If x, y, z are non-zero real numbers, then the inverse of matrix

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Determinants

Long Answer Type

221.

Compute (AB)¹ where:
$straight A space equals space open square brackets table row 1 cell space space space space 1 end cell cell space space space 2 end cell row 0 cell space space space 2 end cell cell space minus 3 end cell row 3 cell space minus 2 end cell cell space space space space 4 end cell end table close square brackets space space space space space space space straight B to the power of negative 1 end exponent space equals space open square brackets table row 1 cell space space space space space 2 end cell cell space space space space space 0 end cell row 0 cell space space space 3 end cell cell space space minus 1 end cell row 1 cell space space 0 end cell cell space space space space 2 end cell end table close square brackets$

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222.

If $straight A space equals space open square brackets table row 2 cell space space space 2 end cell cell space space space 1 end cell row cell negative 2 end cell cell space space space 1 end cell cell space space 2 end cell row 1 cell space minus 2 end cell cell space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row 1 cell space space 3 end cell cell space space 2 end cell row 1 cell space space space 1 end cell cell space 1 end cell row 2 cell negative 3 end cell cell space 1 end cell end table close square brackets comma$ verify that (AB)^–1 = B"^–1 A^–1.

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223.

If $straight A to the power of negative 1 end exponent space equals space open square brackets table row 3 cell space space minus 1 end cell cell space space space space space 1 end cell row cell negative 15 end cell cell space space space space space 6 end cell cell space minus 5 end cell row 5 cell space minus 2 end cell cell space space space space 2 end cell end table close square brackets space space space and space straight B space equals space open square brackets table row cell space space space 1 end cell cell space space space space 2 end cell cell space space space minus 2 end cell row cell negative 1 end cell cell space space space space space 3 end cell cell space space space space space 0 end cell row cell space space 0 end cell cell space space minus 2 end cell cell space space space space space 1 end cell end table close square brackets comma$ find (AB)^–1.

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Multiple Choice Questions

224. Let A be a non-singular square matrix of order 3 × 3. Then | adj A| j is equal to

| A |
| A |²
| A |³
| A |³

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225.

If A is an invertible matrix of order 2, then det (A^–1) is equal to

det (A)
$fraction numerator 1 over denominator det space left parenthesis straight A right parenthesis end fraction$
1
1

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226. If x, y, z are non-zero real numbers, then the inverse of matrix $straight A space equals open square brackets table row straight x cell space space space space 0 end cell cell space space space 0 end cell row 0 cell space space space straight y end cell cell space space space 0 end cell row 0 cell space space space 0 end cell cell space space space straight z end cell end table close square brackets space is$
$open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$xyz open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space space 0 end cell row 0 cell space space straight y to the power of negative 1 end exponent space end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$
$1 over xyz open square brackets table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space straight z end cell end table close square brackets$

A.

open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space straight z to the power of negative 1 end exponent end cell end table close square brackets

straight A space equals space open square brackets table row straight x cell space space 0 end cell cell space 0 end cell row 0 cell space space straight y end cell 0 row 0 cell space 0 end cell cell space straight z end cell end table close square brackets

therefore space space space open vertical bar straight A close vertical bar space equals space open vertical bar table row straight x cell space space 0 end cell cell space space 0 end cell row 0 cell space space straight y end cell cell space space 0 end cell row 0 cell space space 0 end cell cell space space straight z end cell end table close vertical bar space equals space xyz

[Product of diagonal elements]
Co-factors of the elements of first row of | A | are

open vertical bar table row straight y cell space space 0 end cell row 0 cell space straight z end cell end table close vertical bar comma space minus open vertical bar table row 0 cell space space 0 end cell row 0 cell space space straight z end cell end table close vertical bar comma space space open vertical bar table row 0 cell space space space straight y end cell row 0 cell space space 0 end cell end table close vertical bar

i.e. yz, 0, 0 respectively.
Co-factors of the elements of second row of | A | are

negative open vertical bar table row 0 cell space space space 0 end cell row 0 cell space space space straight z end cell end table close vertical bar comma space space open vertical bar table row straight x cell space space space 0 end cell row 0 cell space space space straight z end cell end table close vertical bar comma space space minus open vertical bar table row straight x cell space space space 0 end cell row 0 cell space space space 0 end cell end table close vertical bar

i.e. 0, zx, 0 respectively.
Cofactors of the elements of third row of | A | are

open vertical bar table row 0 cell space space 0 end cell row straight y cell space space 0 end cell end table close vertical bar comma space space minus open vertical bar table row straight x cell space space 0 end cell row 0 cell space space 0 end cell end table close vertical bar comma space space open vertical bar table row straight x cell space space 0 end cell row 0 cell space space straight y end cell end table close vertical bar

i.e. 0, 0, xy respectively.

therefore space space space adj space straight A space equals space open square brackets table row yz cell space 0 end cell cell space space 0 end cell row 0 cell space zx end cell cell space space 0 end cell row 0 0 cell space space xy end cell end table close square brackets to the power of apostrophe space equals space open square brackets table row yz cell space 0 end cell cell space 0 end cell row 0 cell space zx end cell cell space 0 end cell row 0 0 cell space xy end cell end table close square brackets space therefore space space space straight A to the power of negative 1 end exponent space equals space fraction numerator adj over denominator open vertical bar straight A close vertical bar end fraction space equals space 1 over xyz open square brackets table row yz cell space space space 0 end cell cell space space 0 end cell row 0 cell space space space zx end cell cell space space 0 end cell row 0 cell space 0 end cell cell space space xy end cell end table close square brackets space space space space space space space space space space space space space space equals space open square brackets table row cell 1 over straight x end cell 0 0 row 0 cell 1 over straight y end cell 0 row 0 0 cell 1 over straight z end cell end table close square brackets space equals space open square brackets table row cell straight x to the power of negative 1 end exponent end cell cell space space 0 end cell cell space space 0 end cell row 0 cell space straight y to the power of negative 1 space end exponent end cell cell space space 0 end cell row 0 0 cell space space straight z to the power of negative 1 end exponent end cell end table close square brackets therefore space space space left parenthesis straight A right parenthesis space is space correct space answer.

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Short Answer Type

227.

Examine the consistency of the system of equations:
x + 2y = 2
2x + 3y = 3

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228.

Examine the consistency of the system of equations:
2x – y = 5
x + y = 4

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229.

Examine the consistency of the system of equations:
x + 3y = 5
2x + 6y = 8

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230.

Examine the consistency of the system of equations:
x+y+z = 1
2x + 3y + 2z = 2
ax+ay+2az = 4

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