ - y}&z; \end{array}} \right]
We know that transpose of a matrix A = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right] is given by A' = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{21}}}&{{a_{31}}} \\\ {{a_{12}}}&{{a_{22}}}&{{a_{32}}} \\\ {{a_{13}}}&{{a_{23}}}&{{a_{33}}} \end{array}} \right]
So, transpose of A is given by,
A' = \left[ {\begin{array}{*{20}{c}} 0&x;&x; \\\ {2y}&y;&{ - y} \\\ z&{ - z}&z; \end{array}} \right]
We know that a matrix multiplied with its inverse will give the identity matrix.
$\Rightarrow {A^{ - 1}}A = I$
It is given that $A' = {A^{ - 1}}$
$\Rightarrow A'A = I$
On substituting the matrices, we get,
\Rightarrow \left[ {\begin{array}{*{20}{c}} 0&x;&x; \\\ {2y}&y;&{ - y} \\\ z&{ - z}&z; \end{array}} \right]\left[ {\begin{array}{*{20}{c}} 0&{2y}&z; \\\ x&y;&{ - z} \\\ x&{ - y}&z; \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]
On doing the matrix multiplication, we get,
\Rightarrow \left[ {\begin{array}{*{20}{c}} {0 + {x^2} + {x^2}}&{0 + xy - xy}&{0 - xz + xz} \\\ {0 + xy - xy}&{4{y^2} + {y^2} + {y^2}}&{2yz - yz - yz} \\\ {0 - zx + zx}&{2yz - yz - yz}&{{z^2} + {z^2} + {z^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]
On further simplification, we get,
\Rightarrow \left[ {\begin{array}{*{20}{c}} {2{x^2}}&0&0 \\\ 0&{6{y^2}}&0 \\\ 0&0&{3{z^2}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&0&0 \\\ 0&1&0 \\\ 0&0&1 \end{array}} \right]
On equating the diagonal terms, we get,
$2{x^2} = 1$
On dividing throughout with 2, we get,
$\Rightarrow {x^2} = \dfrac{1}{2}$
On taking the square root, we get,
$\Rightarrow x = \pm \dfrac{1}{{\sqrt 2 }}$
Equating the terms with y, we get,
$6{y^2} = 1$
On dividing throughout with 6, we get,
$\Rightarrow {y^2} = \dfrac{1}{6}$
On taking the square root, we get,
$\Rightarrow y = \pm \dfrac{1}{{\sqrt 6 }}$
Equating the terms with z, we get,
$3{z^2} = 1$
On dividing throughout with 6, we get,
$\Rightarrow {z^2} = \dfrac{1}{3}$
On taking the square root, we get,
$\Rightarrow z = \pm \dfrac{1}{{\sqrt 3 }}$
Therefore, the required values are $x = \pm \dfrac{1}{{\sqrt 2 }}$ , $y = \pm \dfrac{1}{{\sqrt 6 }}$ and $z = \pm \dfrac{1}{{\sqrt 3 }}$.

Note: Inverse of a matrix exists only for square matrix. The inverse of a matrix multiplied with the matrix gives an identity matrix. The identity matrix is the square matrix with all the diagonal entries equal to 1 and all other entries equal to 0. Matrix multiplication is defined only when the number of columns of the 1st matrix is equal to the number of the rows of the 2nd matrix. The resultant matrix will have the number of rows of 1st matrix and the number of columns of the 2nd matrix. Matrix multiplication is not commutative. So the order is important for matrix multiplication.