MyQuestionIcon

1

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Finding Inverse Using Elementary Transformations

1 & 1 & 1 1 &...

Question

$If A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦, find A^{- 1} .$ Hence, solve the system of equations
$x + y + z = 6, x + 2 z = 7, 3 x + y + z = 12.$

Open in App

Solution

$| A | = 1 (0 - 2) - 1 (1 - 6) + 1 (1 - 0) = - 2 + 5 + 1 = 4$

Cofactors of the elements of matrix $A$ are
$C_{11}$ = $(- 1)^{1 + 1} [\begin{matrix} 0 & 2 1 & 1 \end{matrix}]$ = $1 (0 - 2) = - 2$

$C_{12}$ = $(- 1)^{1 + 2} [\begin{matrix} 1 & 2 3 & 1 \end{matrix}]$ = $(- 1) (1 - 6) = 5$

$C_{13}$ = $(- 1)^{1 + 3} [\begin{matrix} 1 & 0 3 & 1 \end{matrix}]$ = $1 (1 - 0) = 1$

$C_{21}$ = $(- 1)^{2 + 1} [\begin{matrix} 1 & 1 1 & 1 \end{matrix}]$ = $(- 1) (0 - 0) = 0$

$C_{22}$ = $(- 1)^{2 + 2} [\begin{matrix} 1 & 1 3 & 1 \end{matrix}]$ = $1 (1 - 3) = - 2$

$C_{23}$ = $(- 1)^{2 + 3} [\begin{matrix} 1 & 1 3 & 1 \end{matrix}]$ = $(- 1) (1 - 3) = 2$

$C_{31}$ = $(- 1)^{3 + 1} [\begin{matrix} 1 & 1 0 & 2 \end{matrix}]$ = $(2 - 0) = 2$

$C_{32}$ = $(- 1)^{3 + 2} [\begin{matrix} 1 & 1 1 & 2 \end{matrix}]$ = $(- 1) (2 - 1) = - 1$

$C_{33}$ = $(- 1)^{3 + 3} [\begin{matrix} 1 & 1 1 & 0 \end{matrix}]$ = $(0 - 1) = - 1$

So, cofactor matrix $C = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 5 & 1 0 & - 2 & 2 2 & - 1 & - 1 \end{matrix} ⎤ ⎥ ⎦$

$Adj A = C^{T} = ⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 2 5 & - 2 & - 1 1 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦$

$∴ A^{- 1} = \frac{Adj A}{| A |} = \frac{1}{4}$ $⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 2 5 & - 2 & - 1 1 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦$

The given system of equations can be expressed as

$⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦$ $⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ = $⎡ ⎢ ⎣ \begin{matrix} 6712 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow A X = B$
$where A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦, X = ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ $and B = ⎡ ⎢ ⎣ \begin{matrix} 6712 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow X = A^{- 1} B$

$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ = $\frac{1}{4}$ $⎡ ⎢ ⎣ \begin{matrix} - 2 & 0 & 2 5 & - 2 & - 1 1 & 2 & - 1 \end{matrix} ⎤ ⎥ ⎦$ $⎡ ⎢ ⎣ \begin{matrix} 6712 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ = $\frac{1}{4} ⎡ ⎢ ⎣ \begin{matrix} 1248 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow ⎡ ⎢ ⎣ \begin{matrix} x y z \end{matrix} ⎤ ⎥ ⎦$ = $⎡ ⎢ ⎣ \begin{matrix} 312 \end{matrix} ⎤ ⎥ ⎦$

$\Rightarrow x = 3; y = 1; z = 2$

Suggest Corrections

thumbs-up

0

Similar questions

If A = ⎡ ⎢ ⎣ \begin{matrix} 1 & 1 & 1 1 & 0 & 2 3 & 1 & 1 \end{matrix} ⎤ ⎥ ⎦, find A^{- 1} .

Hence, solve the system of equations

x + y + z = 6, x + 2 z = 7, 3 x + y + z = 12.

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x + y + 3z = 8, −2y + z = 7

(ii)

A = [\begin{matrix} 3 & - 4 & 2 \\ 2 & 3 & 5 \\ 1 & 0 & 1 \end{matrix}]

, find A⁻¹ and hence solve the following system of equations:
3x − 4y + 2z = −1, 2x + 3y + 5z = 7, x + z = 2

(iii)

A = [\begin{matrix} 1 & - 2 & 0 \\ 2 & 1 & 3 \\ 0 & - 2 & 1 \end{matrix}] and B = [\begin{matrix} 7 & 2 & - 6 \\ - 2 & 1 & - 3 \\ - 4 & 2 & 5 \end{matrix}]

, find AB. Hence, solve the system of equations:
x − 2y = 10, 2x + y + 3z = 8 and −2y + z = 7

(iv) If

A = [\begin{matrix} 1 & 2 & 0 \\ - 2 & - 1 & - 2 \\ 0 & - 1 & 1 \end{matrix}]

, find A⁻¹. Using A⁻¹, solve the system of linear equations
x − 2y = 10, 2x − y − z = 8, −2y + z = 7

(v) Given

A = [\begin{matrix} 2 & 2 & - 4 \\ - 4 & 2 & - 4 \\ 2 & - 1 & 5 \end{matrix}], B = [\begin{matrix} 1 & - 1 & 0 \\ 2 & 3 & 4 \\ 0 & 1 & 2 \end{matrix}]

, find BA and use this to solve the system of equations
y + 2z = 7, x − y = 3, 2x + 3y + 4z = 17

(vi) If

A = (\begin{array}{rrr} 2 & 3 & 1 \\ 1 & 2 & 2 \\ - 3 & 1 & - 1 \end{array})

, find A^–1 and hence solve the system of equations 2x + y – 3z = 13, 3x + 2y + z = 4, x + 2y – z = 8.
(vii) Use product

[\begin{matrix} 1 & - 1 & 2 \\ 0 & 2 & - 3 \\ 3 & - 2 & 4 \end{matrix}] [\begin{matrix} - 2 & 0 & 1 \\ 9 & 2 & - 3 \\ 6 & 1 & - 2 \end{matrix}]

to solve the system of equations x + 3z = 9, −x + 2y − 2z = 4, 2x − 3y + 4z = −3.

Q. Find A⁻¹, if

A = [\begin{matrix} 1 & 2 & 5 \\ 1 & - 1 & - 1 \\ 2 & 3 & - 1 \end{matrix}]

. Hence solve the following system of linear equations:
x + 2y + 5z = 10, x − y − z = −2, 2x + 3y − z = −11

A^{- 1}

A = ω ⎡ ⎢ ⎣ \begin{matrix} 4 & 2 & 3 1 & 1 & 1 3 & 1 & - 2 \end{matrix} ⎤ ⎥ ⎦

Hence, solve the following system of linear equations

4 x + 2 y + 3 z = 2

x + y + z = 1

3 x + y - 2 z = 5

Q. Solve the following system of equations by matrix method

x + y + z = 6

x - y - z = - 4

x + 2 y - 2 z = - 1

View More

Join BYJU'S Learning Program

similar_icon

Related Videos

lock

Finding Inverse Using Elementary Transformations

MATHEMATICS

Watch in App

explore_more_icon

Explore more

Finding Inverse Using Elementary Transformations

Standard XII Mathematics

Join BYJU'S Learning Program

CrossIcon