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Maths-

General

Easy

Question

If A= $open square brackets table row 1 2 3 row 3 1 2 row 2 3 1 end table close square brackets$ , B = $open square brackets table row cell negative 5 end cell 7 1 row 1 cell negative 5 end cell 7 row 7 1 cell negative 5 end cell end table close square brackets$ then AB =

I₃
2I₃
4I₃
18I₃

The correct answer is: 18I₃

Related Questions to study

If A = $open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets$ , B = $open square brackets table row 0 cell negative 1 end cell row 1 0 end table close square brackets$ and C = $open square brackets table row 0 i row i 0 end table close square brackets$ , then A² = B²=C² =

If A = $open square brackets table row i 0 row 0 cell negative i end cell end table close square brackets$ , B = $open square brackets table row 0 cell negative 1 end cell row 1 0 end table close square brackets$ and C = $open square brackets table row 0 i row i 0 end table close square brackets$ , then A² = B²=C² =

If Ai = $open square brackets table row cell a to the power of i end exponent end cell cell b to the power of i end exponent end cell row cell b to the power of i end exponent end cell cell a to the power of i end exponent end cell end table close square brackets$ and if |a|<1, |b|<1, then $not stretchy sum subscript i equals 1 end subscript superscript infinity end superscript blank$ det (A_i) is equal to-

If Ai = $open square brackets table row cell a to the power of i end exponent end cell cell b to the power of i end exponent end cell row cell b to the power of i end exponent end cell cell a to the power of i end exponent end cell end table close square brackets$ and if |a|<1, |b|<1, then $not stretchy sum subscript i equals 1 end subscript superscript infinity end superscript blank$ det (A_i) is equal to-

If matrix A = $open square brackets table row 1 0 cell negative 1 end cell row 3 4 5 row 0 6 7 end table close square brackets$ and its inverse is denoted by $A to the power of negative 1 end exponent equals open square brackets table row cell a subscript 11 end subscript end cell cell a subscript 12 end subscript end cell cell a subscript 13 end subscript end cell row cell a subscript 21 end subscript end cell cell a subscript 22 end subscript end cell cell a subscript 23 end subscript end cell row cell a subscript 31 end subscript end cell cell a subscript 32 end subscript end cell cell a subscript 33 end subscript end cell end table close square brackets$ , then the value of a₂₃is equal to-

If matrix A = $open square brackets table row 1 0 cell negative 1 end cell row 3 4 5 row 0 6 7 end table close square brackets$ and its inverse is denoted by $A to the power of negative 1 end exponent equals open square brackets table row cell a subscript 11 end subscript end cell cell a subscript 12 end subscript end cell cell a subscript 13 end subscript end cell row cell a subscript 21 end subscript end cell cell a subscript 22 end subscript end cell cell a subscript 23 end subscript end cell row cell a subscript 31 end subscript end cell cell a subscript 32 end subscript end cell cell a subscript 33 end subscript end cell end table close square brackets$ , then the value of a₂₃is equal to-

parallel

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which of the following is correct

For the matrix A = $open square brackets table row 1 1 0 row 1 2 1 row 2 1 0 end table close square brackets$ , which of the following is correct

If a matrix A is such that3A³+ 2A²+ 5A + I = 0, then A^–1is equal to-

If a matrix A is such that3A³+ 2A²+ 5A + I = 0, then A^–1is equal to-

Let F( $alpha$ ) = $open square brackets table row cell cos invisible function application alpha end cell cell negative sin invisible function application alpha end cell 0 row cell sin invisible function application alpha end cell cell cos invisible function application alpha end cell 0 row 0 0 1 end table close square brackets$ , where $alpha$ $element of$ R. Then (F( $alpha$ ))–1 is equal to

Let F( $alpha$ ) = $open square brackets table row cell cos invisible function application alpha end cell cell negative sin invisible function application alpha end cell 0 row cell sin invisible function application alpha end cell cell cos invisible function application alpha end cell 0 row 0 0 1 end table close square brackets$ , where $alpha$ $element of$ R. Then (F( $alpha$ ))–1 is equal to

parallel

If A= $open square brackets table row 1 3 row 3 4 end table close square brackets$ and A²– kA – 5I₂= O, then the value of k is-

If A= $open square brackets table row 1 3 row 3 4 end table close square brackets$ and A²– kA – 5I₂= O, then the value of k is-

If M = $open square brackets table row 1 2 row 2 3 end table close square brackets$ and M²– $lambda$ M – $straight I subscript 2$ = 0, then $lambda$ equals -

If M = $open square brackets table row 1 2 row 2 3 end table close square brackets$ and M²– $lambda$ M – $straight I subscript 2$ = 0, then $lambda$ equals -

If $open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 cell negative tan theta end cell row cell tan theta end cell 1 end table close square brackets$ $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets to the power of negative 1 end exponent$ = $open square brackets table row a cell negative b end cell row b a end table close square brackets$ , then -

If $open square brackets table attributes columnalign center center columnspacing 1em end attributes row 1 cell negative tan theta end cell row cell tan theta end cell 1 end table close square brackets$ $open square brackets table row 1 cell tan invisible function application theta end cell row cell negative tan invisible function application theta end cell 1 end table close square brackets to the power of negative 1 end exponent$ = $open square brackets table row a cell negative b end cell row b a end table close square brackets$ , then -

parallel

If A = $open square brackets table row 0 1 row cell negative 1 end cell 0 end table close square brackets$ and (aI₂ + bA)² = A, then -

If A = $open square brackets table row 0 1 row cell negative 1 end cell 0 end table close square brackets$ and (aI₂ + bA)² = A, then -

If A is invertible matrix, then det (A^–1) equals-

If A is invertible matrix, then det (A^–1) equals-

If D = diag (d₁, d₂, d₃, ...,d_n), where d₁ $not equal to$ 0 for all i = 1,2,....,n; then D^–1 is equal to-

If D = diag (d₁, d₂, d₃, ...,d_n), where d₁ $not equal to$ 0 for all i = 1,2,....,n; then D^–1 is equal to-

parallel

If A and B are square matrices such that AB = B and BA = A, then A² + B² is equal to-

If A and B are square matrices such that AB = B and BA = A, then A² + B² is equal to-

If $straight A subscript alpha$ = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ , then which of following statement is true -

If $straight A subscript alpha$ = $open square brackets table row cell cos invisible function application alpha end cell cell sin invisible function application alpha end cell row cell negative sin invisible function application alpha end cell cell cos invisible function application alpha end cell end table close square brackets$ , then which of following statement is true -

If A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell 0 row cell sin invisible function application theta end cell cell cos invisible function application theta end cell 0 row 0 0 1 end table close square brackets$ , then -

If A = $open square brackets table row cell cos invisible function application theta end cell cell negative sin invisible function application theta end cell 0 row cell sin invisible function application theta end cell cell cos invisible function application theta end cell 0 row 0 0 1 end table close square brackets$ , then -

parallel

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